Representing a Signal

1

Continuous-Time Fourier Methods

M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl

1

Representing a Signal •  The convolution method for finding the

response of a system to an excitation takes advantage of the linearity and time-invariance of the system and represents the excitation as a linear combination of impulses and the response as a linear combination of impulse responses

•  The Fourier series represents a signal as a linear combination of complex sinusoids

M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 2

Linearity and Superposition If an excitation can be expressed as a sum of complex sinusoids the response of an LTI system can be expressed as the sum of responses to complex sinusoids.


Real and Complex Sinusoids

�

cos x( ) =e jx + e− jx

2

�

sin x( ) =e jx − e− jx

j2


Jean Baptiste Joseph Fourier

3/21/1768 - 5/16/1830


Conceptual Overview The Fourier series represents a signal as a sum of sinusoids. The best approximation to the dashed-line signal below using only a constant is the solid line. (A constant is a cosine of zero frequency.)

t-4 10

Constant

-0.6

0.6

t-4 10

x(t)1.6 Exact x(t)

1Approximation of x(t) by a constant

t0 t0 +T


2

Conceptual Overview The best approximation to the dashed-line signal using a constant plus one real sinusoid of the same fundamental frequency as the dashed-line signal is the solid line.

t-4 10

Sinusoid 1

-0.6

0.6

t-4 10

x(t)

1.6 Exact x(t)

1Approximation of x(t) through 1 sinusoid

t0 t0 +T


Conceptual Overview The best approximation to the dashed-line signal using a constant plus one sinusoid of the same fundamental frequency as the dashed-line signal plus another sinusoid of twice the fundamental frequency of the dashed-line signal is the solid line.


Conceptual Overview The best approximation to the dashed-line signal using a constant plus three sinusoids is the solid line. In this case (but not in general), the third sinusoid has zero amplitude. This means that no sinusoid of three times the fundamental frequency improves the approximation.

t-4 10

Sinusoid 3

-0.6

0.6

t-4 10

x(t) Exact x(t)

1

Approximation of x(t) through 3 sinusoids

t0 t0 +T


Conceptual Overview The best approximation to the dashed-line signal using a constant plus four sinusoids is the solid line. This is a good approximation which gets better with the addition of more sinusoids at higher integer multiples of the fundamental frequency.


Continuous-Time Fourier Series Definition

The Fourier series representation of a signal x(t)over a time t0 < t < t0 +T is

x t( ) = cx k[ ]e j2πkt /Tk=−∞

∞

∑where cx[k] is the harmonic function and k is the harmonicnumber. The harmonic function can be found from the signalusing the princple of orthogonality.


Orthogonality


3

Using Euler's identity

e j2πkt /T ,e j2πqt /T( ) = cos 2π k − qT

t⎛⎝⎜

⎞⎠⎟ + j sin 2π k − q

Tt⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥dt

t0

t0 +T

∫If k = q,

e j2πkt /T ,e j2πqt /T( ) = cos 0( ) + j sin 0( )⎡⎣ ⎤⎦dtt0

t0 +T

∫ = dtt0

t0 +T

∫ = T .

If k ≠ q, the integral

e j2πkt /T ,e j2πqt /T( ) = cos 2π k − qT

t⎛⎝⎜

⎞⎠⎟ + j sin 2π k − q

Tt⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥dt

t0

t0 +T

∫is over a non-zero integer number of cycles of a cosine and a sine and is therefore zero.

Orthogonality


Therefore e j2πkt /T and e j2πqt /T are orthogonal if k and q are not equal.

Now multiply the Fourier series expression x t( ) = cx k[ ]e j2πkt /Tk=−∞

∞

∑by e− j2πqt /T (q an integer)

x t( )e− j2πqt /T = cx k[ ]e j2π k−q( )t /T

k=−∞

∞

∑and integrate both sides over the interval t0 ≤ t < t0 +T

x t( )e− j2πqt /T dtt0

t0 +T

∫ = cx k[ ]e j2π k−q( )t /T

k=−∞

∞

∑⎡⎣⎢

⎤⎦⎥dt

t0

t0 +T

∫ .

Orthogonality


Orthogonality


Summarizing

x t( ) = cx k[ ]e j2πkt /Tk=−∞

∞

∑ and cx k[ ] = 1T

x t( )e− j2πkt /T dtt0

t0 +T

∫ .

The signal and its harmonic function form a Fourier seriespair x t( ) F S

T← →⎯⎯ cx k[ ] where T is the representation time and, therefore, the fundamental period of the CTFS representation of x t( ). If T is also period of x t( ), the CTFS representation of x t( ) is valid for all time. This is, by far, the most common use of the CTFS in engineering applications. If T is not a period of x t( ), the CTFS representation is generally valid only in the interval t0 ≤ t < t0 +T .

Continuous-Time Fourier Series Definition


CTFS of a Real Function

It can be shown that the continuous-time Fourier series (CTFS) harmonic function of any real-valued function x t( ) has the property that cx k[ ] = cx

* −k[ ].

One implication of this fact is that, for real-valued functions,the magnitude of the harmonic function is an even function and the phase is an odd function.


The Trigonometric CTFS The fact that, for a real-valued function x t( ) cx k[ ] = cx

* −k[ ]also leads to the definition of an alternate form of the CTFS, the so-called trigonometric form.

x t( ) = ax 0[ ]+ ax k[ ]cos 2πkt /T( ) + bx k[ ]sin 2πkt /T( ){ }k=1

∞

∑where

ax k[ ] = 2T

x t( )cos 2πkt /T( )dtt0

t0 +T

∫

bx k[ ] = 2T

x t( )sin 2πkt /T( )dtt0

t0 +T

∫


4

The Trigonometric CTFS

Since both the complex and trigonometric forms of the CTFS represent a signal, there must be relationships between the harmonic functions. Those relationships are

ax 0[ ] = cx 0[ ]bx 0[ ] = 0

ax k[ ] = cx k[ ]+ cx* k[ ]

bx k[ ] = j cx k[ ]− cx* k[ ]( )

⎧

⎨

⎪⎪

⎩

⎪⎪

⎫

⎬

⎪⎪

⎭

⎪⎪

, k = 1,2,3,…

cx 0[ ] = ax 0[ ]cx k[ ] = ax k[ ]− j bx k[ ]

2

cx −k[ ] = cx* k[ ] = ax k[ ]+ j bx k[ ]

2

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

⎫

⎬

⎪⎪⎪

⎭

⎪⎪⎪

, k = 1,2,3,…


CTFS Example #1


CTFS Example #1


CTFS Example #2 Let a signal be defined by x t( ) = 2cos 400πt( ) and letT = 10 ms which is 2T0 .


CTFS Example #2


CTFS Example #3 Let x t( ) = 1 / 2 − 3 / 4( )cos 20πt( ) + 1 / 2( )sin 30πt( ) and let T = 200 ms.


5

CTFS Example #3


CTFS Example #3


CTFS Example #3


Linearity of the CTFS

These relations hold only if the harmonic functions of all the component functions are based on the same representation time T.


CTFS Example #4 Let the signal be a 50% duty-cycle square wave with an amplitude of one and a fundamental period T0 = 1. x t( ) = rect 2t( )∗δ1 t( )


CTFS Example #4


6

CTFS Example #4


CTFS Example #4


CTFS Example #4 A graph of the magnitude and phase of the harmonic function as a function of harmonic number is a good way of illustrating it.


The Sinc Function

Let x t( ) = A rect t /w( )∗δT0t( ) , w < T0 . Then

x t( ) = A rect t /w( )∗δT0t( ) F S

T0← →⎯⎯ cx k[ ] = A sin πkw /T0( )

πk

The mathematical form sin π x( )π x

arises frequently enough

to be given its own name "sinc". That is sinc t( ) = sin πt( )πt

.


CTFS Example #5 Let x t( ) = 2cos 400πt( ) and let T = 7.5 ms which is1.5 fundamental periods of this signal.


CTFS Example #5


7

CTFS Example #5


CTFS Example #5 The CTFS representation of this cosine is the signal below, which is an odd function, and the discontinuities make the representation have significant higher harmonic content. This is a very inelegant representation.


CTFS of Even and Odd Functions

For an even function, the complex CTFS harmonic function cx k[ ] is purely real and the sine harmonic function ax k[ ] is zero.

For an odd function, the complex CTFS harmonic functioncx k[ ] is purely imaginary and the cosine harmonic functionbx k[ ] is zero.


Convergence of the CTFS

For continuous signals, convergence is exact at every point.

A Continuous Signal

Partial CTFS Sums xN t( ) = cx k[ ]e j2πkt /T0

k=−N

N

∑



For discontinuous signals, convergence is exact at every point of continuity.

Discontinuous Signal

Partial CTFS Sums



At points of discontinuity the Fourier series representation converges to the mid-point of the discontinuity.


8

Numerical Computation of the CTFS How could we find the CTFS of a signal which has no known functional description? Numerically.

cx k[ ] = 1

Tx t( )e− j2πkt /T dt

T∫Unknown


Numerical Computation of the CTFS

cx k[ ] ≅ 1T

x nTs( )e− j2πknTs /T dtnTs

n+1( )Ts

∫⎡

⎣⎢⎢

⎤

⎦⎥⎥n=0

N−1

∑Samples from x(t)



It can be shown (Web Appendix F) that, for harmonic numbersk << N

cx k[ ] ≅ 1 / N( )D F T x nTs( )( ) , k << Nwhere

D F T x nTs( )( ) = x nTs( )e- j2πnk /N

n=0

N -1

∑



The Discrete Fourier Transform

D F T x nTs( )( ) = x nTs( )e− j2πnk /N

n=0

N−1

∑is an intrinsic function in most modern high-level computerlanguages.


CTFS Properties

Linearity

α x t( ) + β y t( ) F ST← →⎯⎯α cx k[ ]+ β cy k[ ]

Let a signal x(t) have a fundamental period T0 x and let asignal y(t) have a fundamental period T0 y . Let the CTFSharmonic functions, each using a common period T as the representation time, be cx[k] and cy[k]. Then the following properties apply.


CTFS Properties Time Shifting x t − t0( ) F S

T← →⎯ e− j2πkt0 /T cx k[ ]


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CTFS Properties

Frequency Shifting (Harmonic Number

Shifting) ej2πk0t /T x t( ) F S

T← →⎯ cx k − k0[ ]

A shift in frequency (harmonic number) corresponds to multiplication of the time function by a complex exponential.

Time Reversal x −t( ) F ST← →⎯ cx −k[ ]


CTFS Properties

Time Scaling Let z t( ) = x at( ) , a > 0Case 1. T = T0 x / a = T0z for z t( ) cz k[ ] = cx k[ ]Case 2. T = T0 x for z t( ) If a is an integer,

cz k[ ] = cx k / a[ ] , k / a an integer0 , otherwise

⎧⎨⎩


CTFS Properties Time Scaling (continued)


CTFS Properties

Change of Representation TimeWith T = T0 x , x t( ) F S

T← →⎯ cx k[ ]With T = mT0 x , x t( ) F S

T← →⎯ cx,m k[ ]

cx,m k[ ] = cx k /m[ ] , k /m an integer0 , otherwise

⎧⎨⎩

(m is any positive integer)


CTFS Properties Change of Representation Time


CTFS Properties

Time Differentiation

ddtx t( )( ) F S

T← →⎯ j2πk cx k[ ] /T

F ST← →⎯⎯

F ST← →⎯⎯


10

CTFS Properties Time Integration

Case 1. cx 0[ ] = 0

x λ( )dλ

−∞

t

∫ F ST← →⎯

cx k[ ]j2πk /T

, k ≠ 0

Case 2. cx 0[ ] ≠ 0

x λ( )dλ−∞

t

∫ is not periodic

Case 1 Case 2


CTFS Properties

Multiplication - Convolution Duality x t( )y t( ) F S

T← →⎯ cx k[ ]∗cy k[ ](The harmonic functions cx[k] and cy[k] must be basedon the same representation time T .)

x t( )y t( ) F ST← →⎯ T cx k[ ]cy k[ ]

The symbol indicates periodic convolution.Periodic convolution is defined mathematically by

x t( )y t( ) = x τ( )y t −τ( )dτT∫

x t( )y t( ) = xap t( )∗y t( ) where xap t( ) is any single period of x t( )


CTFS Properties


CTFS Properties

Conjugation x* t( ) F S

T← →⎯ cx* −k[ ]

Parseval’s Theorem

1T

x t( ) 2 dtT∫ = cx k[ ] 2

k=−∞

∞

∑The average power of a periodic signal is the sum of theaverage powers in its harmonic components.


Some Common CTFS Pairs

1 F ST← →⎯ δ k[ ] , T arbitrary

δT0t( ) F S

mT0← →⎯⎯

1 /T0( ) , k /m an integer0 , otherwise

⎧⎨⎩

e j2πqt /T0 F SmT0

← →⎯⎯ δ k − mq[ ]sin 2πqt /T0( ) F S

mT0← →⎯⎯ j / 2( ) δ k + mq[ ]−δ k − mq[ ]( )

cos 2πqt /T0( ) F SmT0

← →⎯⎯ 1 / 2( ) δ k − mq[ ]+δ k + mq[ ]( )rect t /w( )∗δT0

t( ) F SmT0

← →⎯⎯ w /T0( )sinc wk /mT0( )δm k[ ]tri t /w( )∗δT0

t( ) F SmT0

← →⎯⎯ w /T0( )sinc2 wk /mT0( )δm k[ ]


CTFS Examples


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CTFS Examples

x t( ) = 12sin 2πt / 0.01( ) rect t / 0.01( )∗δ0.02 t( )⎡⎣ ⎤⎦x t( ) = 12sin 200πt( ) rect 100t( )∗δ0.02 t( )⎡⎣ ⎤⎦Find the CTFS harmonic function of x t( ) with T = 20 ms.

sin 2πqt /T0( ) F SmT0

← →⎯⎯ j / 2( ) δ k + mq[ ]−δ k − mq[ ]( )sin 200πt( ) F S

2×0.01← →⎯⎯ j / 2( ) δ k + 2 ×1[ ]−δ k − 2 ×1[ ]( )rect t /w( )∗δT0

t( ) F SmT0

← →⎯⎯ w /T0( )sinc wk /mT0( )δm k[ ]rect 100t( )∗δ0.02 t( ) F S

1×0.02← →⎯⎯ 1 / 2( )sinc k / 2( )Using x t( )y t( ) F S

T← →⎯ cx k[ ]∗cy k[ ],12sin 200πt( ) rect 100t( )∗δ0.02 t( )⎡⎣ ⎤⎦

F S0.02← →⎯⎯ 12 j / 2( ) δ k + 2[ ]−δ k − 2[ ]( )∗ 1 / 2( )sinc k / 2( )

12sin 200πt( ) rect 100t( )∗δ0.02 t( )⎡⎣ ⎤⎦F S0.02← →⎯⎯ j3 sinc k + 2( ) / 2( )− sinc k − 2( ) / 2( )( )


CTFS Examples Find the CTFS harmonic function of x t( ) with T = 10−8.

cx k[ ] = 1 /T( ) x t( )e− j2πkt /T dtT∫ ⇒ cx 0[ ] = 108 35 ×108 t( )dt

0

10−8

∫ = 35 / 2

cx k[ ] = 108 35 ×108 t( )e− j2π×108 kt dt0

10−8

∫ = 35 ×1016 te− j2π×108 kt dt0

10−8

∫

cx k[ ] = 35 ×1016 t e− j2π×108 kt

− j2π ×108 k⎡

⎣⎢

⎤

⎦⎥

0

10−8

− e− j2π×108 kt

− j2π ×108 k⎛

⎝⎜⎞

⎠⎟dt

0

10−8

∫⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

cx k[ ] = 35 ×1016 − e− j2πk ×10−8

j2π ×108 k− e− j2π×108 kt

j2π ×108 k( )2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

0

10−8⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

cx k[ ] = 35 ×1016 −10−16e− j2πk

j2πk+ 1− e− j2πk

j2πk( )2 ×1016

⎡

⎣⎢

⎤

⎦⎥ = 351− e− j2πk − j2πke− j2πk

j2πk( )2

cx k[ ] = 351 / 2 , k = 0j

2πk , k ≠ 0

⎧⎨⎪

⎩⎪


LTI Systems with Periodic Excitation

The differential equation describing an RC lowpass filter is RC ′vout t( ) + vout t( ) = vin t( )If the excitation vin t( ) is periodic it can be expressed as aCTFS,

vin t( ) = cin k[ ]e j2πkt /Tk=−∞

∞

∑The equation for the kth harmonic alone is RC ′vout,k t( ) + vout,k t( ) = vin,k t( ) = cin k[ ]e j2πkt /T



If the excitation is periodic, the response is also, with thesame fundamental period. Therefore the response can beexpressed as a CTFS also. vout,k t( ) = cout e

j2πkt /T

Then the equation for the kth harmonic becomesj2kπRC /T( )cout k[ ]e j2πkt /T + cout k[ ]e j2πkt /T = cin k[ ]e j2πkt /T

Notice that what was once a differential equation is nowan algebraic equation.



Solving the kth-harmonic equation,

cout k[ ] = cin k[ ]j2kπRC /T +1

Then the response can be written as

vout t( ) = cout k[ ]e j2πkt /Tk=−∞

∞

∑ =cin k[ ]

j2kπRC /T +1e j2πkt /T

k=−∞

∞

∑



The ratio cout k[ ]cin k[ ] is the

harmonic response of the system.


12

Continuous-Time Fourier Methods

M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl

67

Extending the CTFS

•  The CTFS is a good analysis tool for systems with periodic excitation but the CTFS cannot represent an aperiodic signal for all time

•  The continuous-time Fourier transform (CTFT) can represent an aperiodic signal for all time


CTFS-to-CTFT Transition

Its CTFS harmonic function is cx k[ ] = AwT0

sinc kwT0

⎛⎝⎜

⎞⎠⎟

As the period T0 is increased, holding w constant, the duty cycle is decreased. When the period becomes infinite (and the duty cycle becomes zero) x t( ) is no longer periodic.

Consider a periodic pulse-train signal x t( ) with duty cycle w /T0



�

w =T02

�

w =T010

Below are graphs of the magnitude of cx k[ ] for 50% and 10% duty cycles. As the period increases the sinc function widens and its magnitude falls. As the period approaches infinity, the CTFS harmonic function becomes an infinitely-wide sinc function with zero amplitude.


CTFS-to-CTFT Transition This infinity-and-zero problem can be solved by normalizing the CTFS harmonic function. Define a new “modified” CTFS harmonic function T0 cx k[ ] = Awsinc wkf0( ) and graph it

versus kf0 instead of versus k. f0 = 1 /T0( )


CTFS-to-CTFT Transition In the limit as the period approaches infinity, the modifiedCTFS harmonic function approaches a function of continuousfrequency f (kf0 ).


13



Definition of the CTFT

Forward f form Inverse

X f( ) = F x t( )( ) = x t( )e− j2π ft dt−∞

∞

∫ x t( ) = F -1 X f( )( ) = X f( )e+ j2π ft df−∞

∞

∫

Forward ω form Inverse

X jω( ) = F x t( )( ) = x t( )e− jωt dt−∞

∞

∫ x t( ) = F -1 X jω( )( ) = 12π

X jω( )e+ jωt dω−∞

∞

∫Commonly-used notation:

x t( ) F← →⎯ X f( ) or x t( ) F← →⎯ X jω( )


Some Remarkable Implications of the Fourier Transform

The CTFT expresses a finite-amplitude, real-valued, aperiodic signal which can also, in general, be time-limited, as a summation (an integral) of an infinite continuum of weighted, infinitesimal- amplitude, complex sinusoids, each of which is unlimited in time. (Time limited means “having non-zero values only for a finite time.”)


Frequency Content

Lowpass Highpass

Bandpass


δ t( ) F← →⎯ 1

e−αt u t( ) F← →⎯ 1/ jω +α( ) , α > 0 − e−αt u −t( ) F← →⎯ 1/ jω +α( ) , α < 0

te−αt u t( ) F← →⎯ 1/ jω +α( )2, α > 0 − te−αt u −t( ) F← →⎯ 1/ jω +α( )2

, α < 0

t ne−αt u t( ) F← →⎯ n!

jω +α( )n+1 , α > 0 − t ne−αt u −t( ) F← →⎯ n!

jω +α( )n+1 , α < 0

e−αt sin ω0t( )u t( ) F← →⎯ω0

jω +α( )2+ω0

2, α > 0 − e−αt sin ω0t( )u −t( ) F← →⎯

ω0

jω +α( )2+ω0

2, α < 0

e−αt cos ω0t( )u t( ) F← →⎯ jω +α

jω +α( )2+ω0

2, α > 0 − e−αt cos ω0t( )u −t( ) F← →⎯ jω +α

jω +α( )2+ω0

2, α < 0

e−α t F← →⎯ 2αω 2 +α 2 , α > 0

Some CTFT Pairs


Convergence and the Generalized Fourier Transform

Let x t( ) = A. Then from the definition of the CTFT,

X f( ) = Ae− j2π ft dt−∞

∞

∫ = A e− j2π ft dt−∞

∞

∫

This integral does not converge so, strictly speaking, the CTFT does not exist.


14


But consider a similar function,

xσ t( ) = Ae−σ t , σ > 0Its CTFT integral

Xσ f( ) = Ae−σ t e− j2π ft dt−∞

∞

∫does converge.



Carrying out the integral, Xσ f( ) = A 2σσ 2 + 2π f( )2 .

Now let σ approach zero.

If f ≠ 0 then limσ→0

A 2σσ 2 + 2π f( )2 = 0. The area under this

function is A 2σσ 2 + 2π f( )2 df

−∞

∞

∫ which is A, independent of

the value of σ . So, in the limit as σ approaches zero, the CTFT has an area of A and is zero unless f = 0. This exactly

defines an impulse of strength A. Therefore A F← →⎯ Aδ f( ).M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 80


By a similar process it can be shown that

cos 2π f0t( ) F← →⎯ 12

δ f − f0( ) +δ f + f0( )⎡⎣ ⎤⎦

and

sin 2π f0t( ) F← →⎯ j2

δ f + f0( )−δ f − f0( )⎡⎣ ⎤⎦

These CTFT’s which involve impulses are called generalized Fourier transforms (probably becausethe impulse is a generalized function).




Negative Frequency This signal is obviously a sinusoid. How is it described mathematically?

It could be described by x t( ) = Acos 2πt /T0( ) = Acos 2π f0t( )But it could also be described by x t( ) = Acos 2π − f0( )t( )


Negative Frequency

x(t) could also be described by

x t( ) = A ej2π f0t + e− j2π f0t

2or

x t( ) = A1 cos 2π f0t( ) + A2 cos 2π − f0( )t( ) , A1 + A2 = Aand probably in a few other different-looking ways. So who isto say whether the frequency is positive or negative? For thepurposes of signal analysis, it does not matter.


15

Negative Frequency Consider an experiment in which we multiply two sinusoidalsignals x1 t( ) = cos 2π f1t( ) and x2 t( ) = cos 200πt( ) to form

x t( ) = x1 t( )x2 t( ). x t( ) can be expressed using a trigonometricidentity as

x t( ) = 1 / 2( ) cos 2π f1 −100( )t( ) + cos 2π f1 +100( )t( )⎡⎣ ⎤⎦Now imagine that we continuously change f1 from a frequency above100 to a frequency below 100. f1 −100 becomesnegative.


More CTFT Pairs

The generalization of the CTFT allows us to extend the tableof CTFT pairs to some very useful functions.

δ t( ) F← →⎯ 1 1 F← →⎯ δ f( ) sgn t( ) F← →⎯ 1/ jπ f u t( ) F← →⎯ 1/ 2( )δ f( ) +1/ j2π f

rect t( ) F← →⎯ sinc f( ) sinc t( ) F← →⎯ rect f( ) tri t( ) F← →⎯ sinc2 f( ) sinc2 t( ) F← →⎯ tri f( ) δT0

t( ) F← →⎯ f0δ f0f( ) , f0 = 1/ T0 T0δT0

t( ) F← →⎯ δ f0f( ) , T0 = 1/ f0

cos 2π f0t( ) F← →⎯ 1/ 2( ) δ f − f0( ) +δ f + f0( )⎡⎣ ⎤⎦ sin 2π f0t( ) F← →⎯ j / 2( ) δ f + f0( )−δ f − f0( )⎡⎣ ⎤⎦


CTFT Properties

If F x t( )( ) = X f( ) or X jω( ) and F y t( )( ) = Y f( ) or Y jω( )then the following properties can be proven.

α x t( ) + β y t( ) F← →⎯ α X f( ) + β Y f( ) α x t( ) + β y t( ) F← →⎯ α X jω( ) + β Y jω( )Linearity


CTFT Properties

Time Shiftingx t − t0( ) F← →⎯ X f( )e− j2π ft0x t − t0( ) F← →⎯ X jω( )e− jωt0


CTFT Properties

Frequency Shiftingx t( )e+ j2π f0t F← →⎯ X f − f0( )x t( )e+ jω0t F← →⎯ X ω −ω0( )


CTFT Properties

Time Scaling x at( ) F← →⎯ 1

aX f

a⎛⎝⎜

⎞⎠⎟

x at( ) F← →⎯ 1a

X jωa

⎛⎝⎜

⎞⎠⎟

Frequency Scaling

1a

x ta

⎛⎝⎜

⎞⎠⎟

F← →⎯ X af( )

1a

x ta

⎛⎝⎜

⎞⎠⎟

F← →⎯ X jaω( )


16

The “Uncertainty” Principle The time and frequency scaling properties indicate that if a signal is expanded in one domain it is compressed in the other domain. This is called the “uncertainty principle” of Fourier analysis.

e−π t /2( )2 F← →⎯ 2e−π 2 f( )2

e−π t2 F← →⎯ e−π f2


CTFT Properties

Transform ofa Conjugate

x* t( ) F← →⎯ X* − f( )x* t( ) F← →⎯ X* − jω( )

MultiplicationConvolution

Duality

x t( )∗y t( ) F← →⎯ X f( )Y f( )x t( )∗y t( ) F← →⎯ X jω( )Y jω( )

x t( )y t( ) F← →⎯ X f( )∗Y f( )x t( )y t( ) F← →⎯ 1 / 2π( )X jω( )∗Y jω( )


CTFT Properties


CTFT Properties

In the frequency domain, the cascade connection multiplies the frequency responses instead of convolving the impulse responses.


CTFT Properties

Time Differentiation

ddt

x t( )( ) F← →⎯ j2π f X f( )ddt

x t( )( ) F← →⎯ jω X jω( )

Modulation x t( )cos 2π f0t( ) F← →⎯ 1

2X f − f0( ) + X f + f0( )⎡⎣ ⎤⎦

x t( )cos ω0t( ) F← →⎯ 12

X j ω −ω0( )( ) + X j ω +ω0( )( )⎡⎣ ⎤⎦

Transforms ofPeriodic Signals

x t( ) = X k[ ]e− j2πkfFtk=−∞

∞

∑ F← →⎯ X f( ) = X k[ ]δ f − kf0( )k=−∞

∞

∑

x t( ) = X k[ ]e− jkωFt

k=−∞

∞

∑ F← →⎯ X jω( ) = 2π X k[ ]δ ω − kω0( )k=−∞

∞

∑


CTFT Properties


17

CTFT Properties

Parseval’sTheorem

x t( ) 2 dt

−∞

∞

∫ = X f( ) 2 df−∞

∞

∫

x t( ) 2 dt−∞

∞

∫ = 12π

X jω( ) 2 df−∞

∞

∫

Integral Definitionof an Impulse

e− j2π xy−∞

∞

∫ dy = δ x( )

Duality X t( ) F← →⎯ x − f( ) and X −t( ) F← →⎯ x f( )

X jt( ) F← →⎯ 2π x −ω( ) and X − jt( ) F← →⎯ 2π x ω( )


CTFT Properties

Total -AreaIntegral

X 0( ) = x t( )e− j2π ft dt−∞

∞

∫⎡

⎣⎢

⎤

⎦⎥f→0

= x t( )dt−∞

∞

∫

x 0( ) = X f( )e+ j2π ft df−∞

∞

∫⎡

⎣⎢

⎤

⎦⎥t→0

= X f( )df−∞

∞

∫

X 0( ) = x t( )e− jωt dt−∞

∞

∫⎡

⎣⎢

⎤

⎦⎥ω→0

= x t( )dt−∞

∞

∫

x 0( ) = 12π

X jω( )e+ jωt dω−∞

∞

∫⎡

⎣⎢

⎤

⎦⎥t→0

= 12π

X jω( )dω−∞

∞

∫

Integration x λ( )dλ

−∞

t

∫ F← →⎯X f( )j2π f

+ 12

X 0( )δ f( )

x λ( )dλ−∞

t

∫ F← →⎯X jω( )jω

+π X 0( )δ ω( )


CTFT Properties

�

x 0( ) = X f( )df−∞

∞

∫

�

X 0( ) = x t( )dt−∞

∞

∫


CTFT Properties


Numerical Computation of the CTFT It can be shown (Web Appendix G) that the DFT can be used to approximate samples from the CTFT. If the signal x t( ) is a causal energy signal and N samples are taken from it over a finite time beginning at t = 0, at a rate fs then the relationship between the CTFT of x t( ) and the DFT of the samples taken from it is X kfs / N( ) ≅ Tse− jπk /N sinc k / N( )XDFT k[ ]For those harmonic numbers k for which k << N X kfs / N( ) ≅ Ts XDFT k[ ]As the sampling rate and number of samples are increased, this approximation is improved.


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