EE-3424, Mathematics in Signals and Systemsengineering.utsa.edu/agrigoryan/wp-content/uploads/sites/... · 2019-01-12 · EE-3424, MATHEMATICS IN SIGNALS & SYSTEMS, ART GRIGORYAN

EE-3424, MATHEMATICS IN SIGNALS & SYSTEMS, ART GRIGORYAN 1

.

EE-3424, Mathematics inSignals and Systems

Instructor: Dr. Artyom M. Grigoryan

ECE Department, The University of Texas at San Antonio, 2015 – 2018

EE-3424, MATHEMATICS IN SIGNALS & SYSTEMS, ART GRIGORYAN -2

CONTENTS

PART I: LECTURES

I. Preface, Introduction 2

A. Modeling of signals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

II. Elementary Continuous-Time Signals 4

A. Continuous-time and discrete-time signals . . . . . . . . . . . . . . . . . . . . . . . . 5

III. Transformations of signals 11

IV. Signal Characteristics 19

V. Periodic Signals 22

A. Discrete-time periodic signals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27B. Time constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

VI. Sinusoidal signals in engineering 28

A. Complex arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28A1. Geometry of complex arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

B. Simple Differential equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34B1. Complex exponential signals with complex amplitude . . . . . . . . . . . . . . . . . 36

B2. Complex exponential signals with exponential amplitude . . . . . . . . . . . . . . . 36

VII.Unit Step Function 37

A. Useful properties of the unit function . . . . . . . . . . . . . . . . . . . . . . . . . . 38B. The unit impulse function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

B.1. Properties of the δ function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

VIII.Systems 46

A. Properties of systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

IX. Impulse Representation of Signals 51

A. Properties of linear convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54A.1. Causality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54A.2. Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

X. LTI Systems and Differential Equations 56

A. First-order LTI systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57B. General solution of equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

C. Transfer function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59D. n-order LTI systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

XI. Block diagrams for LTI systems 63


XII.The Fourier series: Periodic Signals 70

A. The concept of the Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

B. Main formulas for coefficients of Fourier series . . . . . . . . . . . . . . . . . . . . . 75

XIII. Basis functions and finite Fourier series 75

A. Finite Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

XIV. Exponential Fourier series representation 88

XV. The Fourier series and FFT with MATLAB 94

A. Full Script of Program with Description and Explanation . . . . . . . . . . . . . . . 96

XVI. The Fourier series Application: Ideal low pass filter with MATLAB 102

XVII. Fourier transform 107

A. Definition of the Fourier transform (FT) . . . . . . . . . . . . . . . . . . . . . . . . 108B. Properties of the FT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

B. Fourier transform of periodic functions . . . . . . . . . . . . . . . . . . . . . . . . . 119C. Frequency characteristics of LTI systems . . . . . . . . . . . . . . . . . . . . . . . . 121

XVIII. The Laplace transform 122

A. Definition of the Laplace transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 123B. Properties of the Laplace transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

XIX. Brief Notes in Sampling Theorem 141

PART II: MATLAB Project

Convolution and Fast Fourier Transform 149-1


P R E F A C E

Digital signal processing (DSP) is an area of science and engineering that has developed rapidly

over the past thirty years. The rapid developed DSP is a result of the significant advances indigital computer technology and integrated-circuit fabrication. These lecture-notes are based on

the instructor notes, the material of many text-books, and among them, the following books shouldbe mentioned:

1. Signals, Systems, and Transforms, (5th Ed., 2014), by Charles L. Philips, John M.Parr, and Eve A. Riskin, Chapters 7, 8, 10-13.

2. Signals and Systems: Analysis using Transform methods and MATLAB (2nd Ed.,2004) by M.J. Roberts.

3. Fundamentals of Signals and Systems: with MATLAB Examples (2000) by EdwardKamen and Bonnie Heck.

4. Signals and Systems, (2nd Ed., 2002) by Simon Haykin and Barry Van Veen.5. ...

I. Introduction in Mathematics in Signals and Systems

Signal is a real or abstract concept of what carries, represents, or encodes information. Signalscan be manipulated, stored, or transmitted by a physical process. Examples: speech signals, audio

signals, video signals, images, radar signals, biomedical signals. System is a transformation or amore complicated process of transformation of signals, which may include the recording, changing,

and transmitting signals.Examples: 1. Audio compact disk (CD) records music signals on the disk as a sequence of

numbers. 2. A system for converting these numbers to an acoustic signal is the CD player.Mathematics is an appropriate language for describing and understanding many signals and

systems. Mathematical equations are used to describe and represent signals and systems and todesign new systems to achieve desired results. Examples: differential equations of linear systems,Fourier transformation, filters used in signal and image denoising, restoration, and enhancement.

Signal processing plays a central role in modern sciences and technology. Applications are foundedeverywhere, including speech communication, acoustic, biomedical engineering, seismology, and

many others. The theory of signal processing is concerned in representation, transmission, andmanipulations of signals. Until to the 1960s the technology for signal processing has been carried

in general by analog methods. The evolution of digital computer and microprocessors has beenpushed the developing discrete versions of signal processing methods. Digital signal processing

becomes applicable in many areas beginning from medical imaging to radar processing. Thatgrowth has created a massive amount of data, which is important to analyze, process, and transmit

signals. The digital signal and image processing have successful applications in geology, biology,meteorology, astronomy, radio-location, television, medical diagnosis, and many others domains inscience and engineering.

A. Modeling and Signals

We consider two topics of signals and systems as related to engineering:

1. Modeling of physical systems by mathematical equations.2. Modeling of physical signals by mathematical functions.


Model: It is advantageous to represent a device or an entire system by a circuit model. Forinstance, we can consider the 10m of copper wire is wound into the form of a multi-turn coil and

that a voltage of variable frequency is applied. If the ratio of applied voltage V to resulting currentI is measured as a function of frequency. The model of this simple physical system is described by

the equation

Ldi(t)

dt+ Ri(t) +

1

C

∫

τ<t

i(τ)dτ = v(t)

where R is the parameter of resistance, C is capacitance, and L is induction.Signals. Our word is full of signals, both natural and man-made. A signal can be defined as

a function that conveys information about the state or behavior of physical system. Signals arerepresented mathematically as functions of one or more independent variables.

For example, the functions f(t) = 2t + 1 and f(t) = 3t2 + 2t + 1 describe two signals, which varyrespectively linearly and quadratically with variable t.

Example of signals:

– The daily highs and lows in temperature. We can express the temperature at a designatedpoint in the room as a function θ(x, y, z, t) of four independent variables, where (x, y, z) are space

coordinates of the point and t is time.– The periodic electrical signals generated by the heart. Electrocardiogram (ECG) signal pro-

vides information about the activity of the patient’s heart. Electroencephalogram (EEG) signalprovides information about the activity of the brain. The variation is air pressure when we speak.

A speech can be represented mathematically as function of time.– The music we hear from compact disk (CD) player, which is due to changes in the air pressure

caused by the vibration of the speaker diaphragm. The information stored on CD is in a digitalform and, then, is converted to the analog form before we can hear the music.

– Seismic signal is shown in Fig. 1.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

x 104

−5

−4

−3

−2

−1

0

1

2

3

4

5x 10

5

samples (1:50000)

ampl

itute

Example of a seismic signal

x(t)

t

Fig. 1. Seismic signal.

– A photograph image (2-D signal) is represented as a brightness function(s) of two spatialvariables (coordinates in a plane). As an example, Figure 2 shows 506 random balls placed randomly

in the 3-D box 640×520×100. At each point (x, y, z) of those balls, an intensity θ(x, y, z) is definedthat is concentrated in the center of balls (see Fig. 3).

The projection of the intensity along the middle lines of Y-axis and X-axis are shown respectively

on Figs. 4(a) and (b).


Fig. 2. Image with random balls.

020

40

0

500

50

100

430

440

450

255

260

265

270

275

280

68

70

72

74

76

78

Slice of ball 58 on Z−axis 73

Fig. 3. Random balls.

II. Elementary Continuous-time Signals

Signals convey information and include physical quantities such as voltage, current, and inten-

sity. We will consider 1-D signals (analog or discrete) that are functions (continuous or step-wisecontinuous) usually of time or frequency. They are two important variables, time and frequency (t

and ω), used in signal processing.In the one-dimensional case, we used to consider the independent variable of the mathematical

representation of a signal as the time. The independent variable in the mathematical representationof a signal may be either continuous or discrete.


0 100 200 300 400 500 600

50

100

150

200Intensity of balls (projection by y=256)

x

0 50 100 150 200 250 300 350 400 450 500

50

100

150

200Intensity of balls (projection by x=320)

y

Fig. 4. Projections of the intensity of random balls.

A. Continuous-time and discrete-time signals

Signals may have either a continuous or discrete variable representation, and if certain conditions

hold, these representations are entirely equivalent.Continuous-time signals are often refereed to as analog signals and are defined along a continuum

of times. We will notations x(t), y(t), f(t), . . . for such signals.

−100 −50 0 50 100 150 200 250 300−2

−1

0

1

2Continuous−time signal

ma

gn

itu

de

t

y(t)=exp(−αt)cos(βt)

Fig. 5. The signal of finite duration.

Figure 5 shows the graph of a right side signal x(t) with the interval (domain) of definition[0, 260], i.e., when t ∈ [0, 260]. This interval can also be called the time-interval, or time-segment,

or continuous-time segment of the signal. The range of the signal are the interval of numbers wherethe values of the signal lie. In this example, the range is the interval which is smaller than or

inside the interval [−2, 2]. In general, the range of the signal may be real, i.e., all x(t) are realnumbers, or complex, i.e., all x(t) are complex numbers. As a special case, we mention the integer-

valued signals, when the range of the signals is a set of integer numbers, for example, numbers0, 1, 2, · · · , 255.

Discrete-time signals are refereed mathematically as a sequences of numbers and those are defined

at discrete times (the independent variable has discrete values). We will notations x[n], y[n], f [n], . . . ,


or xn, yn, fn, . . . , for such signals. If a discrete signal y[n] was composed (sampled) from a continuous-time signal y(t), i.e., only the values y(nT ) were recorded, where T is a small number (it is called

sampling period), we use to consider this short notation y[n] instead of y(nT ).As an example, Figure 6 shows the original continuous-time signals (modeled by MATLAB) in

part (a) and its discrete version in the interval [0, 128], or the discrete signal when the samplingperiod T = 1 in part (b).

0 50 100 150 200 250−2

−1

0

1

2Continuous function (model)

mag

nit

ude

y(t)=2exp(−αt)cos(βt)

t

0 20 40 60 80 100 120−2

−1

0

1

2Discrete sequence of data

Number of points

mag

nit

ude

t

a)

b)

Fig. 6. Graphic representations of sequences: (a) analog signal y(t) = 2 exp(−αt) sin(βt) for α = 0.02,β = 0.25, and (b) discrete analog of the signal, y[n] = y(nT ).

Many analog signals can be described by a mathematical expression or graphically by a curve,

or by a set of tabulated values. Real signals are not easy to describe quantitatively. They mustoften be approximated by idealized forms or models. Signals can be of finite or infinite duration.

Finite durations signals are called time-limited. Signals of semi-infinite extent may be right-side, ifthey are zero for t < a (a is finite), or left-side, if they are zero for t > a. Signals are causal if they

are zero for t < 0.

A continuous-time signal as a function x(t) may have discontinuity at some points t0, t2, . . . . It

means that at each of such points, for instance, t0, the following holds:

limε→0

x(t0 − ε) 6= limε→0

x(t0 + ε), (ε > 0).

As an example, Figure 10 shows the signal which is discontinuous at points t0 = −1 and t1 = 1.

Indeed, at point the t0 = −1, we have the following:

limε→0

x(t0 − ε) = 0 6= 1 = limε→0

x(t0 + ε),


−10 −8 −6 −4 −2 0 2 4 6 8 10−1

0

1

2

3

4

Fig. 7. The rectangle signal x(t) of finite duration.

and at point t1 = 1, we have the following:

limε→0

x(t1 − ε) = 1 6= 0 = limε→0

x(t1 + ε).

A.1 Simple signals

A Linear functionsx(t) = at + b, t ∈ [t1, t2],

where numbers t1 and t2 are boundary points of the domain of the signal x(t). The number a and

−10 −5 0 5 10 15 20−5

0

5

a=0

a<0

a>0

lines x(t)=at+b

t

Fig. 8. Functions.

b are two parameters of the line. a defines the slop of the signal and b is the point at which the

line intersects with the vertical axis, i.e., b = x(o). We see that, a = x′(t) at each point t ∈ [t1, t2].Figure 8 shows three lines for cases when a > 0, a< 0, and a = 0. When a = 0, the line x(t) ≡ b.

Properties:1. The sum of two different lines x1(t) = a1t + b1 and x2(t) = a2t + b2 is the line

x(t) = x1(t) + x2(t) = (a1 + a2)t + (b1 + b2).

2. The sum of two different lines x1(t) = a1t + b1 and x2(t) = −a1t + b2 is the line parallel to the

horizontal axisx(t) = x1(t) + x2(t) ≡ b1 + b2.

3. Two different lines x1(t) = a1t + b1 and x2(t) = a2t + b2 are parallel, if a1 = a2 and b1 6= b2.


4. The sum of two the lines x1(t) = at + b1 and x2(t) ≡ b2 is the line which is parallel to x1(t),

x(t) = x1(t) + b2 = at + (b1 + b2).

B Piece-wise linear signals (functions)

As an example, Figure 9 shows the continuous-time signal which is continuous.

−10 −5 0 5 10 15 20 25 30 35−15

−10

−5

0

5

10

15

y(−5)=−3, y(2)=11 y(10)=5 y(20)=5 y(30)=−5

Continuous−time piece−wise linear signal

ma

gn

itu

de

t

Fig. 9. The signal y(t) of duration of 35 sec.

The time segment of the signal or domain of definition of the signal is the interval [−5, 35]. To

write analytically this signal, we first consider the equation of the line

x(t) = at + b

passing two points (t1, x(t1)) and (t3, x(t2)) :

x(t) =x(t2) − x(t1)

t2 − t1t +

x(t1)t2 − x(t2)t1t2 − t1

(1)

Here

a =x(t2) − x(t1)

t2 − t1and b =

x(t1)t2 − x(t2)t1t2 − t1

. (2)

We now consider and write separately this signal in intervals [−5, 2], (2, 10], [10, 30), and [30, 35],by using Eqs. 1 and 2. This signal can be written in the following form:

y(t) =

+11 − (−3)

7t +

−3(2) − 11(−5)

7, if t ∈ [−5, 2],

−11 − 5

8t +

11(10)− 5(2)

8, if t ∈ (2, 10],

5, if t ∈ (10, 20],

−5 − (−5)

10t +

5(30)− (−5)20

10, if t ∈ (20, 30].

Therefore,

y(t) =

+2t + 7, if t ∈ [−5, 2],

−3

4t +

25

2, if t ∈ (2, 10],

5, if t ∈ (10, 20],

−t + 25, if t ∈ (20, 30].


C. Sinusoidal waves (sine and cosine signals)We consider two functions, which are defined by the following power series:

cos(t) = 1 −t2

2!+

t4

4!−

t6

6!+

t8

8!−

t10

10!+ . . .

and

sin(t) = t −t3

3!+

t5

5!−

t7

7!+

t9

9!−

t11

11!+ . . .

where the factorial of the integer n is defined as n! = 1 · 2 · 3 . . . · (n − 1) · n if n 6= 0, and 0! = 0.

These functions are cosine and sine functions which are defined for any time-point t, and inaddition

−1 ≤ cos(t) ≤ 1 and − 1 ≤ sin(t) ≤ 1,

for any t ∈ (−∞,∞).Consider the waves defined as

x(t) = 2 sin(2t − 1) and y(t) = −4 cos(3t + 2).

The amplitude of the signal x(t) is 2, and for the signal y(t), it is −4. The magnitude of the signalx(t) is 2, and for the signal y(t), it is 4. The first signal oscillates between −2 and 2, and the second

signal oscillates between −4 and 4.Figure 10 shows the signals x(t) and y(t) in parts (a) and (b), respectively.

−5 0 5 10−3

−2

−1

0

1

2

3x(t)=2sin(2t−1)

mag

nit

ud

e

(a)

−5 0 5 10−5

0

5x(t)=−4cos(3t+2)

mag

nit

ud

e

(b)

Fig. 10. The graphs of the signals (a) x(t) and (b) y(t) in the time-interval [−2π, 3π].

The variable t for cosine and sine functions are usually considered as an angle; we will use the

letter ϑ instead of t. 1

The main properties of the trigonometric functions sin(ϑ) and cos(ϑ) :

1. sin(ϑ) = − cos(ϑ + π/2);

2. cos(ϑ) = sin(ϑ + π/2);

1ϑ – Greek letter ‘vartheta’


3. sin(ϑ + 2πk) = sin(ϑ), for any integer k;

4. cos(ϑ + 2πk) = cos(ϑ), for any integer k;

5. cos(πk) = (−1)k = ±1, and sin(πk) = 0, when k is integer;

6. cos(2π(k + 1/2)) = −1, when k is integer;

7. sin(−ϑ) = − sin(ϑ);

8. cos(−ϑ) = cos(ϑ);

9. cos(π/2) = 0 and sin(0) = 0;

10. cos(0) = 1 and sin(π/2) = 1.

The main equations:

1. sin2(ϑ) + cos2(ϑ) = 1;

2. sin(2ϑ) = 2 sin(ϑ) cos(ϑ);

3. cos(2ϑ) = cos2(ϑ)− sin2(ϑ) = 2 cos2(ϑ)− 1;

4. sin(ϑ1 + ϑ2) = sin(ϑ1) cos(ϑ2) + sin(ϑ2) cos(ϑ1);

5. cos(ϑ1 + ϑ2) = cos(ϑ1) cos(ϑ2) − sin(ϑ2) sin(ϑ1);

and a few more with the derivatives:

6. (sin(ϑ))′ = cos(ϑ);

7. (cos(ϑ))′ = − sin(ϑ);

8. sin(ϑ)dϑ = −d cos(ϑ);

9. cos(ϑ)dϑ = d sin(ϑ).


III. Transformations of signals

Let x(t) be a function of one independent variable (time) defined for all values −∞ < t < +∞,i.e., for all t ∈ R, where R = R1 denotes the real line.

A. Time transformations:

1. Time reversal. Given a signal x(t), a time-reversal transformation of the signal is defined as

y(t) = x(−t), −∞ < t < +∞. (3)

Drawing the graph of the signal, we can see that the time-reversal transformation creates the mirrorimage about the vertical axis.

As an example, Figure 12 shows the signal x(t) defined in the interval [−1, 2] in part a, alongwith the time-reversal transform in b and c, and two signals together in d. Note that in c, the

numbers along the x-axis have been changed only.

−2 −1 0 1 2−2

−1

0

1

2

3

(a)

mag

nit

ud

e

−2 −1 0 1 2−2

−1

0

1

2

3

(b)

2 1 0 −1 −2 −2

−1

0

1

2

3

(c)

mag

nit

ude

−2 −1 0 1 2−2

−1

0

1

2

3

(d)

x(t)

x(t)x(−t)

x(−t)

t

t

t

t

Fig. 12. Signal and its time-reversal transform.


There are many functions which do not change after time-reversal transformation, i.e.,

y(t) = x(t), ∀t.

For example, such are signals x(t) = |t|, x(t) = t2, x(t) = cos(x), which are called symmetric

relative the vertical axis t = 0. These symmetric signals are shown in Fig. 13 along with the nonsymmetric functions t3 and sin(t).

−2 −1 0 1 2−4

−3

−2

−1

0

1

2

3

4

t2

|t|

t3

symmetric functions

(a)

t2

t2

t2

−5 0 5−1.5

−1

−0.5

0

0.5

1

1.5

x(t)=cos(t)

(b)

x(t) x(t)

t t

Fig. 13. Symmetric and non symmetric signals.

Note that the above signal x(t) is defined only on the interval [−1, 2]. We can thus define and

plot y(t) only on the interval [−2, 1], i.e., y(t) is defined by (3) when t ∈ [−2, 1].In practice, working with digital signals which are defined (or can be defined) on a finite interval

[a, b], we used to determine the time-reversal transformation of the signal relative to the verticalline crossing the middle point of [a, b]. In order words, we define the time-reversal transformationsas (see Fig. 14)

y(t) = x(b + a − t), t ∈ [a, b]. (4)

That results in the mirror image about the vertical axis shifted by (b + a)/2 to the right. Indeed,

we can write that

x(b + a − t) = x

(

b + a

2−

[

t −b + a

2

])

.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−2

−1

0

1

2

3

mag

nit

ude

t0=(a+b)/2

a=1, b=4

window

x(t) y(t)

Fig. 14. Time-reversal signals relative to the middle point.

here, we face with a new operation over the function, namely the time-shifting of the signal.


2. Time shifting. Given a signal x(t) and a value t0, a time-shifted version of the signal is definedas

y(t) = xt0(t) = x(t − t0), −∞ < t < +∞. (5)

When we plot the signal, the time-shifting transformations yields the shift of the vertical axis byvalue t0 to the left or right, depending on t0 > 0 or t0 < 0.

As an example, Figure 15 shows the signal x(t) and its shifted transforms by ±3.5

y(t) = x(t − 3.5) (shifted to the right by 3.5)

and

y(t) = x(t + 3.5) (shifted to the left by 3.5).

−5 −4 −3 −2 −1 0 1 2 3 4 5 6−2

−1

0

1

2

3

mag

nit

ude

shifted signals

x(t)

x(t+3.5) x(t−3.5)

t

Fig. 15. Signal x(t) and its time-shift transforms by ±3.5.

Example 1: We consider the cosine signal

x(t) = cos(t), t ∈ R.

Taking values of t0 equal π/2 and −π/2, we obtain respectively (see Fig. 16):

xπ/2(t) = x(t − π/2) = cos(t − π/2) = − sin(t)

x−π/2(t) = x(t + π/2) = cos(t + π/2) = sin(t).

−8 −6 −4 −2 0 2 4 6 8−1.5

−1

−0.5

0

0.5

1

1.5

shifted signals

mag

nit

ude

x(t)=cos(t)

xπ/2

(t)x−π/2

(t)

Fig. 16. The signal x(t) = cos(t).


Example 2: We consider the following signal, which plays important role in digital signal pro-cessing,

x(t) = e−t, t ≥ 0, x(t) = 0, t < 0.

The shift by t0 yields the following amplification of the signal:

xt0(t) = x(t − t0) = e−(t−t0) = et0e−t = ae−t (6)

for t ≥ t0, and xt0(t) = 0, otherwise. The amplitude of the time shifted signal at every point of twill increase or decrease for t ≥ t0 depending on a > 1 or a < 1, respectively (a = et0) (see Fig. 17)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.5

1

1.5

2

2.5

signal

mag

nit

ud

e

Time in µsecs

x1(t)=exp(−(t−1))

x(t)=exp(−t)

x−1

(t)=exp(−(t+1))

t

Fig. 17. The shift of the signal x(t) = e−t, t ≥ 0.

Task 1: For signal which is described as

x(t) = e−t cos(t), t ≥ 0, x(t) = 0, t < 0,

write and plot the time-shifted signals for t0 = π/2, −π/2, π, π/4, and −π/4.

3. Time scaling. Given a signal x(t) and a positive constant a, a time scaled version of the signalis defined as

y(t) = x(at), −∞ < t < +∞. (7)

As an example, Fig. 18 shows the time scaling transformations for a = 2 and a = 1/2. The signalcan be defined on the whole line R = (−∞, +∞) which does not change after multiplying by a 6= 0,i.e., aR = R.

Figure 19 shows the time-scaling of the cosine wave, when the scale factors are 2 and 1/2.


−2 −1 0 1 2−2

−1

0

1

2

3

(a)

mag

nit

ud

e

−2 −1 0 1 2−2

−1

0

1

2

3

(b)

−2 −1 0 1 2−2

−1

0

1

2

3

(c)

mag

nit

ud

e

−2 0 2 4−2

−1

0

1

2

3

(d)

x(t)

x(t)x(t/2)

x(2t)

t

t

t

t

Fig. 18. (a) Signal x(t) and its time-scaling versions for (b) a = 2, (c) a = 1/2, and (d) all signals together.

−6 −4 −2 0 2 4 6

−1

0

1

Time−transformation of the signal

mag

nit

ude

x(t)=cos(t)

−6 −4 −2 0 2 4 6

−1

0

1

mag

nit

ude

y(t)=cos(2t)

−6 −4 −2 0 2 4 6

−1

0

1

mag

nit

ude

y(t)=cos(t/2)

t

Fig. 19. Cosine signal x(t) = cos(t) and time-scaled versions y(t) = cos(2t) and y(t) = cos(t/2).

3. Combination of time transformations.

We consider the general time transformation which is defined as

y(t) = x(at − t0), ∞ < t < +∞, (8)


where a > 0 is a value of time scaling, t0 is a value of time shifting. Denoting the new variable

t′ = at − t0,

we can write thaty(t) = x(t′), where t = (t′ + t0)/a. (9)

Example 3: Let a = −1/2 and t0 = −1,

y(t) = x

(

1 −1

2t

)

= x(t′)

t = 2 − 2t′

The construction of y(t) is shown in Fig. 20(c).

We can also plot y(t) by using the following step-by-step transformations:

x′(t) = x(−t) (10)

x′

1/2(t) = x′(t/2) (11)

x′

1/2(t − 2) = x′([t − 2]/2) = x′(t/2 − 1) (12)

= x(1− t/2) (13)

which is illustrated in Fig. 20.

−2 −1 0 1 2−2

−1

0

1

2

3

(a)

mag

nit

ude

−2 −1 0 1 2−2

−1

0

1

2

3

(b)

−2 0 2 4−2

−1

0

1

2

3

(c)

mag

nit

ude

−4 −2 0 2−2

−1

0

1

2

3

(d)

x(t)

x(−t/2)y(t)

x(−t)

t

t

t

t

Fig. 20. Construction of the time-transformation y(t) of the signal x(t) in Example 3.


B. Signal-amplitude transformation

5. Amplitude transformation. Given a signal x(t), a amplitude amplification (AA) is defined as

y(t) = Ax(t), (14)

where A is a constant. In the general case, the amplitude transformation is defined as

y(t) = Ax(t) + B = ±|A|x(t) + B, (15)

where B is another constant to be added (if B > 0) or subtracted (if B < 0) from the amplifiedsignal in (14). The amplitude transformation signal can also be considered as

y(t) = A

(

x(t) +B

A

)

= ±(

|A|[

x(t) +B

A

])

, A 6= 0,

i.e., the constant B/A is added to the given signal x(t), and then the signal is amplified by A. In

the A < 0 case, we observe amplitude reversal and amplitude scaling |A|, and constant B shifts theamplitude of the amplified signal.

As an example, Figure 21 illustrates the amplitude transformation of the signal y(t) = 2x(t)+1,

by using the following two steps of calculations:

x(t) → 2x(t) → 2x(t) + 1.

−2 −1 0 1 2−2

−1

0

1

2

3

(a)

mag

nit

ude

−2 −1 0 1 2−2

−1

0

1

2

3

4

(b)

−2 −1 0 1 2−2

0

2

4

6

mag

nit

ude

(c)−2 0 2 4

−2

0

2

4

6

(d)

2x(t)+1

x(t) 2x(t)

x(t)2x(t)+1

Fig. 21. Signal and its amplitude transforms.

Figure 22 show the same amplitude transformation of the signal, by using other two steps of

calculationsx(t) → x(t) + 0.5 → 2(x(t) + 0.5).


−2 −1 0 1 2−2

−1

0

1

2

3

(a)

mag

nit

ude

−2 −1 0 1 2−2

−1

0

1

2

3

4

(b)

−2 −1 0 1 2−2

0

2

4

6

mag

nit

ude

(c)−2 0 2 4

−2

0

2

4

6

(d)

2(x(t)+0.5)

x(t) x(t)+0.5

x(t)2x(t)+1

Fig. 22. Signal and its amplitude transforms.

−2 −1 0 1 2−2

−1

0

1

2

3

mag

nit

ude

(a)−2 −1 0 1 2

−3

−1

1

3

5

7

(b)

x(t) 2x(t)+1

Fig. 23. Signal x(t) and the amplitude transform 2x(t) + 1.

To draw the graph of the amplitude transform, we can use the original graph and change onlynumbers along the y-axis. As an example, Figure 23 shows the above amplitude transform of the

signal x(t) → y(t) = 2x(t) + 1, by using this method.Project 1: (Will be given later) Transfer and plot the transient signal x(t) = 2 exp(−αt) sin(βt),

for α = 0.02 and β = 0.25 which is given on the interval [0, 255] into the window [0, 255]× [1, 3].

This signal is illustrated in Fig. 24 (see also Fig. 6). Use MATLAB and print input x(t) and outputy(t) signals.

0 50 100 150 200 250−2

−1

0

1

2

Input signal

mag

nit

ude

2exp(−αt)cos(βt)

t

0 100 200 3000

1

2

3

4

Amplitute transformation

AA

window

Fig. 24. The time and amplitude transformation of the signal x(t) into the window [0, 255]× [1, 3].


IV. Signal Characteristics

We now consider the property of symmetry of the signals. Let x(t) be a signal (function) definedon the real line R (or on a symmetric interval [−1, 1], a > 0).

(A) x(t) is called even, ifx(t) = x(−t)

for all t; The interval of definition of x(t) should be symmetric. For example, the signals describedby the functions x(t) = |t|, x(t) = cos(t), and x(t) = t sin(t) possess the property of even symmetry,

i.e., they have symmetry with respect to the vertical axis t = 0, as shown in Fig. 25. For an evensignal x(t), the following valid:

x(t) =1

2[x(t) + x(−t)] =

1

2[x(t) + x(−t)] +

1

2[x(t) − x(−t)] (16)

−2 −1 0 1 2−1

−0.5

0

0.5

1

1.5

2

x(t)=|t|−0.5

−5 0 5−1.5

−1

−0.5

0

0.5

1

1.5

x(t)=cos(t)

−30 −20 −10 0 10 20 30

−2

−1

0

1

2

x(t)=Atsin(t)

t

t

t

Fig. 25. Even signals.

(B) x(t) is called odd, if

x(t) = −x(−t)

for all t. For example the signals described by the functions x(t) = t, x(t) = sin(t), and x(t) =

t cos(t) possess the property of odd symmetry (see Fig. 26). For an odd signal x(t), the followingvalid:

x(t) =1

2[x(t)− x(−t)] =

1

2[x(t)− x(−t)] +

1

2[x(t) + x(−t)]. (17)

If x(t) is an even (odd) function, then the amplitude transformation

y(t) = Ax(t) + B, A 6= 0, (18)

is even (not odd, if B 6= 0), too. Next, if x(t) is even (odd), then the time reversal signal y(t) = x(−t)is also even (odd). In general, the function y(t) = x(at) has the same evenness as x(t) if a 6= 0, but

y(t) = x(at + b) does not when b 6= 0.


−2 −1 0 1 2−2

−1

0

1

2

x(t)=t

−10 −5 0 5 10−1.5

−1

−0.5

0

0.5

1

1.5

x(t)=sin(t)

−30 −20 −10 0 10 20 30

−2

−1

0

1

2

x(t)=Atcos(t)

Fig. 26. Odd signals.

The following property is important. An arbitrary signal x(t), t ∈ R, can be represented as the

sum of even and odd signals. Indeed, the function x(t) can be written as follows

x(t) =1

2[x(t) + x(−t)] +

1

2[x(t) − x(−t)]

=1

2xe(t) +

1

2xo(t)

where the even component xe(t) = x(t) + x(−t) and the odd component xo(t) = x(t) − x(−t).Indeed

xe(−t) = x(−t) + x(−(−t)) = x(−t) + x(t) = xe(t),

xo(−t) = x(−t) − x(−(−t)) = x(−t) − x(t) = −xo(t).

We denote by xe(t) and xo(t) the even and odd parts of the signal, respectively,

xe(t) =1

2xe(−t) =

1

2[x(t) + x(−t)],

xo(t) =1

2xo(−t) =

1

2[x(t)− x(−t)].

As an example, Figure 28 shows the even and odd components of the considered above signal.If x(t) is even, then xo(−t) = 0 and x(t) = xe(t). If x(t) is odd, then xe(−t) = 0 and x(t) = xo(t).

We will denote xeven(t) = x(t) if x(t) is even and xodd(t) = x(t) if x(t) is odd. Given two signalsx(t) and y(t), the following properties hold:

1) xodd + yodd = (x + y)odd

2) xeven + yeven = (x + y)even

3) xodd + yeven 6= (x + y)even

xodd + yeven 6= (x + y)odd


−2 −1 0 1 2−2

−1

0

1

2

3

(a)

mag

nit

ud

e

−2 −1 0 1 2−2

−1

0

1

2

3

(b)

−2 −1 0 1 2−2

−1

0

1

2

3

mag

nit

ud

e

(c)

−2 −1 0 1 2−2

−1

0

1

2

3

(d)

xe(t)

x(t) x(−t)

xo(t)

Fig. 27. Decomposition of the signal.

For instance, if x(t) = cos(t) and y(t) = sin(t), then x = xeven and y = yodd, and

x(t) + y(t) = cos(t) + sin(t) =√

2 cos(t + π/4)

which is neither even nor odd.4) xeven × yeven = (x × y)even

5) xodd × yodd = (x × y)even

6) xeven × yodd = (x × y)odd

Figure 28 shows the even and odd components of the transient signal x(t) = 2 exp(−0.02t) sin(t/4).

−100 −50 0 50 100−20

−10

0

10

20

(a)

mag

nit

ud

e

transient signal

−100 −50 0 50 100−20

−10

0

10

20

(b)

−100 −50 0 50 100−20

−10

0

10

20

(c)

mag

nit

ud

e

−100 −50 0 50 100−20

−10

0

10

20

(d)

x(t)

xe(t) x

o(t)

x(−t)

Fig. 28. Decomposition of the transient signal.

EE-3424, MATH IN SIGNALS & SYSTEMS, ART GRIGORYAN 22

V. Periodic signals

A signal x(t) is called periodic, if there exist such a constant T > 0 that

x(t) = x(t + T ), ∀t ∈ R. (19)

T is called a period of x(t). By the definition, it is enough to consider the periodic function only

on the interval [0, T ), because any value t can be written as

t = nT + t0, n ∈ Z, t0 ∈ [0, T ),

where Z is the set of all integers, Z = 0,±1,±2, ..., and x(t) = x(t0). Indeed, the following is

valid:

x(t) = x(nT + t0) = x((n− 1)T + t0 + T )

= x((n− 1)T + t0) = ... = x(T + t0) = x(t0).

−T is also period of the signal, since x(t− T ) = x((t− T ) + T ) = x(t). Any integer multiple of the

period, ±kT, where k is an integer, is a period of the signal,

x(t ± T ) = x(t ± 2T ) = · · · = x(t ± kT ) = x(t), ∀t ∈ R.

The smallest period T > 0 of the signal x(t) is called a fundamental period of the signal, x(t) =

x(t + T ). 1

As an example, Figure 29 shows the periodic signal with the fundamental period T = 6 in theinterval [−18, 18], i.e., six periods of the signal.

−15 −10 −5 0 5 10 15−1

−0.5

0

0.5

1x(t)

T=6

t

one period

Fig. 29. Periodic signal and its one period.

We also use the concept of a fundamental frequency in hertz, which shows how many fundamental

periods are placed in the interval of 1 second

f0 =1

T0Hz (20)

and in radians the frequency is defined as

ω = 2πf0 =2π

T0

rad

sec. (21)

1Another definition: The fundamental period is the interval [0, T ).


For the periodic signal of Fig. 29, one can see that there is only 1/6 part of the signal in the intervalof 1 second. Thus the frequency equals

f0 =1

6Hz, or ω =

π

3

rad

sec.

Example 1: Signals x(t) = cos(t) has the fundamental period equal T = 2π, and the signal

x1(t) = cos(4t) has the period T1 = π/2, i.e., T1 = T/4. The time-scaling transformations t → 4ttransfers one period [0, 2π) of x(t) into the interval [0, 2π)/4 = [0, π/2), as shown in Fig. 30.

0 2 4 6 8 10 12

−1

−0.5

0

0.5

1

x(t)=cos(t)

t

0 2 4 6 8 10 12

−1

−0.5

0

0.5

1

t

y(t)=cos(4t)T

T 1

Fig. 30. Two periodic cosine functions (shown only in the interval [0, 12]).

Example 2: We now consider the periodic signal called the sawtooth wave (shown in Figure 31),

which is useful in sweeping a beam of electrons across of face of CRT (cathode ray tube),

x(t) = t0 = t mod 1, (22)

if t = n + t0, where t0 ∈ [0, 1) and n ∈ Z.It is clear, that if

x(t + T ) = x(t), y(t + T ) = y(t)

then

(x + y)(t + T ) = x(t + T ) + y(t + T )

= x(t) + y(t) = (x + y)(t)

i.e., the sum of two periodic signals with the period T is also periodic with the period T.


−4 −3 −2 −1 0 1 2 3 4−0.5

0

0.5

1

1.5x(t)

T=1t

one period

sawtooth wave

Fig. 31. Eight periods of the sawtooth wave.

In the general case, when periods are different,

x1(t + T1) = x1(t), x2(t + T2) = x2(t), T2 6= T1,

the sum of two periodic functions (signals) x(t) = x1(t) + x2(t) not necessary to be periodic. This

function is periodic if only x1(t) and x2(t) have a common period. If these two signals have acommon period at a point T on the line, then

T = n1T1 = n2T2 (23)

for some integers n1 and n2. In this case T is period of x(t). If T is the first positive such point,then T is the fundamental period.

Example 3: Signals x1(t) = cos(t) has the period T1 = 2π, and the signal x2(t) = cos(2t) has theperiod T2 = π. We have 1T1 = 2T2 = 2π, therefore the signal x(t) = cos(t) + cos(2t) has the period

T = 2π, as shown in Fig. 32.Example 4: Consider signals x1(t) = 2 cos(2t) with period T1 = π, and the signal x2(t) = 5 sin(3t)

has period T2 = 2π/3. The signal x(t) = 2 cos(2t)+5 sin(3t) is periodic and its fundamental periodT = 2π, as shown in Fig. 33. Indeed, we have the following

2T1 = 3T2 = 2π → T = 2π,

i.e., we solve (23) with n1 = 2 and n2 = 3.Example 5: We now consider power-supply periodic signals which convert the sinusoidal voltage

(see Fig. 34 in part a) into the constant voltage (in c). For a given constant T0 > 0, we considerthe following non negative signal (called a full-wave rectified signal (shown in b)

x(t) =

∣

∣

∣

∣

sin

(

πt

T0

)∣

∣

∣

∣

, t ∈ R. (24)

The half-wave rectified signal is defined as y(t) = x(2t)u(t), where u(t) is the binary functiondescribed the unit signals (impulse) with period T0/2,

u(t) =

1, if t = 2n(T0/2) + t0, t0 ∈ [0, T0/2);

0, otherwise.(25)

The construction of the full-wave rectified signal is illustrated in Figs. 34 and 35.


0 5 10 15 20 25

−1

0

1

Signals cos(t), cos(2t), and (cos(t)+cos(2t)), t=0:8*3.14.

cos(t)

mag

nit

ude

t

T1

0 5 10 15 20 25

−1

0

1

mag

nit

ude

cos(2t)

T2

t

0 5 10 15 20 25

−1

0

1

2

mag

nit

ude

cos(t)+cos(2t)

t

T=T1

Time in µsecs

Fig. 32. The sum of two periodic functions.

0 5 10 15 20 25

−1

0

1

Signals cos(2t), sin(3t), and cos(2t)+sin(3t), t=0:8*3.14.

cos(2t)

mag

nit

ude

t

T1

0 5 10 15 20 25

−1

0

1 sin(3t)

mag

nit

ude

t

T2

0 5 10 15 20 25

−2

−1

0

1

2

mag

nit

ud

e

t

T=2T1=3T

1

Time in µsecs

cos(2t)+sin(3t)

Fig. 33. Two cosine functions with the common period at point T = 2T1 = 3T2 = 2π.


−10 −5 0 5 10

−1

−0.5

0

0.5

1

full−wave rectified signal

sin(t/T0)

xx

−T0

T0

x

2T0

x

−2T0

mag

nit

ude

ta)

b)

c)

−10 −5 0 5 10

0

0.5

1

mag

nit

ude

−10 −5 0 5 10

0

0.5

1

mag

nit

ude

Fig. 34. Power-supply periodic signal construction (a)-(c).

−10 −5 0 5 10

0

0.5

1

1.5

mag

nit

ud

e

te)

−10 −5 0 5 10

0

0.5

1

mag

nit

ud

e

t

f)

−10 −5 0 5 10

0

0.5

1

mag

nit

ud

e

t

time−scaled version |x(2t)|

d)

Fig. 35. Power-supply periodic signal construction (d)-(f).


A. Discrete-time periodic signals

We consider the process of sampling of a periodic signal

x(t) = x(t + nT ), ∀t ∈ R,

where T is the period of x(t).Let T0 is the sampling interval, i.e., in every T0 seconds, the value xn = x(nT0) will be computed.

In order to reach into one of the periods, rT, after n intervals of time T0, the following equationshould have place

nT0 = rT = r2π

ω0, ∃r ∈ Z,

nω0T0 = 2πr.

If T0 = 1 then nω0 = 2πr, and n = 2π/ω0 = Tr.If T0 is an integer, then one can take r = 1 and N = T.

In the discrete case, the sum of two periodic (discrete-time) signals is a periodic function (or,signal)

xn = xn+N , N is the period

yn = yn+M , M is the period

There exist an integer K such that

xn = xn+K , for ∀n ∈ Z

yn = yn+K , for ∀n ∈ Z

For instance, we can take K = NM.

Indeed, for any integer n,

xn+K = xn+MN = xn+(M−1)N+N = xn+(M−1)N

= xn+(M−2)N+N = xn+(M−2)N = . . .

= xn+N = xn

yn+K = yn+NM = yn+(N−1)M+M = yn+(N−1)M

= yn+(N−2)M+M = yn+(N−2)M = . . .

= yn+M = yn

So, K is a period of the sum of discrete-time signals (but may be not the fundamental period).

B. Time constant

To define the concept of time constant, we consider the solution in the general form x(t) = Ce−at,when a > 0, and draw the tangent to the curve of x(t) at point t = 0 (see Fig. 37). We have

x′(0) = −Ca(e−at)|t=0 = −Ca = −C

τ→ τ =

1

a


0 5 10 15 20 25 300

0.5

1

1.5

2

2.5

exponential signal x(t)=2exp(−0.25t)

ϑ

x(t)=Ce−at

, C=2, a=0.25, τ=1/a=4.

τ 2τ 3τ 4τ 5τ 6τ

C×1/e, 36.8oC (1/e=0.368)

C×1/e2, 13.5

oC (1/e

2=0.135)

C, 100oC

C×1/e4, 2

oC (1/e

3=0.02)

C×1/e3, 5

oC (1/e

3=0.05)

Fig. 36. Exponential and time constant.

Taking value x(t) at time t, the exponential decays to less than 36.8% of its amplitude in 1τ unit

of time, t + τ. That is, x(t + τ) ≈ 0.368x(t), . . . , x(t + 5τ) = 0.00067x(t).In general, we can write that the following take place for any integer n > 1 :

x(t + nτ) ≈ 0.368x(t + (n − 1)τ) ≈ 0.368× [0.368x(t + (n − 2)τ)] ≈ · · · . (26)

VI. Sinusoidal signals in engineering

In many cases, it is useful to transfer our calculations from the real space to complex space,

analyze and solve problems by using methods of the complex analysis (arithmetic), and then transferback the solution to the real space.

Below is a brief introduction to the complex arithmetic which we should know well in order tounderstand better the main concepts of Fourier transforms and the frequency analysis of functions

and signals by the Fourier transforms.

A. Complex Arithmetic

We first go back to XVI century, when the formal and not real solution of the simple quadratic

equationz2 + 1 = 0, or z2 = −1, (27)

was proposed by two Italian mathematicians Rafael Bombelli and Gerolamo Cardano (XVI century)and denoted by z =

√−1. Another solution of this equation is z = −

√−1.


0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.5

1

1.5

2

2.5

signal

mag

nit

ude

Time in µsecs

x(t)=exp(−t)

t

x1(t)=exp(−(t−1))

x−1

(t)=exp(−(t+1))

A

B

C

D

A/B=C/D=2.7183

Fig. 37. Exponential and time constant.

−2 −1 0 1 2−2

0

2

4

6

(a) y=x2+1 and y=x

2−1

−1 −0.5 0 0.5 1

−1

−0.5

0

0.5

1

•+1•−1

•

z1

•z

2

(b) unit circle

Fig. 38. (a) Parabola and (b) the unit circle with four unit points on it.

Figure 38 shows the graph of the parabola y = x2 + 1 in part a, which does not intersect the

horizontal line, i.e., there is no solution of equation (27) in the real arithmetic. Solutions of thisequation should be found in another arithmetic which will be described in a moment; in fact, these

two solutions are on the unit circle shown in part b. There are four unit points on this circle. Twopoints are on the horizontal line, ±1, and they are the solutions of equation z2 − 1 = 0, which can

be written as x2 − 1 with real numbers x. The graph of the parabola y = x2 − 1 is also shown inpart a. Two other points of the circle, which are on the vertical line, are the solutions of equation

z2 + 1 = 0. These points z1 =√−1 and z2 = −

√−1.


Given real number y, the solutions of the equation

z2 + y2 = 0, or

(

z

y

)2

+ 1 = 0, (28)

can be written as z/y = ±√−1, or z = ±y

√−1, or z = ±(

√−1)y. If for a given real number x we

consider the quadratic equation(z − x)2 + y2 = 0, (29)

then, one can see that z − x = ±y√−1 and the solutions of this equation are the numbers z =

x + y√−1 = x + (

√−1)y and z = x − y

√−1 = x − (

√−1)y.

In XVIII century, Euler denoted this imaginary number or symbol√−1 by i, i.e., i =

√−1 and

i2 = −1. This symbol represents an imaginary unit, and in the engineering community it is also

denoted by j, since the letter “i” is used for the electrical current. The number z = x+ iy = x+yiis called a complex number. Numbers x and y are real, x is called the real part of z and y is the

imaginary part of z. The concept of the complex number generalizes the real numbers which canbe considered as the complex numbers with zero imaginary part, i.e., when y = 0. The arithmetical

operations are also generalized in the complex arithmetic and we consider the main operations overcomplex numbers.

Given complex numbers z1 = x1 + iy1 and z2 = x2 + iy2, the following properties for operationsof addition and multiplication are valid:

1. z1 + z2 = [x1 + iy1] + [x2 + iy2] = (x1 + x2) + i(y1 + y2),2. kz1 = k[x1 + iy1] = (kx1) + i(ky1)

for any real number k. One can note, that z1 +z2 = z2 +z1. The operation of subtraction is definedas z1 − z2 = z1 + (−1)z2,

3. z1 − z2 = [x1 + iy1] − [x2 + iy2] = (x1 − x2) + i(y1 − y2).

The most important operation of complex numbers is the multiplication, z = z1z2 which is

calculated directly as

z1z2 = [x1 + iy1][x2 + iy2] = x1x2 + ix1y2 + iy1x2 + i2y1y2.

Considering the definition, i2 = −1, we obtain the following:

4. z1z2 = (x1x2 − y1y2) + i(x1y2 + y1x2).

Thus, the multiplication of two complex numbers z1 and z2 is the complex number z = x + iy with

real and imaginary parts defined by x = x1x2 − y1y2 and y = x1y2 + y1x2, respectively. When thenumbers z1 and z2 are real, i.e., y1 = 0 and y2 = 0, this operation is reduced to the multiplicationof real numbers, z1z2 = x1x2. The set of complex numbers is denoted by C.

Example 6: If z1 = 1 + 2i and z2 = 2 − 3i, then the multiplication

z1z2 = [2 − 2(−3)] + i[−3 + 2(2)] = 8 + i.The operation of multiplication together with operations of addition and subtraction is commu-

tative, i.e.,

5. z2z1 = (x2x1 − y2y1) + i(x2y1 + y2x1) = z1z2.


It is not difficult to see that for any real numbers k1 and k2 and complex numbers z1, z2, and z3,the following holds:

5a. (k1z1)(k2z2) = (k1k2)(z1z2),

5b. z1(z2 + z3) = z1z2 + z1z3.

For a complex number z1 = x1 + iy1, the number z2 = x1 − iy1 is called the complex conjugate

and denoted by z1. It is clear that, ¯z1 = z1, for any complex number, and z1 = z1, if only the

number z1 is real. The operation of conjugation z → z is a linear operation, namely

6. z1 + kz2 = z1 + kz2,

for any real number k and complex numbers z1 and z2.The module |z1| of the complex number z1 is defined as the multiplication z1z1, which according

to multiplication can be written as

6a. z1z1 = (x1x1 + y1y1) + i(−x1y1 + y1x1) = x21 + y2

1

and denoted as |z1|2. Therefore, |z1| =√

x21 + y2

1 and it is positive if z1 6= 0, and |z1| ≥ |x1| and

|z1| ≥ |y1|.In the general case, the following holds for the complex conjugate of the multiplication:

6b. z1z2 = (x1x2 − y1y2) − i(x1y2 + y1x2) = z1z2.

The following equalities hold for the multiplication:

|z1z2|2 = (z1z2)(z1z2) = (z1z2)(z1z2) = (z1z1)(z2z2) = |z1|2|z2|2.

Therefore, the length of the product of two complex numbers equals the product of their lengths

6c. |z1z2| = |z1||z2|.

Example 7: For the complex number z1 = 3 + 4i, the length of the number is

|z1| =√

32 + 42 =√

9 + 16 = 5.

The complex conjugate number z1 = 3 − 4i has the same length,

|z1| =√

32 + (−4)2 =√

9 + 16 = 5.

The property of triangle inequality holds for the complex numbers

6d. |z1 + z2| ≤ |z1| + |z2|.

Th equality holds for the cases one of the complex numbers is zero, or the numbers are real andhave the same sign.

When z1 is an imaginary number, i.e. x1 = 0, the square of the number is z21 = (iy1)

2 = −y21 ,

i.e.,

6e. z21 = −|z1|2.


Example 8: For the complex number z1 = 4i, the length of the number is

|z1| =√

42 = 4, |z1|2 = 16, and z21 = (4i)2 = 16i2 = −16.

From equation z1z1 = |z1|2, the inverse number 1/z1 is defined as

7. (z1)−1 =

1

z1=

1

|z1|2z1, when z1 6= 0.

Indeed, one can verify that (z1)−1(z1) = (z1)(z1)

−1 = 1. Therefore, the operation of division of the

complex numbers z2 and z1, if only z1 6= 0, is defined as

8.z2

z1= z2(z1)

−1 = z21

z1=

1

|z1|2z2z1 =

z2z1

|z1|2.

Example 9: For the complex numbers z1 = 3 − 4i, and z2 = 2 + i, we have the following:

1

z1=

1

3 − 4i=

1

(32 + 42)(3− 4i) =

1

25(3 + 4i) =

3

25+ i

4

25

andz2

z1=

2 + i

3 − 4i=

1

25(2 + i)(3 + 4i) =

1

25(2 + 11i) =

2

25+ i

11

25.

In this complex arithmetic, we can solve equations (27)-(29). For that, we first consider thecomplex numbers z on the imaginary line, i.e., z = iy. The solution of equation z2 + 1 = 0, which

for such complex numbers is −y2 + 1 = 0, are z = i(+1) = i and z = i(−1) = −i. Figure 39 showsthe graph of the parabola z2 + 1 = (iy)2 + 1 in part a, when iy runs the interval of imaginarynumbers [−2i, 2i]. One can see this parabola intersects the horizontal at points +i and −i. Another

−2 −1 0 1 2−4

−3

−2

−1

0

1

2

×i

(a) z2+1

−2 −1 0 1 2−4

−3

−2

−1

0

1

2

×i+1

(b) (z−1)2+2

Fig. 39. Graphs of (a) the parabola z2 + 1 and (b) the parabola (z − 1)2 + 2.

parabola (z −1)2 +2 is also shown in part b. It should be noted, that the graph of this parabola iscalculated for the complex numbers z = 1 + iy when the imaginary components y runs the interval

[−2, 2]. Therefore, the plot is shown versus these complex numbers z = 1 + iy. This parabolaintersects the horizontal line in two points, z1 = 1 + i

√2 and z2 = 1 − i

√2.

A.1 Geometry of complex numbers

Every complex number z = x+ iy can uniquely be presented as the point (x, y) on the real plane

R2. In fact, it is another form of writing the complex number. We denote this point as P = P (z)


which has Cartesian coordinates (x, y). The writing, P = x + jy relates to the complex arithmetic.The distance between the original and P is d(P ) = |z| =

√

x2 + y2. So, the point P lies on the

circle with radius r = d(P ).In particular r = 1 case, when point P lies on the unit circle, we can write P as

P = P (θ) = (x, y) = (cos θ, sin θ) = cos θ + i sin θ

where the imaginary unite i2 = −1. Similarly, the point P = P (−θ) which represent the complex

conjugate number z can be written as

P = P (−θ) = (x,−y) = (cos θ,− sin θ) = cos θ − i sinθ.

Both points P and P are on the same circle of radius r = d(P ) = d(P ). Therefore,

P (θ) + P (−θ) = 2 cosθ,

P (θ) − P (−θ) = i2 sinθ.

The function P (θ) is denoted by eiθ in honor of Euler, who used this function and founded the

above relations for real and imaginary parts of eiθ,

eiθ + e−iθ = 2 cos θ, (30)

eiθ − e−iθ = i2 sinθ. (31)

The point on the unit circle is completely described by the angle θ. In the general case, when

point P lies on the circle of radius r, the point can be represented as (r, θ). It is the so-called polar

Fig. 40. Point P (r, θ) on the complex plane C.

form of representation of P (as shown in Fig. 40) instead of P = (x, y) = (r cos θ, r sin θ),

P = P (r, θ) = (r, θ).

Here the radius and angle are calculated as

r =√

x2 + y2, and θ = arctany

x= tan−1 y

x,


and θ = arccos(y) = ±π/2, if x = 0. This angle θ is also called the argument of z and denotedby θ = arg(z). If the point P lies in the left semi-plane, the calculation of the angle θ should be

corrected by adding the angle ±π depending on the sign of the imaginary part y.Example 10: The complex numbers z = 3 + 4i as point P (z) = (3, 4) is described in the polar

form as (r, θ) = (5, 0.9273). Indeed,

r =√

32 + 42 = 5 and θ = arctan4

3= 0.9273 (in radians),

or

θ =180

π× arctan

4

3= 53.1301 (in degrees).

The polar form for the complex conjugate z = 3 − 4i is P (z) = (5,−0.9273),

r =√

32 + (−4)2 = 5 and θ = arctan−4

3= −0.9273 (in radians),

or θ = −53.1301 in degrees.

B. Simple Differential Equation

The resistance-inductance (R-L) circuit shown in Fig. 41 is described by the following lineardifferential equation

Li′(t) + Ri(t) = 0 (32)

R

i(t)

L

Fig. 41. Symbol of the (R-L) circuit.

Let us consider the general differential equation

x′(t) = ax(t) (33)

The solution of this equation is founded in the exponential form

x(t) = Ceλt (34)

where the constants C and λ are defined from (32), because the derivative of the exponent is alsoexponent (et)′ = et. Therefore,

x′(t) =(

Ceλt)′

= Cλeλt = λ(Ceλt) = λx(t)


which yields λ = a and the solution is

x(t) = Ceat, C = x(0). (35)

In the R-L circuit case, when a = −R/L, the solution is

i(t) = CeR/Lt, C = i(0). (36)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

g(x) =

an; if n ≥ 00; if n < 0

0 100 200 300 400 500 600 700 800 900 10000

5

10

15

20

25Exponential funtion with a>1

Mag

nit

ude

0 20 40 60 80 100 1200

0.2

0.4

0.6

0.8

1Exponential funtion with a<1

Mag

nit

ude

a)

b)

Fig. 42. The exponential sequences g(n), for (a) a = 0.5 and (b) a = 1.025.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Remark 1: The polar form is used (only) for presentation of points on the plane

(x, y) = (r cosϑ, r sinϑ) → (r, ϑ)

x(t) = Ceαt = Cejωt, α = jω

ejωt = cos(ωt) + j sin(ωt), j2 = −1,

only if ω is real.


B.1 Complex exponential signal with complex amplitude

x(t) = Cejω0t, C = ejφA

where φ, ω0, and A are real constants. This complex signal can be written as

x(t) = Aejφejω0t = Aej(ω0t+φ)

x(t) = (A cos(ω0t + φ), A sin(ω0t + φ))

B.2 Complex exponential signal with exponential amplitude

We consider the complex signal with amplitude to be an exponential function of time

x(t) = Cejω0t, C = C(t) = A(t)ejφ, A(t) = A0eσ0t.

The signal x(t) has the form x(t) = A0eσ0tej(ω0t+φ) and can be written as (see Fig. 43)

x(t) =(

Aeσ0t cos(ω0t + φ), Aeσ0t sin(ω0t + φ))

0 5 10 15 20 25−0.5

0

0.5Complex exponential signal

mag

nit

ude

t

Aeσ

0tcos(ω

0t+φ)

0 5 10 15 20 25−0.5

0

0.5

mag

nit

ude

t

a)

b)

Aeσ

0tsin(ω

0t+φ)

Fig. 43. Two complex exponential signals: cosine and sine waves in the exponential envelop.


VII. Unit Step Function

The unit step signal (or the Heaviside function) is defined as

u(t) =

1, t ≥ 0

0, otherwise.(37)

Figure 44 shows this function in the interval [−2, 5].

−2 −1 0 1 2 3 4 5−1

−0.5

0

0.5

1

1.5

2

unit step function

mag

nit

ude

u(t)

t

Fig. 44. The unit step function.

Given t0, the time-shifted unit step signal is

ut0(t) =

1, t ≥ t00, t < 0.

After the time transformationt → at − t0, a 6= 0,

the unit step signal changes as

u(t) → u

(

t −t0a

)

, a > 0

→ u

(

t0a− t

)

, a < 0.

Indeed,1

a) if a > 0

u(at − t0) = u

(

a

[

t −t0a

])

=

1, t −t0a

≥ 0

0, otherwise= u

(

t −t0a

)

.

b) if a < 0

u(at − t0) = u

(

a

[

t −t0a

])

1Correction to (2.34) in the textbook


=

1, t −t0a

≤ 0

0, otherwise

= u

(

−

(

t −t0a

))

= u

(

t0a

− t

)

.

A. Useful Properties of Unit function

a) switching −1/ + 1

x(t) = 2u(t)− 1

which is shown in Fig. 45.

−2 −1 0 1 2 3 4 5−2

−1

0

1

2

3

switching signal

mag

nit

ude

2u(t)−1

t

Fig. 45. The switching signal 2u(t) − 1.

Example 1:

y(t) = x(t) sin(t)

is a symmetric even function (see Fig. 46).

−10 −5 0 5 10

−1

−0.5

0

0.5

1

x(t)=sin(t)

t

−10 −5 0 5 10

−1

−0.5

0

0.5

1

y(t)=x(t)[2u(t)−1]

Fig. 46. The signal x(t) sin(t).


b) Unit pulse signal in [T1, T2] :Let us consider two translations (time-shift transforms) of the unit step function

uT1(t) = u(t − T1),

uT2(t) = u(t − T2).

As an example, Figure 47 shows two such shifts for T1 = −1 and T2 = 2.

−2 0 2 4−1

−0.5

0

0.5

1

1.5

2

mag

nit

ud

e

(a)−2 0 2 4

−1

−0.5

0

0.5

1

1.5

2

(b)

u(t+1)u(t−2)

tt

Fig. 47. The time-shifting transforms of the unit step function.

Assuming T1 < T2, we define

y(t) = uT1(t) − uT2

(t) =

1, t ∈ [T1, T2]0, otherwise.

as shown in Fig. 48, for T1 = −1 and T2 = 2.

−2 0 2 4−1

−0.5

0

0.5

1

1.5

2

mag

nit

ude

Fig. 48. The difference of two unit step functions (non symmetric rectangle).

c) Unit rectangular pulse is defined for the case when T1 = −T2 = −0.5,

rect(t) = u−

1

2

(t) − u 1

2

(t)

=

1, t ∈

[

−1

2,1

2

]

0, otherwise.

The unit rectangular pulse in the symmetric interval [−T/2, T/2] is defined as

rect

(

t

T

)

=

1, −1

2≤

t

T≤

1

20, otherwise


rect

(

t

T

)

=

1, −T

2≤ t ≤

T

20, otherwise

= u−

T

2

(t) − uT

2

(t).

d) Sum of shifted unit step signals.

Given T > 0, the unit step signal on the interval [0, T/2] is defined as

u0, T

2

(t) = u0(t) − uT

2

(t)

and the sum of shifted step signals

u0, T

2

(t) + u0, T

2

(t − T ) + u0, T

2

(t − 2T ) + ... + u0, T

2

(t − nT )

is defined (n + 1) unit step signals.The periodic parts of the considered above half-wave rectified signal can be defined as

v1(t) = vm sin(ω0t)u0, T

2

(t), v1(t − T ), v1(t − 2T ), ...,

where T = T0 = 2π/ω0.

0 5 10 15 20 25

−1

−0.5

0

0.5

1

x(t)=sin(wt), w=3/2.

t

0 5 10 15 20 25

0

0.5

1

sum of unit step signals

t

0 5 10 15 20 25−1

−0.5

0

0.5

1

1.5half−wave rectified signal

t

T 2T 3T 4T 5T

T/2 T+T/2 2T+T/2 3T+T/2 4T+T/2 5T/2

T/2 T

Fig. 49. The half-wave rectified signal.


B. The Unite Impulse Function (δ-function)

The delta function, δ(t), is a very important mathematical concept which is widely used inengineering. This function allows us to represent and describe many properties of physical systems

in digital signal processing, including the filtration and restoration of signals.Definition 1: δ(t) function (or Dirac delta function) is a generalized function, namely a functional

which operates over functions continuous at point 0, x(t), in the following way:

δ : x → x(0), (38)

i.e., δ[x] = x(0), and this functional can be represented in the integral form as

δ[x] =

∞∫

−∞

x(t)δ(t)dt,

where δ(t) is a function of t.

Thus, by the definition, the δ(t) function is a function such that

∞∫

−∞

x(t)δ(t)dt = x(0), (39)

if the function x(t) is continuous at the point t = 0.

The property of the δ function:∞∫

−∞

δ(t)dt = 1. (40)

There are many functions gn(t), n = 1, 2, 3, ..., exist such that the following convergence takesplace

limn→∞

∞∫

−∞

x(t)gn(t)dt = x(0). (41)

In other words, we can say that gn(t) → δ(t). Sequence gn(t) is called delta-sequence.Example 1:

gn(t) =

2n−1, if t ∈[

− 12n , 1

2n

]

0, otherwise.(42)

Figure ?? shows the first six functions gn(t).Example 2: We also can consider the sequence of Gaussian functions

gn(t) =1

σn

√2π

e− t

2

2σ2n , t ∈ (−∞, +∞), (43)

for any sequence σn → 0. For instance, σ1 = 2, σ1 = 1, σ1 = 0.5, σ1 = 0.25, ... .

For all these functions, the integrals

∞∫

−∞

gn(t)dt = 1 (44)


−0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

0

5

10

15

20

25

30

mag

nit

ud

e

g1(t)

g2(t)

g3(t)

g4(t)

g5(t)

g6(t)

∫gn(t)=1

t

Fig. 50. Delta sequence of functions gn(t) → δ(t), as n → ∞.

and (??) holds, since the following known result (Theorem of Mean) can be used

∞∫

−∞

x(t)gn(t)dt = x(tn)

for certain points tn ∈ [−1/2n, 1/2n] which approach to 0, when n → ∞.

B.1 Properties of the δ-function

+∞∫

−∞

f(t)δ(t)dt = f(0).

1. Given t0 6= 0, the shifting δ(t − t0) is defined as follows

∫ +∞

−∞f(t)δ(t − t0)dt = (t − t0 → t) =

∫ +∞

−∞f(t + t0)δ(t)dt

[g(t) = f(t + t0)]

=

∫ +∞

−∞g(t)δ(t)dt = g(0) = f(t0).

So, if the function f(t) is continuous at t0, then

∫ +∞

−∞f(t)δ(t − t0)dt = f(t0).


2. Given t0,f(t)δ(t − t0) = f(t0)δ(t − t0).

To prove this equation, we have to use the correct (full) definition of the δ-function as a generalized

function, i.e., functional - operation over the functions,

δ : f → f(0), δ[f ] = f(0).

Given a function g(t), t ∈ R,

∫ +∞

−∞g(t)f(t)δ(t− t0)dt =

∫ +∞

−∞g(t)f(t)δ(t − t0)dt

[c(t) = g(t)f(t)]∫ +∞

−∞c(t)δ(t− t0)dt = c(t0) = g(t0)f(t0)

=

∫ +∞

−∞g(t)δ(t− t0)dt · f(t0)

=

∫ +∞

−∞g(t)f(t0)δ(t − t0)dt

So, for any function g(t) which is continuous at point t0,

∫ +∞

−∞g(t)f(t)δ(t− t0)dt =

∫ +∞

−∞g(t)f(t0)δ(t − t0)dt

which results in the equalityf(t)δ(t − t0) = f(t0)δ(t − t0).

3. Given t0 6= 0,∫ t

−∞δ(τ − t0)dτ =

∫ +∞

−∞f(τ)δ(τ − t0)dτ

(

f(τ) =

1, τ ≤ t

0, τ > t= χ[−∞,t](τ)

)

= f(t0) =

1, t0 ≤ t0, t0 > t

= u(t − t0)

So,

∫ t

−∞δ(τ − t0)dτ = u(t − t0).

4. The time transformation t → at − t0 of the delta function results

δ(at − t0) =1

|a|δ(

t − t0a

)

, a 6= 0, (45)

which means that∫ +∞

−∞f(t)δ(at − t0)dt =

1

|a|

∫ +∞

−∞f(t)δ

(

t − t0a

)

dt


Indeed,

∫ +∞

−∞f(t)δ(at − t0)dt = (at − t0 → t′, adt → dt′)

(t =t′

a+

t0a

, dt =1

adt′)

=1

a

∫ ∓∞

±∞f

(

t′

a+

t0a

)

δ(t′)dt′

=1

|a|

∫ +∞

−∞f

(

t′

a+

t0a

)

δ(t′)dt′

=1

|a|f(

0

a+

t0a

)

=1

|a|f(

t0a

)

=1

|a|

∫ +∞

−∞f(t)δ(t − t0

a)dt

=

∫ +∞

−∞f(t)

1

|a|δ(t −t0a

)dt

5. Symmetry

δ(−t) = δ(t) [as functional]

Indeed, the time reversal transformation t → −t is a particular case of the time transformationt → at − t0 when a = −1 and t0 = 0. Therefore, by property (??)

δ(−t) = δ(−1 · t − 0) =1

| − 1|δ(

t − 0

a

)

= δ(t).

In other words, for any function f(t) which is continuous at 0

+∞∫

−∞

f(t)δ(−t)dt =

+∞∫

−∞

f(t)δ(t)dt = f(0).

6. Time-scaling

δ(at) =1

|a|δ(t).

By substituting t0 = 0, we obtain from property (??)

δ(at) = δ(at − 0) =1

|a|δ(

t − 0

a

)

=1

|a|δ(t).

In other words, for any function f(t) which is continuous at 0

+∞∫

−∞

f(t)δ(at)dt =1

|a|

+∞∫

−∞

f(t)δ(t)dt =1

|a|f(0).


Example 3:Find the value of the following integral:

+∞∫

−∞

[cos2(t2 + π/4) + 1]δ(t)dt.

Answer:

+∞∫

−∞

[cos2(t2 +π/4)+1]δ(t)dt = [cos2(t2 +π/4)+1]|t=0 = cos2(π/4)+1 =

(

1√2

)2

+1 =1

2+1 = 1.5.

Example 4:Find the value of the following integral:

+∞∫

−∞

[cos2(t2 + π/4) + 1]δ(2t − 1

4)dt.

Answer:

+∞∫

−∞

[cos2(t2 + π/4) + 1]δ(2t− 1

4)dt =

+∞∫

−∞

[cos2(t2 + π/4) + 1]1

2δ(t − 1

8)dt

=1

2[cos2(t2 + π/4) + 1]|t= 1

8

=1

2cos2(

1

64+ π/4) + 1.

Example 5:

Evaluate the following integral:

+∞∫

−∞

[sin(t − π/3)− 1] δ(3t− π

3)dt.

Answer:

+∞∫

−∞

[

sin(t − π

3)− 1

]

1

3δ(t−π

9)dt =

1

3

[

sin(t − π

3) − 1

]

|t=π

9

=1

3

[

sin(−2π

9) − 1

]

= −1

3

[

sin(2π

9) + 1

]

.

Example 6:

Evaluate the following integral:+∞∫

−∞

e−2|t|δ(t − 2)dt.

Answer:+∞∫

−∞

e−2|t|δ(t − 2)dt = e−2|t||t=2 = e−4.


VIII. Systems

A signal is a function that represents the time (or coordinate) variation of a physical variable.When processing signals x(t), we use the general concept of systems, which describe many physical

systems such as a device, algorithm, filter. A system is a relationship, T, between two signals, x(t)and y(t), and can be defined as follows. A system generates a response (output signal y(t)) for a

given input signal x(t). The input represents a physical process that is generated independentlyfrom the system. The output is generated by the system when the input is present.

A system is represented by the mathematical symbol as (see Fig. 46)

y(t) = (T [x])(t), ∀t ∈ R (or t ∈ [a, b]),

which often is written as y(t) = T [x(t)].

Tinput x(t) output y(t)

T:x(t) −−> y(t)

Fig. 46. Diagram of the system.

The process of deriving a system representation is called modeling.A system which describe a relationship between continuous-time input x(t) signals and continuous-

time output y(t) signals is called a continuous-time system.A system which describe a relationship between discrete-time input x(t) signals and discrete-time

output y(t) signals is called a discrete-time system.If a system has one input and one output signal, then we call the system is a single-input-single-

output (SISO) system. If a system has more than one input and/or output signal, then we call thesystem is a multi-input-multi-output (MIMO) system.

We have already considered the following simple systems:

T : x(t) → x(−t), ∀t,T : x(t) → x(t − t0), t0 6= 0, (ideal time delay)

T : x(t) → x(at), a 6= 0, (“compression”)T : x(t) → Ax(t), A 6= 0, (ideal amplifier).

In the A = 1 case, the system T : x(t) → x(t) is called the identity system.

We consider the important case, when a system H is described by the linear convolution whichis calculated by the following integral

y(t) =

∞∫

−∞

h(t − τ)x(τ)dτ, t ∈ R. (46)

h(t) is the impulse response function of the system H. When the input signal is the ideal infiniteimpulse, i.e., ”delta” function δ(t), or unit impulse function the output is

y(t) =

∞∫

−∞

h(t − τ)δ(τ)dτ =

∞∫

−∞

h(τ)δ(t − τ)dτ = h(t). (47)


So, the impulse characteristics h(t) of the system H can be determined as the output of the signalwhen input is the unit impulse function (ideal infinite impulse).

It is common to write (46) in the short form as

y = h ∗ x, (y(t) = (h ∗ x)(t) = h(t) ∗ x(t).

This operation is commutative, h ∗ x = y = x ∗ h. The system H is also linear, which means that

the following two conditions are satisfied for any inputs x1(t) and x2(t) :

H [x1 + x2] = H [x1] + H [x2],

H [Ax1] = AH [x1], for any constant A.(48)

Indeed, taking any two functions x1(t) and x2(t), we can write the following for the linear

convolution:x1(t) → h(t) ∗ x1(t), x2(t) → h(t) ∗ x2(t)

which implies

∞∫

−∞

h(t − τ)[x1(τ) + x2(τ)]dτ =

∞∫

−∞

h(t − τ)x1(τ)dτ +

∞∫

−∞

h(t − τ)x2(τ)dτ

or,(x1(t) + x2(t) → h(t) ∗ (x1(t) + x2(t)) = h(t) ∗ x1(t) + h(t) ∗ x2(t).

Comments: In the discrete-time signal case, the linear convolution of the signal x(n) withsequence h(n) is defined as

y(n) =∞∑

m=−∞

h(n − m)x(m), n ∈ Z = 0,±1,±2, . . .,

=M∑

m=−M

h(m)x(n − m).

(49)

When h(n) = 1/(2M + 1), for n = −M : M, and h(n) = 0 for other points, the linear convolutionresults in the local means of the input signals in window W = −M, ...,−1, 0, 1, ...,M,

y(n) = xmean,W (n) =M∑

m=−M

h(m)x(n− m), n ∈ Z = 0,±1,±2, . . .. (50)

For instance, let the impulse characteristic of the system be is h(m) = 1/3(1, 1, 1). This is so-calledthe mean filter with window 3, for which the output is calculated by

y(n) =1

3[x(n − 1) + x(n) + x(n + 1)].

As an example, Figure 47 shows the original signal degraded by a random impulse noise and theresult of the linear convolution (mean filter). For comparison, the result of filtration with nonlinearmedian filter is also given. In general, the mean filter provides better filtration in the terms of the

root mean square error (RME), not the mean-absolute error (MAE).


0 2 4 6 8 10 12−4

−2

0

2

4m

ag

nitu

de

original signal y(t)

Signal processing of the degraded signal

0 2 4 6 8 10 12−4

−2

0

2

4

Ma

gn

itu

de signal+ impulse noise

0 2 4 6 8 10 12−4

−2

0

2

4

Convolution of the signal

ma

gn

itu

de

RMS error 0.0143

MAE error 0.1013

0 2 4 6 8 10 12−4

−2

0

2

4

Median filtering

Time in secs

Ma

gn

itu

de

RMS error 0.0175

MAE error 0.0795

t

window 3

Fig. 47. Linear convolution of the degraded signal by mean filter with the symmetric window of length 3.

We can also consider other impulse responses. For instance, when h(m) = 1/5(1, 1, 1, 1, 1), theconvolution is calculated by

y(n) =1

5[x(n− 2) + x(n − 1) + x(n) + x(n + 1) + x(n + 2)],

and, in the h(m) = 1/9(1, 2, 3, 2, 1) case, the output is calculated by

y(n) =1

9[x(n− 2) + 2x(n− 1) + 3x(n) + 2x(n + 1) + x(n + 2)],

for n = 0,±1,±2, . . . .

The basic connections for systems are parallel and cascade connections which are shown in Fig-ures 48 and 49:

T1 : x(t) → y(t), T2 : y(t) → z(t), z = T2[T1[x]] = (T2 T1)[x],

T1 : x(t) → y1(t), T2 : x(t) → y2(t), (T1 + T2) : x(t) → (y1 + y2)(t).

A. Properties of systems

1. A system is called memoryless, if an output y(t) at any time t depends only on input of the

signal x(t) at the same time: (T [x])(t) = T (x(t)). Example: y(t) = Ax(t) + B. But the system


T1

x(t) y(t)T2

z(t)

T1oT

2

Fig. 48. Cascade-wise connections of two systems.

T1

x(t)

y2(t)

T2

y1(t)

T1+T

2

y1(t)+y

2(t)

+

Fig. 49. Two systems are connected in parallel.

y(t) = x(at − b) for a 6= 1 is not memoryless, as well as the system y(t) = x(t) − x(t − 1); we say

such systems have memory.

2. A system is called invertible, if any inputs x1(t) 6= x2(t) result in outputs y1(t) 6= y2(t).

3. A system T1 is called inverse to a system T , if T1 T = I, which means that T1 : y(t) → x(t),when T : x(t) → y(t), for any input signals x(t). The inverse system is denoted by T−1. For example,if T : x(t) → y(t) = x(t − 1), then the system T−1 : y(t) → y(t + 1), because y(t + 1) = x(t) for

all t. So T−1 : y(t) → x(t).

4. A system is called causal, if output y(t0) at any time t0 depends only on input of the signal x(t)

at time t ≤ t0. For example, y(t) = x(t)−x(t−1), but the system with outputs y(t) = x(t+1)−x(t)is not causal.

5. A system is called bounded-input bounded-output (BIBO) stable, if exists a real number B > 0such that

|y(t)| ≤ B, when |x(t)| ≤ A,

and A > 0 is a real number. This conditions should hold for any bounded input of the system.

6. A system is called time invariant (TI), if xt0(t) → yt0(t) for any time shift t0, when x(t) → y(t),i.e.

if x(t) → y(t), ∀t ∈ R, then x(t − t0) → y(t − t0).

For example, relationship y(t) = x(at) determines a TI system T : x(t) → y(t), but

T : x(t) → y(t) = e−tx(t), t ∈ R,

is not TI system (take x(t) = sin(t)). Indeed, for this system we can write the following

T : x(t − t0) → y1(t) = e−tx(t − t0), t ∈ R,


but, when t0 6= 0,

y(t − t0) = e−t+t0x(t − t0) = et0 [e−tx(t − t0)] = et0y1(t) 6= y1(t),

which shows thatwhen x(t) → y(t), then x(t − t0) 6→ y(t − t0).

In the case when a system is not TI, we say the system is time varying.Example 1: Consider the following system:

T : x(t) → y(t) = tx(t) + 1.

This system is not TI. To show that, consider the following response

T : x(t − 1) → tx(t − 1) + 1

and that is not y(t − 1) = (t − 1)x(t − 1) + 1.

This system is:

1. not linear2. not invertible

3. not inverse4. has memory

5. causal6. not stable (not BIBO).

EE-3424, MATH IN SIGNALS AND SYSTEMS, ART GRIGORYAN 51

IX. Impulse Representation of Signals

In this section, we consider main properties of linear time-invariant systems. By definition ofthe delta function, δ(t), every continuous function (continuous-time continuous signal) x(t) can be

represented in the integral form as

x(t) =

+∞∫

−∞

δ(t − τ)x(τ)dτ =

+∞∫

−∞

x(t − τ)δ(τ)dτ

=

+∞∫

−∞

δ(τ − t)x(τ)dτ

= (δ ∗ x)(t) = δ(t) ∗ x(t).

(51)

This identity convolution system is shown in Fig. 50.

Iinput x(t) output x(t)

T:x(t)−−>x(t)*δ(t)

Fig. 50. Diagram of the identity system.

In general when a system, H, is linear and time-invariant, the output (response) of the systemwhen input is x(t) is described in the form of the linear convolution of the input with a function,which is called the impulse response function h(t) :

H : x(t) →h(t) y(t) =

+∞∫

−∞

h(t − τ)x(τ)dτ =

+∞∫

−∞

x(t − τ)h(τ)dτ

6=

+∞∫

−∞

h(τ − t)x(τ)dτ (if h(τ) is not symmetric)

= (h ∗ x)(t) = h(t) ∗ x(t) = x(t) ∗ h(t)

and therefore, H : δ(t) → h(t) (see Fig. 51). The impulse response function is the response of the

Hinput x(t) output y(t)=x(t)*h(t)

δ(t) h(t)

Fig. 51. Diagram of the LTI system with the impulse response h(t).

system on the input being δ(t). The identity system of Fig. 50 has the impulse response function

h(t) = δ(t).

Example 1 (Integrator) Consider a causal system H which is described as

H : x(t) → y(t) =

t∫

−∞

x(τ)dτ, t ∈ (−∞, +∞).


H is linear time invariant system, since for any given t0, we have the following

H : x(t − t0) →

t∫

−∞

x(τ − t0)dτ =

t−t0∫

−∞

x(τ ′)dτ ′

butt−t0∫

−∞

x(τ ′)dτ ′ = y(t − t0).

So, H : x(t − t0) → y(t − t0) when x(t) → y(t), for any t0.

Example 2 (Impulse and Signal) Consider the linear convolution system H with the impulse re-sponse h(t). Let y(t) be the response of the system on the input signal

x(t) = δ(t + 3) + 3e−1

2tu(t)

where u(t) is the unit function. Since the system is linear, we can consider the input as the sum oftwo signals as follows:

x(t) = (x1(t) = δ(t + 3)) +(

x2(t) = 3e−1

2tu(t)

)

and, then, compute two outputsH : x1(t) → y1(t)H : x2(t) → y2(t).

The sum y1(t) + y2(t) is the output of the system when input is x(t), i.e.,

H : x(t) → H [x1(t) + x2(t)] = H [x1(t)] + H [x2(t)] = y1(t) + y2(t) = y(t).

It is clear, that

y1(t) =

+∞∫

−∞

h(t − τ)x1(τ)dτ =

+∞∫

−∞

h(t − τ)δ(τ + 3)dτ = h(t − τ)|τ=−3= h(t + 3). (52)

Let h(t) be the rectangle u(t)−u(t−2) on the interval [0, 2]. To compute the output y2(t), we first

note that the area A of Fig. 52 is

A = A(t) = 2(1− e−1

2t).

Indeed+∞∫

0

e−tdt = 1, therefore

+∞∫

0

e−1

2tdt = 2

+∞∫

0

e−tdt = 2.

We now write the convolution in the form

+∞∫

−∞

h(t − τ)x2(τ)dτ =

+∞∫

−∞

h′(τ − t)x2(τ)dτ


e−τ/2

τ t 0

1

A(t)

area

τ

1

h(τ)

0 2

Fig. 52. The real exponential signal e−1

2t.

where h′(τ) = h(−τ).1. For the case when 0 ∈ [t − 2, t], i.e., t ∈ [0, 2], we obtain

y2(t) =

+∞∫

−∞

h′(τ − t)x2(τ)dτ =

t∫

0

3e−1

2τdτ = 3A(t) = 6(1− e−

1

2t).

A(t)

A(t)−A(t−2)

h’(τ+t)

h’(τ−t)

τ

τ

e−τ/2

e−τ/2

t 0

0

t−2

t−2 t

1

1 h’(τ)=h(−τ)

τ 0 −2

1

Fig. 53. The convolution of (x2 ∗ h′)(t).


2. For the case when t > 2, we obtain

y2(t) =

+∞∫

−∞

h′(τ − t)x2(τ)dτ =

t∫

t−2

h′(τ − t)x2(τ)dτ

= A(t) − A(t − 2) = 6(1 − e−1

2t)− 6(1− e−

1

2(t−2)) = 6(e−

1

2(t−2) − e−

1

2t) = 6(e − 1)e−

1

2t.

Therefore, the output of the linear system is

y(t) = y1(t) = [u(t + 3) − u(t + 1)]

+ y2(t) = [u(t) − u(t − 2)]6(1− e−1

2t) + [u(t − 2)]6(e− 1)e−

1

2t.

A. Properties of the Linear Convolution

The following properties of the linear convolution follow directly from the definition of the linear

convolution1. x ∗ h = h ∗ x

2. x ∗ (h1 + h2) = x ∗ h1 + x ∗ h2

3. (x1 + x2) ∗ h = x1 ∗ h + x2 ∗ h

4. (x ∗ h1) ∗ h2 = x ∗ (h1 ∗ h2) = x ∗ (h2 ∗ h1) = (x ∗ h2) ∗ h1

5. x ∗ (kh) = k(x ∗ h) = (kx) ∗ h, k ∈ R.

A.1 Causality

The linear convolution can be written as follows:

y(t) =

+∞∫

−∞

x(t − τ)h(τ)dτ =

+∞∫

0

x(t − τ)h(τ)dτ +

0∫

−∞

x(t − τ)h(τ)dτ

=

+∞∫

0

x(t − τ)h(τ)dτ + 0 (”for system H to be causal”)

=

t∫

−∞

x(τ)h(t − τ)dτ

The second integral should be zero for all input signals x(t), if the system H is causal, since theoutput y(t) is a function of x(t− τ); τ ≥ 0. Therefore, the impulse response function h(−τ) = 0,

for all τ > 0. This is the only condition for a linear system to be causal.

A.2 Stability

In a stable linear system H, every bounded input results in a bounded output, i.e., if

|x(t)| ≤ M, ∀t ∈ R

then, ∃N > 0 such that

|y(t)| ≤ N, ∀t ∈ R, if H : x(t) → y(t).


To show the sufficiency of this condition for stability of the system, let us take the followingfunction

x(t) =

+1; if h(−t) ≥ 0,

−1; if h(−t) < 0.

Then, |x(t)| ≡ 1, i.e., x(t) is bounded, and at point t = 0 we obtain

+∞∫

−∞

x(t − τ)h(τ)dτ =

+∞∫

−∞

x(−τ)h(τ)dτ =

+∞∫

−∞

|h(τ)|dτ

In order that L to be stable, the last integral should exist, i.e.,

+∞∫

−∞

|h(τ)|dτ = B < ∞.

This is the necessary and also sufficient condition for stability of the system. Indeed, if suchcondition takes place, then, for a bounded function |x(t)| ≤ M,

|y(t)| =

∣

∣

∣

∣

∣

∣

+∞∫

−∞

x(τ − t)h(τ)dτ |

∣

∣

∣

∣

∣

∣

≤

+∞∫

0

|x(τ − t)||h(τ)|dτ ≤ M

+∞∫

0

|h(τ)|dτ < MB, ∀ t.

Example 3: Consider the impulse response function h(t) = e−atu(t), a > 0. Then

+∞∫

−∞

|h(t)|dt =

+∞∫

0

h(t)dt =1

a

+∞∫

0

e−atd(at) = −1

ae−t

∣

∣

∣

∞

0=

1

a

and the system is stable.

In the case, when the impulse response function h(t) = eatu(t), a > 0, we obtain

+∞∫

−∞

|h(t)|dt =

+∞∫

0

h(t)dt =1

aet

∣

∣

∣

∞

0= ∞

and the system is not stable.

0 1 2 30

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 30

50

100

150

200

250

300

350

400

450

e−αt

u(t) e−αt

u(t)

α>0α<0

Fig. 54. Two exponential signals eatu(t), when a > 0 and a < 0.


X. LTI Systems and Differential Equations

Many physical systems as linear time-invariant systems are defined by differential equations withconstant coefficients

dny(t)

dtn+ a1

dn−1y(t)

dtn−1+ a2

dn−2y(t)

dtn−2+ · · ·+ any(t) = b0

dmx(t)

dtm+ b1

dm−1x(t)

dtm−1+ · · ·+ bnx(t),

which can also be written as

Ly(1, a1, a2, . . . , an) = Lx(b0, b1, b2, . . . , bm). (53)

Here, integers n ≥ m, and ak, k = 1 : n and bk, k = 0 : m, are constant coefficients; n is the orderof the differential equation.

What we need to know:1. The above equation can be considered as

Ly(1, a1, a2, . . . , an) = Lx(b0, b1, b2, . . . , bm) + 0.

Therefore, the solution of this equation y(t) can be represented as the solution y0(t) of the homo-geneous equation

Ly(1, a1, a2, . . . , an) = 0,

and single solution y1 of the equation

Ly(1, a1, a2, . . . , an) = Lx(b0, b1, b2, . . . , bm).

The general form of the solution of the equation in (53) is

y(t) = y1(t) + y0(t).

To make this solution unique, we need to know n initial values (conditions), for instance values

y(0), y(1)(0), y(2)(0), . . . , y(n−1)(0)

at the point t = 0 or any other point t = t0.

2. For the solution y0(t) of the homogeneous equation

Ly(1, a1, a2, . . . , an) = 0,

we study the corresponding characteristic polynomial

P (λ) = λn + a1λn−1 + a2λ

n−2 + · · ·+ an−1λ + an

and solve the characteristic equation P (λ) = 0. Assuming all roots of this equations are different,

λ1, λ2, . . . , λn, the solution of the homogeneous equation can be written as

y0(t) = C1eλ1t + C2e

λ2t + · · ·+ Cneλnt,

where the coefficients Ck are found from the initial conditions (they should be given).


3. If one of the roots, let say λ1 is complex, then the complex conjugate λ1 is also the root ofthe characteristic polynomial, i.e., P (λ1) = 0. So, λ1 is one of the roots, let say λ3.

It also means that if the equation is real (or system is real, i.e., real input → real output), thenthe solution y0(t) contains the cosine wave,

C1eλ1t + C3e

λ3t = C1eλ1t + C3e

λ1t

and C3 = C1 (since the system is real), therefore,

C1eλ1t + C3e

λ3t = C1eλ1t + C1e

λ1t = C1eλ1t + C1eλ1t.

C1 and λ1 in general is are complex numbers, C1 = |C1|eiϕ and λ1 = σ1 + iω1. Therefore, we can

continue the above equation and write

C1eλ1t + C1eλ1t = |C1|e

iϕe(σ1+iω1)t + |C1|e−iϕe(σ1−iω1)t = 2|C1|e

σ1t cos(ω1t + ϕ).

A. First-order LTI systems

We consider linear time-invariant systems, which are defined by the simple differential equation

dy(t)

dt− ay(t) = x(t), a > 0,

that describe RL circuits (y(t) = i(t), x(t) = v(t)).Assuming the input signal is constant, the above equation takes the form

dy(t)

dt+ ay(t) = const, ∀t. (54)

It is clear that the constant function y0(t) = const/a is a solution (named a particular solution) of

this equation. Therefore, we find the solution of the differential equation

dy1(t)

dt+ ay1(t) = 0. (55)

and define the solution of Eq. 54 as y(t) = y1(t) + y0(t).We will find a solution of Eq. 55 in the form of the exponential function,

y1(t) = Ce−st

where s is real number.Substituting y(t) in Eq. 55, we obtain

−sCe−st + aCe−st = 0, ∀t ∈ (−∞, +∞),

which shows that s = a, and the solution

y(t) = Ce−at + const/a.

The value of C can be found from y(0),

y(0) = Ce−a0 + const/a = C + const/a,

or C = y(0)− const/a.


B. General solution of the 1st order differential equation

In this section, we drive the analytical formula for the solution of the differential equation differ-ential equation

dy(t)

dt+ ay(t) = x(t), (56)

with a constant coefficient a and initial value y(t0).1

For that, first we define the function

f(t) = eaty(t) (57)

and consider the derivative of this (unknown) function

df(t)

dt= eat dy(t)

dt+ aeaty(t) = eat

[

dy(t)

dt+ ay(t)

]

= eatx(t).

Therefore, we can write the following (when t ≥ t0):

f(t) =

t∫

−∞

eaτx(τ)dτ = f(t0) +

t∫

t0

eaτx(τ)dτ.

Also,f(t0) = eat0y(t0)

and

f(t) = eat0y(t0) +

t∫

t0

eaτx(τ)dτ.

From Equation 57 it follows that

y(t) = e−atf(t) = e−at

eat0y(t0) +

t∫

t0

eaτx(τ)dτ

= e−a(t−t0)y(t0) +

t∫

t0

e−a(t−τ )x(τ)dτ

or

y(t) = e−a(t−t0)y(t0) + e−at

t∫

t0

eaτx(τ)dτ. (58)

This is the general solution of the differential equation (57).

Example 1: Consider the RC circuit between two notes across which the voltage potential is x(t).

Let R be of 1Ω and C be of 1/4F. The voltage across the capacitor is described by the followingdifferential equation:

RCdy(t)

dt+ y(t) = x(t)

ordy(t)

dt+

1

RCy(t) =

1

RCx(t)

1You can omit the proof and consider only Eq. 58


We consider that the voltage x(t) = 2V, namely x(t) = 2u(T ) and the initial value at t0 = 0 isy(0) = 0.

As follows from (58), the following calculations hold for the solution of th RC system:

a = 1/RC, and x(t) = x(t)/(RC) = 2u(t)/(RC)

and (when t ≥ 0)

y(t) = e−a(t−0)0 + e−at

t∫

0

eaτ2u(τ)dτ = e−1

RCt

t∫

0

e1

RCτ2u(τ)

1

RCdτ

= 2u(t)e−1

RCt

t∫

0

de1

RCτ = 2u(t)e−

1

RCt[

e1

RCt − 1

]

= 2u(t)[

1− e−1

RCt]

.

The time constant RC = 1/4 and 1/(RC) = 4, and the final answer for the solution is

y(t) = 2[

1− e−4t]

u(t).

C. Transfer function for the 1-O LTIS

We now define a concept of a transfer function for the linear system H that is defined by thedifferential equation

dy(t)

dt− ay(t) = x(t), a > 0, (59)

Given a value s ∈ R, we consider the input x(t) as a exponential function

x(t) = xs(t) = Xest, t ∈ (−∞, +∞), X 6= 0,

and output of the systemy(t) = ys(t) ← xs(t) = x(t).

We now consider the output y(t) in the form

y(t) = ys(t) = H[xs(t)] = H(s)xs(t) = H(s)x(t),

where H(s) 6= 0 is an unknown (for the present) function to be defined.

Owing to Eq. 59, we have the following:

d(H(s)x(t))

dt− a(H(s)x(t)) = x(t),

H(s)dx(t)

dt− (aH(s))x(t) = x(t),

H(s)dx(t)

dt= (aH(s) + 1)x(t).

The last equality can be written as

dx(t)

dt= C(s)x(t), (60)


where C(s) = a + H(s)−1 when H(s) 6= 0.Because of our assumtion x(t) = Xest, the equation

dx(t)

dt= C(s)x(t),

yields

dx(t)

dt= Xsest = C(s)Xest

and for X 6= 0,

C(s) = s, or, a + H(s)−1 = s

Therefore, the function

H(s) =1

s− a, if s 6= a.

We define H(s) to be a transfer function of 1st order of the system H defined by the first-order

differential equation

dy(t)

dt− ay(t) = x(t), a > 0,

and for which the following transformation takes place

x(t) → y(t) =1

s − ax(t). (61)

We now define the same linear system H by using the integral-type linear convolution

y(t) =

+∞∫

−∞

x(t− τ)h(τ)dτ

for a response function h(t) to be found.

According to (61), we will define the response function in a way such that for the input function

x(t) = Xest the output y(t) to be equal1

s− ax(t).

By simple calculations, we have the following:

y(t) =

+∞∫

−∞

Xes(t−τ )h(τ)dτ

= Xest

+∞∫

−∞

e−sτh(τ)dτ

= x(t)

+∞∫

−∞

e−sτh(τ)dτ


Therefore, the transfer function of the system H equals

H(s) =

+∞∫

−∞

e−sτh(τ)dτ. (62)

This equation shows the relation between the response function h(t) of the system H and its

transfer function H(s).

D. n-order LTI systems

We now consider a general nth-order linear system H : x(t) → y(t) that is described by the

n-order linear differential equation with constant coefficients,

a0y(t)+a1y(1)(t)+a2y

(2)(t)+ . . .+any(n)(t) = b0x(t)+ b1x(1)(t)+ b2x

(2)(t)+ . . .+ bmx(m)(t) (63)

with coefficients ak and bl, (k = 0 : n, l = 0 : m). y(k)(t) and x(l)(t) denote respectively thederivatives dky(t)/dtk and dlx(t)/dtl. We consider an exponential input function with amplitude

X,x(t) = Xest, t ∈ (−∞, +∞)

and output in the formy(t) = H(s)x(t), t ∈ (−∞, +∞)

where H(s) is a function of s.Since

x(l)(t) = slx(t) = slest, l = 0, . . . , m,

andy(k)(t) = H(s)x(k)(t) = H(s)skx(t) = sky(t)

for k=0,...,n, we can write Eq. 63 as

a0y(t) + a1sy(t) + a2s2y(t) + . . . + ansny(t) =

= b0x(t) + b1sx(t) + b2s2x(t) + . . . + bmsmx(t)

(64)

or,

[a0 + a1s + a2s2 + . . . + ansn]y(t) = [b0 + b1s + b2s

2 + . . . + bmsm]x(t). (65)

We finally obtain

y(t) =b0 + b1s + b2s

2 + . . . + bmsm

a0 + a1s + a2s2 + . . . + ansnx(t). (66)

Therefore the nth-order linear system H : x(t) → y(t), that has been defined by the n-orderlinear differential equation with constant coefficients, is described by the transform function

H(s) =b0 + b1s + b2s

2 + . . . + bmsm

a0 + a1s + a2s2 + . . . + ansn(67)

when inputs are exponential functions.


We now consider the linear convolution

y(t) =

+∞∫

−∞

x(t− τ)h(τ)dτ

when input is x(t) = est.

y(t) =

+∞∫

−∞

es(t−τ )h(τ)dτ = est

+∞∫

−∞

e−sτh(τ)dτ.

Also, the following holds:

y(t) = H(s)x(t) = estH(s), t ∈ (−∞, +∞).

Therefore, we obtain the following formula for computing the response function of the system bythe impulse function:

H(s) =

+∞∫

−∞

e−sτh(τ)dτ. (68)

Generalizing the above reasoning, we see that the linear convolution system can be representedby two ways:

x(t) −→ h(t) −→ y(t) (integral representation),

x(t) −→ H(s) −→ y(t) (polynomial representation).

and the relations between these two functions is described by (68), which in the general case leads

to the concepts of the integral Fourier and Laplace transforms.


XI. Block diagrams for LTI systems

We have defined the transfer function of the linear time-invariant systems which are describedby differential equations. In particular case of the first-order differential equation

dy(t)

dt− ay(t) = x(t), a > 0, (69)

the function H(s) is defined such that

x(t) = xs(t) = Xest → H(s)Xest = H(s)x(t)

and equals

H(s) =1

s − a.

Since, for a given t,t

∫

−∞

(dy(τ)

dτ

)

dτ = y(t)

Eq. 69, i.e.,dy(t)

dt= ay(t) + x(t)

can be considered as

y(t) =

t∫

−∞

[

ay(τ) + x(τ)]

dτ. (70)

According to Eq. 70, the diagram of realization of the first-order LTI system takes the form whichis shown in Fig. 55.

- l s∫

6x(t) y(t)+ - -

a

Fig. 55. Realization of the first-order LTI system of Eq. 69. (One integrator and one multiplier by factor a

are used in the diagram.)

The block with∫

denotes the integrator,

x(t) → y(t) =

∫

t

−∞x(τ)dτ

The response function, h(t), of the integrator is the unit step function, h(t) = u(t). Indeed, whenthe input function is x(t) = δ(t),

x(t) = δ(t) → h(t)


and

h(t) =

t∫

−∞

δ(τ)dτ =

+∞∫

−∞

[1 − u(τ − t)]δ(τ)dτ

= [1− u(τ − t)]|τ=0= 1 − u(−t) = u(t).

We can use also the known property of the δ function:

t∫

−∞

δ(τ − t0)dτ = u(t − t0)

for t0 = 0.

So, we have three different forms of representation of the system as shown in Fig. 56. The

first two forms hold for any inputs, and the last form of realization holds only for inputs havingexponential form, i.e., for x(t) = xs(t) = Xest, where X and s are constants.

- l s∫

6x(t) y(t)+ - -

h(τ)x(t) y(t)

--

x(t) y(t)-- 1

s−a

Eq. 62

y(t) = x(t) ∗ h(t)

y(t) = H(s)x(t)

H(s) = 1s−a

=∞∫

−∞h(τ)e−sτdτ

a

Fig. 56. Three diagrams of realization of the 1st order LTI system.


Remark: The first two systems result in the same outputs for equal input functions. The thirdsystem gives the similar outputs only if the input function is x(t) = Xest.

In the general case of the first-order differential equation

a1

dy(t)

dt− ay(t) = bx(t) (71)

where a1, a, and b are some coefficients, we can write the realization of the system x(t) → y(t) as

y(t) =1

a1

t∫

−∞

[

ay(τ) + bx(τ)]

dτ (72)

The diagram of realization of the first-order LTI system will take form shown in Fig. 57.

l s

∫

x(t) y(t)+ - -

6

a

1a1-b

Fig. 57. (Is is correct?) Realization of the 1st order LTI system of Eq. 71.

We now consider the second-order differential equation

a2d2y(t)

dt2− a1

dy(t)

dt− ay(t) = bx(t) (73)

with coefficients a2, a1, a, and b. The realization of the system x(t) → y(t) can be represented as

a2

d2y(t)

dt2= a1

dy(t)

dt+ ay(t) + bx(t)

a2

dy(t)

dt=

t∫

−∞

[

a1

dy(τ)

dτ+ ay(τ) + bx(τ)

]

dτ

= a1y(t) +

t∫

−∞

[

ay(τ) + bx(τ)]dτ

(74)


a2y(t) =

t∫

−∞

a1y(τ)d(τ) +

t∫

−∞

(

τ∫

−∞

[

ay(τ1) + bx(τ1)]

dτ1

)

dτ

y(t) =1

a2

a1

t∫

−∞

y(τ)d(τ) +

t∫

−∞

(

τ∫

−∞

[

ay(τ1) + bx(τ1)]

dτ1

)

dτ

1

a2

b

t∫

−∞

(

τ∫

−∞

x(τ1)dτ1

)

dτ +

+ a1

t∫

−∞

y(τ)dτ + a

t∫

−∞

(

τ∫

−∞

y(τ1)dτ1

)

dτ

(75)

The diagram of realization of the first-order LTI system will take form shown in Fig. 58, wheretwo integrators are used when input is x(t) and two integrators are used when input is y(t).

l s

∫

x(t) y(t)+ - -

a1

1a2

b

s

∫

a

CCCCCO6CCC

s

∫

∫

l+

6

-

-

Fig. 58. Realization of the 2nd order LTI system of Eq. 75.

We consider the first-order differential equation of the form

a1

dy(t)

dt− ay(t) = bx(t) + b1

dx(t)

dt(76)

with coefficients a1, a, and b, b1


Similar to Eq. 72, we can write the realization of the system x(t) → y(t) as

y(t) =1

a1

t∫

−∞

[

ay(τ) + bx(τ) + b1dx(τ)

dτ

]

dτ

=1

a1

(

t∫

−∞

[

ay(τ) + bx(τ)]

dτ + b1x(t))

=1

a1

(

t∫

−∞

ay(τ)dτ + b

t∫

−∞

x(τ)dτ + b1x(t))

(77)

The diagram of realization of that kind of first-order LTI system will take form shown in Fig. 59.

l s

∫

x(t) y(t)+ - -

a

1a1

b

s

∫

l+

-

-

6

w(t)-b1

6

w(t) = bt∫

−∞x(τ)dτ + b1x(t)

Fig. 59. Realization of the 1st order LTI system of Eq. 76.

In the case of the second-order differential equation of the form

a2d2y(t)

dt2− a1

dy(t)

dt− ay(t) = bx(t) + b1

dx(t)

dt, (78)

the realization of the system x(t) → y(t) can be represented as

y(t) =1

a2

t∫

−∞

a1y(τ)d(τ) (a2 6= 0)

+

t∫

−∞

(

τ∫

−∞

[

ay(τ1) + bx(τ1) + b1

dx(τ1)

dτ1

]

dτ1

)

dτ

=1

a2

t∫

−∞

(

a1y(τ) +

τ∫

−∞

ay(τ1)dτ1+

+b1x(τ) +

τ∫

−∞

bx(τ1)dτ1

)

dτ

(79)


l s

∫

x(t) y(t)+ - -

a1

1a2

b1 s

∫

a

CCCCCO6CCC

s

∫

l+

-

-w(t)

b

s

∫

-

6

-

Fig. 60. Realization of the 2nd order LTI system of Eq. 78. Three integrators are required for this realization,but it will be shown later that the realization of this LTI system can be reduced to a diagram with twointegrators. All described above diagrams are so-called diagram of realization of form I.

The diagram of realization of that kind of second-order LTI system will take form shown in

Fig. 60.The function ω(t) = b(x ∗ u)(t) + b1x(t), i.e.,

ω(t) = b

t∫

−∞

x(τ)dτ + b1x(t),

because the unit step function u(t) is the impulse response function of the integrator.Eq. 79 can be written in the following compact form:

y(t) =1

a2

[a1(y ∗ u)(t) + a((y ∗ u) ∗ u)(t) + b1(x ∗ u)(t) + b((x ∗ u) ∗ u)(t)] . (80)

This diagram can be transfered to the diagram with only two integrators as shown in Fig. 61.

∫

∫ a

b1

b

0 + + •

•

•

a1

x(t) y(t)

a2y

(2)(t)−a

1y’(t)−ay(t)=bx’(t)+b

1x(t)

1/a2

Fig. 61. Diagram of realization of the system in Form II.


This diagram of realization in Form II can be obtained from the diagram of Fig. 60 by replacingthe left and right parts of the diagram as shown in Fig. 62

l s

∫

x(t) y(t)+ -

a1

1a2

b1s

∫

a

CCCCCO6CCC

s

∫

l+

-

b

s

∫

-

6

--

Fig. 62. (Step 1) Realization of the 2nd order LTI system of Eq. 78.

Then, since each pair of integrators have the same inputs and outputs, the diagram of Fig. 62

can be simplified as shown in Fig. 63.

l s

∫

x(t) y(t)+ -

a1

1a2

b1s

∫

a

CCCCCO6CCC

l+

-

b-

6

--

s

Fig. 63. (Step 2) Realization of the 2nd order LTI system of Eq. 78.

And this diagram is equals to the diagram of Fig. 61.


XII. The Fourier series: Periodic signals

In this section, we consider the representation of the time signals (functions) in another domain,

the frequency domain. Illustration: Figures 65 and 66 show the signals (1-D and 2-D signals) and

their representation in the frequency domain, which is called the Fourier series (or the discrete

Fourier transform (DFT)). The representation in the frequency domain is unique.

0 50 100 150 200 2500

100

200

0 50 100 150 200 2500

1000

2000

3000

4000

5000

−100 −50 0 50 1000

1000

2000

3000

4000

5000

DFT(0)=48233

DFT(0)=48233

Fig. 64. (top) 1-D signal, (middle) coefficients of the Fourier series (FS), and (bottom) shifted FS to the

middle. (FS is shown in the absolute scale)

Art

Fig. 65. (left) Two two-dimensional signals (images), (middle) coefficients of the Fourier series of images,

and (right) shifted FSs to the middle. (FSs are shown in the absolute scale)


A. The concept of the Fourier series

We define the important concept of the Fourier transform, which is defined in a way similar to

the transfer function H(s) in the integral form of Eq. 50 (or Eq. 56), when the input function is

the complex exponential function

x(t) = xjs(t) = ejst, j2 = −1,

which results in the transform function

H(js) =

+∞∫

−∞

e−jsth(t)dt. (81)

Let us consider the following periodic signal

x(t) = 10 + 3 cos(ω0t) + 5 cos(2ω0t − π/6) + 4 sin(3ω0t)

= 10 + 3 · cos(ω0t) + 5A · cos(2ω0t) + 0 · cos(3ω0t)

+ 0 · sin(ω0t) + 5B · sin(2ω0t) + 4 · sin(3ω0t)

where A = cos(π/6) and B = sin(π/6). The fundamental period of this signal is T = 2π/ω0. The

illustration of the decomposition of this signal x(t) is given in Figure 67.

Since, for each natural n,

cos(nω0t) =ejnω0t + e−jnω0t

2

sin(nω0t) =ejnω0t − e−jnω0t

2j

we can write the above signal in the form

x(t) = c0 + c1ejω0t + c−1e

−jω0t

+ c2ej2ω0t + c−2e

−2jω0t

+ c3ej3ω0t + c−3e

−3jω0t,

where the real or complex coefficients cn, n = 0,±1,±2, ±3, can be easily defined.

Indeed,

x(t) = 10 + 3 cos(ω0t) + 5A cos(2ω0t)

+5B sin(2ω0t) + 4 sin(3ω0t)

= 10 + 3ejω0t + e−jω0t

2+ 5A

ej2ω0t + e−j2ω0t

2

+5Be2jω0t − e−j2ω0t

2j+ 4

ej3ω0t − e−j3ω0t

2j

= 10 +3

2ejω0t +

3

2e−jω0t

+5

2(A − jB)ej2ω0t +

5

2(A + jB)e−j2ω0t

−2jej3ω0t + 2je−j3ω0t


and c0 = 10, c1 = c−1 = 3/2, c2 = 5/2(A − jB), c−2 = 5/2(A + jB), and c3 = −2j, c−3 = 2j.

So, the continuous-time periodic function x(t) with period T = 2π/ω0 = 2π is completely defined

by seven coefficients cn, i.e.,

x(t) ↔ (c0, c1, c−1, c2, c−2, c3, c−3)

↔ (10, 3/2, 3/2, 5/2(A− jB), 5/2(A + jB),−2j, 2j).

This statement is very important: the function x(t) defined (and/or periodic) in the interval

[0, 2π] is completely defined by seven coefficients.

Moreover, the following property of the complex conjugation takes place for the coefficients c±n,

n = 1, 2, 3,

c−1 = c1, c−2 = c2, c−3 = c3,

which means that it is enough to know only four coefficients to define the periodic signal x(t) in

the interval [0, 2π],

x(t) ↔ (c0, c1, c2, c3)

↔ (10, 3/2, 5/2(A− jB),−2j)

↔ (c0, c−1, c−2, c−3)

↔ (10, 3/2, 5/2(A + jB), 2j).

Since ω = 1 and T = 2π, we can write the signal x(t) as follows:

x(t) = 10 + 3 cos(t) + 5A cos(2t)

+5B sin(2t) + 4 sin(3t)

= 10 + 3 cos(2π

Tt) + 5A cos(

2π

T2t)

+5B sin(2π

T2t) + 4 sin(

2π

T3t)

= c0 + c1ej 2π

Tt + c−1e

−j 2π

Tt

+ c2ej 2π

T2t + c−2e

−j 2π

T2t

+ c3ej 2π

T3t + c−3e

−j 2π

T3t.

The general result comes from 1800’s and belongs to Jean Baptiste Joseph Fourier who has used

for a periodic function f(t) on an interval [0, T ] the expansion in the trigonometrical series (the

Fourier series)

f(t) =a0

2+

+∞∑

n=1

an cos2πnt

T+

+∞∑

p=1

bn sin2πnt

T(82)

for t ∈ [0, T ].


The coefficients an and bn, n = 1, 2, ..., are calculated as

an =2

T

T∫

0

f(t) cos

(

2πn

Tt

)

dt

bn =2

T

T∫

0

f(t) sin(2πn

Tt) dt.

(83)

This Fourier series of the function f(t) can also be written in the complex form

f(t) =+∞∑

n=−∞

cnej 2πt

Tn, t ∈ [0, T ], (84)

with the coefficients

c±n =1

T

T∫

0

f(t)e∓j 2πn

Tt dt = (an ∓ jbn)/2,

n = 1, 2, . . . , (c0 = a0/2).

(85)

Thus, the set of the coefficients cn (or an, bn) can be compared to the continuous function f(t) and

present the latter as a linear superposition of the harmonic oscillations (cosine and sine functions).

In the general f(t) case, the function

(F f)(ω) = F (ω) =

+∞∫

−∞

f(t)ejωt dt, (86)

where ω ∈ (−∞, +∞), which is called the Fourier transform of the function f, can be compared to

the absolute integrable function f, determined in the real line R1,

+∞∫

−∞

|f(t)|dt < ∞.

The operation F is called an 1-D Fourier transformation. The module of the function |F f | is

called an energy Fourier spectrum of the function f.

At that, the following inverse formula takes place:

f(t) =1

2π

+∞∫

−∞

F (ω)e−jωt dω, t ∈ (−∞, +∞), (87)

if the function f satisfies additional conditions (Dini), for instance, if f is continuous and has the

finite derivative.

One can consider (86) as a transition from the time description of the function, f, to the descrip-

tion in the frequency domain, F = (F f).


Figure 67 shows the decomposition of the signals

x(t) = 10 + 3 cos(ω0t) + 5 cos(2ω0t − π/6) + 4 sin(3ω0t)

and

y(t) = 10 + 3 cos(ω0t) + 5 cos(2ω0t) + 4 cos(3ω0t)

in the interval [0, 3π]. The phase φ = π/6) and the fundamental frequency ω2 = 2.

5 cos(2ω0t − π/6) = 5 cos(2ω0t) cos(π/6) + 5 sin(2ω0t) sin(π/6)

0 2 4 6 85

10

1510+3cos(wt)

Mag

nit

ude

0 2 4 6 8−10

−5

0

5

105Acos(2wt)

0 2 4 6 8−10

−5

0

5

10

5Bsin(2wt)

Mag

nit

ude

0 2 4 6 8−10

−5

0

5

10

4sin(3wt)

0 2 4 6 8

0

5

10

15

20

25sum of 4 signals

Mag

nit

ude

tT

0 2 4 6 8

0

5

10

15

20

25sum of 3 signals

tT

The decomposition of the signal x(t)

Fig. 66. Decomposition of the signal x(t).


B. Main formulas for coefficients of Fourier Series

Given a periodic function x(t) with period T can be represented in the three different ways which

are trigonometrical, exponential, and cosine representation (t ∈ [0, T ]):

x(t) =a0

2+

+∞∑

n=1

an cos(nωt) ++∞∑

n=1

bn sin(nωt), (88)

x(t) =+∞∑

n=−∞

cnej(nωt), (89)

x(t) = d0 ++∞∑

n=1

dn cos(nωt + ϑn), (90)

where

dn =√

a2n + b2

n = 2|cn| = 2|c−n|, n = 1, . . . , d0 = a0/2,

cn =an − jbn

2

c−n =an + jbn

2, and c0 = a0/2.

(91)

Indeed, we can write the following in (88):

an cos(nωt) + bn sin(nωt) =√

a2n + b2

n

(

an√

a2n + b2

n

cos(nωt) +bn

√

a2n + b2

n

sin(nωt)

)

=√

a2n + b2

n (cos(ϑn) cos(nωt) + sin(ϑn) sin(nωt))

=√

a2n + b2

n cos(nωt − ϕn),

(92)

where

cos(ϕn) =an

√

a2n + b2

n

, sin(ϕn) =bn

√

a2n + b2

n

. (93)

Therefore if an 6= 0, then the phase is calculated as follows:

tan(ϕn) = bn/an, ϕn = −ϑn = tan−1(bn/an), (94)

and ϕn = π/2 and dn = bn, if an = 0.

XIII. Basis Functions and Finite Fourier series

For a periodic function f(t) defined on an interval [0, T ] the expansion in the trigonometrical

series (the Fourier series) is

f(t) =a0

2+

+∞∑

n=1

an cos2πnt

T+

+∞∑

n=1

bn sin2πnt

T, t ∈ [0, T ], (95)


and this series of the function f(t) is written in the complex form as

f(t) =+∞∑

n=−∞

cnej 2πt

Tn, t ∈ [0, T ]. (96)

The coefficients an and bn, n = 1, 2, ..., and cn, n = 0,±1,±2, ..., respectively are calculated by

formulae (76) and (78). The set of the coefficients cn (or an, bn) can be compared to the continuous

function f(t). The function f(t) is a linear superposition of the harmonic oscillations (cosine and

sine functions), and coefficients an, bn show the amplitudes of such oscillations.

Consider first the case of periodic functions f(t) on the interval [0, 2π], i.e., when the period

T = 2π. We consider the linear space L2[0, 2π] of square-integrable functions on the interval [0, 2π],

2π∫

0

|f(t)|2dt < ∞.

In the space L2[0, 2π], the set of functions

1, cos t, cos 2t, cos 3t, . . . cos nt, . . . ,

sin t, sin 2t, sin 3t, . . . sinnt, . . .(97)

compose a complete orthogonal system (trigonometric system) of functions.

In L2 space, we define the inner product of two real functions φk(t) and φn(t) as

(φk, φn) =

2π∫

0

φk(t)φn(t)dt

and the orthogonality means that (φ1, φ2) = 0.

The functions of (97) are orthogonal.

Indeed, if k 6= n = 0,

(cos kt, 1) =

2π∫

0

cos kt dt =

2π∫

0

sin kt dt = 0.

If k 6= n 6= 0,

(cos kt, cosnt) =

2π∫

0

cos kt cosnt dt =1

2

2π∫

0

[cos(k + n)t + cos(k − n)t] dt = 0,

(sinkt, sinnt) =

2π∫

0

sin kt sinnt dt =1

2

2π∫

0

[cos (k − n)t − cos(k + n)t] dt = 0.

(sinkt, cosnt) =

2π∫

0

sin kt cosnt dt =1

2

2π∫

0

[sin(k + n)t + sin(k − n)t] dt = 0.


0 2 4 6 8 10 12

−1

−0.5

0

0.5

1

Orthogonality of cosine and sine functions

0 2 4 6 8 10 12

−1

−0.5

0

0.5

1

time

•

•

•

A

B

−B

A sin(t)

cos(t)

cos(2t)

•

•

•

•

A

•

•

•

•

sin(t)

B

B

A A

•

Fig. 67. Orthogonality of two pair of sinusoidal waves.

As an example, Figure 68 illustrates the orthogonality of functions cos(t) and sin(t) as well sin(t)

and cos(2t).

If k = n 6= 0,

(cos kt, cosnt) =

2π∫

0

cos2 nt dt = π,

(sinkt, sinnt) =

2π∫

0

sin2 nt dt = π,

(1, 1) =

2π∫

0

1 dt = 2π.

Therefore, the set of functions

1√2π

,cos t√

π,

cos 2t√π

,cos 3t√

π, . . . ,

sin t√π

,sin 2t√

π,

sin 3t√π

, . . .

(98)

compose a complete orthonormal system of functions in L2[0, 2π].


Note, that when we write the Fourier expansion for the function f ∈ L2[0, 2π],

f(t) =a0

2+

+∞∑

k=1

ak cos kt ++∞∑

k=1

bk sin kt, t ∈ [0, 2π]. (99)

we understand that the partial sums, Sn converge to f,

Sn(t) =a0

2+

n∑

k=1

ak cos kt +n∑

k=1

bk sinkt → f(t), ∀t ∈ [0, 2π] (100)

when n → +∞, and the convergence is in the metric of the space L2, i.e.,

2π∫

0

|Sn(t)− f(t)|2 dt → 0

when n → +∞.

Moreover, for a given number n, the partial sum Sn gives the best approximation of the function

f among all trigonometric polynomials,

Tn(t) =α0

2+

n∑

k=1

αk cos kt +n∑

k=1

βk sin kt.

Remark: System of functions in (97) compose the complete set of functions on the interval [0, 2π].

But, each of the following systems

1, cos t, cos 2t, cos 3t, . . . cos nt, . . .

and

sin t, sin 2t, sin 3t, . . . sinnt, . . .

compose the complete system of function only on the space L2[0, π], i.e., for functions defined (or

periodic) on the interval [0, π].

In the general case of the space L2[0, T ] of square-integrable functions on the interval [0, T,

we can use all mentioned above formulas changing 2π by T and basis functions cosnt and sin nt

respectively by cos2π

Tnt and sin

2π

Tnt.

Denoting the frequency ω0 =2π

T, we obtain the complete orthogonal system of functions,

1, cosω0t, cos 2ω0t, cos 3ω0t, . . . cos nω0t, . . . ,

sin ω0t, sin 2ω0t, sin 3ω0t, . . . sinnω0t, . . .(101)

in the space of functions L2[0, T ].

Every function f(t) of L2[0, T ], can be described by its Fourier series

f(t) =a0

2+

+∞∑

n=1

an cos(nω0t) ++∞∑

n=1

bn sin(nω0t), t ∈ [0, T ]. (102)


The coefficients an and bn, n = 1, 2, ..., are calculated as

an =2

T

T∫

0

f(t) cos(nω0t) dt, bn =2

T

T∫

0

f(t) sin(nω0t) dt. (103)

In the comples form, the Fourier series of the function f(t) is written as

f(t) =+∞∑

n=−∞

cnejnω0t, t ∈ [0, T ], (104)


c±n =1

T

T∫

0

f(t)e∓jnω0t dt = (an ∓ jbn)/2, n = 1, 2, . . . , (c0 = a0/2). (105)

The Fourier series of Eq. 104 is defined for real and complex functions of the space L2[0, T ]. The

inner product of two functions φk(t) and φn(t) of L2[0, T ] is defined as

(φk, φn) =

T∫

0

φk(t)φn(t)dt.

The functions

φn(t) = ejnω0t, n ∈ Z,

are orthogonal. Indeed, if k 6= n

(φk, φn) =

T∫

0

ejkω0te−jnω0tdt

=

T∫

0

ej(k−n)ω0tdt

=1

(k − n)ω0ej(k−n)ω0t

∣

∣

∣

T

0

=1

(k − n)ω0

[

ej(k−n)ω0T − 1]

= 0

because ω0T = 2π.

And,

(φn, φn) =

T∫

0

ejnω0te−jnω0tdt =

T∫

0

dt = T.


A. Finite Fourier series

Consider the Fourier series of the function f(t) of the space L2[0, T ],

f(t) =+∞∑

n=−∞

cnejnω0t, t ∈ [0, T ], (106)

or,

f(t) =a0

2+

+∞∑

k=1

ak cos(kω0t) ++∞∑

k=1

bk sin(kω0t), t ∈ [0, T ], (107)

where ω0 = 2π/T.

Consider the discrete-time functions f(t) = f(n), when n = 0 : N −1. In other words, we assume

that the period T is an integer number, i.e., T = N. Then, we can write

f(t) =a0

2+

+∞∑

k=1

ak cos(2π

Nkt) +

+∞∑

k=1

bk sin(2π

Nkt), (108)

which shows that the coefficients

ak = bk = 0, ∀k > N,

and (108) is of the form

f(t) =a0

2+

N−1∑

k=1

ak cos(2π

Nkt) +

N−1∑

k=1

bk sin(2π

Nkt). (109)

Indeed, the functions cos(2π

Nkt) and sin(

2π

Nkt) are periodic with period N as the function of

parameter k,

cos(2π

N[k + N ]t) = cos(

2π

Nkt + 2πt) = cos(

2π

Nkt), t = 0 : N − 1,

sin(2π

N[k + N ]t) = sin(

2π

Nkt + 2πt) = sin(

2π

Nkt), t = 0 : N − 1.

In the comples form, the finite Fourier series takes form

f(t) =N−1∑

n=−N+1

cnej 2π

Nnt, t = 0 : N − 1. (110)

where c0 = a0/2, and cn = (an − jbn)/2, c−n = (an − jbn)/2.

We can note, that the following property takes place for the complex coefficients c±n:

c−n = cN−n, n = 1 : N − 1.

Indeed,

e−j 2π

Nnt = e−j 2π

Nnt+j 2π

NNt = ej 2π

N(N−n)t


and Eq. 110 can be written as

f(t) =N−1∑

n=0

c′nej 2π

Nnt, t = 0 : N − 1, (111)

where

c′n = c−n + cN−n, n = 1 : N − 1.

In the case when the discrete-time function is real, the following take place

c′n = c′N−n, n = 1 : N − 1.

So to define the function f(t) we can use 1 + 2N real coefficients

a0, a1, a2, . . . , aN−1, b1, b2, . . . , bN−1,

or 1 + N/2 (or, 1 + [N + 1]/2) complex coefficients

c0, c1, c2, . . . , cN/2−1, (c[N+1]/2−1).

After changing the notation c′ → cp, we define the finite Fourier series of the sequence f(n) as

f(n) =N−1∑

p=0

cpej 2π

Npn, n = 0 : N − 1. (112)

The coefficients cp compose the 1-D N -point discrete Fourier transform (1-D DFT) of the se-

quence f(n), an can be defined as

cp = F (p) =1

N

N−1∑

n=0

f(n)e−j 2π

Nnp, p = 0 : N − 1. (113)

and Eq. 112 describes the inverse 1-D N -point discrete Fourier transform (IDFT).

In MATLAB, we can use the following functions calculating the DFT:

1. output=fft(input) for the direct 1-D DFT of the sequence f

2. output=ifft(input) for the inverse 1-D DFT of the sequence f

3. fftshift(input) to swap the left and right halves of the 1-D DFT.


XIV. Exponential Fourier series representation

Given a periodic function x(t) with period T, we consider the exponential, representationof x(t), i.e., the trigonometric Fourier series in complex form :

x(t) =+∞∑

n=−∞

cnejnωt, t ∈ [0, T ], (117)


cn =1

T

T∫

0

x(t)e−j 2πn

Tt dt, n = ±1,±2, . . . ,

c0 =1

T

T∫

0

x(t) dt, (ω = 2π/T ).

(118)

Example 1: Let us consider the following periodic ramp function with period T,

x(t) = r(

t

T

)

=t

T, t ∈ [0, T ].

T 2T −T −2T

1

x(t)

t0

Tc0

Fig. 72. Periodic ramp function.

The coefficients of the Fourier series of x(t),

c0 =1

T

T∫

0

r(

t

T

)

dt =

T∫

0

r(

t

T

)

dt

T

=

1∫

0

r (t) dt =

1∫

0

t dt =1

2.

Further, by the direct calculations we obtain in the n 6= 0 case, that

cn =

T∫

0

r(

t

T

)

e−j 2πn

Tt d

t

T

=

1∫

0

r (t) e−j2πnt dt = −1

j2πn

1∫

0

tde−j2πnt


= −1

j2πn

te−j2πnt|10 −

1∫

0

e−j2πntdt

= −1

j2πn

1 − 0 +1

j2πn

1∫

0

de−j2πnt

= −1

j2πn

[

1 +1

j2πne−j2πnt|10

]

= −1

j2πn

[

1 +1

j2πn[1 − 1]

]

we obtain

cn = −1

j2πn=

j

2πn, n 6= 0.

Therefore,

x(t) =1

2+

+∞∑

n=±1

j

2πnejnωt, t ∈ [0, T ].

Owing to symmetry, c−n = −cn, for all integers n > 0, we have the following representation

x(t) =1

2+

+∞∑

n=±1

j

2πnejnωt

=1

2+

+∞∑

n=1

j

2πn

[

ejnωt − e−jnωt]

=1

2+

+∞∑

n=1

j

2πn2j sin(nωt)

=1

2−

+∞∑

n=1

1

πnsin(nωt). (119)

Note, that the function y(t) = x(t) − 0.5 is odd, i.e., y(−t) = −y(t), and y(t) can berepresented by the Fourier series of sin(·) functions.

T 2T −T −2T

1/2

y(t)

t0

Fig. 73. Odd ramp function.

We can also write the cosine representation of x(t),

x(t) =1

2+

+∞∑

n=1

1

πncos(nωt + π/2),


and the coefficients and phases of the cosine Fourier series

d0 =1

2, dn =

1

πn, ϑn =

π

2, n = 1, 2, ... .

0 2 4 6 8 100

0.1

0.2

0.3

0.4

0.5

0 2 4 6 8 10−0.4

−0.2

0

0.2

0.4

0.6

(a) (b)

dn

d0

d1

d2

c0

−|cn|

n

n

Fig. 74. Coefficients of the Fourier series.

Example 2: The similar representation has the following periodic step function, or ”square-wave” function,

xs(t) =

1, t ∈ [0, T/2),−1, t ∈ [T/2, T ),

which shows the sign of the function sin(2πt/T ).

T 2T −T −2T

1

xs(t)

t0

−1

Fig. 75. Square-wave function.

Indeed, we first consider the Fourier series for the periodic pulse function

y(t) = xs(t) + 1 =

2, t ∈ [0, T/2),0, t ∈ [T/2, T ).

Then,

c0 =1

T

T∫

0

y (t) dt =1

T

T/2∫

0

2dt = 1,

and for n 6= 0,

cn =1

T

T/2∫

0

2e−jωnt dt


= −2

T

1

jωne−jωnt|

T/20

=2

T

1

jωn

[

1 − e−jωTn/2]

=1

jπn

[

1 − e−jπn]

, (ωT = 2π)

or,

cn =

2

jπ(2k + 1), n = 2k + 1, k = 0,±1, ...

0, n = 2k, k = 0,±1, ... .

Therefore,

xs(t) + 1 = 1 ++∞∑

k=−∞

2

jπ(2k + 1)ej(2k+1)ωt, t ∈ [0, T ].

and since cn = −c−n,

xs(t) =+∞∑

k=−∞

2

jπ(2k + 1)ej(2k+1)ωt

=+∞∑

k=0

2

jπ(2k + 1)

[

ej(2k+1)ωt − e−j(2k+1)ωt]

=+∞∑

k=0

2

jπ(2k + 1)2j sin [(2k + 1)ωt]

=+∞∑

k=0

4

π(2k + 1)sin [(2k + 1)ωt] .

We can compare the Fourier transforms of functions xs(t) and x(t) of Example 1. Forthat, we take the function 2 − 4x(t), which can be written as (see Eq. 119)

y(t) = 4

1

2− x(t) =

+∞∑

n=±1

1

2πnsin(nωt)

, t ∈ [0, T ].

If we omit each second term from the Fourier series of y(t), we obtain the Fourier series ofthe function xs(t).

Problem 1: (a). Find the coefficients of the periodic function x(t) with period 2T,

x(t) = r(

t

T

)

=

t

T, t ∈ [0, T ),

0, t ∈ [T, 2T ).

(b) Find the coefficients of the time-reversal version of the periodic function x(t),

x(−t) = r(

−t

T

)

=

0, t ∈ [0, T ),

2 −t

T, t ∈ [T, 2T ).


Remark. Note, that for a periodic function x(t) with period T, we use to consider theintervals [0, T ] or [−T/2, T/2]. But, any other interval [a, T + a] of length T can also beconsidered. Indeed, for all cases, the coefficients of the Fourier series cn are the same. i.e.,

cn =1

T

T∫

0

x(t)e−j 2πn

Tt dt, n = 0,±1,±2, . . . ,

=1

T

T/2∫

−T/2

x(t)e−j 2πn

Tt dt (ω = 2π/T )

=1

T

a+T∫

a

x(t)e−j 2πn

Tt dt, a ∈ R.

Indeed,

a+T∫

a

x(t)e−j 2πn

Tt dt =

T∫

a

+

a+T∫

T

x(t)e−j 2πn

Tt dt

=

T∫

a

x(t)e−j 2πn

Tt dt +

a∫

0

x(t + T )e−j 2πn

T(t+T ) dt

=

T∫

a

x(t)e−j 2πn

Ttdt +

a∫

0

x(t)e−j 2πn

Tt dt

=

T∫

0

x(t)e−j 2πn

Ttdt

Example 3: We now consider the periodic function

x(t) =

t

T, t ∈ [0, T ),

2 −t

T, t ∈ [T, 2T ),

for which

cn =1

2

T∫

0

t

Te−j 2πn

2Tt d

t

T+

1

2

2T∫

T

[

2 −t

T

]

e−j 2πn

2Tt d

t

T

=1

2

T∫

0

t

Te−j 2πn

2Tt d

t

T+

1

2

0∫

−T

−t

Te−j 2πn

2Tt d

t

T


=1

2

1∫

0

te−jπnt dt +1

2

0∫

−1

−te−jπnt dt

=1

2

1∫

0

te−jπnt dt +1

2

1∫

0

tejπnt dt

=1

2

1∫

0

t[

e−jπnt + ejπnt]

dt

=

1∫

0

t cos(πnt) dt =1

πn

1∫

0

td sin(πnt)

=1

πn

t sin(πnt)|10 −

1∫

0

sin(πnt)dt

=1

πn

−1

πn

1∫

0

d cos(πnt)

= −(

1

πn

)2

[1 − cos(πn)]

= −(

1

πn

)2

[1 − (−1)n]

So,

cn = −(

1

πn

)2

[1 − (−1)n] = cn =

−2

(πn)2 , n = 2k + 1,

0, n = 2k.

c0 = 1, and the Fourier series of x(t) is

x(t) = 1 −+∞∑

n=−∞

(

1

πn

)2

[1 − (−1)n] ejnωt, t ∈ [0, T ].

Since, c−n = cn, for all n, we can write

x(t) = 1 −∑

n=±1,±2,...

(

1

πn

)2

[1 − (−1)n] 2 cos(nωt),

or

x(t) = 1 −+∞∑

k=0

4

[π(2k + 1)]2cos((2k + 1)ωt).


XV. The Fourier series and FFT with MATLAB

In this section, we consider the simple script of the program to calculate the Fouriertransform of a periodic signal x(t) = x(n) with integer time points. The input signal isconsidered to be the following:

x(t) = 1 + 2 cos(2πf1t) + 4 sin(2πf2t), f1 = 0.02(Hz), f2 = 0.03(Hz). (120)

The signal is periodic with the fundamental frequency f0 = 0.01Hz. where ω0 = 2π · 0.01(in rad/s). The carrier frequencies f1 and f2 are integer multiple to f0, i.e., f1 = 2f0 andf2 = 3f0. Therefore, the fundamental period and frequency ω0 (in rad/s) of this signal are

T = 1/f0 = 100 s, ω0 = 2πf0 = 2π · 0.01 rad/s = 0.02π rad/s.

This signal x(t) in the interval [0, 300] is given in Figure 76. The representation of the firstperiod (we also call it the fundamental period), which we denote by x1(t), in the frequencydomain in the absolute scale also is shown in the figure. This representation is also calledthe 100-point discrete Fourier transform (DFT) of the signal x1(t).

0 50 100 150 200 250 300−6

−4

−2

0

2

4

6

8original signal x(n)

−3 −2 −1 0 1 2 30

50

100

150

200

250

| 100−point DFT of the period x1(n) |

Fig. 76. (top) 1-D signal x(t) and (bottom) the shifted 100-point DFT of the first period x1(t). (DFT isshown in the absolute scale)

Below are the scripts of the program which was written in the class to calculate the signalx(t) and its 100-point DFT. This script (maybe with some small changes) was also postedin the web cite (BB).

% =======================================================================================

% call: work_withFS.m (the first draft)%% for EE 3424 (Math in SS), ECE UTSA

% Artyom Grigoryan (from the class work on October 26, 2015)


% 1. Signal composition% x(t) = 1 + 2cos(2pi0.02t) + 4sin(2pi0.03t)

t_end=300;t=0:t_end;

f0=0.01;

f1=2*f0; % in Hzf2=3*f0;w1=2*pi*f1; w2=2*pi*f2;

N=length(t);x=zeros(1,N);

x = 1 + 2*cos(w1*t) +4*sin(w2*t);

% 2. Display the signalfigure;

subplot(2,1,1);plot(t,x);title(’original signal x(n)’);

% 3. Take the fund. period of the signalT=1/f0;

x1=zeros(1,T);x1=x(1:T);t1=t(1:T);

hold on;plot(t1,x1,’r’);

% 4. Calculate the FS (X1) of the signal x1X1=fft(x1);

X1abs=abs(X1);w=linspace(0,2*pi-2*pi/T,T);

% 5. Plot the DFT of the signalsubplot(2,1,2);stem(w,X1abs);

title(’| Fourier series of x1 |’);axis([ 0,2*pi+0.01, 0,250]);

pause(2)axis([ 0,2*pi/5+0.01, 0,250]);

pause(2)X1abs_shifted=fftshift(X1abs);

w1=w-pi;

stem(w1,X1abs_shifted);title(’| 100-point DFT of the period x_1(n) |’);axis([ -pi-0.01,pi+0.01, 0,250]);

% =======================================================================================


A. Full Script of the Program with Description and Explanation

Now, I will try to explain the above given script and correct it. Namely, we will see therelationship between the 100-point DFT of the period x1(t) and the coefficients Ck of theFourier series of the signal x(t).

1. Note, the periodic signal x(t) is given in the integer interval [0, 300] and the coefficientsCk should be calculated for this signal. The 100-point DFT is calculated on the first periodx1(t) of the signal, not on the entire signal x(t).

2. The period of the signal N = T = 100 is integer. The fundamental frequency ω0 = 2π/N(in rad/s).

3. The finite Fourier series of the sequence x(n) is

x(n) =N−1∑

k=0

Ckej 2π

Nkn, n = 0, 1, 2, 3, . . . . (121)

The coefficients Ck are calculated as

Ck =1

N

N−1∑

n=0

x(n)e−j 2π

Nnk =

1

N

N−1∑

n=0

x1(n)e−j 2π

Nnk, k = 0 : N − 1. (122)

4. The N -point discrete Fourier transform of the sequence x1(n) is calculated by ”fft.m”function in MATLAB as

Xk =N−1∑

n=0

x1(n)e−j 2π

Nnk, k = 0 : N − 1. (123)

Therefore, to calculate the coefficients ck from the values of Fk, the following normalizationof the N -point DFT should be considered:

Ck =1

NXk, k = 0 : N − 1.

Now, we consider the script of the program in more detail. The script of the program“work withFS1.m” is given below; it uses the functions which are in the standard library ofMATLAB:

1. “fft.m” – for the direct 1-D DFT of the signal2. “fftshift.m” to swap the left and right halves of the 1-D DFT,3. functions to display and draw graphics, such as “figure, plot, stem, text, xlabel, ... .”

One can use the command “help” to know more about these functions. For instance, for“stem” on can write in the command window of MATLAB the command

>> help stem

Comments: When start MATLAB, use the command “cd” which means “change direc-tory” to move away from the folder “MATLAB” and work in another folder, as I do in theclass when writing in the command line the following:

>> cd C:/Art/Work inclass


The program “work withFS1.m” calculates the signal, its coefficients of the Fourier series,and prints the results as shown in Figure 77.

0 50 100 150 200 250 300−6

−4

−2

0

2

4

6

8

original signal x(n) and one period x1(n)

(a)

0 1 2 3 4 5 60

0.5

1

1.5

2

2.5

3

|cn| coeff. of the Fourier series of signal x

1(n)

(b) frequency (rad/s)

Fig. 77. (top) 1-D signal x(t) and (bottom) the shifted 100-point DFT of the first period x1(t). (DFT isshown in the absolute scale)

Code Description:

% ========================== comments =========================================

% call: work_withFS1.m (from class work of October 26, 2015)% -----------------------------------------------------------------------------% To demonstrate in detail the example of script written for

% composing the signal and calculating its Fourier series coefficients Cn% by using the fast Fourier transform (FFT) function fft.m from MATLAB.%

% For class EE 3424 Math in Signals and Systems% Artyom Grigoryan, October 26-27, 2015, ECE Dept. UTSA% -----------------------------------------------------------------------------

% --------------------------------- PART 1 ------------------------------------% 1. Signal composition

% x(t) = 1 + 2cos(2pi0.02t) + 4sin(2pi0.03t), [pi stands for 3.1416...]% in the integer interval [0,300], or [0,t_end]

t_end=300;t=0:t_end; % the vector of time points 0,1,2,...,300f0=0.01; % the fundamental frequency of the signal x(t) in Hz

f1=2*f0; f2=3*f0; % two frequencies of the signal x(t) in Hzw1=2*pi*f1; w2=2*pi*f2; % two frequencies of the signal x(t) in rad/s


%N=length(t); % the length of the signal x(t), which is N=301

x=zeros(1,N); % asking memory for the new variable (signal x(t))x = 1 + 2*cos(w1*t) +4*sin(w2*t); % calculation of the signal at all time points%

% 2. Display the signalfigure; % open the window for the figuresubplot(2,1,1); % the window is divided by 2 parts and 1st part will be used

plot(t,x); % plot the signal x(t) versus the time points ttitle(’original signal x(n) with 3 periods’); % add the title to the graphxlabel(’ (a) ’); % label the x-axis as "(a)"

% 2. Calculate select the fundamental period of the signalT=1/f0; % it is 100 for this example

% 3. Calculate and plot the first period of the signal, which is denoted as x1(t)x1=zeros(1,T); % asking memory for x1(t); T zeros will be filled in x1x1=x(1:T); % assign the first T values of the signal x(t) to x1(t)t1=t(1:T); % time points for the period x1(t)

pause(2) % pause for 2 seconds (it is good to use when running the code)% draw or highlight the period x1(t) by the red color on the original signal

hold on; % hold the graph of x(t)plot(t1,x1,’r’); % plot in red the signal x1(t) versus the time points t1title(’original signal x(n) and one period x_1(n)’); % change the title of the graphs

% The result of code on this stage is shown above in Figure 77 (top)

% --------------------------------- PART 2 --------------------------------------------

% 1. Calculate the Fourier series (FS) of the signal x(t) by using the function fft

X1=fft(x1); % or fft(x1,T); calculate the T-point DFT of the signal x1(t)

% according to equation (111). Here X1 stands for F.X1abs=abs(X1); % calculate the DFT in the absolute scale; for plotting only% X1real=real(X1); % calculate the real part of the DFT

% X1imag=imag(X1); % calculate the imaginary part of the DFT% X1ang=phase(X1); % calculate the phase part of the DFT [angle(X1) can be used]

% DFT and FS will be plotted versus the T frequency-points in the interval [0,2*pi).w=linspace(0,2*pi-2*pi/T,T); % T or 100 frequency-points are written in the vector w

% (!!) Consider the relationship between the FFT and Cn coefficients:cn_coefficients=X1/T; % calculation of Cn coefficientscn_coefficients_inabs=abs(cn_coefficients); % Cn in the absolute scale

% The following string variable will be used for the title of FSs_title=’|c_n| coeff. of the Fourier series of signal x_1(n) ’;

subplot(2,1,2); % the 2nd sub-window will be used for the linear spectrum |FS|stem(w,cn_coefficients_inabs); % plot coefficients Cn versus the frequency-points

% one can use plot function as well instead of "stem"

title(s_title); % title will be added to the graphaxis([ 0,2*pi+0.01, 0,3]); % to correct a little the window for the graphicxlabel(’ (b) frequency (rad/s)’); % add the label along x-axis

% -------------- The result of the code on this stage is shown in Figure 78. ------------


0 50 100 150 200 250 300−6

−4

−2

0

2

4

6

8

original signal x(n) and one period x1(n)

(a)

0 1 2 3 4 5 60

0.5

1

1.5

2

2.5

3


1(n)


Fig. 78. The result of the code.

The same linear spectrum can be plotted versus the frequency-points in Hz with thefollowing similar commands. The frequency points will be taken from the interval [0, 1] asthe results of division of the ω-frequency interval [0, 2π] by 2π, since f = ω/(2π).

figure; % open the new figure (Figure 79)

subplot(2,1,2); % the 2nd sub-window will be used for |FS|f=linspace(0,1-1/T,T); % or f=w/(2*pi); frequency-points in Hzstem(f,cn_coefficients_inabs); % plot coefficients Cn versus the frequency-points

axis([ 0-0.02,1+0.02, 0,3]); % to correct the window for the graphictitle(s_title); % add the title to the graphxlabel(’ (b) frequency (Hz)’); % add the label along x-axis

pause(2) % use this line when adding this part to the code

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

2

2.5

3


1(n)

(b) frequency (Hz)

Fig. 79. The Cn coefficients versus the frequency-points in Hz.


The same linear spectrum can shown only in the small sub-interval of the frequency-points.

% -------------------------- Addition commands ------------------------------------------

figure; % open the new figure

subplot(2,1,2); % the 2nd sub-window will be used for |FS|h_st=stem(w,cn_coefficients_inabs); % plot coefficients Cn versus the frequency-points

% by using the pointer h_st, one can change the properties of the graph as shown here:set(h_st, ’MarkerSize’,4, ...

’MarkerEdgeColor’,’b’, ...

’MarkerFaceColor’,’g’);

axis([ 0-0.02,2*pi/5+0.02, 0,3]); % to correct the window for the graphictitle(s_title); % add the title to the graph

xlabel(’ \omega frequency (rad/s)’); % add the label along x-axis

pause(2) % use this line when adding this part to the code

% -------------- The result of the code on this stage is shown in Figure 80. ------------% ----------------------------------------------------------------------------------------

0 0.2 0.4 0.6 0.8 1 1.20

0.5

1

1.5

2

2.5

3


1(n)

ω frequency (rad/s)

Fig. 80. Additional commands: The first 21 Cn coefficients versus the frequency-points in rad/s.

The linear spectrum can be plotted by shifting or cyclicly shifting the coefficients to themiddle, as shown in Figure 81.

% -------------------------- Addition commands ------------------------------------------

figure; % open the new figuresubplot(2,1,2); % the 2nd sub-window will be used for |FS|

% use command fftshift to shift Cn coefficients cyclically the graph to the middle pointcn_coefficients_inabsandshifted=fftshift(cn_coefficients_inabs);

% These coefficients should be plotted versus new frequency-points, namely points in the% interval [-pi,pi]. A simple way is to use the interval for frequency-points w, as shown


w1=w-pi;

h_st2=stem(w1,cn_coefficients_inabsandshifted); % plot shifted coefficients Cn

% Here, we change the properties of the graph as

set(h_st2,’MarkerSize’, 4, ...’MarkerEdgeColor’,’b’, ...’MarkerFaceColor’,’b’);

title(s_title); % add the title to the graphaxis([ -pi-0.05,pi+0.05, 0,3]); % to correct the window for the graphic

xlabel(’ \omega frequency (rad/s)’); % add the label along x-axis

text(-2.75,2.5,’ |c_-n|=|c_100-n| (T=100)’) % add text at points (-2.75,2.5)

% We can save this figure as the eps-file with the name "figure_fs5.eps" asprint -depsc figure_fs5.eps;


−3 −2 −1 0 1 2 30

0.5

1

1.5

2

2.5

3


1(n)

ω frequency (rad/s)

|c_−n|=|c_100−n| (T=100)

Fig. 81. The linear spectrum of the signal x(t), or Cn coefficients versus the frequency-points in rad/s.

% ----------------------------------------------------------------------------------------

I hope you will run and analyze all lines of these scripts and it will help you in writing yourown codes in DSP.

Instructor: Dr. Artyom Grigoryan10/28/2015


XVI. The Fourier series application: Ideal low pass filter with MATLAB

In this section, we consider the simple application of the concept of the Fourier series,or discrete Fourier transform in de-noising of signal, by illuminating the coefficients of theFourier series.

Ideal Low-Pass Filter: Assume, that the coefficients of the Fourier series of the periodicsignal x(t) are plotted versus the frequencies, i.e., consider

Cn = C(ωn), n = 0, 1, 2, ..., (N − 1),

whereωn = nω0, n = 0, 1, 2, ..., (N − 1).

Here, ω0 = 2π/N is the fundamental frequency (in rad/s) of the signal, and the integer Nits period. Given a frequency ωcutoff = ωn0

, the frequencies

0, ω1, ω2, · · · , ωn0−1, ωn0

are considered are low frequencies together with the frequencies

ωn0+1, ωn0+2, · · · , ωN−2, ωN−1,

since the coefficients of the Fourier series are periodic, i.e.,

CN−n = C−n, or C(2π − ωn) = C(−ωn).

Figure 82 shows such a low-pass filter (LPF) with the cut-off frequency

0 2 4 60

0.5

1

1.5

2

Coefficients H(n)=H(ωn) of LPF

(a) frequency (rad/s)

wcutoff

= 0.7793 rad/s

−2 0 20

0.5

1

1.5

2

Coefficients H(n)=H(ωn) of LPF


wcutoff

= 0.7793 rad/s

Fig. 82. Frequency characteristic of the LPF (a) before and (b) after the shifting.

ωcutoff = ω127 = 2π/N · 127 = 0.7988 rad/s.

We denote the coefficients of this LPF by H0, H1, H2, . . . , HN−2, HN−1.


The operation of signal filtration by this filter is described as

Cn → Cn = H(n)Cn, n = 0, 1, 2, ..., (N − 1).

A new signal with the Fourier series coefficients Cn is the filtered signal x, i.e.,

x(n) =N−1∑

k=0

Ckej 2π

Nkn, n = 0, 1, 2, 3, . . . (N − 1). (124)

Below is the scripts of the program which was written in the class to calculate the operationof filtration signal

x(n) → x(n)

over the signal of length N = 512. This script (maybe with some small changes) will also beposted in the web cite (BB).

% ============================== comments =========================================% call: work_lpf_inMSS.m (from class work of October 28, 2015)% ---------------------------------------------------------------------------------

% Example of application of the Fourier series (or discrete Fourier transform):% Ideal filter in the frequency domain with a given cut-off frequency% The coefficients of this filter take values 1 and 0, namely

% 1; in the region of low-frequencies (around zero)% 0; outside this region% This filter is used in the code to filter the noise on the signal%

% For class EE 3424 Math in Signals and Systems% Artyom Grigoryan, October 26-27, 2015, ECE Dept. UTSA% ----------------------------------------------------------------------------------

clear all; close all; % clear all data used before and close all figures% ------------------------------------ part 1 --------------------------------------

% 1. Signal composition from the binary fileS1_name=’data2.8’; % file name with the signal datafid=fopen(S1_name,’rb’); % open to r(read) b(inary) file with name

X=fread(fid); fclose(fid); clear fid; % read the file, close it and remove fidN=length(X); % the length of this signal is largeN=512; % only the first 512 values will be used

x=X(1:N); clear X; % x(n) is the new signal, clear X-data

figure; % open the window for the new figure

subplot(2,2,1); % the window is divided by 2x2 parts, use the 1stplot(x); % plot x(n) versus the time points n=1,2,3,..., Naxis([0,512,0,40]); % to correct a little the window for the graph

S_title=’Original signal x(1:)’; % title for the graphh_title=title(S_title); % add the title to the graphset(h_title,’Color’,’b’,’FontName’,’Times’,’FontSize’,10); % change its properties

xlabel(’ (a) ’); % label the x-axis as "(a)"pause(2) % use this line when running the code

% 2. 512-point DFT calculation (which is the Fourier series up to the constant N)X=zeros(1,N); % memory for the new variable for the DFT, X


X=fft(x); % or fft(x,N); % calculation of the 512-point FDT of x(n)Xabs=fftshift(abs(X)); % cyclically shift of the magnitude of the DFT

subplot(2,2,2); % the 2nd sub-window will be used for the |DFT|plot(fftshift(Xabs)); % plot the original (not shifted) DFT in abs. scale

axis([0,512,0,1000]); % to correct a little the window for the graphtitle(’512-point |DFT| and Ideal LPF’); % add the title to the graphxlabel(’ (b) ’); % label the x-axis as "(b)"

pause(2) % use this line when running the code

0 100 200 300 400 5000

10

20

30

40Original signal x(1:)

(a) 0 100 200 300 400 500

0

200

400

600

800

1000512−point |DFT| and Ideal LPF

(b)

−200 −100 0 100 2000

200

400

600

800

1000the shifted |512−point DFT|

(d) 0 100 200 300 400 500

0

10

20

30

40

(c)

the filtered signal

Fig. 83. (a) The original signal, (b) the 512-point DFT in absolute scale and the LPF, (c) the result offiltration, and (d) the filtered spectrum of the signal.

% ----------------------------------------------------------------------------------------

%3. calculate the low-pass filter (LPF)wcut=64; % cut-off point (integer, not frequency) for the LPFH=ones(N,1); % memory for the new variable with all 1s for the DFT, H

H(wcut+1:N-wcut+1)=0; % fill hight frequencies with zero valueshold on; % hold on the current active part - subplot(2,2,2)plot(H*200,’r’); % plot the LP Filter after amplifying by the factor of 400


%4. Filtration

X1=H.*X; % the operation is reduced to the multiplicationsubplot(2,2,4); % the 4th sub-window will be used for the |X1|


w1=-256:255; % new frequency-points for the shifted spectrumplot(w1,fftshift(abs(X1))); % plot the result of multiplication of spectra

axis([-260,260,0,1000]); % to correct a little the window for the graphtitle(’the original and filtered signals’); % add the title to the graphxlabel(’ (d) ’); % label the x-axis as "(d)"


%4. Calculate and plot the result of filtration of high frequencies

subplot(2,2,3); % the 3rd sub-window will be used for signalx1=real(ifft(X1)); % the filtered signal x1(n)plot(real(ifft(X1))); % plot the signal x1(n)

axis([0,512,0,40]); % to correct a little the window for the graphtitle(’the filtered signal’); % add the title to the graphxlabel(’ (c) ’); % label the x-axis as "(c)"

pause(2) % use this line when running the code% -------------- The result of the code on this stage is shown in Figure 83. ------------% ----------------------------------------------------------------------------------------

% ------------------------------ continuation ------------------------------------------figure; % open the window for the new figuret=1:512; % define the time-points for signals

subplot(2,1,2); % the 2nd sub-window will be used for signalsplot(t,x,’b’, t,real(ifft(X1)),’r’); % plot the signal x(n) in blue, x1(n) in redaxis([0,512,0,40]); % to correct a little the window for the graph

title(’the original and filtered signals’); % add the title to the graph% -------------- The result of the code on this stage is shown in Figure 84. ------------

0 50 100 150 200 250 300 350 400 450 5000

10

20

30

40the original and filtered signals

Fig. 84. The original signal together with the filtered signal.

% ------------------------------ continuation ------------------------------------------

%5. The inverse DFT (h(n)) of the ideal filter; h(n) is called the impulse function of LPFh=zeros(1,N); % memory for the new variable h(k)h=real(ifft(H)); % calculate the inverse 512-DFT of the LPF H(n)

figure; % open the window for the new figuresubplot(2,1,1); % the 1st sub-window will be used for h(k)

plot(h); % plot the impulse function of the LPF H(n)axis([-10,520, -0.1, 0.3]); % to correct a little the window for the graphtitle(’inverse DFT h(n) of the Ideal LPF’); % add the title to the graph


xlabel(’ (a) ’); % label the x-axis as "(a)"

t1=-N/2:N/2-1; % calculate new time-point for the shifted h(k)subplot(2,1,2); % the 2nd sub-window will be used for shifted h(k)plot(t1,fftshift(h)); % plot the impulse function h(k) after shifting

axis([-270,270, -0.1, 0.3]); % to correct a little the window for the graphtitle(’shifted inverse DFT h(n) of the Ideal LPF’); % add the title to the graphxlabel(’ (b) ’); % label the x-axis as "(a)"


0 50 100 150 200 250 300 350 400 450 500−0.1

0

0.1

0.2

0.3inverse DFT h(n) of the Ideal LPF

(a)

−250 −200 −150 −100 −50 0 50 100 150 200 250−0.1

0

0.1

0.2

0.3shifted inverse DFT h(n) of the Ideal LPF

(b)

Fig. 85. The impulse function of the LPF (a) before and (b) after shifting.

% =================================== end of the code ====================================


XVII. Fourier transform

In this section, we discuss the general concept of the Fourier transform, when not onlyperiodic function can be uniquely represented in the frequency-domain.

It was mentioned in the lectures, that the coefficients of the Fourier series of the commonperiodic function f(t) are not functions of the period T or fundamental frequency ω0, i.e.,

C0 6= C0(T ), Cn 6= Cn(T ), n = ±1,±2, . . . ,

andC0 6= C0(ω0), Cn 6= Cn(ω0), n = ±1,±2, . . . .

The scaling of the function f(t)→ f(αt), when α 6= 0, does not change the coefficients, i.e.,

Cn(f(t)) = Cn(f(αt)), n = ±1,±2, . . . .

For generalized functions like the delta function, this fact does not hold.

Example 1: Consider the periodic function x(t) with the period T = 2π; thus ω0 = 2π/T =1 rad/s. We assume that x(t) = δ(t) in the interval [−T/2, T/2) = [−π, π). Then, the Fouriercoefficients of x(t) can be calculated as follows:

cn =1

2π

+π∫

−π

δ(t)e−jnω0t dt =1

2π=

1

T, (125)

for all n = 0,±1,±2, . . . .So, the delta function can be represented by the exponential basis functions as

δ(t) =+∞∑

n=−∞

cnejnω0t =

1

2π

+∞∑

n=−∞

ejnt

=1

2π+

1

π

+∞∑

n=1

cos(nt) =1

T+

2

T

+∞∑

n=1

cos(nω0t) =1

T+

2

T

+∞∑

n=1

cos(

2πnt

T

)

.

Thus, the coefficients, dn and phases, ϑn, of the cosine Fourier series for the periodic deltafunction are defined as

d0 = 1/(2π), dn = 1/π, and ϑn = 0, n = 1, 2, . . . .Note, that

1

2π

+π∫

−π

e−jntejmtdt = δn,m

where δn,m is the Dirac symbol, δn,m = 1 if n = m, and δn,m = 0 if n 6= m.


A. Definition of the Fourier transform (FT)

For not periodic function, one can imagine that the function is considered on the largeinterval [−T/2, T/2) and periodically extended on the real line. The fundamental frequencyω0 = 2π/T becomes small when T grows. In the limit when T → ∞, this period will coverthe entire real line R and the concept of the Fourier series will be reduced to the new conceptwhich we call the Fourier transform.

For an absolute (and square) integrable function x(t) defined on the numeral line R,

+∞∫

−∞

|x(t)|pdt <∞, p = 1, 2

we consider the Fourier transform,

(F x)(ω) = X(ω) =

+∞∫

−∞

x(t)e−jωt dt, ω ∈ (−∞, +∞). (126)

The inverse Fourier transform is described as

x(t) =1

2π

+∞∫

−∞

X(ω)ejωt dω, t ∈ (−∞, +∞). (127)

B. Properties of the FT

1. Duality takes place:x(t) → X(ω)↓ ↓

X(t) → 2πx(−ω)

2. If for a function x(t),+∞∫

−∞

x(t)e−jωtdt = 0,

then x(t) ≡ 0 almost everywhere, namely it follows that the zero response of the integrator

x(t)→

t∫

−∞

x(τ )dτ = 0, ∀t ∈ (−∞, +∞),

yields the input x(τ ) = 0 a.e. In other words, if (x ∗ u)(t) = 0 then x = 0 a.e.

3.

|X(ω)| ≤

+∞∫

−∞

|x(t)|dt, ∀ω ∈ (−∞, +∞).


4. For k-order derivative, f (k)(t), of the function f,

F [f (k)](ω) = (F f (k))(ω) = (jω)kF (ω), k = 1, 2, 3, . . . .

Indeed, using the rule of integration by parts, we obtain the following:

F [f ′](ω) =

+∞∫

−∞

f ′(t)e−jωtdt

= f(t)e−jωt|+∞−∞ + jω

+∞∫

−∞

f(t)e−jωtdt

(f(t)→ 0, t→ ±∞, for absolute integrable function)

= jω

+∞∫

−∞

f(t)e−jωtdt = jωF (ω).

F [f (k)](ω) = jωF [f (k−1)](ω) = (jω)2F [f (k−2)](ω) = · · · = (jω)kF [f ](ω).

Application 1: We can use this property when solving the linear differential equations,for instance,

y(k)(t) + bk−1y(k−1)(t) + · · ·+ b1y

(1)(t) + by(t) = ax(t) + a1x(1)(t).

Indeed, taking the Fourier transform of both sides, we obtain

(jω)kF [y] + bk−1(jω)k−1F [y] + · · · + b1(jω)F [y] + bF [y] = aF [x] + a1(jω)F [x],

F [y](

(jω)k + bk−1(jω)k−1 + · · · + b1(jω) + b)

= F [x] (a + a1(jω)) .

Thus,Y (ω) = F [y](ω) = H(ω)F [x](ω) = H(ω)X(ω),

H(ω) =a + a1(jω)

b + b1(jω) + · · ·+ bk−1(jω)k−1 + (jω)k

And the impulse response function of the linear time-invariant system described by the abovedifferential equation is

h(t) =1

2π

+∞∫

−∞

H(ω)ejωtdω.

5. For a periodic and square integrable function x(t) on the interval [0, 2π] the followingholds:

+∞∑

−∞

|cn|2 =

1

2π

2π∫

0

|x(t)|2dt


where, the coefficients

cn =1

2π

2π∫

0

x(t)e−jnt dt, n = 0,±1, . . . .

In the general case, when the function x(t) is defined on the real line R

+∞∫

−∞

|X(ω)|2 dω = 2π

+∞∫

−∞

|x(t)|2 dt.

The following general formula holds. For given two functions x1(t) and x2(t)

+∞∫

−∞

X1(ω)X2(ω) dω = 2π

+∞∫

−∞

x1(t)x2(t) dt.

As a particular case, the distance between two functions f(t) and g(t) can be measuredin the time domain as well as in the frequency domain (Parseval’s theorem):

+∞∫

−∞

|f(t)− g(t)|2 dt =1

2π

+∞∫

−∞

|F (ω)−G(ω)|2 dω.

if use the previous result for x1(t) = x2(t) = f(t)− g(t).

6. Consider the time-scaled version x(at) of the signal

x(t) → x(at)↓ F ↓ F

X(ω) →1

|a|X(

ω

a)

Indeed, if a > 0+∞∫

−∞

x(at)e−jωtdt =

+∞∫

−∞

x(at)e−j ω

aatdat ·

1

a

=1

a

+∞∫

−∞

x(t)e−j ω

atdt

=1

aX(

ω

a)

In particularx(t) → x(−t)↓ F ↓ F

X(ω) → X(−ω)


x(−1

2t)← x(t) → x(2t)

↓ F ↓ F ↓ F

2X(−2ω)← X(ω) →1

2X(

ω

2)

7. Consider the time-shifted version x(t− t0) of the signal

x(t) → x(t− t0)↓ F ↓ F

X(ω) → e−jωt0X(ω) = (|X(ω)|, ϑ(ω)− ωt0)

where X(ω) = (|X(ω)|, ϑ(ω)) .Indeed,

+∞∫

−∞

x(t− t0)e−jωtdt =

+∞∫

−∞

x(t− t0)e−jω(t−t0+t0)dt·

=

+∞∫

−∞

x(t− t0)e−jω(t−t0)e−jωt0dt

= e−jωt0

+∞∫

−∞

x(t− t0)e−jω(t−t0)d(t− t0)

= e−jωt0X(ω)

In particularx(t− 1) ← x(t) → x(t + 1)↓ F ↓ F ↓ F

e−jωX(ω)← X(ω) → ejωX(ω)

Due to the principle of duality, the following diagram holds

ejtω0x(t) . x(t− t0) → e−jt0ωX(ω)↓ . F

X(ω − ω0) . e−jtω0X(t) → 2πx(ω − ω0)

8. Using property 7, we can find the Fourier transform of functions

y(t) = x(t) sin(w0t), y(t) = x(t) cos(w0t).

Indeed

x(t) sin(w0t) = x(t)ejw0t − e−jw0t

2j

=1

2j

[

x(t)ejw0t − x(t)e−jw0t]


It directly results the diagram

x(t) → x(t) sin(w0t)↓ F ↓ F

X(ω) →1

2j[X(ω − ω0)−X(ω + ω0)]

Similarly, the following diagram is valid for function x(t) cos(w0t) :

x(t) → x(t) cos(w0t)↓ F ↓ F

X(ω) →1

2[X(ω − ω0) + X(ω + ω0)]

9. For amplitude amplification, we have the following diagram for the Fourier transform:

x(t) → Ax(t) + B↓ F ↓ F

X(ω) → AX(ω) + B · 2πδ(ω).

10. The theorem of multiplication of spectra:

x(t), h(t) =⇒ (x ∗ h)(t)↓ ↓ ↓

X(ω), H(ω) =⇒ X(ω)H(ω)

Indeed,+∞∫

−∞

(x ∗ h)(t)e−jωtdt =

+∞∫

−∞

+∞∫

−∞

x(t− τ )h(τ ) dτ

e−jωtdt

In the integral, exponential function can be written as

e−jωt = e−jω(t−τ+τ ) = e−jω(t−τ )e−jωτ

and therefore the last integral equals

+∞∫

−∞

+∞∫

−∞

x(t− τ )h(τ ) dτ

e−jω(t−τ )e−jωτdt

=

+∞∫

−∞

x(t− τ )e−jω(t−τ ) dt

+∞∫

−∞

h(τ )e−jωτdτ

= X(ω)H(ω).


In particular, when h(t) is the delta function,

x(t), δ(t) =⇒ (x ∗ δ)(t) = x(t)↓ ↓ ↓ ↓

X(ω), D(ω) =⇒ X(ω)D(ω) = X(ω)

Therefore, D(ω) = (F δ)(ω) ≡ 1.And, due to the principle of duality

δ(t) → D(ω) ≡ 1↓ ↓

D(t) ≡ 1 → 2πδ(−ω) = 2πδ(ω)

Example 2: Consider the exponential function

x(t) = e−a|t|, a > 0.

The Fourier transform of this functions can be computed as follows:

X(ω) =

+∞∫

−∞

x(t)e−jωtdt =

+∞∫

−∞

e−a|t|[cos ωt− j sinωt]dt

=

+∞∫

−∞

e−a|t| cosωtdt

= 2

+∞∫

0

e−at cosωtdt =2a

a2 + ω2

Due to the duality

x(t) =2a

a2 + t2←→ 2πe−a|ω|.

In particular a = 1 case,

e−|t| ←→2

1 + ω2,

2

1 + t2←→ 2πe−|ω|.

Example 3: Given a > 0, consider the exponential function

x(t) = e−atu(t), t ∈ (−∞, +∞).

Then

X(ω) =

+∞∫

0

e−ate−jωtdt

=

+∞∫

0

e−(a+jω)tdt =1

a + jω


and due to the duality,

x(t) =1

a + jt←→ 2πeaωu(−ω).

Example 4: Consider the rectangle function on the interval [−T0, T0],

x(t) = rect(

t

2T0

)

= u(t + T0)− u(t− T0).

Then,

X(ω) =

+∞∫

−∞

x(t)e−jωtdt =

T0∫

−T0

e−jωtdt

=ejωT0 − e−jωT0

jω

=2j sin ωT0

jω

= 2T0sin ωT0

ωT0= 2T0 sinc(ωT0)

where we denote

sinc(t) =sin(t)

t, t 6= 0, sinc(0) = 1.

As a particular case, we obtain

rect(t)←→ sinc(ω/2),

Note that sinc(ω) function is not absolute integrable, i.e.,

+∞∫

−∞

|sinc(ω)| dω =∞.

Figure 86 shows the sinc(ω) function.Example 5: Consider the integrator

x(t)→ y(t) =

t∫

−∞

x(τ ) dτ.

The response function of the integrator is u(t), and the following diagram valid

x(t), u(t) =⇒ (x ∗ u)(t)↓ ↓ ↓

X(ω), U(ω) =⇒ X(ω)U(ω)


−6 −4 −2 0 2 4 60

0.5

1

1.5Rectangle and DFT

t

Am

plit

ud

e

−0.3 −0.2 −0.1 0 0.1 0.2 0.3

−500

0

500

Re

al(F

FT

[x])

w

−0.3 −0.2 −0.1 0 0.1 0.2 0.30

200

400

600

ab

s(F

FT

[x])

w

Fig. 86. The rectangle function, its Fourier transform, sinc(ω), and |sinc(ω)| funstion.

The Fourier transform of the unit step function is

(F u)(ω) = U(ω) =1

jω+ πδ(ω).

Therefore,

X(ω)U(ω) = X(ω)

[

1

jω+ πδ(ω)

]

=1

jωX(ω) + πX(0)δ(ω).


We now consider the Fourier transform over functions (non periodic) defined on the realline R = (−∞, +∞).

Example 6: Consider first the periodic pulse function defined on the interval [0, T ] as

x(t) = rect(t) → z(t) = x(t −

1

2

)

z(t) → y(t) = z(

t

T

)= x

(t

T−

1

2

)

−1/2 1/2

1

x(t)

1

z(t)

4

y(t)

t t t

1 1

Fig. 87. Transformation of the unit rectangle.

The Fourier transforms of above functions are described by the diagram as

x(t) → z(t) = x(t − 12) → z( t

T) = x

(tT− 1

2

)

↓ F ↓ F ↓ F ↓ F ↓ F

X(ω) → Z(ω) = X(ω)e−j 1

2ω → TZ(Tω) = TX(Tω)e−j 1

2ωT

We also know that the Fourier transform of rect(t) is the sinc function

X(ω) = sinc(

ω

2

)

therefore, we obtain

F : x(

t

T−

1

2

)→ T sinc

(Tω

2

)e−j 1

2ωT .

Example 7: Consider function

x(t) = e−a|t|, a > 0.

To find the Fourier transform of x(t), we perform the following calculations

X(ω) =

∞∫

−∞

e−a|t|e−jωt dt


=

∞∫

−∞

e−a|t|[cosωt − j sinωt] dt

=

∞∫

−∞

e−a|t| cos ωt dt = 2

∞∫

0

e−at cos ωt dt

Therefore, by using two times the integration by parts we find the following:

∞∫

0

e−at cos ωt dt = −1

a

∞∫

0

cos(ωt)de−at

= −1

a

cos(ωt)e−at |∞0 −

∞∫

0

e−atd cos ωt

= −1

a

0 − 1 − ω

∞∫

0

e−at sin(ωt)dt

= −1

a

−1 − ω

−1

a

∞∫

0

sin(ωt)de−at

= −1

a

−1 +ω

a

sin(ωt)e−at |∞0 −

∞∫

0

e−atd sin(ωt)

= −1

a

−1 +ω2

a

∞∫

0

e−at cos(ωt)dt

Denoting by

A =

∞∫

0

e−at cosωt dt

we obtain

A = −1

a

[−1 +

ω2

aA

]→ aA = 1 −

ω2

aA,

which results inA =

a

a2 + ω2.

Finally, we obtain

X(ω) =

∞∫

−∞

e−a|t|e−jωt dt = 2A =2a

a2 + ω2.

x(t) is symmetric and its Fourier transform is real.

Example 8: For signalx(t) = e−atu(t), a > 0,


−10 −8 −6 −4 −2 0 2 4 6 8 100

0.5

1

1.5

2

2.5Symmetric exponential function and FT (a=2)

t w

x(t)

X(w)

Art

−10 −8 −6 −4 −2 0 2 4 6 8 100

1

2

3

4

Symmetric exponential function and FT (a=1/2)

t w

x(t)

X(w)

Fig. 88. Laplacian functions x(t) = 2e−a|t| and their Fourier transforms for a = 2 and a = 0.5.

the Fourier transform can be calculated as follows

X(ω) =

∞∫

−∞

(e−atu(t)

)e−jωtdt

=∫ ∞

0e−ate−jωtdt =

∞∫

0

e−at−jωtdt

=

∞∫

0

e−(a+jω)tdt = −1

a + jωe−(a+jω)t|∞o

=1

a + jω

So, the power spectrum of the signal

|X(ω)|2 =1

a + jω·

1

a − jω=

1

a2 + ω2

if Re a 6= 0.


Example 9: Consider the Fourier transform of the Fourier series of the periodic functionx(t) with period T. Taking the Fourier series of the function

x(t) =∑

n∈Z

cnejω0nt, ω0 = 2π/T,

we can write the Fourier transform as

X(ω) =

∞∫

−∞

x(t)e−jωtdt

=

∞∫

−∞

[∑

n∈Z

cnejω0nt

]e−jωtdt

=∑

n∈Z

cn

∞∫

−∞

ejω0nte−jωtdt

=∑

n∈Z

cn

∞∫

−∞

e−jt(ω−ω0n)dt

=∑

n∈Z

cn2πδ(ω − ω0n)

= 2π∑

n∈Z

cnδ(ω − nω0)

B. Fourier transform of periodic functions

We now consider main properties of the Fourier transform of a periodic function, namely,the representation of the Fourier transform. Let T be a fundamental period of a functionx(t),

x(t) = x(t + T ), t ∈ R.

Then, defining the function (one period)

x(t) =

x(t), t ∈ [0, T )0, t /∈ [0, T )

we can write function x(t) as the following set of shifted versions of x(t), i.e.

x(t) = ..., x(t + T ), x(t), x(t − T ), ...

or

x(t) =∑

k∈Z

x(t − kT ) =∑

k∈Z

x(t) ∗ δ(t − kT )

= x(t) ∗∑

k∈Z

δ(t− kT ).


We know, the Fourier series of the periodic infinite-pulse (delta) function can be written as

∑

k∈Z

δ(t− kT ) =∑

k∈Z

1

Te−jkω0t, ω0 = 2π/T.

Therefore

x(t) = x(t) ∗∑

k∈Z

1

Te−jkω0t.

Then, denoting by

X(ω) =

∞∫

−∞

x(t)e−jωt dt,

X(ω) =

∞∫

−∞

x(t)e−jωt dt,

and using the fact that the convolution is a operation of multiplication in the Fourier domain,we obtain the following:

x(t) = x(t) ∗1

T

∑

k∈Z

e−jkω0t

↓ ↓ ↓ ↓

X(ω) = X(ω) ·1

T2π

∑

k∈Z

δ(ω − kω0)

So

X(ω) =2π

T

∑

k∈Z

X(ω)δ(ω − kω0)

= ω0

∑

k∈Z

X(kω0)δ(ω − kω0)

and the Fourier transform X(ω) of the periodic function has nonzero values only at integerfrequencies kω0.

Example 10: In the case, when x(t) = δ(t), we have

x(t) = δ(t)↓ ↓

X(ω) = 1

x(t) =∑

n∈Z

x(t − nT )

↓ ↓

X(ω) = ω0

∑

n∈Z

δ(ω − nω0)

where the fundamental frequency ω0 = 2π/T.


C. Frequency characteristics of a LTI system

Let H be a linear time-invariant system with the impulse response function h(t). Weconsider the input of the system, that have the form

x(t) = ejωt, given ω ∈ R.

Then, by definition of the linear convolution,

x(t) → (x ∗ h)(t) =

∞∫

−∞

h(τ )x(t − τ ) dτ

=

∞∫

−∞

h(τ )ejω(t−τ ) dτ

=

∞∫

−∞

h(τ )ejωte−jωτ dτ

= ejωt

∞∫

−∞

h(τ )e−jωτ dτ

= ejωtH(ω)

(Y (ω)

X(ω)= H(ω)

)

So

ejωt → ejωtH(ω)

a1ejω1t + a2e

jω2t → a1ejω1tH(ω1) + a2e

jω2tH(ω2)

and using the Fourier transform representation, we obtain

x(t) =

∞∫

−∞

X(ω)ejωt dω

→

∞∫

−∞

X(ω)H[ejωt] dω

y(t) =

∞∫

−∞

X(ω)H(ω)ejωt dω

which is the theorem of multiplication of spectra,

y(t) = x(t) ∗ h(t) → X(ω)H(ω) = Y (ω).


XVIII. The Laplace transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EE-3424

The most common system in electrical engineering is circuit analysis that can be done in thetime domain or/and in the frequency domain. Because of powerful methods of the linear algebra,

the circuit analysis is usually done in the frequency domain. The analysis of differential equationsthat describe the relations between output and input as well as between different components of

the circuit is reduced to algebraic equations. Many simple circuit elements such as resistors, con-ductors, capacitors, diodes, voltage and current sources can be modeled as linear devices (systems)

and described by transfer functions. The frequency analysis uses mainly the Laplace transform.The technique of the Fourier transform cannot be applied in many cases for signals or system

characteristics for which the integral does not converge.In many practical applications, we need solve differential equations with functions f(t) that grow

fast when t tends to ±∞. The functions are not absolute integrable, and therefore, the Fouriertransform cannot be applied directly to simplify the differential equations in the frequency domain.But if functions do not grow faster than exponential functions, then a new Fourier integral can be

evaluated. We face with a situation when a function f(t) is not integrable but becomes such aftermultiplying by exponential function e−αt, for a certain real (or complex) number a. For example, the

function f(t) = e2tu(t) is not absolute integrable, but the function f(t)e−3t = e2te−3tu(t) = e−tu(t)is integrable. In other words, for such case the following integral exists

F[f(t)e−αt

]=

∞∫

−∞

[f(t)e−αt

]e−jωtdt =

∞∫

−∞

f(t)e−(α+jω)tdt (128)

we can even derive the domain of α for which this integral is defined. Indeed, let α = α1 + jα2,then the integral in the above equation can be written as

∞∫

−∞

f(t)e−(α1+jα2+jω)tdt =

∞∫

−∞

f(t)e−α1te−j(α2+ω)tdt

and if the function f(t)e−α1t is absolute integrable then the transform F[f(t)e−αt

]exists.

Example 1: Given number p (on the imaginary vertical p = jω), we consider the following trans-

form for the unit function u(t):

f(t) = u(t)→

∫ ∞

−∞u(t)e−ptdt =

∫ ∞

0e−ptdt

=

∫ ∞

0

1

−pde−pt =? −

1

pe−pt |∞0 → (p = jω)→? 1

jω−

1

jωe−jωt|t=∞

with no result, because the last integral cannot be evaluated. Let us now multiply the function f(t)

by a factor e−αt, where α = α1 + jα2 is a complex number to be defined. The Fourier transform ofthe new function g(t) = f(t)e−αt is defined as

g(t) → G(ω) =

∫ ∞

−∞g(t)e−jωtdt =

∫ ∞

−∞u(t)e−αte−jωtdt

=

∫ ∞

0e−(α1+jα2)te−jωtdt =

∫ ∞

0e−j(α2+ω)te−α1tdt ∃


because ∣∣∣e−j(α2+ω)te−α1t∣∣∣ = e−α1t → 0, t→∞,

when α1 > 0. We can therefore define integral G(ω) for any α1 > 0, or Re α > 0.To understand the concept of the Laplace transform, we first describe the response of a linear

time invariant system L to a complex exponential signal

x(t) = xs(t) = Xest, t ∈ R, (s ∈ R or C2)

where s is a real or complex number. The response is defined as follows

xs(t) → y(t) = xs(t) ∗ h(t) =

∞∫

−∞

h(τ)xs(t− τ)dτ

=

∞∫

−∞

h(τ)Xes(t−τ )dτ = Xest

∞∫

−∞

h(τ)e−sτdτ = XestH(s) = xs(t)H(s)

where the coefficient H(s) is defined by

H(s) =

∞∫

−∞

h(τ)e−sτdτ. (129)

Similarly, the response of the system to a linear combination of complex exponentials xs1(t) and

xs2(t) is expressed as a linear combination of exponentials with the coefficients multiplied by H(s1)

and H(s2). In other words,

X1xs1(t) + X2xs2

(t)→ H(s1)X1xs1(t) + H(s2)X2xs2

(t).

We recall here that the Fourier transform of a time-continuous function is a linear combination(finite or infinite) of a complex sinusoidal functions of the form ejωt, where ω is a frequency (real

variable)

F (ω) =

∞∫

−∞

f(t)e−jωtdt.

A. Definition of the Laplace transform

Consider the function f(t) of the real variable t ∈ (−∞, +∞). The transform

F (p) =

∞∫

0

f(t)e−ptdt, p = p1 + jp2, p1, p2 ∈ (−∞, +∞), (130)

is called the Laplace transform (or unilateral Laplace transform). The set of p for which the

Laplace transform can be defined is called a region of convergence for which we will use notationROC(F ) or (ROC).

The transformationL : f(t)→ F (p)


of the function of real variable t to the complex function of complex variable p, is defined forfunctions f(t) of exponential order on [0,∞) , i.e., for which there exist positive number M and a

such that|f(t)| ≤Meatu(t), t ∈ (−∞, +∞).

The integral in (130) can be defined if Re p = p1 > a.

Indeed, the following calculations hold:

∣∣∣∣∣∣

∞∫

0

f(t)e−ptdt

∣∣∣∣∣∣≤ M

∞∫

0

eate−p1tdt = M

∞∫

0

e−(p1−a)tdt

= M1

−(p1 − a)

∞∫

0

de−(p1−a)t = M1

p1 − a

when p1 > a.The Laplace transform exists for any piecewise continuous function of exponential order.Example 2: For the unit function u(t), the Laplace transform is

U(p) =

∞∫

0

u(t)e−ptdt =

∞∫

0

e−ptdt =1

−p

∞∫

0

de−pt =1

p

when Re p > 0.

Example 3: For the exponential function

f(t) = eαtu(t),

the Laplace transform is calculated as follows:

F (p) =

∞∫

0

f(t)e−ptdt =

∞∫

0

eαte−ptdt =1

p− α

when Re p > Re α.

As a result, we obtain the following

L : u(t)→1

p, Re p > 0,

and for the cosine function the Laplace transform is defined as follows

cos(ωt)u(t) =1

2[ejωt + e−jωt]u(t)

↓ L ↓ L

L[cos(ωt)u(t)] =1

2

[1

p− jω+

1

p + jω

]=

[p

p2 + ω2

]

if Re p > |Re (jω)|= 0.


We obtain a similar result for the sine function

sin(ωt)u(t) =1

2j[ejωt − e−jωt]u(t)

↓ L ↓ L

L[sin(ωt)u(t)] =1

2j

[1

p− jω−

1

p + jω

]=

[ω

p2 + ω2

].

B. Properties of the Laplacian transform

1. (Linearity)If

x(t)→ X(p), y(t)→ Y (p),

thenx(t) + ky(t)→ X(p) + kY (p),

for any constant k ∈ R, but p from the intersection of the ROCs.

2. (Delay)

Let us define the shift function

xt0(t) = x(t− t0)u(t− t0) =

0, t < t0,x(t− t0), t ≥ t0.

Then, the following calculations hold:

x(t) → xt0(t)↓ L ↓ L

X(p) → X(p)e−pt0

for any real t0.

Indeed,

Xt0(p) =

∞∫

0

xt0(t)e−ptdt =

∞∫

t0

x(t− t0)e−ptdt =

∞∫

0

x(t)e−p(t+t0)dt = X(p)e−pt0

for p such that Re p > α.

Example 4: The piecewise function

f(t) =

0, t < a,

n, na ≤ t < (n + 1)a, n = 1, 2, 3, ...

can be written as

f(t) = [u(t− a) + u(t− 2a) + u(t− 3a) + ...]

The following diagram holds for the Laplace transforms

f(t) = u(t− a) + u(t− 2a) + u(t− 3a) + ...↓ L ↓ L ↓ L ↓ L

F (p) = U(p)e−pa + U(p)e−2pa + U(p)e−3pa + ...


and since U(p) = 1/p, we obtain

F (p) =1

p[e−pa + e−2pa + e−3pa + ...]

=1

p[(e−pa) + (e−pa)2 + (e−pa)3 + ...] =

1

p

e−pa

1− e−pa, if p ∈ ROC(F ).

Region of convergence of the above geometrical progression is defined as the set of numbers p for

which |e−pa| < 1. Since p = p1 + jp2, then |e−pa| = |e−p1a| · |e−jp2a| = |e−p1a| < 1. Since a > 0, thep1 > 0 and ROC(F ) is the set p; Re p > 0.

3. (Derivative)

x(t) → x′(t)↓ L ↓ L

X(p) → pX(p)− x(0)(Re p > a)

Indeed,

L[x′](p) =

∞∫

0

x′(t)e−ptdt = x(t)e−pt|∞0 + p

∞∫

0

e−ptx(t)dt = −x(0) + pX(p).

Similarly, in the general n ≥ 1 case, we obtain

L : x(n)(t)→ pn

[

X(p)−x(0)

p−

x′(0)

p2−

x(2)(0)

p3− ...−

x(n−1)(0)

pn

]

.

Application

Consider the solution of the differential equation with constant coefficients

y(n)(t) + a1y(n−1)(t) + a2y

(n−2)(t) + ... + any(t) = bx(t),

y(0) = y′(0) = y(2)(0) = ... = y(n−1)(0) = 0.

Using the Laplace transform, we obtain

Y (p)[pn + a1pn−1 + a2p

n−2 + · · ·+ an−1p′ + an] = bX(p),

or

Y (p) =bX(p)

Pn(p)=

bX(p)

pn + a1pn−1 + a2pn−2 + ... + an

.

4. (Integrator)

x(t) →

t∫

0

x(τ)dτ

↓ L ↓ L

X(p) →X(p)

p

(Re p > a)


Indeed, if x(t)→ y(t), then

y′(t) = x(t), y(0) = 0

↓ L ↓ LpY (p)− y(0) = X(p)

4a. (n Integrators)

x(t) →

t∫

0

dt1

t1∫

0

dt2 . . .

tn−1∫

0

x(tn)dtn

↓ L ↓ L

X(p) →X(p)

pn

(Re p > a)

5. (Convolution)

x(t), y(t), x(t) ∗ y(t)

↓ L ↓ L ↓ LX(p), Y (p), X(p)Y (p)

(Re p > maxa, b)

if

|x(t)| ≤Meatu(t), |y(t)| ≤ Nebtu(t), t > 0.

Example 5: Let us find the original function of the following Laplace transform

F (p) =wp

(p2 + ω2)2.

Then, we can compose the following diagram:

cos(ωt)u(t), sin(ωt)u(t), [cos(ωt)u(t)] ∗ [sin(ωt)u(t)] =t

2sin(ωt)

↓ L ↓ L ↓ L[p

p2 + ω2

],

[ω

p2 + ω2

],

wp

(p2 + ω2)2= F (p)

if Re p > |Im ω|, or Re p > 0 if ω is real.

Therefore, the convolution of these cosine and sine waves can be calculated as follows:

f(t) =

t∫

0

sin(ω[t− τ ]) cos(ωτ)dτ =1

2

t∫

0

[sin(ωt) + sin(ω[t− 2τ ])]dτ

=1

2

t∫

0

sin(ωt)dτ +1

2

t∫

−t

sin(ωτ)dτ =t

2sin(ωt).


6. (Inverse formula)

f(t) =1

2πj

x+j∞∫

x−j∞

F (p)eptdp, x > a.

where x is real. The vertical line Re p = x is from the region of the existence of the Laplace

transform of f(t).

7. (Delta function)

For any real number t0 > 0,δ(t) → δ(t− t0)

↓ L ↓ L1 → e−pt0 .

Indeed,∞∫

0

δ(t− t0)e−ptdt = e−pt|t=t0 = e−pt0.

Example 6: For the ramp function

x(t) = r(t)u(t),

the following diagram takes place

e−atu(t) = 1u(t) = x′(t) ← x(t)

↓ L (a = 0) ↓ L ↓ L ↓ L1

p + a=

1

p= pX(p) ← X(p)

and, therefore,

X(p) =1

p2, Re p > 0.

8. (Time scaling)

For any real number k > 0,

f(t) → f(kt)↓ L ↓ L

F (p) →1

kF (

p

k)

Indeed,

∞∫

0

f(kt)e−tpdt

=

∞∫

0

f(kt)e−p 1

kktdkt

1

k


=1

k

∞∫

0

f(t)e−p

ktdt

=1

kF (

p

k).

Remark that k > 0 and we do not define the Laplace transform over the function f(−t), when t ispositive.

Example 7: Let us consider the time-transformation version of the function sin(t)u(t)

sin(t)→ sin(ωt − t0)u(ωt− t0) = sin

(ω

[t−

t0ω

])u

(ω

[t−

t0ω

]).

Then, the Laplace transform can be calculated as

sin(ωt)u(t) → sin

(ω(t−

t0ω

)

)u

(ω(t−

t0ω

)

)

↓ L ↓ L

ω

p2 + ω2→

ω

p2 + ω2e−p

t0ω

or, we can use the calculation by the following diagram:

sin(t)u(t) → sin (t− t0) u (t− t0) → sin (tω − t0) u (tω − t0)↓ L ↓ L ↓ L

1

p2 + 1→

1

p2 + 1e−pt0 →

1

ω

1( p

ω

)2+ 1

e− p

ωt0

9. (Derivative of the Laplace image)

The following diagram takes place

F (p) → −F ′(p)

↑ L ↑ Lf(t) → tf(t)

Indeed,

F ′(p) =d

dp

∞∫

0

f(t)e−ptdt

= −

∞∫

0

f(t)te−ptdt

= −

∞∫

0

[tf(t)]e−ptdt

= −L[tf(t)](p).


Example 8: Let us consider the function f(t) = sin(t), then

sin(ωt) → cos(ωt) ∗ sin(ωt) = t2 sin(ωt)

↓ L ↓ L ↓ Lω

p2 + ω2

wp

(p2 + ω2)2= −1

2

(ω

p2 + ω2

)′

if Re p > 0.

Similarly, we obtain

cos(ωt)u(t) → t · cos(ωt)u(t)

↓ L ↓ Lp

p2 + ω2→ −

(p

p2 + ω2

)′

=ω2 − p2

(ω2 + p2)2

since

−

(p

p2 + ω2

)′

=1 · (p2 + ω2)− 2p2

(p2 + ω2)2.

Another example of using this property is described by the following diagram:

tu(t) → t2u(t) t3u(t) → t4u(t)↓ L ↓ L ↓ L ↓ L1

p2→ −

(1

p2

)′

=2

p3→ −

(2

p3

)′

=2 · 3

p4→ −

(2 · 3

p4

)′

=2 · 3 · 4

p5

Thereforetu(t) → tnu(t)↓ L ↓ L1

p2→

2 · 3 · 4 · · ·n

pn+1=

n!

pn+1

when Re p > 0.

10. (Periodic function)

Let f(t) be a periodic function with fundamental period T,

f(t) = f(t + T ), t ∈ R.

Denoting byf(t) = f(t), t ∈ [0, T ],

we obtain the following:

∞∫

0

f(t)e−ptdt

=

T∫

0

+

2T∫

T

+

3T∫

2T

+ · · ·

f(t)e−ptdt


=

T∫

0

f(t)e−ptdt +

+

2T∫

T

f(t)e−ptdt =

T∫

0

f(t + T )e−p(t+T )dt

+

3T∫

2T

f(t)e−ptdt =

T∫

0

f(t + 2T )e−p(t+2T )dt

+ · · ·

=

T∫

0

f(t)e−pt[1 + e−pT + e−p2T + · · ·

]dt

=

T∫

0

f(t)e−pt 1

1− e−pTdt

=1

1− e−pT

T∫

0

f(t)e−ptdt =1

1− e−pT

∞∫

0

f(t)e−ptdt

=1

1− e−pT· F (p)

when |e−pT | < 1 (which means that p1 > 0). We define by F (p) the Laplace transform of theperiodic part f (t) of f(t).

12. (Fourier and Laplace transforms)

Consider a function f(t) of exponential order, t is the real variable from (−∞, +∞).The transform

L : f(t)→ F (p) =

∞∫

0

f(t)e−ptdt, p ∈ C2, (131)

is the Laplace transform.The transform

F : f(t)→ F (ω) =

∞∫

−∞

f(t)e−jωtdt, ω ∈ R, (132)

is the Fourier transform, which can be written as

L : f(t)→ F (jω) =

∞∫

0

f(t)e−(jω)tdt, ω ∈ R, (133)

if f(t) = 0, for all t < 0.So, for functions f(t) = 0, t < 0, the Fourier transform F (ω) is the part of the Laplace transform

F (p), i.e., the Laplace transform on the line p = jω, ω ∈ R. It is assumed that the Laplace transformexists on the vertical line p = jω.


For example,

f(t) = e−αtu(t), α > 0,

Lf(t)(p) = F (p) =1

p + α, p ∈ C2, (Re p > 0)

and

Ff(t)(ω) = F (ω) =1

jω + αω ∈ R.

We have defined the Laplace transform of f(t) for p such that Re p > 0, and not for Re p = 0.

F (p)|p=jω = F (ω).

But, for the function

f(t) = e−α|t|, α > 0,

the Fourier image is

Ff(t)(ω) = F (ω) =2α

ω2 + α2, ω ∈ R

but this function is not zero for negative ω, and the Laplace transform cannot be defined.

The following example, when these two transform are ”not equal”:

f(t) = u(t).

Indeed,

Lf(t)(p) = U(p) =1

p, Re p > 0, (134)

and

Ff(t)(ω) = U(ω) = πδ(ω) +1

jω, ω ∈ R.

In this case also, we have defined the Laplace transform of u(t) for p such that Re p > 0, andformula (134) cannot be used for vertical line Re p = 0. Indeed,

1

p = jw=

1

jω6=

1

jω+ πδ(ω) = U(ω).

So, we extend the Laplace transform on the vertical Re p = 0, defining the Laplace transform as

the Fourier transformU(p)|p=jω = U(ω).


13. (Stability of causal systems)Let h(t) be an impulse response function such that h(t) = 0, t < 0, and let H(p) be the Laplace

transform of h(t). We consider the transfer function of an nth-order system x(t)→ y(t),

H(p) =Y (p)

X(p)=

bn−1pn−1 + ... + b1p + b0

pn + an−1pn−1 + ... + a1p + a0

where bn−1 6= 0.

We know, that the transfer function can be represented as

H(p) =b′1

p− p1+

b′2p− p2

+ ... +b′n

p− pn

where p1, p2, ..., pn are zeros of the denominator (or, poles of the H(p)), i.e.

pn + an−1pn−1 + ... + a1p + a0 = (p− p1)(p− p2) · · · (p− pn).

The output transform can be represented as

Y (p) =k1

p− p1+

k2

p− p2+ ... +

kn

p− pn

+ YX(p)

where YX(p) has the same poles as X(p).

Therefore,y(t) = k1e

p1t + k2ep2t + ... + knepnt + yx(t).

The system H is unstable when the output is unbounded for bounded input x(t), or when thereexists

pm > 0, m ∈ [1, n].

For example,

H(p) =1

p + 2+

2

p− 3+

3

p + 4←− e−2t + 2e3t + 3e−4t

is for not stable system (p2 = 3 > 0), but

H(p) =1

p + 2+

2

p + 3+

3

p + 4←− e−2t + 2e−3t + 3e−4t

is the transfer function for the stable system.

We remain here, that a complex function

X(p) : C2 → C2, (C2 = R× jR)

has a pole at point p = p0 iflim

p→p0

X(p) = ±∞,

and we say this function has a zero at point p = p0 if

limp→p0

X(p) = 0.


For example, functions

X(p) =1

p + 2, X(p) = (p− 2), X(p) =

p− 2

p + 2

have poles or/and zeros respectively at points p = −2 and p = 2.

A rational function is a ratio of two (numerator and denominator) polynomials with real coeffi-cients

X(p) =Pm(p)

Pn(p)=

bmpm + bm−1pm−1 + ... + b1p + b0

pn + an−1pn−1 + ... + a1p + a0,

and n > m is order of the function X(p).We consider X(p) in the form

X(p) =Pm(p)

pn + an−1pn−1 + ... + a1p + a0

and assume that it has n distinct poles, p1, p2, ..., pn.Then, there exist coefficients b1, b2, ..., bn such that

X(p) =b1

p− p1+

b2

p− p2+ ... +

bn

p− pn

(this is so-called a partial fraction representation of X(p)).

Example 9:

X(p) =p + 3

p(p + 2)=

p + 2 + 1

p(p + 2)=

1

p+

1

p(p + 2)

=1

p+

1

2

[1

p−

1

p + 2

]

=3

2·1

p−

1

2·

1

p + 2=

32

p−

12

p + 2

This function has two poles at points p1 = 0, and p2 = −2, and b1 = 3/2, and b2 = −1/2. There

is another simple way to find the coefficients b1 and b2.Indeed,

pX(p)|p=0 =

[p + 3

p + 2

]

p=0

=3

2= b1,

(p + 2)X(p)|p=−2 =

[p + 3

p

]

p=−2

= −1

2= b2.

So, the function with the image equal X(p) can be computed as

X(p) =3

2·1

p−

1

2·

1

p + 2↑ L ↑ L ↑ L

x(t) =3

2· u(t) −

1

2· e−2tu(t)


so

x(t) =3

2· u(t)−

1

2· e−2tu(t) =

1

2

[3− e−2t

]u(t).

In the general case, to find the coefficient bk, we calculate

(p− pk)X(p)|p=pk= bk, k = 1, 2, ..., n.

Example 10:

X(p) =1

p2(p + 2)=

d1

p+

b1

p2+

b2

p + 2.

We have

p2X(p)|p=0 =

[1

p + 2

]

p=0

=1

2= b1,

(p + 2)X(p)|p=−2 =

[1

p2

]

p=−2

=1

4= b2,

therefore

X(p) =d1

p+

12

p2+

14

p + 2

and we can find that d1 = −1/4.The function x(t) with that image can be computed from the diagram:

X(p) = −1

4·1

p+

1

2·

1

p2+

1

4·

1

p + 2↑ L ↑ L ↑ L ↑ L

x(t) = −1

4· u(t) +

1

2· tu(t) +

1

4· e2tu(t)

which results

x(t) =

[−

1

4+

1

2t−

1

4e2t

]u(t).

14. Uniqueness

Let f(t) 6= g(t) be two continuous functions with an exponential order, and let their Laplacetransforms are the same

F (p) = G(p), p ∈ ROC(F ) ∩ ROC(G).

Then, for any interval [0, T ], where T > 0, the functions coincide, f(t) = g(t), except maybe for

a finite number of points.


14. Limits of pF (p)

A. Initial value theorem:

f(0) = limp→∞

pF (p). (135)

B. Final value theorem: If the limit limt→∞ f(t) exists, then

limt→∞

f(t) = limp→0

pF (p). (136)

The following example shows the necessity of condition for this property. Consider the function

F (p) =p

p2 + 4

ThenF (p) =

p

p2 + 4L←− cos(2t)u(t)→ lim

t→∞cos(2t)u(t) does not exist.

Example 11:

F (p) =p + 2

p2 + 4p + 5.

Then

f(0) = limp→∞

pF (p) = limp→∞

p2 + 2p

p2 + 4p + 5= 1

limt→∞

f(t) = limp→0

pF (p) = limp→0

p ·p + 2

p2 + 4p + 5= 0.

Example 12:

F (p) =p + 2

p2 + 4p− 5=

12

p− 1+

12

p + 5

Then

f(0) = limp→∞

pF (p) = limp→∞

[1

2

p

p− 1+

1

2

p

p + 5

]= 1

limp→0

pF (p) = limp→0

[1

2

p

p− 1+

1

2

p

p + 5

]= 0 =? = lim

t→∞f(t).

The property of FVT cannot be applied. Indeed

F (p) =1

2

1

p− 1+

1

2

1

p + 5→ f(t) =

1

2etu(t) +

1

2e−5tu(t)

and the first component of f(t) does not have a finite limit when t tends to the positive infinity.


15. Using MATLABConsider the following function as the Laplace transform

F (p) =p + 2

p3 + 2p2 + 4p + 5.

The following next show a way to calculate perform the partial fraction expansion in MATLAB.

(1) Define two vectors with coefficients of numerator and denominator num=[1 2]; den=[1 2 4 5];

(2) Use ”residue” function to finds the residues, poles, and direct term of the partial fraction

expansion of F (p). [r,p,k]=residue(num,den);

As a result, we obtain the following data:

r =–0.0486 – 0.3135i–0.0486 + 0.3135i

0.0971

p =

–0.2370 + 1.7946i–0.2370 – 1.7946i

-1.5260

k =[ ]

(3) Therefore, the Laplace transform can be written as follows

F (p) =−0.0486− 0.3135i

p− (−0.2370 + 1.7946i)+

0.0486 + 0.3135i

p− (−0.2370− 1.7946i)+

0.0971

p− (−1.5260)

or

F (p) =−0.0486− 0.3135i

p + 0.2370− 1.7946i+

0.0486 + 0.3135i

p + 0.2370 + 1.7946i+

0.0971

p + 1.5260.

16. Laplace transform in Circuit Analysis

Resistance R:

vR(t) = RiR(t)L−→ VR(p) = RIR(p).

Inductor L:

vL(t) = LdiL(t)

dtL−→ VL(p) = L [IL(p)− iL(0)] .

Capacitor C:

vC(t) =1

C

t∫

0

iC(τ)dτ + vC(0)L−→ VC(p) =

1

C·1

pIC(p) +

vC(0)

p.


17. Forced and Natural ResponsesWe consider a linear system described by the following differential equation

y′′(t) + 7y′(t) + 12y(t) = x′(t) + 3x(t) + 0, t > 0 (137)

with the initial conditions y′(0) = −4, y(0) = 2.

We now find the response of the system to the input x(t) = u(t). For that, we recall that we have

y′′(t)L−→ p2Y (p)− py(0)− y′(0) = p2Y (p)− 2p + 4

y′(t)L−→ pY (p)− y(0) = pY (p)− 2

x′(t)L−→ pX(p)− x(0) = pX(p)−

1, if u(0) = 10.5, if u(0) = 0.5

and we consider that u(0) = 1, then pX(p)− 1 = pU(p)− 1 = 0 for all p such that Rep > 0.

Applying the Laplace transform over both parts of the differential equation in (137), we obtain[p2Y (p)− 2p + 4

]+ 7 [pY (p)− 2] + 12Y (p) = 3X(p)

Y (p)[p2 + 7p + 12]− 2p− 10 = 31

p, Rep > 0

Y (p) =2p + 10

p2 + 7p + 12︸︷︷︸NR

+3

p(p2 + 7p + 12)︸︷︷︸FR

.

The Laplace transform is composed from the transforms of the natural response (NR) and forced

response (FR) Y (p) = YN (p) + YF (p), where

YN (p) =2p + 10

p2 + 7p + 12=

A1

p + 3+

B1

p + 4

YF (p) =31

p

p2 + 7p + 12=

A2

p + 3+

B2

p + 4+

C2

p

where it is not difficult to find that A1 = 4, B1 = −2, and A2 = −1, B2 = 3/4, C2 = 1/4. The

Laplace transform of the natural response have two poles −3 and −4, and the Laplace transformof the forced response has one pole 0 more.

We now consider the inverse Laplace transforms of YN (p) and YF (p) :

YN (p) =4

p + 3+−2

p + 4L−1

−→ yN (t) = 4e−3tu(t)− 2e−4tu(t)

YF (p) =−1

p + 3+

3

4·

1

p + 4+

1

4·1

pL−1

−→ yF (t) = −e−3tu(t) +3

4e−4tu(t) +

1

4u(t).

The response of the system (as the inverse Laplace transform of Y (p)) to the input u(t) equals

y(t) = yN (t) + yF (t)

= 4e−3tu(t)− 2e−4tu(t)− e−3tu(t) +3

4e−4tu(t) +

1

4u(t)

=

[3e−3t −

5

4e−4t +

1

4

]u(t).


We can check the initial conditions for the obtained response

y(0) = 3−5

4+

1

4= 2, y′(0) = (−3)3− (−4)

5

4= −9 + 5 = −4.

Bilateral Laplace Transform

F (p) =

∞∫

−∞

f(t)e−ptdt, p ∈ C2.

For the function of exponential order the following holds

f(t) = Aeαtu(t)L−→

A

p− α, Rep > α.

We now consider the time-inverted function

g(t) = f(−t) = Ae−αtu(−t).

The Laplace transform of g(t) is calculated as follows

G(p) =

∞∫

−∞

f(−t)e−ptdt = −

−∞∫

+∞

f(−t)e−(−p)(−t)d(−t)

=

∞∫

−∞

f(t)e−(−p)tdt = F (−p) = −A

p + α

for any p such that Re(−p) > α ⇒ Rep < −α.

Thus we haveAeαtu(t) Ae−αtu(−t)↓ L ↓ LA

p− α−

A

p + α

Re p > α Re p < −α

This example illustrates also the importance of the ROC for the Laplace transform. Indeed, thefollowing diagram holds

Ae−αtu(t) −Ae−αtu(−t)

L LA

p + αRe p > −α Rep < −α

Thus for the Laplace transform, the ROC should be specified.

Properties of the bilateral Laplace transform are similar to unilateral Laplace transform, exceptthe following two.

1. (Time scaling)

For any real number k 6= 0,


f(t) → g(t) = f(kt)↓ L ↓ L

F (p) → G(p) =1

|k|X(

p

k)

and ROC(G) =ROC(F/k).

2. (Derivative)

f(t) → f ′(t)↓ L ↓ L

F (p) → pF (p)

(Rep > a)

Consider the following non-causal signal

x(t) = 3e−4tu(t) + 2e−2tu(−t)

and the impulse response function

h(t) = 2δ(t)− e−3|t|

of a non-causal system.The following calculations hold for the Laplace transform

3e−4tu(t) + 2e−2tu(−t)

↓ L ↓ L3

p + 4+ −

1

p + 2

Rep > −4 ∩ Re p < −2

and ROC(X)=p; −4 < Re p < −2.For the impulse response

2δ(t) − e−3|t|

↓ L ↓ L

2 –−2 · 3

p2 − 9

∀p ∩ |Rep| < 3

and ROC(H) = p; −3 < Re < 3.We can define the Laplace transform of the convolution in the region

ROC(Y ) = p; −3 < Re p < −2,

where

Y (p) = H(p)X(p) =

[3

p + 4−

2

p + 2

]·

[2 +

6

p2 − 9

]

=A1

p + 4+

A2

p + 2+

A3

p− 3+

A4

p + 3

for some numbers A1, ..., A4.


XIX. Brief Notes in Sampling Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EE-3424

I. Digital signal processing provides an alternative method for processing the analog signals. Toperform processing digitally, there is a need for an interface between the analog signal and digital

processor. This interface is called an analog-to-digital (A-D) converter. The output of A-D is adigital signal to be processed in the digital processor.

A / D DSP x(t) x(nT) y(n)

Fig. 89. Diagram of digital signal processing.

We consider the sampling process, where the analog signal x(t) (or continuous-time signal) is mea-sured periodically every T seconds, as x(nT ). T is the sampling time, or period, and time is counted

at points multiple to T, i.e. t = 0, T, 2T, 3T, . . . , (t = nT, n=0,... ), or t = 0,±T,±2T,±3T, . . . .The following problems are considered:

1. How does the sampling effect on the Fourier spectrum of the original signal?2. What is the maximal value of sampling interval T ?3. Is it possible to reconstruct the original continuous-time signal x(t) from its sampled version,

by interpolating between the samples x(nT ).

We consider an absolute (or square) integrable function x(t) defined on the real line R for which

the Fourier transform exists

X(ω) = (F x)(ω) =

+∞∫

−∞

x(t)e−jωt dt, ω ∈ (−∞, +∞). (138)

The inverse Fourier transform is described as

x(t) =1

2π

+∞∫

−∞

X(ω)ejωt dω, t ∈ (−∞, +∞). (139)

A.

Let T be an interval of sampling the function x(t), that results in the sequence

x(t) → Analog/Digital Converter → x(nT )

x(nT ) = x(t)|t=nT , n = 0,±1,±2, . . . .

We consider the relation between the spectrum of the sequence x(nT ) with the Fourier transform

of the continuous-time representation of this sequence, or sampled signal,

xs(t) = x(t)s(t) = x(t)∑

n=0,±1,±2,...

δ(t − nT ) =∑

n=0,±1,±2,...

x(nT )δ(t − nT )


We need to remember that the train of delta functions is represented by the train of delta functionsin the frequency domain as

∑

n=0,±1,±2,...

δ(t − nT ) → ω0

∑

n=0,±1,±2,...

δ(ω − nω0)

where ω0 = 2π/T. Therefore, in the frequency domain, this equation can be written as (see moredetails in pages 106-107)

Xs(ω) =1

T

∑

n=0,±1,±2,...

X(ω − n2π

T) =

1

T

∑

n=0,±1,±2,...

X(ω − n2πfs). (140)

The sampling process generates high frequency components and every frequency component of

the original signal is periodically replicated over entire frequency axis with period

fs =1

T(sampling rate (frequency))

in units of samples per second. 1

B. Let us assume that the signal has bounded spectrum, i.e. X (ω) = 0 for all |ω| > Ω, and

π/T > Ω for a positive number Ω. In this case, for all integers m 6= 0, we have

X

(ω + m

2π

T

)∣∣∣(−Ω,Ω)

= 0.

This means that if a frequency ω ∈ (−Ω, Ω), then X(ω + m2π

T

)= 0 and the replicas of X(ω) do

not overlap (see Fig. 90(a)).

Then, we can state that the Fourier transform of the continuous-time signal can be defined fromthe discrete Fourier transform of the sampled signal as

1

TX (ω) = Xs(ω), or X (ω) = TXs(ω) (141)

for all frequencies |ω| ≤ Ω.And opposite, if period of sampling T such that π/T < Ω, then at least two additional terms

X(ω + 2π

T

)and X

(ω − 2π

T

)will contribute to form the spectrum of the sampled signal in the

interval [−Ω, Ω] as shown in Fig. 90(b), and the Fourier transform X(ω) in the interval [−Ω, Ω]

cannot be defined asX (ω) = TXs(ω), ω ∈ [−Ω, Ω],

from the spectrum of the sampled signal. The reason is the small sampling rate fs (or, not small

sampling time T ), which results in the overlapping (aliasing) condition

π

T< Ω.

1The periodicity is in term of Hz and means that Xs(ω) = Xs(2πf) is periodic when shifting f → f + fs.


−10 −5 0 5 100

0.1

0.2

0.3

0.4

0.5

2π/T=6

Ω−Ω π/T 2π/T

2π/T

(a)

(b)

X(ejω

)

ω

X(ω)/T

−10 −5 0 5 100

0.1

0.2

0.3

0.4

0.5

2π/T=2.5

Ω−Ω

π/T

2π/T

2π/TX(e

jω)

ω

Fig. 90. (a) Effect of sampling when π/T > Ω, and (b) effect of sampling when π/T < Ω.

To determine the Fourier transform of the function x(t), the period of sampling should be taken as

a period T1 which provides the condition

−π

T1

< −Ω and Ω <π

T1

and, therefore T1 ≤ T.

C. From the first example of Fig. 90 (part a), one can see that it is possible to reconstruct theoriginal spectrum from the continuous-discrete Fourier transform, taking

X(ω) = TXs(ω), |ω| ≤π

T> Ω (142)

since X(ω) ≡ 0 when ω lies outside the interval (−π/T, π/T ). From the second example, we observethat one can never reconstruct the original spectrum from X(ejωT ). Namely, the original Fourier

transform can be reconstructed partially (reconstruction for low frequencies).


It should be noted that the sampling condition in terms of frequencies in Hz is derived as follows

π

T≥ Ω ⇔

1

2T=

fs

2≥ fb =

Ω

2π.

(see illustration of this condition in Figure 91). In other words, the sampling rate fs ≥ 2fb.

Ω π/T=πfsampling

ω=2πf (rad/sec)

f (Hz)fb=Ω/2π 1/(2T)=f

sampling/2

Nyquist interval

Nyquist frequency

0

Fig. 91. Condition of the proper sampling the signal bounded by Ω.

D. The following statement holds.Theorem 1: Let x(t) be a function with the bounded spectrum, i.e.

X(ω) = 0, |ω| > Ω (143)

and let such value Ω > 0 exists. Then x(t) can be described uniquely by its samples taken at

discrete time with the sampling interval

T <π

ωs

for a frequency ωs such that Ω ≤ ωs ≤ π/T .This statement has been proved first by Kotelnikov [V.A. Kotelnikov, Theory of potential noise

stability, Moscow, Nauka, 1956] but is often called Uttekker and Shannon’s sampling theorem.Example 1: Consider the signal x(t) that is bounded in the spectral domain of 50 · 103rad/sec.

Then, Ω = 25 · 103rad/sec. Let us assume that T = 10−4sec. Checking the condition

ωperiod/2 =π

T= 3.14 · 104 > Ω

we see that T can be considered as a good sampling period.

If we take T = 1.5 · 10−4sec, then after checking the condition

ωperiod/2 =π

T= 3.14/1.5 · 104 ≈ 2.28 · 104 < Ω

we can state that such T cannot be used for sampling a signal with bounded spectrum of 50 ·103rad/sec. We need reduce by ∆T the sampling period for signal reconstruction. The minimum

change in sampling time is defined as

∆T = T −π

Ω= 1.5 · 10−4 −

π

2.5 · 104= 10−4(1.5−

π

2.5) = 2.4336 · 10−5sec.


In terms of hertz, the spectrum is bounded by the frequency

fb =Ω

2π=

25 · 103

2πHz = 3.9789 · 103Hz

the sampling rate fs therefore should be greater or equal 2fb = 7.9578 · 103Hz.

Owing to the Kotelnikov theorem (given bellow), each function with the bounded spectrum can

be restored from its discrete values chosen in the defined way. So, the one-dimensional signal x(t),that is given in the finite interval [0, L] and satisfies the condition of spectrum to be band limited:

X(ω) = 0 for all |ω| > Ω, can be represented one-to-one by the discrete sequence of its values

xn = x(nT ), n = 0 : (N − 1) (144)

taken with a sampling interval T ≤ π/Ω = 1/(2fb) such that N = L/T is an integer. The

restoration of the assumed continuous-time function from its discrete values (144) is described bythe expansion formula of the function:

x(t) =N−1∑

n=0

xnsincΩ(t− nT ), t ∈ [0, L] (145)

by the basis functions

sincΩ(t − nT ) =sinΩ(t − nT )

Ω(t − nT )(146)

that are the shifted by nT and time-scaled by Ω versions of sinc(t) functions

sinc(t) → sinc(Ωt) → sinc(Ω(t− nT )).

Thus, a “signal” with the bounded spectrum (143) is defined completely by the finite number ofits values at points situated uniformity on the segment [0, L].

In the proof the statement of the Kotelnikov theorem in (145), the condition of not aliasing isused,

X(ω) = TXs(ω), |ω| < Ω <π

T. (147)

In the general case, Eq. 145 has the form

x(t) =TΩ

π

∑

n∈Z

x(nT )sinc [Ω(t − nT )]

=∑

n∈Z

x(nT )sinc [Ω(t − nT )] ( ifπ

T= Ω).

E. In the real word, the signals are not bound limited. They are limited in time but not in thefrequency domain.

Example 2: Consider the rectangle function on the interval [−T0, T0]

x(t) = rect

(t

2T0

)= u(t + T0)− u(t − T0) → 2T0 sinc(ωT0).

In the particular T0 = 1/2 case, we obtain rect(t)F−→ sinc(ω/2). Note that sinc(ω) is the function

that is not limited neither absolute integrable.


−10 −5 0 5 10−2

−1

0

1

2

3

4

5

6

7

8

9sum of three sinc functions

1 2 3

Fig. 92. Elements of the decomposition of the reconstructed signal by basis sinc(t) functions.

D. We consider a continuous-in-time signal that itself represents a step-function with values of

the sampled sequence

x(t) = x(nT ), t ∈ An = (nT − T/2, nT + T/2], n ∈ Z. (148)

The function x(t) can be represented as the infinite sum of the pulse signals

x(t) =∑

n∈Z

x(nT )rect

(t

T− n

).

Each function r(t) = rect(

tT − n

)represents the performance of the following time-transforma-

tions t :→ t/T and t → t − nT of the rect(t) function. Therefore, the Fourier transform is defined

by these transformation as it shown in the diagram:

rect(t) → z(t) = rect

(t

T

)→ z(t − nT ) = rect

(t

T− n

)

↓ F ↓ F ↓ F ↓ F = ↓ F

sinc

(ω

2

)→ Z(ω) = T sinc

(ωT

2

)→ Z(ω)e−jωnT = T sinc

(ωT

2

)e−jωnT

Therefore, the Fourier transform of the function x(t) is calculated as

X(ω) =∑

n∈Z

x(nT )T sinc

(ωT

2

)e−jωnT

=∑

n∈Z

x(nT )e−jωnT ·

[T sinc

(ωT

2

)]= TXs(ω)sinc

(ωT

2

).


On the other hand, if x(nT ) is the sampled signal x(t) with the sampling period T , then accordingto (140)

Xs(ω) =1

TX (ω) +

1

T

∑

m=±1,±2,...

X

(ω + m

2π

T

).

Therefore,

X(ω) = TXs(ω)sinc

(ωT

2

)

= X (ω) sinc

(ωT

2

)+

∑

m6=0

X

(ω + m

2π

T

)

sinc

(ωT

2

).

and we can see that in generalEven in the case when x(t) has a spectrum bounded by Ω < π/T , the following holds for

ω ∈ (−Ω, Ω)

X(ω) = X(ω)sinc

(ωT

2

)6= X(ω) (if ω 6= 0)

which shows the error of the digital-to-analog convertor.F. (Periodic sampling step function)We consider the mathematical representation of the sampling

x(t) → xs(t) = x(t) · s(t) =

x(nT ), if t = nT

0, otherwise

when the continuous-time signal is transformed into a continuous-time signal with values of theoriginal signal at points nT. The modulated signal s(t) is periodic impulse function with period T

s(t) = . . . , δ(t + 2T ), δ(t + T ), δ(t), δ(t− T ), δ(t− 2T ), . . .

=∑

n∈Z

δ(t − nT ) =1

T

∑

n∈Z

e−jnωst

where ωs = 2π/T is the sampling frequency.

In the Fourier domain, we following diagram holds for the function s(t)

s(t) =1

T

∑

n∈Z

e−jnωst

↓ F ↓ F

S(ω) =1

T

∑

n∈Z

2πδ(ω − nωs)

Next, because of the property of the convolution in the frequency domain, we obtain the following

xs(t) = x(t) · s(t)↓ F = ↓ F

Xs(ω) =1

2π

∫ ∞

−∞X(ω − τ)S(τ)dτ


and

Xs(ω) =1

2π

∫ ∞

−∞X(ω − τ)

[1

T

∑

n∈Z

2πδ(τ − nωs)

]dτ =

1

T

∑

n∈Z

∫ ∞

−∞X(ω − τ)δ(τ − nωs)dτ

=1

T

∑

n∈Z

X(ω − nωs) =1

TX(ω) +

1

T

∑

n∈Z\0

X(ω − nωs).

We can see again that the Fourier transform of the function xs(t) consists of periodic copies (withe

period ωs) of the scaled Fourier transform of the original signal x(t). If the Fourier transform ofx(t) is bound limited, X(ω) = 0 for ω /∈ (−Ω, Ω), then these copies do not overlap if ωs > 2Ω. In

this case, we haveX(ω) = TXs(ω), ∀ω ∈ (−Ω, Ω).

In other words, the ”lowpass” filter

Hlp(ω) =

T, if |ω| < ωcut

0, otherwiseF−1

−→ hlp(t) =Tωcut

πsinc(ωcutt)

with a cutoff frequency ωcut = ωcutoff from the interval (Ω, ωs/2) can be used to recover the Fouriertransform

X(ω) = Hlp(ω)Xs(ω).

−6 −4 −2 0 2 4 60

0.5

1

1.5

2

2.5

3

Y(ω)

Xs(ω) = X(e

jω)

X(ω−2π/T)

2π/T −2π/T

Ω−Ω ω

X(ω+2π/T)

Fig. 93. Fourier transform calculation by the lowpass filter H(ω).

In the case when ωs < 2Ω (or T > π/Ω), the Fourier transform of the signal x(t) cannot be

recovered from the Fourier transform of the signal xs(t), because of overlapping of copies X(ω−nωs).Thus, X(ω) 6= TXs(ω), or X(ω) 6= Hlp(ω)Xs(ω) for any lowpass filter.

EE-3424, MATH IN SIGNALS AND SYSTEMS, ART GRIGORYAN, UTSA-2016 1

MATLAB-Based Project: The Fourier series and convolution of signals

Fast Fourier Series - or - the N-point Discrete Fourier Transform 1

Time: April 6 - April 29 (by 5:00pm)

Instructor: Dr. Artyom Grigoryan

Student name:

Section:

1For this project you will be given maximum of 10 points which will be added to your final score.

EE-3424, MATH IN SIGNALS AND SYSTEMS, ART GRIGORYAN, UTSA-2016 2

Speech signal and linear convolution

The speech signal for this project was posted on our web page in BB. Write your code2 to calculate theconvolution of the speech signal by performing the following steps.

1. Read the speech signal fn from the file ’mike.wav’ (see Figure 1); for that the command “wavread”can be used. (To sound this signal f , use the command “sound(f,Fs)”. This signal is of 38.7360 seconds

0 5 10 15 20 25 30 35−0.5

0

0.5speech signal

time (seconds)

Fig. 1. Original 1-D signal.

duration.

2. Consider the triangle filter

hn = 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 8, 7, 7, 6, 6, 5, 5, 4, 4, 3, 3, 2, 2/K

with the center h0 = 8; note the filter is not symmetric. Define value of K from the condition

12∑

n=−12

hn = 1.

3. Calculate and sketch the convolution of the signal with this filter.

4. Divide the signal fn by parts of 7 seconds each; add zeros to the last part to complete it to 7 seconds.

5. Perform the linear convolution of each part of the signal by this filter, by using the direct method ofconvolution.

6. Compose one new signal from these convoluted parts; this signal, gn, should be of the same length asthe original speech signal.

7. Repeat steps 3-6 by calculating the convolutions in steps 3 and 5 by the method of the Fourier seriesby using the MATLAB code of the fast Fourier transform (fft).

Provide your code(s) and report for this project in hard copy with illustrations, as well as in electronicform (in one zip-file). Functions from MATLAB which calculate the convolution and filters cannot be used,only “fft” and “ifft” and “fftshift’ can be used for the N -point DFT (Fourier series). Plot the signals,convolutions, Fourier transforms of the signals and filter in absolute scale. Explain clearly your work in theproject, and write how fast your program(s) work.

A.M.G. (e-mail: [email protected])

2No MATLAB commands “conv”, “filter”,... can be used.

Documents

EE-3424, Mathematics in Signals and Systemsengineering.utsa.edu/agrigoryan/wp-content/uploads/sites/... · 2019-01-12 · EE-3424, MATHEMATICS IN SIGNALS & SYSTEMS, ART GRIGORYAN