TE2035

Analysis of Signals and Systems

Class Notes

The Department of Electrical and Computer EngineeringInstituto Tecnologico y de Estudios Superiores de Monterrey

Campus Monterrey

byCesar Vargas-Rosales

B.S., Universidad Nacional Autonoma de Mexico, June 1988M.S. in E.E., Louisiana State University, December 1992Ph.D. in E.E., Louisiana State University, August 1996

.

c©Copyright 2014Cesar Vargas RosalesAll rights reserved

ii

Contents

List of Tables v

List of Figures vi

Chapter

I Signals and System Analysis 1

1 Signals 31.1 Important Signals and some Properties . . . . . . . . . . . . . . . . . . . . 6

1.1.1 Cosenoidal Signal . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 Time Averages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2.1 Mean Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2.2 Mean Squared Value . . . . . . . . . . . . . . . . . . . . . . . . . 81.2.3 rms Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 RC Low-Pass Filter Time Domain Analysis 112.1 Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Impulse Response and Convolution Integral . . . . . . . . . . . . . . . . . 13

3 Geometric Interpretation of Signals 193.1 Orthogonal Signal Expansion . . . . . . . . . . . . . . . . . . . . . . . . . 193.2 Complex Exponential Fourier Series . . . . . . . . . . . . . . . . . . . . . 203.3 Trigonometric Fourier Series (TFS) . . . . . . . . . . . . . . . . . . . . . 22

4 Examples Fourier Series 254.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

5 CEFS of Cosine Functions 335.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

iii

II Frequency Domain Analysis 37

6 Fourier Transform 396.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406.2 Fourier Transform of a Periodic Signal . . . . . . . . . . . . . . . . . . . . 426.3 Example of the FT of a Periodic Signal . . . . . . . . . . . . . . . . . . . . 43

7 Frequency Domain Analysis 457.1 Transfer Function and Impulse Response . . . . . . . . . . . . . . . . . . . 467.2 Frequency Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

8 Homework 4: Solutions 518.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 518.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 568.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

9 Partial Exam 2: Solutions 659.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 659.2 Problem 2: RLC circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

9.2.1 Output from Inductor . . . . . . . . . . . . . . . . . . . . . . . . . 679.2.2 Output from Capacitor . . . . . . . . . . . . . . . . . . . . . . . . 689.2.3 Output from Resistor . . . . . . . . . . . . . . . . . . . . . . . . . 69

9.3 Problem 3: RL Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 699.4 Problem 4: Triangular Pulse . . . . . . . . . . . . . . . . . . . . . . . . . 71

Appendix A. Complex Analysis 76A.1 Complex Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

A.1.1 The Real numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 77A.1.2 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 79

Appendix B. Complex Functions 85

Appendix C. Sequences and Series of Functions 91

Appendix D. Power Series 96

Bibliography 97

iv

List of Tables

v

vi

List of Figures

1.1 Different signals conveying the same information. . . . . . . . . . . . . . . 31.2 Different signals conveying the same information. . . . . . . . . . . . . . . 41.3 Periodic signals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Aperiodic signals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.5 Different representations of a cosenoidal signal. . . . . . . . . . . . . . . . 7

2.1 RC low-pass circuit. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Case 1 of convolution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3 Case 2 of convolution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.4 Case 3 of convolution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.5 Result of the convolution example. . . . . . . . . . . . . . . . . . . . . . . 172.6 Result of the convolution example. . . . . . . . . . . . . . . . . . . . . . . 17

4.1 Triangular Waveform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.2 Discrete Magnitude spectrum of coefficients Cn. . . . . . . . . . . . . . . . 304.3 Shifted Triangular Waveform. . . . . . . . . . . . . . . . . . . . . . . . . 31

6.1 Rectangular pulse signal A = 3, T = 10. . . . . . . . . . . . . . . . . . . . 416.2 Fourier transfrom of a rectangular pulse signal A = 3, T = 2. . . . . . . . . 42

7.1 RC low-pass circuit. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457.2 Magnitude response of RC low-pass circuit. . . . . . . . . . . . . . . . . . 487.3 Magnitude squared response of RC low-pass circuit. . . . . . . . . . . . . 487.4 Magnitude response in dBs of RC low-pass circuit. . . . . . . . . . . . . . 49

8.1 Signal gp(t) for A = 3, φ = π/4, and f0 = 2Hz. . . . . . . . . . . . . . . . 518.2 Magnitude of coefficients Cn for A = 3. . . . . . . . . . . . . . . . . . . . 538.3 Phase of coefficients Cn for A = 3. . . . . . . . . . . . . . . . . . . . . . . 548.4 Three periods of signal xp(t). . . . . . . . . . . . . . . . . . . . . . . . . . 568.5 Circuit for problem 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 598.6 Frequency response of low-pass filter. . . . . . . . . . . . . . . . . . . . . 628.7 Square of frequency response of low-pass filter. . . . . . . . . . . . . . . . 628.8 dB of frequency response of low-pass filter. . . . . . . . . . . . . . . . . . 63

vii

9.1 Signal x(t) with a = 2 and k = 2. . . . . . . . . . . . . . . . . . . . . . . 659.2 RLC circuit in series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 679.3 RL circuit in series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 709.4 Triangular pulse for A = 2 and T = 3. . . . . . . . . . . . . . . . . . . . . 729.5 X(f) for T = 2 and A = 1. . . . . . . . . . . . . . . . . . . . . . . . . . . 749.6 X(f) for T = 2 and A = 1. . . . . . . . . . . . . . . . . . . . . . . . . . . 75A.7 Geometric interpretation of a complex number. . . . . . . . . . . . . . . . 82A.8 Polar coordinates of a complex number. . . . . . . . . . . . . . . . . . . . 82A.9 Circle of radius ρ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85A.10 Mapping of complex function. . . . . . . . . . . . . . . . . . . . . . . . . 87A.11 Mapping of complex function f = z2 with domain B = z = a + jb ∈ C :

<z = 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88A.12 Plot of the real and imaginary parts of the complex-valued function f (x) =

(3 + j2x)2 /2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90A.13 Convergence of the sequence in graphical form. . . . . . . . . . . . . . . . 94

viii

Part I

Signals and System Analysis

1

Chapter 1

Signals

A signal is a representation of a message or information. It carries information in someformat and is represented mathematically as a function. Signals can be produced by manydifferent methods, for example an audio signal coming from a microphone, or an imagecoming from a camera, temperature or pressure from a sensor, etc. Different signals canrepresent the same information conveying the same message, see Figure 1.1.

Figure 1.1: Different signals conveying the same information.

A signal needs to have an emitter or transmitter and an interpreter or receiver, otherwiseit would be information that does not make any sense. A signal describes a physically re-alizable process and its variability and special features such as amplitude, frequency, phase,range, location, power received measured in a space (see Figure 1.2,) etc.

Definition 1 Signal. A signal is a measurement or observation that contains informationdescribing some phenomenon such as the time history of an economic process, chemicalsignals of fragrances, variation of light intensity, among many more.

The representation of a signal can be in continuous and discrete time, and can take formsthat can be described mathematically or graphically. Then a different definition of a signal,and perhaps a formal one would be the following

Definition 2 Signal. A signal is a real or complex valued function of one or more realvariables such as time (continuous or discrete) or space.

3

4

Figure 1.2: Different signals conveying the same information.

A signal may depend on one variable that would make it a one-dimensional signal, e.g.,speech vs. time, temperature vs. time, etc. Dependence on two or more variables gives amultidimensional signal, e.g., an image, power received at certain coordinates (see Figure1.2), etc. As one can see from definition 2, a signal is defined by the concept of a function.

Recall that a function is a rule that maps a domain set A into a range set or image set B,we denote this as f : A −→ B, and f is a rule that assigns, to each element of a set A anelement of a set B, see [4]. The rule defines that for every element a ∈ A, there exists anelement f(a) ∈ B called the image of a under the function f . For any subset D ⊂ A, f(D)will be the set of all image points of D under the function f , i.e., the image of D under f . Incontrast, for a set E ⊂ B, denote f−1(E) the set of all points of A whose images under thefunction f line in E, i.e., the preimage or inverse image of E under f . Functions can be, see[4],

• Injective or one-to-one: If for every pair of elements a1, a2 ∈ A, a1 6= a2, their imagesare different, in other words, f(a1) 6= f(a2). Injectivity depends on the rule of thefunction.

• Surjective or f maps A onto B: If for every element of the range set b ∈ B, b = f(a)for at least one element a ∈ A. Surjectivity depends on the rule and on the range setB.

• Bijective or one-to-one correspondence: If the function f is both injective and surjec-tive. A bijective function f has an inverse f−1, where for each b ∈ B, f−1(b) is thatunique element a ∈ A for which f(a) = b.

In general, the sets involved in the mapping of a function can be specified or can be under-stood from the context of the analysis, for example, in the signal analysis part, it will beunderstood that the domain set A will be related to time in any of its forms.

Signals can be classified in different ways, in the following, a list of contrasting charac-teristics is presented.

5

1. Continuous vs. Discrete: A signal s(t) is continuous if at any point t0 of its domain,if it satisfies that limt↑t0s(t) = limt↓t0s(t) for every point t0 in its domain. A signal isdiscrete if its range or image takes only some countable set of values.

2. Periodic vs. aperiodic: A signal s(t) is periodic if for a positive constant number T0called the period, the following is satisfied

s (t) = s (t+ kT0) , ∀t ∈ R, T0 > 0, k = 0,±1,±2, . . . . (1.1)

A convention is that a periodic signal will always have infinite duration. An aperiodicsignal does not satisfy the relation in Equation (1.1) for any value of T0. Also, anaperiodic signal can be of infinite or finite duration.

3. Deterministic vs. Random: A deterministic signal can be represented by explicit math-ematical rules, e.g., completely specified time functions such as s(t) = cos(2πft),t ∈ R. A random signal can take a random value for any element of its domain set,and can be modeled probabilistically with distributions and density functions. Figures1.3 and 1.4 show two periodic signals and two aperiodic signals, respectively, a cosinefunction and a rectangular pulse train in the periodic signal example, and a truncatedcosine and a rectangular pulse for the aperiodic signal example. Note that the trun-cated cosine signal can be seen as the product of the aperiodic rectangular pulse signalin the figure and the periodic cosine signal of Figure 1.3.

Figure 1.3: Periodic signals.

4. Time limited vs. Band limited: The signals in Figure 1.4 can be seen or classifiedas time-limited signals since their duration or existence is for a limited set of values(interval) of the domain or time variable, outside this interval the signal is zero or non-existent. A band-limited signal is zero everywhere in the frequency domain, except fora limited set or range of values of frequency.

6

Figure 1.4: Aperiodic signals.

5. Even vs. Odd symmetry: A signal s(t) has even symmetry if it satisfies

s (t) = s (−t) , ∀t, (1.2)

on the other hand, a signal s(t) has odd symmetry if it satisfies

s (t) = s (−t) , ∀t. (1.3)

A signal can be the combination of an even and an odd symmetric signals, then theeven and odd parts of a signal can be used to form the signal in the following way

s (t) = seven (t) + sodd (t) , (1.4)

and we can also see that

seven (t) =1

2[s (t) + s (−t)] , (1.5)

and that

sodd (t) =1

2[s (t)− s (−t)] , (1.6)

1.1 Important Signals and some Properties

In this section, we present some of the most important signals because they represent relevantprocesses in communication systems. Also, some properties of functions will be introducedwith the examples.

7

1.1.1 Cosenoidal Signal

Figure 1.5 shows two representations of a cosenoidal signal. The signal is deterministic andit is defined by s(t) = 3 cos(2πt). Note that it is a cosine of peak amplitude of 3 V andfrequency f = 1 Hz. Recall that frequency f will be defined in Hz, and that it is related toω = 2πf . In the same figure, we can see that there is a shifted version of s(t). The shiftis to the right hand side of the original signal by an amount of 1/4 so that the shifted signalis s(t − 1

4) = 3 cos

[2π(t− 1

4)]. Note that the same signal can be represented by another

function that depends on the shift of the phase of the original signal, the phase of the originalsignal is zero. In the same figure, we can also see that the original signal s(t) has evensymmetry, and that the shifted version has odd symmetry.

Figure 1.5: Different representations of a cosenoidal signal.

Property 3 Shifting: A representation s(t − τ) of signal s(t) implies that s(t) is shifted tothe right τ time units for τ > 0. For s(t+ τ), the signal s(t) is shifted to the left τ time units.

1.2 Time Averages

The time averages of a signal are parameters used for design of systems. These time averagesare obtained by the use of a linear operator and the type of signal (periodic signal, aperiodicpower signal and aperiodic energy signal). The time averages are the mean value, the meansquared value and the root mean square (rms) value.

8

1.2.1 Mean Value

The mean value of a signal represents the Direct Current (DC) component of the signal. Ifone could see the frequency components of the signal, the mean value would be situated atzero frequency. The mean value is obtained as follows

1. Periodic signal xp(t) with period T0. Recall that this signal exists for all the time, i.e.,t ∈ R.

〈xp(t)〉 =1

T0

T02∫

−T02

xp(t)dt. (1.7)

2. Aperiodic power signal x(t). Recall that this signal has infinite duration, but it couldbe for all t ∈ R or for example it could be for t ≥ 0. Define an interval of length T ,then

〈x(t)〉 =1

T

T2∫

−T2

x(t)dt. (1.8)

3. Aperiodic energy signal x(t). Recall that this signal exists only for a finite durationof time, for example T . The integral must be calculated only on the interval where thesignal exists, i.e.,

〈x(t)〉 =

∫T

x(t)dt. (1.9)

1.2.2 Mean Squared Value

The mean squared value of a signal represents the power or energy of the signal. In this case(communications and signal processing) we talk about the normalized average total power(energy) of a signal since we consider that the power or energy is dissipated in a 1Ω resistor.This time average is given by

1. Periodic signal xp(t) with period T0. Recall that this signal exists for all the time, i.e.,t ∈ R.

〈x2p(t)〉 =1

T0

T02∫

−T02

x2p(t)dt. (1.10)

2. Aperiodic power signal x(t). Recall that this signal has infinite duration, but it couldbe for all t ∈ R or for example it could be for t ≥ 0. Define an interval of length T ,

9

then

〈x2(t)〉 =1

T

T2∫

−T2

x2(t)dt. (1.11)

3. Aperiodic energy signal x(t). Recall that this signal exists only for a finite durationof time, for example T . The integral must be calculated only on the interval where thesignal exists, i.e.,

〈x2(t)〉 =

∫T

x2(t)dt. (1.12)

1.2.3 rms ValueThe rms value is obtained by calculating the square root of the mean squared value, hencewe have

1. Periodic signal xp(t) with period T0. Recall that this signal exists for all the time, i.e.,t ∈ R.

√〈x2p(t)〉 =

√√√√√√√ 1

T0

T02∫

−T02

x2p(t)dt. (1.13)

2. Aperiodic power signal x(t). Recall that this signal has infinite duration, but it couldbe for all t ∈ R or for example it could be for t ≥ 0. Define an interval of length T ,then

√〈x2(t)〉 =

√√√√√√ 1

T

T2∫

−T2

x2(t)dt. (1.14)

3. Aperiodic energy signal x(t). Recall that this signal exists only for a finite durationof time, for example T . The integral must be calculated only on the interval where thesignal exists, i.e., √

〈x2(t)〉 =

√√√√∫T

x2(t)dt. (1.15)

10

Chapter 2

RC Low-Pass Filter Time DomainAnalysis

2.1 Differential Equation

Consider the RC low-pass circuit shown in Figure 4.1, and solve the following: We can see

Figure 2.1: RC low-pass circuit.

that the output y(t) is given by

y(t) =1

C

t∫−∞

i(τ)dτ. (2.1)

We can also see from the circuit that

i(t) =x(t)− y(t)

R. (2.2)

11

12

By substituting Equation (4.2) into (4.1) we get the following

y(t) =1

C

t∫−∞

x(τ)− y(τ)

Rdτ

=1

RC

t∫−∞

x(τ)dτ − 1

RC

t∫−∞

y(τ)dτ. (2.3)

From Equation (2.3) we get the following

y(t) +1

RC

t∫−∞

y(τ)dτ =1

RC

t∫−∞

x(τ)dτ. (2.4)

Now, taking the derivative with respect to time t in Equation (2.4), we get the followingdifferential equation

dy(t)

dt+

1

RCy(t) =

1

RCx(t). (2.5)

Now, Equation (2.5) can be recognized as a first order differential equation written inthe standard form. See that in order to be of the standard form, the right hand side of theequation only contains x(t), i.e., no derivative or integral of x(t) exists in this equation.Another condition for the equation to be in the standard form is that the coefficient of thefirst order derivative of y(t) is one. Now, we can solve the differential equation in (2.5) byusing the following procedure

1. Obtain the result ofe∫

1RC

dt = etRC . (2.6)

2. Multiply both sides of Equation (2.5) by the expression obtained in (2.6) to get

etRCdy(t)

dt+ e

tRC

1

RCy(t) = e

tRC

1

RCx(t). (2.7)

3. See that the left hand side of (2.7) can be changed so that we get from it the following

d

dt

[e

tRC y(t)

]= e

tRC

1

RCx(t). (2.8)

4. Now, we can integrate both sides of Equation (2.8), see that on the right hand side wechange the integration variable from t to τ to obtain

etRC y(t) =

1

RC

t∫−∞

eτRC x(τ)dτ. (2.9)

13

5. Then the solution to the differential equation in (2.5) is given by

y(t) =1

RCe−

tRC

t∫−∞

eτRC x(τ)dτ. (2.10)

2.2 Impulse Response and Convolution IntegralNow, consider that the input is an impulse (Dirac’s delta), so that x(τ) = δ(τ) and y(t)becomes the impulse response of the circuit which is given by h(t). Since δ(τ) exists onlyfor τ = 0, then the integral in Equation (2.10) is zero for all time t < 0. Now, for times t ≥ 0we use the sifting property of the Dirac’s delta function which for any signal g(t) is given by

t∫−∞

g(τ)δ(τ − k)dτ = g(k), t ≥ 0. (2.11)

Now, using the property in (2.11) with k = 0 and g(τ) = eτRC , we get g(0) = 1, then the

impulse response is obtained as

h(t) =1

RCe−t/RC , t ≥ 0. (2.12)

The impulse response h(t) of any system is used to obtain the output of the system y(t)to any input x(t) by using the convolution integral given by any of the two definitions (recallthat convolution is commutative and that we represent it using the operator asterisk ∗)

y(t) = h(t) ∗ x(t)

=

∫x(τ)h(t− τ)dτ

=

∫h(τ)x(t− τ)dτ. (2.13)

As an example, consider that the input is a rectangular pulse of amplitude A, centered atthe origin with duration T given by

x(t) = AΠ

(t

T

), (2.14)

First recall that either x(t) or h(t) must be chosen to be changed to x(−t) or h(−t). Afterobtaining the mirror image of one of them, you must recognize the part of the mirror imagethat is at the vertical axis because it will indicate you the time t to be used whenever youshift in time this signal in order to obtain the convolution integral. After finding such part

14

of the signal that corresponds to time t, now see if it applies that when the signal is moved,there is a time t for which no overlap of both signals exists.

In this example, the mirror of signal x(t) will be considered, and since it has even sym-metry, its mirror is exactly the same as the original rectangular pulse. Now, once we get themirror, we must see that the vertical axis is located right in the middle of the signal, hencethe middle of the mirrored rectangular pulse will define the time t. You can also see that theright hand edge of the rectangular pulse will define time t + T/2 and the left hand edge thetime t− T/2.

In our example of a rectangular pulse with amplitude A = 1.2, duration T = 2, we cansee CASE 1 when no overlap is obtained in Figure 2.2. You need to confirm that CASE 1will exist for all times that satisfy t < −T

2

Figure 2.2: Case 1 of convolution.

Since no overlap exists for t < −T2

in signals x(t − τ) and h(τ), we get that the outputfor CASE 1 is given by y1(t) = 0 for these times.

CASE 2, when partial overlap exists, is shown in Figure 2.3. You need to confirm thatCASE 2 exists for times −T

2≤ t < T

2.

For the partial overlap case, we can see that the overlap is only between time 0 and time

15

Figure 2.3: Case 2 of convolution.

t+ T/2, then we need to solve the following convolution integral

y2(t) =

t+T2∫

0

A

RCe−τ/RCdτ

=A

RC

(1

− 1RC

)e−τ/RC

∣∣∣∣t+T/20

= −A(e−(t+T/2)/RC − 1

)= A

(1− e−t/RC−T/2RC

)= Ae−T/2RC

(eT/2RC − e−t/RC

). (2.15)

If we consider the same values of A = 1.2 and T = 2, we get that the output in CASE 2is given by

y2(t) = 1.2e−1/RC(e1/RC − e−t/RC

). (2.16)

CASE 3, when total overlap exists, is shown in Figure 2.4. You need to confirm thatCASE 3 exists for times T

2≤ t, i.e., the mirrored signal was shifted to the right T/2 or more

time units. In this case, wherever we move the mirrored signal to times above T/2 seconds,the overlap of both signals will be only for times that are in the interval [t − T

2, t + T

2].

Therefore, the integral will have limits defined by such an interval.Hence, the convolution integral in this CASE 3 can be calculated by the following pro-

16

cedure

y3(t) =

t+T2∫

t−T2

A

RCe−τ/RCdτ

=A

RC

(1

− 1RC

)e−τ/RC

∣∣∣∣t+T/2t−T

2

= −A(e−(t+T/2)/RC − e−(t−T/2)/RC

)= A

(e−t/RC+T/2RC − e−t/RC−T/2RC

)= Ae−t/2RCeT/2RC

(1− e−T/RC

). (2.17)

Figure 2.4: Case 3 of convolution.

Now getting all three cases together, we get the following output for the rectangular pulsex(t)

y(t) =

0, t ≤ −T

2,

Ae−T/2RC(eT/2RC − e−t/RC

), − T

2≤ t < T

2,

Ae−t/RCeT/2RC(1− e−T/RC

), t ≥ T

2.

(2.18)

For the example with A = 1.2 and T = 2 we get

y(t) =

0, t ≤ −1,

1.2e−1/RC(e1/RC − e−t/RC

), − 1 ≤ t < 1,

Ae−t/RCe1/RC(1− e−2/RC

), t ≥ 1.

(2.19)

17

Assuming that R = 1 MΩ and that C = 1µF, we get the output y(t) shown in Figure 2.5.As you can see, the input signal is also included in the figure so that we can see the effectsof distortion caused by the RC filter.

Figure 2.5: Result of the convolution example.

Now, calculating the convolution using the command conv in MATLAB, and after scalingand plotting, we get Figure 2.6. It can be seen when compared to Figure 2.5 that the outputobtained by convolving x(t) with h(t) with the command conv in MATLAB is exactly thesame as that obtained by the solution in Equation (2.18)

Figure 2.6: Result of the convolution example.

18

Chapter 3

Geometric Interpretation of Signals

The analysis of every digital communications system considers the transmitter, the receiver,and the channel. The transmitter and receiver know the signals that are being sent and re-ceived.

The problem of the geometric interpretation of signals consist of finding a set of functionsϕn (t)∞n=−∞ defined in an interval [a, b], such that the following conditions are satisfied:

1. The functions are orthogonal, i.e.∫ b

a

ϕn (t)ϕ∗m (t) dt =

0, n 6= m,

‖ ϕn (t) ‖2, n = m.(3.1)

2. Every signal f(t) defined in the same interval [a, b], can be expressed as a linear com-bination of ϕn (t)∞n=−∞, in other words

f (t) =∞∑

k=−∞

Ckϕk (t) , Ck ∈ C. (3.2)

3.1 Orthogonal Signal Expansion

Assume that the set of functions ϕn (t)∞n=−∞ are orthogonal and complex on an interval[a, b], then if f(t) is also defined on the same interval [a, b], it is possible to have coefficientsCk, k = −∞, . . . ,−1, 0, 1, . . . ,∞, such that

f(t) = · · ·+ C−1ϕ−1 (t) + C0ϕ0 (t) + C1ϕ1 (t) + C2ϕ2 (t) + · · ·+ Cnϕn (t) + · · · (3.3)

Assume that the representation through an infinite series as that in Equation (3.3) ispossible, then multiply both sides of such equation by ϕ∗n (t) and integrate on the interval

19

20

[a, b] to obtain ∫ b

a

f(t)ϕ∗n (t) dt = · · ·+∫ b

a

C−1ϕ−1 (t)ϕ∗n (t) dt

+

∫ b

a

C0ϕ0 (t)ϕ∗n (t) dt

+

∫ b

a

C1ϕ1 (t)ϕ∗n (t) dt

+

∫ b

a

C2ϕ2 (t)ϕ∗n (t) dt

+ · · ·

+

∫ b

a

Cnϕn (t)ϕ∗n (t) dt+ · · · (3.4)

Now, applying the orthogonality conditions of ϕn (t)∞n=−∞ given in Equation (3.1), weget that ∫ b

a

f(t)ϕ∗n (t) dt =

∫ b

a

Cnϕn (t)ϕ∗n (t) dt

=

∫ b

a

Cn|ϕn (t) |2dt

= Cn〈ϕn (t) , ϕn (t)〉= Cn‖ ϕn (t) ‖2 (3.5)

Therefore the coefficients Ck of the series in Equation (3.3) will be given by

Cn =

∫ baf(t)ϕ∗n (t) dt∫ ba|ϕn (t) |2dt

=〈f(t), ϕn (t)〉‖ ϕn (t) ‖2

, ∀n. (3.6)

Then, the signal representation through a series is given by

f(t) =∞∑

n=−∞

Cnϕn (t) =∞∑

n=−∞

〈f(t), ϕn (t)〉‖ ϕn (t) ‖2

ϕn (t) (3.7)

3.2 Complex Exponential Fourier SeriesThe infinite series in Equation (3.7) is called the Generalized Fourier Series (GFS). TheGFS can be used to define other signal representations. For example, we can choose the setof periodic functions with period T0 to be

ϕn (t)∞n=−∞ =ej2πf0tn

∞n=−∞ , (3.8)

Recall that for a periodic signal with period T0, its fundamental frequency f0 is related to its period asf0 = 1/T0, in other words we have that the product of period and frequency is f0T0 = 1

21

and we can see that the set contains orthogonal functions since

∫ T0/2

−T0/2ϕn (t)ϕ∗m (t) dt =

∫ T0/2

−T0/2ej2πf0tne−j2πf0tmdt

=

∫ T0/2

−T0/2ej2πf0t(n−m)dt

=1

j2πf0(n−m)ej2πf0t(n−m)

∣∣∣∣T0/2−T0/2

=1

j2πf0(n−m)

[ej2πf0

T02(n−m) − e−j2πf0

T02(n−m)

]=

1

j2πf0(n−m)

[ejπ(n−m) − e−jπ(n−m)

]=

1

πf0(n−m)sin [π(n−m)] . (3.9)

Recall that in Equation (3.9), the term (n−m) will be an integer number since n and m areintegers, therefore, we can see that for n 6= mwe have the sine function of an integer numberk = n−m times π, which is always zero for any integer number k. We can also realize thatfor n = m we have an indetermination that needs to be solved using L’Hopital’s rule. Theapplication of this rule gives that for n = m Equation (3.9) gives the result of 1/f0 which isthe same as T0. In summary, we get the following result for this set of orthogonal functions

∫ T0/2

−T0/2ϕn (t)ϕ∗m (t) dt =

∫ T0/2

−T0/2ej2πf0tne−j2πf0tndt =

0, n 6= m,T0, n = m.

(3.10)

Now, using the set of orthogonal functions ϕn (t)∞n=−∞ =ej2πf0tn

∞n=−∞, and the

results in (3.10) in the GFS of Equation (3.7), we get that the Complex Exponential FourierSeries (CEFS) of a periodic signal f(t) is given by

f(t) =∞∑

n=−∞

Cnϕn (t)

=∞∑

n=−∞

Cnej2πf0tn

=∞∑

n=−∞

〈f(t), ϕn (t)〉‖ ϕn (t) ‖2

ej2πf0tn

=∞∑

n=−∞

〈f(t), ej2πf0tn〉T0

ej2πf0tn, (3.11)

22

where the coefficients are given by

Cn =〈f(t), ϕn (t)〉‖ ϕn (t) ‖2

=

∫ T0/2−T0/2 f(t)ϕ∗n (t) dt∫ T0/2−T0/2 |ϕn (t) |2dt

=

∫ T0/2−T0/2 f(t)e−j2πf0tndt

T0

=1

T0

∫ T0/2

−T0/2f(t)e−j2πf0tndt, ∀n. (3.12)

3.3 Trigonometric Fourier Series (TFS)

Consider the definition of the CEFS for a periodic signal s(t), with period T0 or fundamentalfrequency f0, given by

s(t) =∞∑

n=−∞

Cnej2πf0tn, (3.13)

With the definition of the CEFS in Equation (3.13), we can separate the summation intotwo parts to obtain an alternate expression

s(t) =∞∑

n=−∞

Cnej2πf0tn

=−1∑

n=−∞

Cnej2πf0tn + C0 +

∞∑n=1

Cnej2πf0tn

=∞∑n=1

C−ne−j2πf0tn + C0 +

∞∑n=1

Cnej2πf0tn

= C0 +∞∑n=1

Cne

j2πf0tn + C−ne−j2πf0tn

. (3.14)

Now, in Equation (3.14) we can use the trigonometric identities of the complex exponen-tials in terms of the sine and cosine functions, i.e., the trigonometric identities ej2πf0tn =

23

cos(j2πf0tn) + j sin(j2πf0tn) and e−j2πf0tn = cos(j2πf0tn)− j sin(j2πf0tn), to obtain

s(t) = C0 +∞∑n=1

Cn [cos(j2πf0tn) + j sin(j2πf0tn)]

+ C−n [cos(j2πf0tn)− j sin(j2πf0tn)]

= C0 +∞∑n=1

[Cn + C−n] cos(j2πf0tn) + j [Cn − C−n] sin(j2πf0tn)

= C0 +∞∑n=1

An cos(j2πf0tn) +Bn sin(j2πf0tn) . (3.15)

The last row of Equation (3.15) is known as the Trigonometric Fourier Series repre-sentation of the periodic signal s(t). In Equation (3.15), we use the following substitution

An = Cn + C−n, Bn = j [Cn − C−n] . (3.16)

Note that the coefficients An are associated to the real part of the CEFS, whereas thecoefficients Bn are associated to the complex part of the CEFS. This also means that thecoefficients An will be different from zero whenever the signal s(t) has even symmetry orno symmetry. In contrast, the coefficients Bn will be different from zero whenever the signals(t) has odd symmetry o no symmetry. In other words, if the signal s(t) has even symmetry,the coefficients Bn will be zero, and whenever s(t) has odd symmetry, the coefficients Anwill be zero.

The coefficients An and Bn from Equation (3.16) can be obtained using the definitionintegral of the coefficients Cn of the CEFS, this definition is given in Equation (3.12). Oncewe use Equation (3.12) in Equation (3.16), we get for the coefficients An the following

An =1

T0

T0/2∫−T0/2

s(t)e−j2πf0tndt+1

T0

T0/2∫−T0/2

s(t)ej2πf0tndt

=1

T0

T0/2∫−T0/2

s(t)

e−j2πf0tn + ej2πf0tn

dt

=2

T0

T0/2∫−T0/2

s(t) cos(2πf0tn)dt, n = 1, 2, . . . , (3.17)

24

and for the coefficients Bn in Equation(3.16) we obtain

Bn = j

1

T0

T0/2∫−T0/2

s(t)e−j2πf0tndt− 1

T0

T0/2∫−T0/2

s(t)ej2πf0tndt

= j

1

T0

T0/2∫−T0/2

s(t)

e−j2πf0tn − ej2πf0tn

dt

= −j

1

T0

T0/2∫−T0/2

s(t)

−e−j2πf0tn + ej2πf0tn

dt

= −j

2

2T0

T0/2∫−T0/2

s(t)

ej2πf0tn − e−j2πf0tn

dt

=

2

j2T0

T0/2∫−T0/2

s(t)

ej2πf0tn − e−j2πf0tn

dt

=

2

T0

T0/2∫−T0/2

s(t) sin(2πf0tn)dt, n = 1, 2, . . . , (3.18)

Chapter 4

Examples Fourier Series

4.1 Example 1Consider the triangular waveform of Figure 4.1, and solve the following:

Figure 4.1: Triangular Waveform.

1. How would you classify the signal? The signal is continuous, with odd symmetry,periodic with period T0 = π, and deterministic. Since it is periodic, it exists for alltimes, i.e., t ∈ <, which makes it a power signal because it has infinite duration.

2. Find the Complex Exponential Fourier Series (CEFS) representation of the wave-form in Figure 4.1. Provide the general expression for the coefficients Cn.

(a) First, define function f(t). To do that, you need to see that it is made out of twostraight lines, one with positive slope with value of 2

π/2= 4

πcrossing through

the origin (y-intercept is zero), and another line with the “same” slope value butnegative − 4

πand crossing the vertical axis at 2 (y-intercept is two, you need to

see that this is true!). Thus, the function that defines the signal is given by

f(t) =

4πt, −π

4< t < π

4,

2− 4πt, π

4≤ t ≤ 3π

4.

(4.1)

25

26

(b) The second step is to obtain the information from the signal such as period andfundamental frequency, which in this problem we have from Figure 4.1 the fol-lowing

T0 = π, f0 =1

T0=

1

π, then we have 2πf0n = 2n. (4.2)

(c) The third step is to obtain the complex exponential Fourier series coefficients, i.e.,calculate Cn. To do this, we have to consider the definition of these coefficientsand work the integrals from there on substituting the functions obtained in thefirst step and shown in Equation (4.1). So we start the process from the definitionof the coefficients and keep working on it using Equation (4.1) and Equation (4.2)as follows.

Cn =1

T0

∫T0

f(t)e−j2πf0nt dt

=1

π

∫ π4

−π4

4

πte−j2nt dt+

1

π

∫ 3π4

π4

2e−j2nt dt+1

π

∫ 3π4

π4

4

πte−j2nt dt.(4.3)

Now, we work with Equation (4.3) by solving the first and the third integrals (wesolve the integrals by parts or we can use a table of integrals which is what wehave here). From tables of integrals, we have

∫teat dt =

(t

a− 1

a2

)eat. (4.4)

For the specific case of the integrals to solve in Equation (4.3), we have that theparameter a in Equation (5.3) must take the value a = −j2n. Then using thisformula, we solve the integrals in (4.3) to obtain the following

Cn =1

π

∫ π4

−π4

4

πte−j2nt dt+

1

π

∫ 3π4

π4

2e−j2nt dt+1

π

∫ 3π4

π4

4

πte−j2nt dt

=4

π2

(t

−j2n− 1

(−j2n)2

)e−j2nt

]π4

−π4

+2

π

(1

−j2n

)e−j2nt

] 3π4

π4

− 4

π2

(t

−j2n− 1

(−j2n)2

)e−j2nt

] 3π4

π4

. (4.5)

27

We can see that Equation (4.5) can be written as follows

Cn =4

π2

(− t

j2n+

1

4n2

)e−j2nt

]π4

−π4

+2

π

(1

−j2n

)e−j2nt

] 3π4

π4

− 4

π2

(− t

j2n+

1

4n2

)e−j2nt

] 3π4

π4

. (4.6)

Now, we substitute the initial and final values of each of the three terms in Equa-tion (4.6) to get

Cn =4

π2

[(− π

j8n+

1

4n2

)e−jn

π2 −

(π

j8n+

1

4n2

)ejn

π2

]− 1

jnπe−jn

3π2 +

1

jnπe−jn

π2

− 4

π2

[(− 3π

j8n+

1

4n2

)e−jn

3π2 −

(− π

j8n+

1

4n2

)e−jn

π2

]. (4.7)

Now expand each of the terms in brackets shown in Equation (4.7) to obtain thefollowing

Cn =4

π2

[− π

j8ne−jn

π2 +

1

4n2e−jn

π2 − π

j8nejn

π2 − 1

4n2ejn

π2

]− 1

jnπe−jn

3π2 +

1

jnπe−jn

π2

− 4

π2

[− 3π

j8ne−jn

3π2 +

1

4n2e−jn

3π2 +

π

j8ne−jn

π2 − 1

4n2e−jn

π2

].(4.8)

Now, we keep working on Equation (4.8) to expand each of the terms by mul-tiplying the constants 4/π2 that are outside the brackets with each of the termsinside the brackets to obtain the following.

Cn = − 1

j2nπe−jn

π2 +

1

n2π2e−jn

π2 − 1

j2nπejn

π2 − 1

n2π2ejn

π2

− 1

jnπe−jn

3π2 +

1

jnπe−jn

π2

+3

j2nπe−jn

3π2 − 1

n2π2e−jn

3π2 − 1

j2nπe−jn

π2 +

1

n2π2e−jn

π2 . (4.9)

From Equation (4.9), we can see that the first term and the ninth term add up andeliminate the sixth term. Then, see that the second term and the tenth term can

28

be put together. Also, we can put together the fifth and seventh terms to obtain

Cn =2

n2π2e−jn

π2 − 1

j2nπejn

π2 − 1

n2π2ejn

π2

+1

jnπe−jn

3π2

(3

2− 1

)− 1

n2π2e−jn

3π2

=2

n2π2e−jn

π2 − 1

j2nπejn

π2 − 1

n2π2ejn

π2

+1

j2nπe−jn

3π2 − 1

n2π2e−jn

3π2 . (4.10)

Now, in Equation (4.10), we can put together the first, third and fifth terms. Wecan do the same with the second and fourth terms, then we obtain

Cn =1

n2π2e−jn

π2

(2− ejnπ − e−jnπ

)+

1

j2nπejn

π2

(e−j2nπ − 1

). (4.11)

Now, recall that we have the following identities

ejnπ = cos(nπ) + j sin(nπ) = cos(nπ) = (−1)n

e−jnπ = cos(nπ)− j sin(nπ) = cos(nπ) = (−1)n

e−j2nπ = cos(2nπ)− j sin(2nπ) = cos(2nπ) = 1. (4.12)

Now we use the identities in Equation (4.12) and substitute them into Equation(4.11) to obtain

Cn =1

n2π2e−jn

π2 (2− (−1)n − (−1)n)

+1

j2nπejn

π2 (1− 1)

=1

n2π2e−jn

π2 (2− 2(−1)n)

=2

n2π2e−jn

π2 (1− (−1)n)

=2

n2π2[1− (−1)n] e−jn

π2 . (4.13)

See that when n = 0 in Equation (4.13), we have an indetermination, i.e., 00,

hence we need to treat the coefficient C0 separately. So, Equation (4.13) is onlyvalid for n = ±1,±2, . . .. Verify the following result!!!.

C0 =1

T0

∫T0

f(t) dt

=1

π

∫ π4

−π4

4

πt dt+

1

π

∫ 3π4

π4

2 dt+1

π

∫ 3π4

π4

4

πt dt = 0. (4.14)

29

Now, with the result in equations (4.13) and (4.14), we can write the CEFS of thesignal in Equation (4.1), shown in Figure 4.1. This CEFS is given by

f(t) = C0 +∞∑

n=−∞n 6=0

Cnej2πf0nt

=∞∑

n=−∞n 6=0

2

n2π2[1− (−1)n] e−jn

π2 ej2πf0nt. (4.15)

3. Provide the expression for the magnitude |Cn| and the phase ∠Cn. The magnitudeof the coefficients Cn, i.e., |Cn|, gives what is called the magnitude spectrum, i.e.,the information of the frequency content of the signal. For n = 0, we have that C0

gives the DC component or mean value of the signal because from the definitionof the coefficients Cn in Equation (3.12), we have that substituting n = 0 gives theexpression for the mean value of a signal (since C0 = 1

T0

∫T0f(t) dt, which is the same

as the time average of mean value of the signal). The DC component of the signal willalways be related to the value of the signal given at zero frequency. The remainingcomponents (coefficients Cn for n 6= 0) will provide the frequency content at pointsnf0, i.e., integer multiples of the fundamental frequency f0.

(a) Magnitude. The coefficients Cn are given by the last row of Equation (4.13),and their magnitude is given by

|Cn| =

∣∣∣∣ 2

n2π2[1− (−1)n] e−jn

π2

∣∣∣∣=

∣∣∣∣ 2

n2π2[1− (−1)n]

∣∣∣∣ ∣∣e−jnπ2 ∣∣=

∣∣∣∣ 2

n2π2[1− (−1)n]

∣∣∣∣ (4.16)

(b) Phase. The phase of the coefficients is obtained by considering the last row ofEquation (4.13) together with the identity: e−jn

π2 = cos(nπ/2) − j sin(nπ/2).

In this identity, see that whenever n is even or zero, i.e., n = 0,±2,±4, . . .the cosine function takes a value of +1 or −1 and the sine function is alwayszero, but see that when n is even, the whole coefficient Cn is zero because 1 −(−1)n = 0. On the other hand, see that when n is odd, i.e., n = ±1,±3, . . .,the cosine function takes the value zero and the sine function takes the value of+1 or −1, and at the same time we have 1 − (−1)n = 2, hence making Cn apurely complex coefficient for n odd, and zero for n even. Therefore, we havethat the coefficients Cn are purely complex (you should also see that this resultis consistent with the symmetry of the signal, which is odd symmetry) and

30

that take a positive and negative values depending on the positive and negativevalues of n. In conclusion, the phase will be π/2 for n > 0 and odd, and thephase will be −π/2 for n < 0 and odd.

4. Plot the discrete amplitude spectrum. You should obtain the plots using Matlab.The plot for the magnitude is shown in Figure 4.2.

Figure 4.2: Discrete Magnitude spectrum of coefficients Cn.

4.2 Example 2Answer the same questions as in previous example, but for the waveform shown in Figure4.3. Comment the differences that the coefficients Cn have between the two representations.

1. We follow the same steps as those for the first example. Just for this time, considerthat the function given in Figure 4.3 is called g(t). Then we have that the function isgiven by

g(t) =

4πt+ 1, −π

2< t < 0,

1− 4πt, 0 ≤ t ≤ π

2.

(4.17)

2. See that the function g(t) in Equation (4.17) can be obtained by substituting in Equa-tion (4.1) t by t + π

4. The second step is to obtain the information from the signal

such as period and fundamental frequency, which in this problem we have the sameinformation as that for Problem 1, in other words,

T0 = π, f0 =1

T0=

1

π, then we have 2πf0n = 2n. (4.18)

31

Figure 4.3: Shifted Triangular Waveform.

3. The third step is to obtain the complex exponential Fourier series coefficients, i.e.,calculate Cn. We start the process from the definition of the coefficients and keepworking on it using Equation (4.17) and Equation (4.18) as follows.

Cn =1

T0

∫T0

f(t)e−j2πf0nt dt

=1

π

∫ 0

−π2

(4

πt+ 1

)e−j2nt dt+

1

π

∫ π2

0

(1− 4

πt

)e−j2nt dt

=4

π2

∫ 0

−π2

te−j2nt dt+1

π

∫ 0

−π2

e−j2nt dt

+1

π

∫ π2

0

e−j2nt dt− 4

π2

∫ π2

0

te−j2nt dt. (4.19)

Now, we work with Equation (4.19) by solving the first and fourth integrals usingEquation (5.3). You must complete all the steps to get what it is shown here. Thenwe have (after solving and substituting the initial and final values)

Cn =4

π2

[1

4n2− π

j4nejnπ − 1

4n2ejnπ

]− 1

j2nπ+

1

j2nπejnπ − 1

j2nπe−jnπ +

1

j2nπ

− 4

π2

[− 1

4n2− π

j4ne−jnπ +

1

4n2e−jnπ

]. (4.20)

32

By regrouping and eliminating terms in Equation (4.20) we obtain

Cn =2

n2π2− 1

n2π2

(ejnπ + e−jnπ

)+

1

jnπejnπ

(1

2− 1

)+

1

jnπe−jnπ

(1− 1

2

)=

2

n2π2− 1

n2π2cos(nπ)− 1

j2nπejnπ +

1

j2nπe−jnπ

=2

n2π2− 1

n2π2cos(nπ)− 1

nπ

1

2j

(ejnπ − e−jnπ

)=

2

n2π2− 1

n2π2cos(nπ)− 1

nπsin(nπ)

=2

n2π2(1− cos(nπ))

=2

n2π2[1− (−1)n] . (4.21)

Note that this result gives coefficients Cn that are purely real because the signal haseven symmetry. With the result in Equation (4.21) just obtained, you can see thesimilarity to Equation (4.13) where you can find the extra factor e−jn

π2 . So you can

see that with respect to the previous example, we have in this example, a time shift ofg(t) by π/4 time units to the right to obtain f(t) (see functions in figures 4.1 and 4.3),in other words, f(t) = g(t− π

4). Whenever you have a time shift of k time units to the

right of a function, the coefficients Cn will be affected by the multiplication of a factor

e−j2πf0nk (4.22)

In Example 1 we have a signal f(t) with a time shift of k = π/4 time units to the rightof the signal of problem 2, then (recall that in this problem f0 = 1/π) we have

e−j2πf0nk = e−j2π1πnπ

4 = e−jnπ2 . (4.23)

Therefore, the factor obtained in Equation (4.23) is multiplied by the coefficient Cn ofEquation (4.21) to get the coefficients of problem 1.

Chapter 5

CEFS of Cosine Functions

5.1 Example 1Consider the periodic function f(t) with period T0 given by

f(t) = cos(2πf0t). (5.1)

We are interested in obtaining the CEFS representation of the signal. The CEFS of the signalin (5.1) can be obtained by three methods. We show those three methods in the following,beginning with the most elaborate.

1. Method 1: Using the cosine function.In order to obtain the CEFS of function f(t) in Equation (5.1), first determine itsperiod, which in this case is T0. Then calculate the CEFS coefficients Cn which willbe given by

Cn =1

T0

∫ T0/2

−T0/2f(t)e−j2πf0ntdt

=1

T0

∫ T0/2

−T0/2cos(2πf0t)e

−j2πf0ntdt. (5.2)

As one can see, Equation (5.2) is an integral that has the following form∫eατ cos(βτ)dτ =

1

α2 + β2eατ [α cos(βτ) + β sin(βτ)] . (5.3)

To use the solution of the integral, we just see that we have for our example the fol-lowing

α = −j2πf0n, α2 = −4π2f 20n

2,

β = 2πf0, β2 = 4π2f 20 . (5.4)

33

34

Using the parameters defined in (5.4), we get

Cn =1

T0

∫ T0/2

−T0/2cos(2πf0t)e

−j2πf0ntdt

=e−j2πf0nt

4π2f 20 (1− n2)T0

−j2πf0n cos(2πf0t) + 2πf0 sin(2πf0t)]T0/2−T0/2

=e−j2πf0nt

2πf0(1− n2)T0sin(2πf0t)− jn cos(2πf0t)

]T0/2−T0/2

=e−j2πf0nt

2π(1− n2)

ej2πf0t − e−j2πf0t

2j− jne

j2πf0t + e−j2πf0t

2

]T0/2−T0/2

=e−j2πf0nt

2π(1− n2)

ej2πf0t − e−j2πf0t

2j+ n

ej2πf0t + e−j2πf0t

2j

]T0/2−T0/2

=e−j2πf0nt

j4π(1− n2)

ej2πf0t − e−j2πf0t + nej2πf0t + ne−j2πf0t

]T0/2−T0/2

=1

j4π(1− n2)

(1 + n)ej2πf0t(1−n) − (1− n)e−j2πf0t(1+n)

]T0/2−T0/2

=1

j4π

ej2πf0t(1−n)

1− n− e−j2πf0t(1+n)

1 + n

]T0/2−T0/2

=1

j4π

ej2πf0

T02(1−n)

1− n− e−j2πf0

T02(1−n)

1− n− e−j2πf0

T02(1+n)

1 + n+ej2πf0

T02(1+n)

1 + n

=1

j4π

ejπ(1−n)

1− n− e−jπ(1−n)

1− n− e−jπ(1+n)

1 + n+ejπ(1+n)

1 + n

=

1

2j 2π

ejπ(1−n) − e−jπ(1−n)

1− n+ejπ(1+n) − e−jπ(1+n)

1 + n

=

1

2π

sin(π(1− n))

1− n+

sin(π(1 + n))

1 + n

. (5.5)

Now, see that there are two critical values of n in Equation (5.5) which are n = +1and n = −1. These are critical values because they make the denominator zero in oneof the terms in the equation. For n = +1 we obtain C1 using L’Hopital’s rule you mustshow that C1 = 1/2. In a similar way, see that C−1 = 1/2. Show that Cn = 0 forn 6= ±1. Therefore we have

C0 = 0, C1 =1

2, C−1 =

1

2, Cm = 0, m = ±2,±3, . . . . (5.6)

The cosenoidal signal has only two coefficients, namely C1 and C−1 both with value

35

of 1/2. Therefore, the CEFS of the function f(t) = cos(2πf0t) is given by

f(t) =∞∑

n=−∞

Cnej2πf0tn

= C1ej2πf0t + C−1e

−j2πf0t

=1

2ej2πf0t +

1

2e−j2πf0t

= cos(2πf0t). (5.7)

In conclusion, we only need to use the trigonometric identity of the cosine function.

2. Method 2: Using exponentials instead of cosine function. In this part, we alsocompute the CEFS coefficients, but instead of using the integral in Equation (5.3), wefirst substitute the trigonometric identity of the cosine to solve for the coefficients Cnas follows

Cn =1

T0

∫ T0/2

−T0/2cos(2πf0t)e

−j2πf0ntdt

=1

2T0

∫ T0/2

−T0/2

[ej2πf0t + e−j2πf0t

]e−j2πf0ntdt

=1

2T0

∫ T0/2

−T0/2ej2πf0te−j2πf0ntdt+

1

2T0

∫ T0/2

−T0/2e−j2πf0te−j2πf0ntdt

=1

2T0

∫ T0/2

−T0/2ej2πf0t(1−n)dt+

1

2T0

∫ T0/2

−T0/2e−j2πf0t(1+n)dt

=1

j4π(1− n)ej2πf0t(1−n)

]T0/2−T0/2

− 1

j4π(1 + n)e−j2πf0t(1+n)

]T0/2−T0/2

=1

j4π

ej2πf0t(1−n)

1− n− e−j2πf0t(1+n)

1 + n

]T0/2−T0/2

. (5.8)

The last row of Equation (5.8) is exactly the same as the eighth row in Equation (5.5),hence the same procedure and result as that will be obtained.

3. Method 3: Using the trigonometric identity of the cosine function.

f(t) =∞∑

n=−∞

Cnej2πf0tn

= C1ej2πf0t + C−1e

−j2πf0t

=1

2ej2πf0t +

1

2e−j2πf0t

= cos(2πf0t). (5.9)

36

5.2 Example 2Obtain the CEFS representation of the following signal

g(t) = cos2(2π200t)− sin(2π400t). (5.10)

For this problem, recall to use the trigonometric identities for the cosine and sine func-tions. First use the trigonometric identity cos2(α) = 1

2+ 1

2cos(2α) to get

g(t) =1

2+

1

2cos(2π400t)− sin(2π400t). (5.11)

We can see in Equation (5.11) that the fundamental frequency of the signal is f0 = 400Hz, then its period would be T0 = 1/400 sec.

Now you can apply the trigonometric identities of the sine and cosine functions in termsof the exponentials in Equation (5.11) to obtain the CEFS.

g(t) =1

2+

1

4

(ej2π400t + e−j2π400t

)− 1

2j

(ej2π400t − e−j2π400t

)=

1

2+

1

4ej2π400t +

1

4e−j2π400t − 1

2jej2π400t +

1

2je−j2π400t

=1

2+

1

4ej2π400t +

1

4e−j2π400t + j

1

2ej2π400t − j 1

2e−j2π400t. (5.12)

Then the coefficients of this CEFS are

C0 =1

2, C1 =

1

4+ j

1

2, C−1 =

1

4− j 1

2. (5.13)

Part II

Frequency Domain Analysis

37

Chapter 6

Fourier Transform

The Fourier transform is applied in general for aperiodic signals that could be of infinite(power signals) or finite duration (energy signals). Consider the aperiodic signal x(t), itsFourier transform is defined to be

X(f) = F x(t)

=

∞∫−∞

x(t)e−j2πftdt. (6.1)

The inverse Fourier Transform is defined as

x(t) = F−1 X(f)

=

∞∫−∞

X(f)ej2πftdf. (6.2)

The Fourier transform has several properties that could be used to obtain the transformof signals. In this document, you can find some of the properties and applications of them asneeded throughout the examples. One of the properties is the time-shift property. Considera signal s(t) with Fourier transform S(f), now if we need to obtain the Fourier transform ofthe shifted version s(t− t0), we can define first the following x(t) = s(t− t0) and apply the

39

40

FT definition in Equation (6.1) to obtain

X(f) = F x(t)

=

∞∫−∞

x(t)e−j2πftdt

=

∞∫−∞

s(t− t0)e−j2πftdt

=

∞∫−∞

s(z)e−j2πf(z+t0)dz

=

∞∫−∞

s(z)e−j2πfze−j2πft0dz

= e−j2πft0∞∫

−∞

s(z)e−j2πfzdz

= e−j2πft0S(f). (6.3)

Equation (6.3) tells us that the Fourier transform of a shifted version of a signal s(t) byt0 seconds is given by the product of the Fourier transform of the signal (un-shifted version)s(t) multiplied by the complex exponential e−j2πft0 .

6.1 Example

In this example, we analyze the rectangular pulse signal centered at the origin and of durationT given by

s(t) = A∏(

t

T

). (6.4)

A rectangular pulse signal with A = 3 and T = 10 is shown in Figure 6.1.

41

Figure 6.1: Rectangular pulse signal A = 3, T = 10.

The Fourier transform of signal s(t) is given by

X(f) = F x(t)

=

∞∫−∞

s(t)e−j2πftdt

=

∞∫−∞

A∏(

t

T

)e−j2πftdt

=

T/2∫−T/2

A e−j2πftdt

= A

T/2∫−T/2

e−j2πftdt

=A

−j2πfe−j2πft

∣∣∣∣T/2−T/2

=A

−j2πf

e−j2πfT/2 − ej2πfT/2

=

A

j2πf

ejπfT − e−jπfT

=

A

πfsin(πfT )

=AT

πfTsin(πfT )

= AT sinc(fT ). (6.5)

42

Now, the magnitude of the Fourier transform of a rectangular pulse with A = 3 andT = 2 is shown in Figure 6.2. Note in the figure how the magnitude of the sinc functiontouches the horizontal axis in integer multiples of the inverse of the duration, i.e., k/T fork = 0,±1,±2, . . .. The continuous line in the figure is the Fourier transform obtained byusing the command fft in MATLAB, whereas the red dots correspond to the evaluation of thesinc function in Equation (6.5).

Figure 6.2: Fourier transfrom of a rectangular pulse signal A = 3, T = 2.

6.2 Fourier Transform of a Periodic SignalEven if the FT is said to apply to aperiodic signals, we have that it can be applied to periodicsignals as well. Consider a periodic signal xp(t) with period T0, its Fourier transform is givenby the following four-step method.

1. Step 1: Truncation. Take the periodic signal xp(t) and truncate one period, consideralways as the best option to truncate the period centered at the origin, i.e., from−T0/2to T0/2. In other words, you need to obtain the truncated signal x(t) as

x(t) = xp(t)∏(

t

T0

). (6.6)

2. Step 2: Fourier Transform. Obtain the Fourier transform of the truncated signal x(t)by using the definition in Equation (6.1). In other words, obtain X(f).

3. Step 3: “Sampling”. In the Fourier transform just obtained, change every instance ofthe variable f by nf0 where f0 = 1/T0 to obtain X(nf0).

43

4. Step 4: Scale. The Fourier transform of the periodic signal will be given by a scaledversion of X(nf0) as follows

Xp(f) =1

T0X(nf0) = Cn. (6.7)

6.3 Example of the FT of a Periodic SignalConsider the periodic signal given by a rectangular pulse train with period T0, amplitude Aand the duration of each rectangular pulse being T where T ≤ T0/2. The signal can berepresented mathematically by

xp(t) =∞∑

n=−∞

A∏(

t− nT0T

). (6.8)

See that the signal in Equation (6.8) has even symmetry, and hence the rectangular pulsecorresponding to n = 0 is centered at the origin.

Now we can apply the four-step method

1. Step 1: Truncation. Consider the case of n = 0 to obtain

x(t) = A∏(

t

T

). (6.9)

2. Step 2: Fourier Transform. The FT was just obtained in the previous example givenin Equation (6.5), hence we get

X(f) = AT sinc(fT ). (6.10)

3. Step 3: “Sampling”.X(nf0) = AT sinc(nf0T ). (6.11)

4. Step 4: Scale.

Xp(f) =AT

T0sinc

(nT

T0

)= Ad sinc(nd). (6.12)

In Equation (6.12) d is the duty cycle and is given by d = T/T0. We generally areinterested in duty cycles of the form 1/2, 1/3, 1/4, . . ..

Check by yourself that the Fourier transform of a periodic signal gives the CEFS coef-ficients Cn, i.e., Equation (6.12) is exactly the same as the coefficients of the CEFS Cn for arectangular pulse train.

44

Chapter 7

Frequency Domain Analysis

In Chapter 2 you can find the time domain analysis of a passive low-pass filter, where weobtained the differential equation, its solution depending on the input and then obtained theimpulse response h(t) of the circuit. We also applied convolution to obtain the output ofthe filter when a rectangular pulse was the input. In this chapter, we consider the frequencydomain analysis through Laplace of the same circuit that is shown in Figure 7.1.

Figure 7.1: RC low-pass circuit.

In order to perform the frequency domain analysis of the circuit, we need to consider thecomplex impedances of the components, in this case the complex impedance of the resistorR is R because from Ohm’s law we have that v(t) = Ri(t) and in the frequency domainthis equation is V (s) = RI(s) hence the quotient satisfies ZR = V (s)/I(s) = R. For thecapacitor the complex impedance comes from the voltage in the capacitor, i.e.,

v(t) =1

C

t∫−∞

i(τ)dτ, (7.1)

which in the frequency domain is

V (s) =1

CsI(s), (7.2)

45

46

and hence we have that the complex impedance of the capacitor ZC is given by the quotient

ZC =V (s)

I(s)=

1

sC. (7.3)

Now, the input in the frequency domain is X(s) and the output is Y (s), hence we canobtain the output of the circuit by using the complex impedances just as if they were resistors,hence Y (s) is given by the voltage divisor of ZR and ZC , then we have

Y (s) =ZC

ZC + ZRX(S)

=1/(Cs)1Cs

+RX(s)

=1

1 +RCsX(s)

=

[1/RC

s+ 1RC

]X(s). (7.4)

7.1 Transfer Function and Impulse ResponseThe term within the brackets in the last row of Equation (8.27) is the frequency relationshipbetween the output Y (s) and the input X(s) in the frequency domain. From this equationwe can obtain the Transfer Function H(s) as the quotient of the output to the input to getthe following

H(s) =Y (s)

X(s)=

1

RC

1

s+ 1RC

. (7.5)

From the transfer function H(s), we can obtain two important things, one is the impulseresponse and the other is the differential equation of the circuit. To obtain the impulseresponse we need to apply the inverse Laplace transform to H(s). Recall that in this courseyou already must know how to use the Laplace transform. For this example we have

h(t) =1

RCe−

tRC , t ≥ 0. (7.6)

See that Equation (7.6) is exactly the same as the impulse response obtained in Equation(2.12) in Chapter 2 from the solution of the differential equation with input an impulse.

The differential equation of the circuit can be obtained from the transfer function H(s)as follows

H(s) =Y (s)

X(s)

=1

RC

1

s+ 1RC

, (7.7)

47

From (7.7) we can get

Y (s)

(s+

1

RC

)=

1

RCX(s), (7.8)

and using the inverse Laplace transform we get

dy(t)

dt+

1

RCy(t) =

1

RCx(t), (7.9)

see that Equation (7.9) is exactly the same as Equation (2.5) from Chapter 2 which is thedifferential equation of the circuit.

7.2 Frequency ResponseThe frequency response H(f) of a system will be obtained from the transfer function H(s)by making the substitution s = j2πf . Hence in this example we get from Equation (7.5) thefollowing

H(f) = H(s)|s=j2πf

=1

RC

1

j2πf + 1RC

(7.10)

The magnitude response of the circuit is given by

|H(f)| = 1

RC

1√4π2f 2 +

(1RC

)2 . (7.11)

The magnitude square of the system is given by

|H(f)|2 =1

(RC)21

4π2f 2 +(

1RC

)2 . (7.12)

Another important function of the frequency domain is the amount of dBs which is ob-tained as follows

|H(f)|dB = 10 log10 |H(f)|2. (7.13)

As an example, the magnitude response of theRC low pass filter given in Equation (7.11)is shown in Figure 7.2 for R = 10 kΩ and C = 0.1 µF.

The magnitude squared of the same example of RC low pass filter is shown in Figure7.3.

The dB of the frequency response is shown in Figure 7.4. See how in the figure we havethe −3 dB horizontal line indicating the frequency f0 = 1/(2πRC) which is the cut-offfrequency of the filter that defines its bandwidth.

48

Figure 7.2: Magnitude response of RC low-pass circuit.

Figure 7.3: Magnitude squared response of RC low-pass circuit.

49

Figure 7.4: Magnitude response in dBs of RC low-pass circuit.

50

Chapter 8

Homework 4: Solutions

8.1 Problem 1

Consider the periodic signal gp(t) = A cos(2πf0t + φ) with period T0 = 1/f0. An exampleof a realization of this signal is shown in Figure 8.1 for the specific case of frequency f0 = 2Hz, phase shift of φ = π/4, and amplitude A = 3V. Note that since the phase shift ispositive, the signal is shifted to the left of the vertical axis as shown in the figure. Also, thisshift produces a signal that has no symmetry, therefore, you should expect to get that theCEFS coefficients Cn be complex with real and imaginary parts different from zero.

Figure 8.1: Signal gp(t) for A = 3, φ = π/4, and f0 = 2Hz.

1. Obtain the coefficients cn of the CEFS. You need to use the definition of the coefficientswith the integral and show all your work.

51

52

Cn =1

T0

T0/2∫−T0/2

gp(t)e−j2πf0ntdt

=1

T0

T0/2∫−T0/2

A cos(2πf0t+ φ)e−j2πf0ntdt

=A

2T0

T0/2∫−T0/2

[ej2πf0t+jφ + e−j2πf0t−jφ

]e−j2πf0ntdt

=A

2T0

T0/2∫−T0/2

[ejφej2πf0t + e−jφe−j2πf0t

]e−j2πf0ntdt

=A

2T0

T0/2∫−T0/2

[ejφe−j2πf0(n−1)t + e−jφe−j2πf0(n+1)t

]dt

=A

2T0

ejφ

1

−j2πf0(n− 1)e−j2πf0(n−1)t

+e−jφ1

−j2πf0(n+ 1)e−j2πf0(n+1)t

]T0/2−T0/2

=A

2T0

−ejφ

j2πf0(n− 1)

[e−jπ(n−1) − ejπ(n−1)

]− e−jφ

j2πf0(n+ 1)

[e−jπ(n+1) − ejπ(n+1)

]=

A

2T0

ejφ

πf0(n− 1)sin(π(n− 1)) +

e−jφ

πf0(n+ 1)sin(π(n+ 1))

=

Aejφ

2

sin(π(n− 1))

π(n− 1)+Ae−jφ

2

sin(π(n+ 1))

π(n+ 1)

=

Aejφ

2, n = 1,

Ae−jφ

2n = −1,

0, n 6= ±1.

(8.1)

53

2. Provide an expression of the CEFS for gp(t)

gp(t) =∞∑

n=−∞

Cnej2πf0nt

=A

2ejφej2πf0t +

A

2e−jφe−j2πf0t

=A

2ej(2πf0t+φ) +

A

2e−j(2πf0t+φ)

= A cos(2πf0t+ φ). (8.2)

Recall that the CEFS of a cosine function is obtained using the trigonometric identity.

3. Give an expression for the magnitude of the coefficients, i.e., |Cn|

|C+1| =A

2, |C−1| =

A

2. (8.3)

This can also be expressed as |Cn| = A2δ(n− 1) + A

2δ(n+ 1).

4. Give an expression for the phase of the coefficients, i.e., ∠Cn. Since C−1 = Ae−jφ

2=

A2

cos(φ)− jA2

sin(φ) and C+1 = Aejφ

2= A

2cos(φ)+ jA

2sin(φ), we have that the phase

of these is given by

∠C+1 = tan−1

A2

sin(φ)A2

cos(φ)

= tan−1(tan(φ)) = φ,

∠C−1 = tan−1

−A

2sin(φ)

A2

cos(φ)

= tan−1(− tan(φ)) = −φ. (8.4)

5. Plot the magnitude spectrum of the coefficients, |Cn| Figure 8.2 shows a plot of themagnitude of the coefficients when A = 3.

Figure 8.2: Magnitude of coefficients Cn for A = 3.

54

Figure 8.3: Phase of coefficients Cn for A = 3.

6. Plot the phase spectrum of the coefficients, |Cn| Figure 8.3 shows a plot of the phaseof the coefficients when φ = π/4.

7. Obtain the coefficients C0, An andBn of the TFS. From the solution of the coefficientsCn, we have that C0 = 0 and that we only have coefficients for n = 1 and n = −1.Then we can apply the identities An = Cn + C−n and Bn = j(Cn − C−n) to obtain

A1 =Aejφ

2+Ae−jφ

2= A cos(φ). (8.5)

and

B1 = j

(Aejφ

2− Ae−jφ

2

)= −A sin(φ). (8.6)

8. Obtain the Fourier transform of gp(t) using the method seen in class given by

(a) Truncate gp(t), give a sketch or an expression of this signal, call it x(t). Weprovide an expression of this signal which is

x(t) = A cos(2πf0t+ φ)∏(

t

T0

). (8.7)

55

(b) Obtain the Fourier transform X(f) of x(t)

X(f) =

T0/2∫−T0/2

x(t)e−j2πftdt

=

T0/2∫−T0/2

A cos(2πf0t+ φ)e−j2πftdt

=A

2

T0/2∫−T0/2

[ej2πf0t+jφ + e−j2πf0t−jφ

]e−j2πftdt

=A

2

T0/2∫−T0/2

[ejφej2πf0t + e−jφe−j2πf0t

]e−j2πftdt

=A

2

T0/2∫−T0/2

[ejφe−j2π(f−f0)t + e−jφe−j2π(f+f0)t

]dt

=A

2

ejφ

1

−j2π(f − f0)e−j2π(f−f0)t

+e−jφ1

−j2π(f + f0)e−j2π(f+f0)t

]T0/2−T0/2

=A

2

−ejφ

j2π(f − f0)[e−jπ(f−f0)T0 − ejπ(f−f0)T0

]− e−jφ

j2π(f + f0)

[e−jπ(f+f0)T0 − ejπ(f+f0)T0

]=

A

2

ejφ

π(f − f0)sin(π(f − f0)T0) +

e−jφ

π(f + f0)sin(π(f + f0)T0)

= ejφ

AT02sinc((f − f0)T0) + e−jφ

AT02sinc((f + f0)T0). (8.8)

(c) “Sample” X(f) by changing f for nf0, i.e., give X(nf0). To do this we use theexpression (f ± f0) = (nf0 ± f0) = (n± 1)f0 and the identity f0T0 = 1 to get

X(nf0) = ejφAT0

2sinc((nf0 − f0)T0) + e−jφ

AT02sinc((nf0 + f0)T0)

= ejφAT0

2sinc((n− 1)) + e−jφ

AT02sinc((n+ 1))

= ejφAT0

2

sin(π(n− 1))

π(n− 1)+ e−jφ

AT02

sin(π(n+ 1))

π(n+ 1). (8.9)

56

The last row of Equation (8.9) is close to the penultimate row in Equation (8.1)with he only difference of a scaling factor of T0.

(d) Scale your result, i.e., provide 1T0X(nf0)

Gp(f) =1

T0X(nf0) = ejφ

A

2δ(f − f0) + e−jφ

A

2δ(f + f0). (8.10)

(e) Compare this result to the coefficient Cn obtained in part (a), conclude. Compar-ison of equations (8.9) and (8.1) shows that the scaling of X(nf0) gives the sameresult as that of the coefficients of the CEFS Cn.

9. Provide the transform pair of gp(t)

gp(t) = A cos(2πf0t+ φ)F⇐⇒

Gp(f) = ejφA

2δ(f − f0) + e−jφ

A

2δ(f + f0). (8.11)

8.2 Problem 2Define a periodic signal xp(t) with period T0 = 2π. One period of the signal is given by

xp(t) =

t, 0 ≤ t ≤ π,0, −π < t < 0.

(8.12)

1. Sketch the signal with three periods. Figure 8.4 shows three periods of the signal.

Figure 8.4: Three periods of signal xp(t).

2. Obtain the CEFS coefficients Cn and show that they are given by

Cn =

1

2πn2 [(−1)n − 1] + j 12n

(−1)n, n = ±1, ± 2, . . . ,

π4, n = 0.

(8.13)

57

First note that T0 = 2π, then we start with the definition of the CEFS coefficients Cnand work by integrating by parts to obtain them as follows

Cn =1

T0

T0/2∫−T0/2

xp(t)e−j2πf0ntdt

=1

2π

π∫0

te−j2πf0ntdt

=1

2π

π∫0

te−jntdt

=−1

2π

t

jne−jnt

]π0

+1

2π

1

jn

π∫0

e−jntdt

=−1

j2πnπe−jnπ +

1

j2πn

(−1

jn

)e−jnt

]π0

=j

2ne−jnπ +

1

2πn2

(e−jnπ − 1

)=

1

2πn2[(−1)n − 1] + j

1

2n(−1)n. (8.14)

The coefficients in Equation (8.14) are valid for negative and positive n 6= 0. To obtainC0 we also apply the definition of the coefficients to obtain the following

C0 =1

T0

T0/2∫−T0/2

xp(t)dt

=1

2π

π∫0

tdt

=1

2π

t2

2

]π0

=1

2π

(π2

2− 0

)=

π

4. (8.15)

3. Provide the magnitude |Cn|. To calculate this, we use the magnitude square first, i.e.,|Cn|2. See that |C0| = π/4. We also use the fact that (−1)2n = ((−1)2)

n= 1n = 1.

Then for n 6= 0 we have

58

|Cn|2 =

1

2πn2[(−1)n − 1]

2

+

1

2n(−1)n

2

=1

4π2n4

[(−1)2n − 2(−1)n + 1

]+

1

4n2(−1)2n

=1

4π2n4[1− 2(−1)n + 1] +

1

4n2

=1

4π2n4[2− 2(−1)n] +

1

4n2

=1

2π2n4[1− (−1)n] +

1

4n2, |n| > 0. (8.16)

4. Give terms n = 0,±1,±2 of the CEFS.

The CEFS would be given by

xp(t) =∞∑

n=−∞

Cnej2πf0nt (8.17)

For n = 0 we have C0 = π/4, for n = 1, we have C1 from Equation (8.14) and soon. We show the first term for n = 0, the second term for n = 1, the third term forn = −1, the fourth term for n = 2 and the fifth term for n = −2.

xp(t) =1

4

=

(− 1

π− j 1

2

)ejt

=

(− 1

π+ j

1

2

)e−jt

= j1

4ej2t − j 1

4e−j2t. (8.18)

5. Give the coefficients C0, An and Bn of the TFS.

C0 = π/4. Use identities An = Cn + C−n and Bn = j(Cn − C−n) to obtain thecoefficients for the TFS.

An =1

n2π2[(−1)n − 1] ,

Bn = − 1

nπ(−1)n . (8.19)

6. Give terms n = 0, 1, 2, 3 of the TFS. Similar to the CEFS terms.

59

7. Give the Fourier transform of the truncated signal. Show that it is given by

X(f) =1

4π2f 2

(e−j2π

2f − 1)

+ j1

2fe−j2π

2f . (8.20)

Consider the truncated signal as x(t) = xp(t)∏(

t−π2π

). Using the definition and inte-

grating by parts we get

X(f) =

π∫0

te−j2πftdt

= − t

j2πfe−j2πft

]π0

+1

j2πf

π∫0

e−j2πftdt

= − 1

j2fe−j2π

2f +1

j2πf

(− 1

j2πf

)e−j2πft

]π0

= j1

2fe−j2π

2f +1

4π2f 2

(e−j2π

2f − 1). (8.21)

8.3 Problem 3For the circuit in Figure 8.5.

Figure 8.5: Circuit for problem 3.

1. Provide the transfer function H(s).

First name the node that joins R1, R2 and capacitor C1 as node z, consider that thecurrent in the first circuit is given by i1(t) or I1(s) and the current in the second circuitis I2(s), then we can see that

Vz(s) =1

sC1

[I1(s)− I2(s)] , (8.22)

60

where

I1(s) =X(s)− Vz(s)

R1

, (8.23)

and

I2(s) =Vz(s)− Y (s)

R2

. (8.24)

Then, we obtain Vz(s) as follows

Vz(s) =1

sC1

[I1(s)− I2(s)]

=1

sC1

[(X(s)− Vz(s)

R1

)−(Vz(s)− Y (s)

R2

)]=

1

sC1

X(s)

R1

− Vz(s)

sC1R1

− Vz(s)

sC1R2

+Y (s)

sC1R2

. (8.25)

With this, we can obtain the following

Vz(s)

(1 +

1

sC1R1

+1

sC1R2

)=

1

sC1

X(s)

R1

+Y (s)

sC1R2

. (8.26)

We also have that the output Y (s) together with Equation (8.23) is given by

Y (s) =1

sC2

I2(s)

=Vz(s)

sC2R2

− Y (s)

sC2R2

Y (s)

(1 +

1

sC2R2

)=

Vz(s)

sC2R2

Y (s) (1 + sC2R2) = Vz(s). (8.27)

Now we substitute this last result of Equation (8.27) into Equation (8.26) to get

Y (s) (1 + sC2R2)

(1 +

1

sC1R1

+1

sC1R2

)=

1

sC1

X(s)

R1

+Y (s)

sC1R2

Y (s) (1 + sC2R2)

(sC1R1R2 +R2 +R1

sC1R1R2

)=

X(s)

sC1R1

+Y (s)

sC1R2

. (8.28)

From Equation (8.28), we can solve for Y (s) in terms of X(s) to obtain the transferfunction H(s) as follows

Y (s) (1 + sC2R2)

(sC1R1R2 +R2 +R1

sC1R1R2

)− Y (s)

sC1R2

=X(s)

sC1R1

Y (s)

[(1 + sC2R2) (sC1R1R2 +R2 +R1)−R1

sC1R1R2

]=

X(s)

sC1R1

. (8.29)

61

With the result in Equation (8.29) we can obtain the following

H(s) =Y (s)

X(s)

=R2

(1 + sC2R2) (sC1R1R2 +R2 +R1)−R1

=R2

sC1R1R2 +R2 +R1 + s2C1C2R1R22 + sC2R2

2 + sC2R1R2 −R1

=R2

s2C1C2R1R22 + s (C1R1R2 + C2R2

2 + C2R1R2) +R2

=R2/ (C1C2R1R

22)

s2 + s(

C1R1R2

C1C2R1R22

+C2R2

2

C1C2R1R22

+ C2R1R2

C1C2R1R22

)+ R2

(C1C2R1R22)

=1

C1C2R1R2

s2 + s(

1C2R2

+ 1C1R1

+ 1C1R2

)+ 1

C1C2R1R2

. (8.30)

The transfer function H(s) given in Equation (8.30) corresponds to a second orderlow-pass filter with cutoff frequency given by

fc =1

2πC1C2R1R2

. (8.31)

2. Give the frequency response H(f). The frequency response H(f) is obtained fromEquation (8.30) by substituting s = j2πf .

3. Give the differential equation of the circuit. The differential equation is obtained alsofrom Equation (8.30) by using the inverse Laplace transform and considering nullinitial conditions as follows

Y (s)

[s2 + s

(1

C2R2

+1

C1R1

+1

C1R2

)+

1

C1C2R1R2

]= X(s)

1

C1C2R1R2

d2y(t)

dt2+

(1

C2R2

+1

C1R1

+1

C1R2

)dy(t)

dt+

y(t)

C1C2R1R2

=x(t)

C1C2R1R2

. (8.32)

4. For R = 1MΩ = R2 and C1 = C2 = 0.1µF, plot the magnitude response |H(f)|Using the command freqs in Matlab, you can get the magnitude response and thephase response of the filter. This is shown in Figure 8.6

5. Plot |H(f)|2 and |H(f)|dBWe can use the same Matlab command to obtain the plot of the magnitude square andthe dB plot. Figure 8.7 shows the magnitude square of the filter.

Figure 8.8 shows the frequency response in dBs. .

62

Figure 8.6: Frequency response of low-pass filter.

Figure 8.7: Square of frequency response of low-pass filter.

63

Figure 8.8: dB of frequency response of low-pass filter.

64

Chapter 9

Partial Exam 2: Solutions

This chapter presents the solution to the problems of the second Partial Exam. The solutionswill be given in a general form so that any specific case given by the ID numbers used, canbe derived.

9.1 Problem 1Consider the definition of the Fourier transform and the signal x(t) given by

x(t) = u(t− k)e−at, (9.1)

where k is the least significant digit of your student ID number (e.g., if your ID number isA01234567, then k = 7. If your digit is zero then k = 0).

In this case, the problem will be solved using delay k.

1. Sketch the signal x(t).

Figure 9.1: Signal x(t) with a = 2 and k = 2.

65

66

2. Obtain the Fourier transform of x(t), i.e., X(f).

From the sketch shown in Figure 9.1, we can see that the time where the signal existsis limited to the interval from k up to infinity. We can also see that the exponentialis the only part that we need to consider since the unit step function u(t − k) is onlyaffecting the interval where the signal exists, then we have that the Fourier transformis given by

X(f) =

∞∫k

x(t)e−j2πftdt

=

∞∫k

e−ate−j2πftdt

=

∞∫k

e−t(a+j2πf)dt

=1

−(a+ j2πf)e−t(a+j2πf)

∣∣∣∣∞k

=1

−(a+ j2πf)

(0− e−k(a+j2πf)

)=

e−ka

a+ j2πfe−j2πfk. (9.2)

Note that 1a+j2πf

is the Fourier transform of an exponential signal e−at, also note thatsince we are considering the initial time to be k, the amplitude of the exponential ise−ka. Also, since the signal is considering a time-delay of k units, we have the complexexponential term e−j2πfk.

9.2 Problem 2: RLC circuit

Consider the RLC circuit in series as that shown in Figure 9.2

67

Figure 9.2: RLC circuit in series

9.2.1 Output from Inductor

1. Provide the transfer function H(s) when the output of the circuit is taken from thevoltage of the inductor L.

H(s) =sL

sL+R + 1sC

=sL

sL+R + 1sC

sC

sC

=s2LC

s2LC +RCs+ 1

=s2LC

s2LC +RCs+ 1

1LC1LC

=s2

s2 + RLs+ 1

LC

. (9.3)

This transfer function corresponds to a second order filter of the type High-pass. Thecutoff frequency is fc = 1√

2πLC.

2. Give the differential equation of the system obtained from H(s).

Since we have that

H(s) =Y (s)

X(s), (9.4)

then we have that from the transfer function we get

Y (s)

[s2 +

R

Ls+

1

LC

]= X(s)s2. (9.5)

68

hence the differential equation is obtained by applying the inverse Laplace transformwith null initial conditions to get

d2y(t)

dt2+R

L

dy(t)

dt+

1

LCy(t) =

d2x(t)

dt2. (9.6)

9.2.2 Output from Capacitor

1. Provide the transfer function H(s) when the output of the circuit is taken from thevoltage of the capacitor C.

H(s) =1sC

sL+R + 1sC

=1sC

sL+R + 1sC

sC

sC

=1

s2LC +RCs+ 1

=1

s2LC +RCs+ 1

1LC1LC

=1LC

s2 + RLs+ 1

LC

. (9.7)

This transfer function corresponds to a second order filter of the type Low-pass. Thecutoff frequency is fc = 1√

2πLC.

2. Give the differential equation of the system obtained from H(s).

Since we have that

H(s) =Y (s)

X(s), (9.8)

then we have that from the transfer function we get

Y (s)

[s2 +

R

Ls+

1

LC

]= X(s)

1

LC. (9.9)

hence the differential equation is obtained by applying the inverse Laplace transformwith null initial conditions to get

d2y(t)

dt2+R

L

dy(t)

dt+

1

LCy(t) =

1

LCx(t). (9.10)

69

9.2.3 Output from Resistor1. Provide the transfer function H(s) when the output of the circuit is taken from the

voltage of the resistor R.

H(s) =R

sL+R + 1sC

=R

sL+R + 1sC

sC

sC

=RCs

s2LC +RCs+ 1

=RCs

s2LC +RCs+ 1

1LC1LC

=RLs

s2 + RLs+ 1

LC

. (9.11)

This transfer function corresponds to a second order filter of the type Band-pass.

2. Give the differential equation of the system obtained from H(s).

Since we have that

H(s) =Y (s)

X(s), (9.12)

then we have that from the transfer function we get

Y (s)

[s2 +

R

Ls+

1

LC

]= X(s)

R

Ls. (9.13)

hence the differential equation is obtained by applying the inverse Laplace transformwith null initial conditions to get

d2y(t)

dt2+R

L

dy(t)

dt+

1

LCy(t) =

R

L

dx(t)

dt. (9.14)

9.3 Problem 3: RL CircuitConsider an RL circuit in series as that shown in Figure 9.3

1. If the output is the voltage of the inductor L, provide the transfer function H(s).

H(s) =sL

sL+R

=sL

sL+R

1L1L

=s

s+ RL

. (9.15)

70

Figure 9.3: RL circuit in series.

2. Provide the frequency response of the filter H(f).

H(f) = H(s)

]s=j2πf

=s

s+ RL

]s=j2πf

=j2πf

j2πf + RL

=j2πf

j2πf + RL

(−j2πf + R

L

−j2πf + RL

)

=4π2f 2 + j2πf R

L(RL

)2+ 4π2f 2

. (9.16)

3. Provide the differential equation of the filter.

Since we have that

H(s) =Y (s)

X(s), (9.17)

then we have that from the transfer function we get

Y (s)

[s+

R

L

]= X(s)s. (9.18)

hence the differential equation is obtained by applying the inverse Laplace transformwith null initial conditions to get

dy(t)

dt+R

Ly(t) =

dx(t)

dt. (9.19)

71

4. If the input to the circuit is x(t) = cos(2πf0t), obtain the output Y (f)

First recall that the Fourier transform of a cosine signal with fundamental frequencyf0 is given by

x(t) = cos(2πf0t)F⇐⇒

X(f) =1

2δ(f − f0) +

1

2δ(f + f0). (9.20)

Now, we multiply the frequency response in Equation (9.16) and the Fourier transformin Equation (9.20) to obtain the output Y (f) as follows

Y (f) = H(f)X(f)

=

(4π2f 2 + j2πf R

L(RL

)2+ 4π2f 2

)(1

2δ(f − f0) +

1

2δ(f + f0)

)

=

(4π2f 2(

RL

)2+ 4π2f 2

+ j2πf R

L(RL

)2+ 4π2f 2

)(1

2δ(f − f0) +

1

2δ(f + f0)

)

=2π2f 2

0(RL

)2+ 4π2f 2

0

δ(f − f0) + δ(f + f0)

+jπf0

RL(

RL

)2+ 4π2f 2

0

δ(f − f0)− δ(f + f0)

. (9.21)

9.4 Problem 4: Triangular Pulse

Consider the signal x(t) defined as

x(t) =

ATt+ A, −T ≤ t < 0,

A− ATt, 0 ≤ t ≤ T,

0, elsewhere.(9.22)

Here, we also use the general form of a triangular pulse of amplitude A and limited in timefrom −T to T .

1. Sketch the signal x(t).

2. Obtain the Fourier transform of x(t), i.e., X(f).

Considering the triangular pulse in Figure 9.4, we can see that the signal is time lim-ited to the interval [−T, T ], then we proceed applying the definition of the Fourier

72

Figure 9.4: Triangular pulse for A = 2 and T = 3.

transform as follows

X(f) =

T∫−T

x(t)e−j2πftdt

=

0∫−T

x(t)e−j2πftdt+

T∫0

x(t)e−j2πftdt

=

0∫−T

(A

Tt+ A

)e−j2πftdt+

T∫0

(A− A

Tt

)e−j2πftdt. (9.23)

In this case, we are going to need the following integral

b∫a

tectdt =

(t

c− 1

c2

)ect

]ba

(9.24)

Note that in our case, c = −j2πf and that c2 = −4π2f 2. Thus we have the following

73

X(f) =

0∫−T

(A

Tt+ A

)e−j2πftdt+

T∫0

(A− A

Tt

)e−j2πftdt

=A

T

0∫−T

te−j2πftdt+ A

0∫−T

e−j2πftdt+ A

T∫0

e−j2πftdt− A

T

T∫0

te−j2πftdt

=A

T

(t

−j2πf− 1

−4π2f 2

)e−j2πft

]0−T

+A

−j2πfe−j2πft

]0−T

+A

−j2πfe−j2πft

]T0

−AT

(t

−j2πf− 1

−4π2f 2

)e−j2πft

]T0

=A

T

(1

4π2f 2−(−T−j2πf

+1

4π2f 2

)ej2πfT

)− A

j2πf

(1− ej2πfT

)− A

j2πf

(e−j2πfT − 1

)−AT

((T

−j2πf+

1

4π2f 2

)e−j2πfT − 1

4π2f 2

)=

2A

4π2f 2T+

A

j2πf

(ej2πfT − e−j2πfT

)−AT

(T

j2πf+

1

4π2f 2

)ej2πfT − A

T

(−Tj2πf

+1

4π2f 2

)e−j2πfT

=2A

4π2f 2T+

A

πfsin(2πfT )− A

j2πfej2πfT +

A

j2πfe−j2πfT

− A

4π2f 2Tej2πfT − A

4π2f 2Te−j2πfT

=2A

4π2f 2T+

A

πfsin(2πfT )− A

j2πf

(ej2πfT − e−j2πfT

)− A

4π2f 2T

(ej2πfT + e−j2πfT

)=

2A

4π2f 2T+

A

πfsin(2πfT )− A

πfsin(2πfT )− A

2π2f 2Tcos(2πfT )

=A

2π2f 2T− A

2π2f 2Tcos(2πfT ). (9.25)

74

In Equation (9.25), we can see that

X(f) =A

2π2f 2T

1− cos(2πfT )

, (9.26)

and recall that there is a trigonometric identity as follows sin2(α2

)= 1−cos(α)

2which

can be used in Equation (9.26) to get

X(f) =A

2π2f 2T

1− cos(2πfT )

=

A

2π2f 2T2 sin2(πfT )

=A

π2f 2T

T

Tsin2(πfT )

=AT

π2f 2T 2sin2(πfT )

= ATsin(πfT )

πfT

sin(πfT )

πfT

= ATsinc(fT ) sinc(fT )

= ATsinc2(fT ). (9.27)

Figure 9.5 shows the Fourier transformX(f) of the triangular pulse signal x(t). Figure9.6 shows the same figure but with logarithmic vertical axis.

Figure 9.5: X(f) for T = 2 and A = 1.

75

Figure 9.6: X(f) for T = 2 and A = 1.

76

Appendix A. Complex Analysis

A.1 Complex VariableThis chapter is a review of the basic concepts of complex numbers and their operations andfunctions. Before proceeding to the complex number context, we recall some fundamentalconcepts of number theory.

Recall that numbers are organized in different sets, such as the set of natural numbersN = 1, 2, 3, . . ., the set of integers Z = . . . ,−2,−1, 0, 1, 2, . . ., the set of rationalnumbers Q = p/q : p, q ∈ Z, q 6= 0, the set of real numbers R, which sometimes anddepending on the context, the set will contain the −∞ and +∞ symbols. The definitionsand axioms in the following sections are based on [1].

A.1.1 The Real numbersDefinition 4 An ordered set S, is a set with elements satisfying a relation such as <.

Definition 5 A field is a set F with two operations, addition and multiplication which satisfythe following axioms.

• Addition

– For any two numbers x, y ∈ F , we have x+ y ∈ F .

– Commutative property, i.e., ∀x, y ∈ F , x+ y = y + x.

– Associativity property, i.e., ∀x, y, z ∈ F , (x+ y) + z = x+ (y + z).

– Null element, or additive identity, ∀x ∈ F , there exists and element 0 ∈ F suchthat x+ 0 = x.

– Additive inverse, ∀x ∈ F , there is an element −x ∈ F such that x+ (−x) = 0.

• Multiplication

– For any two numbers x, y ∈ F , we have xy ∈ F .

– Commutative property, i.e., ∀x, y ∈ F , xy = yx.

– Associativity property, i.e., ∀x, y, z ∈ F , (xy) z = x (yz).

77

78

– Multiplicative identity, ∀x ∈ F , there exists and element 1 ∈ F such that x1 = x.

– Multiplicative inverse, ∀x ∈ F , x 6= 0, there is an element 1/x ∈ F such thatx(1/x) = 1.

• Distributive Law, ∀x, y, z ∈ F , x (y + z) = xy + xz.

Definition 6 An ordered field is a field F which is also an ordered set, such that

• ∀x, y, z ∈ F , x+ y < x+ z implies y < z.

• ∀x, y ∈ F , xy > 0 implies x > 0 and y > 0, or x < 0 and y < 0.

Definition 7 Consider an ordered set S, and a subset B ⊂ S. The subset B is boundedabove if ∃δ ∈ S such that ∀x ∈ B, x ≤ δ. δ is the upper bound of B.

Definition 8 Let B ⊂ S be bounded above. Assume ∃α ∈ S such that

• α is an upper bound of B

• if β < α, then β is not an upper bound of B.

Then α is the least upper bound of the set B or the supremum of B, i.e., α = sup B.

Theorem 9 The set R = (−∞, ∞) of real numbers together with the addition and multi-plication operations is an ordered field with the least upper bound property, i.e. (R, ·,+) isa field. The set of rational numbers Q with the same operations is a subfield of (R, ·,+).

Theorem 10 For every positive x ∈ R and every integer n > 0, there is a unique realnumber y such that yn = x, or y = n

√x = x1/n.

Definition 11 The real field (R, ·,+) together with the symbols −∞ and +∞ form the Ex-tended Real Number System (ERNS).

∀x ∈ R, some implications of the ERNS are the following

• x/∞ = x/(−∞) = 0

• x+∞ = +∞.

• x+ (−∞) = −∞.

• if x > 0, x(+∞) = +∞, x(−∞) = −∞

• if x < 0, x(+∞) = −∞, x(−∞) = +∞

79

A.1.2 Complex NumbersRecall that there are different sets of numbers such as the natural numbers, the rationalnumbers, the irrational numbers, the integer numbers, the real numbers and the complexnumbers. It is common to see that the real numbers are extended by the ser of complexnumbers, specially when solutions to equations such as x2 = −2 need to be obtained. Inorder to work with the complex numbers, we need the definition i2 = −1, sometimes insteadof the symbol i,the symbol is changed to j2 = −1.We denote that z is a complex number asz ∈ C, where C is the set of complex numbers. A complex number z is written in a formsuch as

z = a+ jb, (A.28)

where a, b ∈ R, and a is called the real part of z, and b is called the imaginary part of z, i.e.,a = <z , and b = =z . The complex conjugate of the complex number z is denoted as

z∗ = a− jb. (A.29)

Consider the complex number in (A.28) and another complex number w = c + jd. Wecan perform the following operations with these two numbers

1. Addition: z+w = (a+ c)+j (b+ d) .This operation satisfies the laws of commutativ-ity (z + w = w + z), and associativity (z + (w + x) = (z + w) + x) , where z, w,andx are complex numbers. The conjugate of a sum is the sum of the conjugates, i.e.,(z + w)∗ = z∗ + w∗.

2. Difference: z − w = (a− c) + j (b− d) .This operation does not satisfy the laws ofcommutativity and associativity. The conjugate of a difference is the difference of theconjugates, i.e., (z − w)∗ = z∗ − w∗.

3. Multiplication: zw = (ac− bd) + j (ad+ bc) . Multiplication also satisfies the com-mutative law since zw = wz, and the associativity law, z (wx) = (zw)x.The conju-gate of a multiplication is the multiplication of the conjugates, i.e., (zw)∗ = z∗w∗.

The multiplication and addition operation also satisfy the distributive law since z (w + x) =zw+zx.When a complex number is multiplied by its complex conjugate, the result is alwaysa purely positive real number (when z 6= 0), since

zz∗ = (a+ jb) (a− jb) (A.30)= a2 − jab+ jba− j2b2

= a2 + b2.

In Equation (A.30), the condition j2 = −1 was used. Also, with the complex conjugateof z we have the following result

z + z∗ = (a+ jb) + (a− jb) (A.31)= 2a+ j0

= 2<z,

80

similarly, we have the following

z − z∗ = (a+ jb)− (a− jb) (A.32)= 0 + j2b

= j2=z,

A real number m can be written as a complex number y with zero imaginary part, i.e.,y = m+ j0. the product of a real number and a complex number is obtained as follows

mz = yz (A.33)= (m+ j0) (a+ jb)

= m (a+ jb)

= am+ jbm.

For complex numbers z and w, with w 6= 0, the quotient z/w is obtained by multi-plication and division with the complex conjugate of the denominator w∗ as shown in thefollowing expression

z

w=

a+ jb

c+ jd(A.34)

=

(a+ jb

c+ jd

)(c− jdc− jd

)=

(ac+ bd) + j (cb− ad)

c2 + d2.

The square of a complex number z is given by

z2 = zz (A.35)= (a+ jb) (a+ jb)

= a2 + jab+ jba+ j2b2

=(a2 − b2

)+ j2ab.

The reciprocal of a complex number z is given by

1

z=

1

(a+ jb)(A.36)

=1

(a+ jb)

(a− jb)(a− jb)

=a− jba2 + b2

=z∗

zz∗.

81

From Equation (A.36) and the previous results, suggest that every operation performedwith complex numbers must be algebraically handled in order to obtain a final expression inthe form f + jg, where the real and imaginary parts can be numbers, or rational expressions.With all the discussion carried out so far, it can be concluded that the set of complex numbersC together with the addition and multiplication operations, forms an ordered field. Thecomplex numbers 0 + j0 and 1 + j0 are the additive and multiplicative identity elements,respectively.

Definition 12 The modulus or absolute value of a complex number z is

|z| = (zz∗)1/2 =√a2 + b2. (A.37)

Some of the properties of the absolute value of complex numbers z, w ∈ C are thefollowing

1. |z| > 0, unless z = 0.

2. |z∗| = |z|.

3. |zw| = |z||w|.

4. |< z | ≤ |z|.

5. Triangular inequality. |z + w| ≤ |z|+ |w|.

Theorem 13 Schwartz Inequality. Let a1, a2, . . . , an ∈ C and b1, b2, . . . , bn ∈ C, then∣∣∣∣∣n∑j=1

ajb∗j

∣∣∣∣∣2

≤n∑j=1

|aj|2n∑j=1

|bj|2 . (A.38)

Geometric Interpretation

A complex number z = a + jb is an ordered pair (a, b) of real numbers, where the firstcomponent of the ordered pair corresponds to the real part <z and the second componentof the ordered pair corresponds to the imaginary part =z. In this way, the ordered pairrepresenting a complex number z can be associated to a point (a, b) in a coordinate planeor in rectangular coordinates. Since a point (a, b) can be interpreted as a vector with initialpoint the origin and end point (a, b), then a complex number can also be seen as a vector inthe complex plane where the horizontal axis corresponds to the real part, and the verticalaxis to the imaginary part. This is shown in Figure A.7

Within this context, we have a new way to represent the complex number z = a + jb asz = (a, b). Recall that a point (a, b) in rectangular coordinates as that shown in Figure A.7can be expressed in terms of polar coordinates (r, θ) as follows

a = r cos θ, b = r sin θ, (A.39)

82

Figure A.7: Geometric interpretation of a complex number.

with the relationship of

r =√a2 + b2, θ = tan−1

(b

a

), (A.40)

the coordinate transformations can be seen in Figure A.8.

Figure A.8: Polar coordinates of a complex number.

With the polar coordinate representation in (A.40), the complex number z = a + jb canbe represented as z = r cos θ + jr sin θ or in Euler’s notation as z = rejθ, with r = |z| themagnitude of the complex number and θ its phase. As exercise, derive the following resultsfor the multiplication and division of two complex numbers z = a + jb or in polar formz = rz (cos θz + j sin θz) and w = c+ jd or w = rw (cos θw + j sin θw), see [3],

zw = rzrw [cos (θz + θw) + j sin (θz + θw)] , (A.41)

83

z

w=rzrw

[cos (θz − θw) + j sin (θz − θw)] . (A.42)

Powers of Complex Numbers

With the results in equations (A.41) and (A.42), integer powers of complex numbers can befound, for example, with w = z in (A.41), we can get

z2 = r2z [cos (2θz) + j sin (2θz)] . (A.43)

To calculate z3, we can use (A.43) with z3 = z2z to obtain

z3 = r3z [cos (3θz) + j sin (3θz)] , (A.44)

and in general, for n > 0 the n-th power of a complex number z is given by

zn = rnz [cos (nθz) + j sin (nθz)] (A.45)

In Figure A.8, the angle θ is also known as the argument of the complex number z, i.e.arg(z) = θ. We can follow the same inductive argument as that in equations (A.43) - (A.45)for the case when n < 0, by noting first that 1 = 1 + j0 and that 1 = (1, 0) which impliesthat arg(1) = 0, then from (A.42) we get

1

z=

1

rz[cos (−θz) + j sin (−θz)] , (A.46)

and then we can obtain from (A.43) and (A.46) the following

1

z2=

1

r2z[cos (−2θz) + j sin (−2θz)] , (A.47)

which in general takes us as well to Equation (A.45).DeMoivre’s Formula. For a normalized complex number z = cos θz + j sin θz, i.e., a

complex number with rz = 1, Equation (A.45) gives

zn = [cos (nθz) + j sin (nθz)] = (cos θz + j sin θz)n . (A.48)

Roots of Complex Numbers

In order to obtain the n-th root of a complex number we can consider that we have two com-plex numbers in polar form, e.g., z = rz (cos θz + j sin θz) and w = rw (cos θw + j sin θw),and that we have the equality wn = z, which becomes from (A.45) the following

rnw (cosnθw + j sinnθw) = rz (cos θz + j sin θz) . (A.49)

From (A.49), it can be concluded that rnw = rz, hence rw = r1/nz , and we also have that

cosnθw + j sinnθw = cos θz + j sin θz. (A.50)

84

From (A.50), we can get

cosnθw = cos θz, and sinnθw = sin θz. (A.51)

Equality in expressions shown in (A.51) will be obtained for k integer as follows

nθw = θz + 2πk, (A.52)

which gives

θw =θz + 2πk

n, k = 0, 1, 2, . . . , n− 1. (A.53)

Note that from Equation (A.53), we will obtain n different roots for the complex numberz. Therefore, the n roots of complex number z = rz (cos θz + j sin θz), i.e., w = z1/n isgiven by

z1/n = r1/nz

[cos

(θz + 2πk

n

)+ j sin

(θz + 2πk

n

)], k = 0, 1, . . . , n− 1. (A.54)

From Equation (A.54), it can be seen that the n roots of the complex number z arelocated on the circumference of the circle of radius rz, at angles defined by (θz + 2πk) /n,for k = 0, 1, . . . , n − 1. The concept of the circle defined by the complex number z isimportant since it defines regions on the complex plane and will be useful to define thoseregions where functions converge.

Complex Plane Sets

In the complex plane as that in Figure A.9, different sets can be defined, for example, theset of point indicated by the circle in the figure are all those points w with a distance tothe complex number z that is less than or equal to ρ, i.e., the distance between the complexnumbers z = a+ jb and w = c+ jd is given by

|z − w| =√

(a− c)2 + (b− d)2, (A.55)

and the circle in Figure A.9 of radius ρ, including the circumference is given by the followingset of points

Bz(ρ) = w ∈ C : |z − w| ≤ ρ . (A.56)

Different sets in the complex plane can be defined, for example, the circle in Equation(A.56) is a closed circle, the open circle or disc of radius ρ is defined by excluding thecircumference to get

Bz(ρ) = w ∈ C : |z − w| < ρ , (A.57)

the right half plane of the complex plane including the imaginary axis is defined as

E = w ∈ C : <w ≥ 0 . (A.58)

85

Figure A.9: Circle of radius ρ.

A ring centered at the origin in the complex plane can be defined by two radii ρ1 and ρ2,where ρ1 < ρ2, hence the closed set of points in the ring is defined as

E(ρ1, ρ2) = w ∈ C : ρ1 ≤ |w| ≤ ρ2 , (A.59)

and the open ring set is defined as

E(ρ1, ρ2) = w ∈ C : ρ1 < |w| < ρ2 . (A.60)

A closed ring centered at a point z = (a, b), with two radii ρ1 and ρ2, where ρ1 < ρ2, isdefined by

Ez(ρ1, ρ2) = w ∈ C : ρ1 ≤ |z − w| ≤ ρ2 . (A.61)

86

Appendix B. Complex Functions

Recall that a function in general is a mapping from a set B to a set E that assigns to eachelement of B a unique element in E, i.e., f : B −→ E. The set B is called the domain ofthe function f , and the set E the range. Our interest is in functions with a complex domainset, i.e., B ⊂ C, then f is a complex function.

Consider a complex number z = a+jb ∈ B and a complex function such that w = f(z),then since w = c+ jd, we actually have that the real and imaginary parts of w are functionsof a and b, i.e., we have

w = f(z) = c(a, b) + jd(a, b). (A.62)

The mapping of a complex function can be seen in Figure A.10, where a complex numberz in the domain B is mapped to a complex number w in the range E of the complex functionf .

Figure A.10: Mapping of complex function.

Example 14 Consider a complex number z = a + jb ∈ C, and let the complex function fbe f(z) = z + <z, then we can express the function using the standard notation as in thefollowing equation

f(z) = a+ jb+ a = 2a+ jb, (A.63)

87

88

then, we obtain thatc(a, b) = 2a, d(a, b) = b. (A.64)

Example 15 Consider the function f : B −→ E, where the domain is the set B = z =a + jb ∈ C : <z = 1, and the mapping is given by w = c + jd = f(z) = z2. The rangeof the function is obtained as follows. From Equation (A.35) we can see that

f(z) =(a2 − b2

)+ j2ab, (A.65)

and then we have thatc(a, b) =

(a2 − b2

), d(a, b) = 2ab. (A.66)

Since <z = 1, we obtain by substituting this in (A.66) that

c(1, b) =(1− b2

), d(1, b) = 2b. (A.67)

From (A.67) we can see that b = d/2 and that c = 1− d2/4. This last expression can beused to plot the result of the mapping as shown in Figure A.11.

Figure A.11: Mapping of complex function f = z2 with domain B = z = a + jb ∈ C :<z = 1.

Definition 16 A complex function f is continuous at a point z0 if

limz→z0

f (z) = f (z0) (A.68)

Consider the complex constant coefficients ai, i = 0, 1, . . . , n and n a nonnegative inte-ger, then define the polynomimal of degree n as

f(z) = anzn + an−1z

n−1 + · · ·+ a2z2 + a1z + a0, an 6= 0. (A.69)

89

Definition 17 A complex function w = f(z) is analytical in a point z0 if its derivative existsin such point and in a neighborhood of the point. A complex function is analytical in adomain B if it is analytical in all the points of the domain.

A complex number c is a zero of a polynommial if and only if z − c is a factor of f(z).For example, we can have

f(z) = z4 + 5z2 + 4 (A.70)=

(z2 + 1

) (z2 + 4

)= (z + j) (z − j) (z + j2) (z − j2) .

From the second expression in (A.70), the roots are obtained and then the last expressiongives the factors of the polynomial.

Recall that a complex number z = a+ jb is considered in a complex domain D ⊆ C. Acomplex-valued function is a function of the form

f (z) = u (a, b) + jv (a, b) , (A.71)

where the real part of the function is<f (z) = u (a, b), and the imaginary part=f (z) =v (a, b) . The complex conjugate of the function f (z) is f ∗ (z) = u (a, b)− jv (a, b), and themagnitude or absolute value is given by

|f (z)| =√u2 (a, b) + v2 (a, b) =

√f (z) f ∗ (z). (A.72)

The values of the complex-valued function f (z) correspond to points on the complexplane, i.e., points with coordinates (u (a, b) , v (a, b)) . The function maps the complex do-main D to a range or image E ⊂ C, where (u (a, b) , v (a, b)) ∈ E, this is denoted byf : D −→ E. The domain of the function can be many different things, including other num-ber sets such as the real or the rational numbers for example D = R ⊂ C, or D = N ⊂ C.

Example 18 Reduce the function, obtain the standard form representation, and plot the realand imaginary parts as a function of the domain. Let the domain of the complex-valuedfunction f be the real numbers, i.e., D = x : x ∈ (−∞,∞). There can also be otherdefinitions of the domain, for example, D = z = a+ jb : b = 0, a ∈ (−∞,∞). Definethe function as f (x) = (3 + j2x)2 /2. To solve this, develop the function as follows

f (x) =1

2(3 + j2x)2 (A.73)

=1

2

(9 + j12x− 4x2

)= 4.5− 2x2 + j6x.

The real part of the function is u (x) = (4.5− 2x2), and the imaginary part is v (x) = 6x.In Figure A.12, the real and imaginary parts of the function are shown.

90

Figure A.12: Plot of the real and imaginary parts of the complex-valued function f (x) =(3 + j2x)2 /2.

At this moment, consider a complex -valued function whose domain are the real numbers,in other words f : R −→ E, let the domain or independent variable be x, then the followingpoints are satisfied

1. The derivative of the complex-valued function exists on an interval if and only if thederivatives of the real and imaginary parts of the function exist on that interval, in otherwords, for x ∈ [x0, x1],

df (x)

dx=du (x)

dx+ j

dv (x)

dx. (A.74)

2. The complex-valued function will be differentiable if it is continuous. Then, f (x)is continuous at a point x0 if and only if u (x) and v (x) are continuous at the samepoint x0. This is extended for intervals, i.e., the function is continuous on an interval ifand only if the real and imaginary parts of the function are continuous on the interval.Continuity can also be replaced for boundedness, smooth, even, periodic, etc. as thecharacteristic of the function that needs to be verified.

3. The integral of the complex-valued function exists on an interval if and only if theintegrals of the real and imaginary parts of the function exist on that interval, in otherwords, for x ∈ [x0, x1],

x1∫x0

f (x) dx =

x1∫x0

u (x) dx+ j

x1∫x0

v (x) dx. (A.75)

4. The complex conjugate of the integral of the complex-valued function is the integral

91

of the complex conjugate of the function, in other words x1∫x0

f (x) dx

∗ =

x1∫x0

u (x) dx+ j

x1∫x0

v (x) dx

∗ (A.76)

=

x1∫x0

u (x) dx− jx1∫x0

v (x) dx

=

x1∫x0

[u (x)− jv (x)] dx

=

x1∫x0

f ∗ (x) dx.

5. Also, the integrals satisfy ∣∣∣∣∣∣x1∫x0

f (x) dx

∣∣∣∣∣∣ ≤x1∫x0

|f (x)| dx. (A.77)

92

Appendix C. Sequences and Series ofFunctions

We briefly discuss about convergence of series and sequences of complex functions. Resultsand facts about these kinds of series and sequences is similar to those of series and sequencesof real-valued functions.

Definition 19 A sequence of complex numbers is a function with domain set the positiveintegers n = 1, 2, 3..., where for each n, a complex number zn is assigned. We denote suchsequence as zn.

Example 20 Consider the sequence zn = 1n2 + jn 1

2n, n = 1, 2, ... then the sequence takes

the values1 + j

1

2,

1

4− 1

4= 0,

1

9− j 1

8, . . .

Definition 21 The sequence zn converges to a number L, if for any small positive numberε, there is an integer number N so that for n > N , |zn − L| < ε.

In Figure A.13, we can see how the sequence of the previous example converges to zn = 0as n grows.

Theorem 22 The sequence zn converges to a complex number L, if and only if,

<zn −→n→∞

<L , and =zn −→n→∞

=L . (A.78)

Example 23 Consider the sequence zn =

3+jnn+j

and obtain the convergence and the

number to which it converges. One way to proceed in this example is to obtain the sequencein standard form and then reduce it algebraically, in other words,

zn =3 + jn

n+ j· n− jn− j

(A.79)

=4n+ j (n2 − 3)

n2 + 1·

1n2

1n2

=4n

1 + 1n2

+ j

(1− 3

n2

)1 + 1

n2

.

93

94

Figure A.13: Convergence of the sequence in graphical form.

From Equation (A.79), the limits can be taken separately to the real and imaginary parts ofthe result. this gives

limn→∞<zn = 0, lim

n→∞=zn = j.

Hence, the sequence converges to L = 0 + j.

Definition 24 An infinite series of complex numbers takes the form

∞∑k=1

zk = z1 + z2 + z3 + · · · (A.80)

Consider the partial sums defined as follows

Sn =n∑k=1

zk = z1 + z2 + z3 + · · ·+ zn−1 + zn, (A.81)

and see the sequence of partial sums Sn , then the infinite series in Equation (A.80) con-verges to a complex number L when

Sn −→n→∞

L.

Example 25 Geometric complex series. Consider a constant number r, and the infinitecomplex series defined as

∞∑k=1

rzk−1 = r + rz + rz2 + · · · (A.82)

95

The n-th partial sum of the infinite series in (A.82) is given by

Sn =n∑k=1

rzk−1 = r + rz + rz2 + · · ·+ rzn−1. (A.83)

In order to obtain the convergence of (A.82) and the conditions under which it occurs, con-sider the following

zSn =n∑k=1

rzk = rz + rz2 + · · ·+ rzn−1 + rzn. (A.84)

Now, subtract (A.84) from (A.83) to obtain

Sn − zSn = r − rzn. (A.85)

Solve for Sn in (A.85) and we get

Sn =r (1− zn)

1− z. (A.86)

In (A.86), as n goes to infinity, the term that dominates convergende is zn, and it is knownthat zn −→ 0 as n −→ ∞ when |z| < 1, otherwise, i.e., when |z| ≥ 1, zn diverges withoutlimit. Therefore, the geometric series in (A.82) converges as n −→∞ when |z| < 1, and thelimit is

∞∑k=1

rzk−1 =r

1− z. (A.87)

It is left to the reader to show the following results

∞∑k=1

zk−1 =1

1− z, (A.88)

∞∑k=1

r (−z)k−1 = r − rz + rz2 − rz3 + · · · (A.89)

=r

1 + z.

Also, see that the following relations are satisfied

1− zn

1− z= 1 + z + z2 + z3 + · · ·+ zn−1, (A.90)

and1

1− z= 1 + z + z2 + z3 + · · ·+ zn−1 +

zn

1− z(A.91)

96

Theorem 26 If the infinite complex series in Equation (A.80) converges, then

limn→∞

zn = 0. (A.92)

The opposite, i.e., limn→∞

zn 6= 0, will make the infinite complex series diverge.

Definition 27 Absolute Convergence. An infinite complex series converges absolutely if∞∑k=1

|zk| converges as well.

Theorem 28 Ratio Test. Given an infinite complex series∞∑k=1

zk

1. it converges if

limn→∞

∣∣∣∣zn+1

zn

∣∣∣∣ < 1, (A.93)

2. it diverges if limn→∞

∣∣∣ zn+1

zn

∣∣∣ > 1,for n ≥ n0 for some fixed integer n0.

3. If limn→∞

∣∣∣ zn+1

zn

∣∣∣ = 1, the test is inconclusive.

Theorem 29 Root Test. Given a series∞∑k=1

zk, define α = limn→∞

n√|zn|, then

1. if α < 1,∞∑k=1

zk converges absolutely;

2. if α > 1,∞∑k=1

zk diverges;

3. if α = 1, there is no conclusion as if the series converges or diverges.

Appendix D. Power Series

Definition 30 Power Series. Consider a sequence cn of complex numbers that are thecoefficients of the series, then the power series is defined as

∞∑n=0

cnzn. (A.94)

Depending on the choice of z in (A.94), the series will converge or diverge. With everypower series, there is a circle of convergence such that Equation (A.94) converges if z isinside that circle, and the series diverges otherwise. The following two theorems are tests forconvergence of series. These theorems will be useful to determine convergence or divergenceof series.

Theorem 31 Root Test. Given a series∑

n an, define α = lim supn→∞

n√|an|, then

1. if α < 1,∑

n an converges;

2. if α > 1,∑

n an diverges;

3. if α = 1, there is no conclusion as if the series converges or diverges.

Theorem 32 Ratio Test. Given a series∑

n an,

1. it converges if

lim supn→∞

∣∣∣∣an+1

an

∣∣∣∣ < 1, (A.95)

2. it diverges if

lim supn→∞

∣∣∣∣an+1

an

∣∣∣∣ ≥ 1, (A.96)

for n ≥ n0 for some fixed integer n0.

97

98

Bibliography

[1] Rudin, W., Principles of Mathematical Analysis, McGraw-Hill Publishing Company,Third edition, New York, 1976.

[2] Sarason, D., Notes on Complex Function Theory, Department of Mathematics, Univer-sity of California Berkeley, California, 1994.

[3] Zill, D.G., Wright, W.S., and Cullen, M.R., Advanced Engineering Mathematics,McGraw-Hill, third edition, 2006.

[4] Munkres, J.R., Topology: A First Course, Englewood Cliffs, New Jersey, Prentice Hall,1975.

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