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Chapter 3 Fourier Representations of Signals

and Linear Time-Invariant Systems

Introduction

Complex Sinusoids and Frequency Response of LTI

Systems.

Fourier Representations for Four Classes of Signals

Discrete-Time Periodic Signals

Continuous-Time Periodic Signals

Discrete-Time Nonperiodic Signals

Continuous-Time Nonperiodic Signals.

Properties of Fourier Representation.

Linearity and Symmetry Properties.

1

Chapter 3 Fourier Representations of Signals

and Linear Time-Invariant Systems

Convolution Property

Differentiation and Integration Properties

Time- and Frequency-Shift Properties

Finding Inverse Fourier Transforms by Using Partial-Fraction

Expansions.

Multiplication Property

Scaling Properties

Parseval Relationships

Time-Bandwidth Product

Duality

2

Introduction

Fourier Analysis

Representing Signals a Superposition of Complex Sinusoids

Four Fourier Representations

DT periodic signals: DT Fourier series (DTFS)

CT periodic signals: CT Fourier series (CTFS)

DT nonperiodic signals: DT Fourier transform (DTFT)

CT nonperiodic signals: CT Fourier transform (CTFT)

3

2. Complex Sinusoids and Frequency

Response of LTI Systems.

The input-output relation of an LTI system is

characterized by convolution with the system’s

impulse response.

By representing the input as a sum of complex

sinusoids, we obtain a more straightforward

understanding of the input-output relation of an LTI

system

4


Response of LTI Systems

Discrete-Time LTI System

The output of a complex sinusoidal input to an LTI system is a

complex sinusoid of the same frequency as the input,

multiplied by the frequency response of the system.

5

LTI Systemx(t) or

x[n]

y(t) or

y[n]

k k

knjekhknxkhny

j n j k j j n

k

y n e h k e H e e

( ) [ ]j j

k

H e h k e

Frequency Response



Continuous-Time LTI System

6

( )( ) ( ) ( )

( )

j t j t j

j t

y t h e d e h e d

H j e

( ) ( ) jH j h e d

Frequency response:

Polar form complex number c = a + jb:

arg{ }j cc c e where 2 2 1and arg{ } tan b

ac a b c

Polar form for H (j):

arg{ ( )}( ) ( ) j H jH j H j e

where ( ) Magnitude response and arg ( ) Phase respnseH j H j

argj t H j

y t H j e



RC Circuit: Frequency response

7

/1( ) ( )t RCh t e u t

RC

deueRC

jH jRC

1

deRC

RCj

0

11

1

0

1 1

1

jRCe

RCj

RC

101

11

RCj

RC

RCj

RC1

1

Frequency response:

Magnitude response:

2

2 1

1

RC

RCjH

Phase response:

RCjH arctanarg


Response of LTI Systems8

2

4

2

4

arg H j

H j

[ ]nH

( )t

[ ]n

( )t ( ) j tH j e H

j te

H

j ne j j nH e e



Eigenvalues and Eigenfunctions of a LTI System

9

Eigenfunction ( ) j tt e

Eigenvalue ( )H j

If ek is an eigenvector of a matrix A with eigenvalue k, then

k k kAe e

Arbitrary input = weighted superpositions of eigenfunctions

Convolution operation Multiplication

Ex. Input:

M

k

tj

kkeatx

1

Output:

M

k

tj

kkejHaty

1

1

[ ] k

Mj n

k

k

x n a e

1

[ ] k

Mj nj

k

k

y n a H e e

[ ] j nn e

( )jH e

3. Fourier Representations for Four

Classes of Signals

Periodic Signals: Fourier Series Representations

10

0 0is the th harmonic of .jk t j t

e k e

1. x[n] = discrete-time signal with fundamental period N. DTFS of x[n] is

0[ ] [ ]jk n

k

x n A k e (3.4)

o = 2/N Fundamental

frequency of x[n]

2. x(t) = continuous-time signal with fundamental period T. FS of x(t) is

0( ) [ ]jk t

k

x t A k e

(3.5) o = 2/T Fundamental

frequency of x(t)

“^” denotes approximate value. A[k] = the weight applied to the kth harmonic.


Classes of Signals 11

<pf.> njknjNnKNjeee 000

njNnj ee 02

njke 0

There are only N distinct complex sinusoids of the form exp(jk0n)

should be used in Eq. (3.4).

Symmetries in x[k]:

k = (N 1)/2 to (N 1)/2

3. The complex sinusoids exp(jkon) are N-periodics in the frequency index k.

4. Continuous-time complex sinusoid exp(jk0t) with distinct frequencies k0

are always distinct. FS of continuous-time periodic signal x(t) becomes

0( ) [ ]jk t

k

x t A k e

Mean-square error (MSE) between the signal and its series representation:

Discrete-time case:

1 2

0

1[ ] [ ]

N

n

MSE x n x n dtN

Continuous-time case:

2

0

1( ) ( )

T

MSE x t x t dtT

1

0

0][][ˆN

k

njkekAnx


Classes of Signals

Nonperiodic Signals: Fourier-Transform

Representations

12

1. FT of continuous-time signal:

X(j)/(2) = the weight applied to a sinusoid

of frequency in the FT representation.

2. DTFT of discrete-time signal:X(e j )/(2) = the weight applied to the

sinusoid ej n in the DTFT representation.

deeXnx njj )(

2

1][ˆ

dejXnx tj)(2

1][ˆ


Classes of Signals 13


Classes of Signals

Periodic Discrete

14

4. Discrete-Time Periodic Signals:

The Discrete-Time Fourier Series15

1. DTFS pair of periodic signal x[n]:

0

1

0

[ ] [ ]N

jk n

k

x n X k e

(3.10)

0

1

0

1[ ] [ ]

Njk n

n

X k x n eN

(3.11)

Fundamental period = N;

Fundamental frequency = o = 2/N

Notation:

0;DTFSx n X k

Fourier coefficients; Frequency domain

representation


The Discrete-Time Fourier Series

Example 3.6

16

MNnM

MnMnx

,0

,1

1. Period of DTFS coefficients = N o = 2/N

2. It is convenient to evaluate Eq.(3.11) over the interval n = M to n = N M 1.

0

0

11[ ] [ ]

1

N Mjk n

n M

Mjk n

n M

X k x n eN

eN

0

0 0

2( )

0

2

0

1[ ]

1

Mjk m M

m

Mjk M jk m

m

X k eN

e eN

3. For k = 0, N, 2N, …, we have

1o ojk jke e

2

0

1 2 11 , 0, , 2 ,

M

m

MX k k N N

N N



3. For k 0, N, 2N, …, we may sum the geometric series in Eq. (3.15) to

obtain 0 0

0

(2 1)1

[ ] , 0, , 2 , ......1

jk M jk M M

jk

e eX k k N N

N e

0 0 0 0

0 0 0 0

2 1 / 2 2 1 2 1 / 2 2 1 / 2

/ 2 / 2 / 2

1 1 1

1

jk M jk M jk M jk M

jk jk jk jk

e e e eX k

N e e N e e

sin 2 1 /1, 0, , 2 ,

sin /

2 1 / , 0, , 2 ,

k M Nk N N

X k N k N

M N k N N

,2/sin

2/12sin1

0

0

k

Mk

NkX

4. Substituting o = 2/N, yields

0, , 2 ,

sin 2 1 /1 2 1lim

sin /k N N

k M N M

N k N N

L’Hôpital’s

Rule

Nk

NMk

NkX

/sin

/12sin1

The value of X[k] for k = 0, N, 2N, …,

is obtained from the limit as k 0.

L'Hôpital's rule

In its simplest form, l'Hôpital's rule states that for

functions ƒ and g:

18



Figure 3.12 (p. 211)

The DTFS coefficients for the square

wave shown in Fig. 3.11, assuming a

period N = 50: (a) M = 4. (b) M = 12.



Symmetry property of DTFS coefficient: X[k] = X[ k].

1. DTFS of Eq. (3.10) can be written as a series involving harmonically related

cosines.2. Assume that N is even and let k range from ( N/2 ) + 1 to N/2. Eq. (3.10)

becomes

2/

12/

0

N

Nk

njkekXnx

12/

1

2/ 0002/0N

m

njknjknjkemXemXeNXXnx

3. Use X[m] = X[ m] and the identity No = 2 to obtain

/ 2 1

0

1

0 / 2 cos 2 cosN

m

X X N n X m m n

0 0/ 2 1

1

0 / 2 22

jm n jm nNj n

m

e ex n X X N e X m

e jn = cos(n)

since sin(n) =

0 for integer n.

4. Define the new set of coefficients

4. Discrete-Time Periodic Signals--


4. Define the new set of coefficients

, 0, / 2

2 , 1, 2, , / 2 1

X k k NB k

X k k N

and write the DTFS for the square wave in terms of a series of harmonically

related cosines as/ 2

0

0

[ ] [ ]cos( )N

k

x n B k k n

(3.17)

A similar expression may be derived for N odd.

Example 3.7 Building a Square Wave From DTFS Coefficients

0

0

[ ] [ ]cos( )J

J

k

x n B k k n

4. Discrete-Time Periodic Signals--


(a) J = 1. (b) J = 3. (c) J = 5. (d) J = 23. (e) J = 25.

Example Figure 3.15

Electrocardiograms for two

different heartbeats and the first

60 coefficients of their

magnitude spectra.

(a) Normal heartbeat.

(b) Ventricular tachycardia.

(c) Magnitude spectrum for the

normal heartbeat.

(d) Magnitude spectrum for

ventricular tachycardia.

心動過速

5. Continuous-Time Periodic Signals–

The Fourier Series24

1. Continuous-time periodic signals are represented by the Fourier series (FS).

2. A signal with fundamental period T and fundamental frequency o = 2/T,

the FS pair of x(t) is

0( ) [ ]jk t

k

x t X k e

(3.19)

0

0

1[ ] ( )

Tjk t

X k x t e dtT

(3.20)

3. Notation:

0;FSx t X k

Frequency domain

representation of x(t)

The variable k determines the

frequency of the complex sinusoid

associated with X[k] in Eq. (3.19).



Mean-square error (MSE)

1. Suppose we define

ˆ ojk t

k

x t X k e

2. If x(t) is square integrable,

T

dttxT 0

21

then the MSE between x(t) and x̂ t is zero, or, mathematically,

T

dttxtxT 0

2

01 at all values of t; it simply implies

that there is zero power in their

difference.

ˆ( )x t x t

3. Pointwise convergence of ˆ( )x t x t is guaranteed at all values of t

except those corresponding to discontinuities if the Dirichlet conditions are

satisfied: x(t) is bounded.

x(t) has a finite number of maximum and minima in one period.

x(t) has a finite number of discontinuities in one period.

If a signal x(t) satisfies the Dirichlet conditions and is not continuous, then

x̂ t converges to the mipoint of the left and right limits of x(t) at each

discontinuity.



1. Fundamental frequency: o = 2/T.

2. X[k]: by integrating over T/2 t T/2

00 0

0

0

0

0

0 0 0 0

/ 2

/ 2

0

0

0 0

0

1 1

1, 0

2, 0

2

2sin, 0

T Tjk t jk t

T T

T

jk t

T

jk T jk T

X k x t e dt e dtT T

e kTjk

e ek

Tk j

k Tk

Tk

For k = 0, we have

0

0

0210

T

T T

Tdt

TX

3. By means of L’Hôpital’s rule, we have



T

T

Tk

Tk

k

0

0

00

0

2sin2lim

0

00sin2

Tk

TkkX

4. X[k] is real valued. Using o = 2/T gives X[k] as a function of the ratio To/T:

02sin( 2 / )[ ]

2

k T TX k

k

(3.23)

5. Fig. 3.22 (a)-(c) depict X[k], 50 k 50, for To/T =1/4, To/T =1/16, and To/T =

1/64, respectively.

Definition of Sinc function:

sin( )sinc( )

uu

u

(3.24) Fig. 3.23.

1) Maximum of sinc function is unity at u = 0, the zero crossing occur at

integer values of u, and the amplitude dies off as 1/u.

2) The portion of this function between the zero crossings at u = 1 is

manilobe of the sinc function.



The FS coefficients, X[k], –50 k 50, for three square waves. (see Fig. 3.21.) (a) To/T

= 1/4 . (b) To/T = 1/16. (c) To/T = 1/64.


The Fourier Series

Example 3.15

29

J

k

J tkkBtx0

0cos

Figure 3.25b-3 (p. 226) (b) J = 3. (c) J = 7. (d) J = 29.

Gibbs phenomenon

Ripple near discontinuiitiies


The Fourier Series

Trigonometric FS for Real Signal

30

0 0

1

( ) [0] [ ]cos( ) [ ]sin( )k

x t B B k k t A k k t

(3.25)

FS coefficients:

0

1[0] ( )

T

B x t dtT

00

2[ ] ( )cos( )

T

B k x t k t dtT

00

2[ ] ( )sin( )

T

A k x t k t dtT

(3.26)

B[0] = X[0] represents the time-

averaged value of the signal.

[ ] [ ] [ ]B k X k X k

[ ] ( [ ] [ ])A k j X k X k for k 0

Relation between exponential FS and trigonometric FS coefficients

6. Discrete-Time Nonperiodic Signals–

The Discrete-Time Fourier Transform31

1. The DTFT is used to represent a discrete-time nonperiodic signal as a

superposition of complex sinusoids.

2. DTFT representation of time-domain signal x[n]:

(3.31) 1

[ ] ( )2

j j nx n X e e d

( ) [ ]j j n

n

X e x n e

(3.32)

DTFT

pair

3. Notation:

Frequency-domain representation x[n]

DTFT jx n X e


The Discrete-Time Fourier Transform

Condition for convergence of DTFT:

32

If x[n] is of infinite duration, then the sum converges only for certain classes

of signals. If x[n] is absolutely summable, i.e.,

n

x n

The sum in Eq. (3.32) converges uniformly to a

continuous function of .

If x[n] is not absolutely summable, but does satisfy (i.e., if x[n] has finite

energy),

n

nx2

It can be shown that the sum in Eq. (3.32)

converges in a mean-square error sense,

but does not converge pointwise.



Let 1,

0,

n Mx n

n M

1. DTFT of x[n]: 1 .

Mj j n

n M

X e e

2 2

( )

0 0

2( 1)1, 0, 2 , 4 ,

1

2 1, =0, 2 , 4 ,

M Mj j m M j M j m

m m

j Mj M

j

X e e e e

ee

e

M

Change of variable

m = n + M

0, 2 , 4 , The expression for X(e j ) when

2 1 / 2 2 1 / 2 2 1 / 2

/ 2 / 2 / 2

2 1 / 2 2 1 / 2

/ 2 / 2

( )

j M j M j M

j j M

j j j

j M j M

j j

e e eX e e

e e e

e e

e e

sin 2 1 / 2( )

sin 2

jM

X e

0, 2 , 4 , ,

sin 2 1 / 2lim 2 1;

sin 2

MM

L’Hôptital’s Rule

With understanding that X(e j ) for

is obtained as limit.0, 2 , 4 ,



Example 3.19 Inverse DTFT of A Rectangular Pulse

34

1,

,0,

jW

X eW n

1

2

W

j n

W

x n e d

sin / ,W

x n c Wn



Example 3.20 DTFT of The Unit Impulse

35

Find the DTFT of x[n] = [n].

<Sol.>1. DTFT of x[n]:

1j j n

n

X e n e

2. This DTFT pair is depicted in Fig. 3. 32.

1.DTFTn

Figure 3.32 (p. 235)

Example 3.20. (a) Unit impulse

in the time domain. (b) DTFT

of unit impulse in the

frequency domain.



Example 3.21 Inverse DTFT of A Unit Impulse Spectrum

1. Inverse DTFT of X(e j ): 1

.2

j nx n e d

1

,2

DTFT

.

2. This DTFT pair is depicted in Fig. 3. 33.

We can define X(e j ) over all by writing it as an infinite sum of delta

functions shifted by integer multiples of 2.

Dilemma: The DTFT of x[n] = 1/(2) does not converge, since it is not a

square summable signal, yet x[n] is a valid inverse DTFT!

2 .j

k

X e k

7. Continuous-Time Nonperiodic

Signals– The Fourier Transform37

1. The Fourier transform (FT) is used to represent a continuous-time non-

periodic signal as a superposition of complex sinusoids.

2. FT representation of time-domain signal x(t):

1( ) ( )

2

j tx t X j e d

(3.35)

( ) ( ) j tX j x t e dt

(3.36)

Frequency-domain representation of

the signal x(t)

3. Notation for FT pair:

.FTx t X j



Convergence condition for FT:

1) Approximation: 1

ˆ ,2

j tx t X j e d

Squared error between x(t) and : the error energy, given by x̂ t

2

ˆ ,x t x t dt

is zero if x(t) is square

integrable, i.e., if

2

.x t dt

Zero squared error does not imply pointwise convergence [i.e., ] ˆx t x tat all values of t.

Pointwise convergence of ˆ( )x t x t is guaranteed at all values of t

except those corresponding to discontinuities if x(t) satisfies the Dirichlet

conditions for nonperiodic signals:

x(t) has a finite number of maximum, minima, and discontinuities in any

finite interval.

The size of each discontinuity is finite.

There is zero energy in the difference of ˆ( ) andx t x t

x(t) is absolutely integrable: .x t dt



Example 3.25 FT of A Rectangular Pulse

Find the FT of x(t).

Consider the rectangular pulse depicted in Fig. 3.40 (a) and defined as

0 0

0

1,

0,

T t Tx t

t T

1. The rectangular pulse x(t) is absolutely integrable, provided that To < .

2. FT of x(t):

0

0

0

0 0

1 2sin , 0

Tj t j t

T

Tj t

T

X j x t e dt e dt

e Tj

3. For = 0, the integral simplifies to 2To.

0 00

2lim sin 2 .T T

0

2sin ,X j T

With understanding that

the value at = 0 is

obtained by evaluating

a limit.



0sin

2 ,T

X j

5. Phase spectrum:

0

0

sin 00,arg

sin 0,

TX j

T

6. X(j) in terms of sinc function:

0 02 sinc .X j T T

Fig. 3. 40 (b).

v ; p

As To increases, the nonzero time

extent of x(t) increases, while X(j)

becomes more concentrated about

the frequency origin.

4. Magnitude spectrum:



Example 3.26 Inverse FT of A Rectangular Pulse

Find the inverse FT of the rectangular spectrum depicted in Fig. 3. 42 (a) and

given by

1,

0,

W WX j

W

<Sol.>

Figure 3.42 (p. 246)

Example 3.26. (a) Rectangular spectrum in

the frequency domain.

(b) Inverse FT in the time domain.

v ; p

1. Inverse FT of x(t):

1 1 1

sin( ), 02

Wj t j t W

WW

x t e d e Wt tj t t



0

1lim sin ,t

Wt Wt

3. Inverse FT is usually written as

1

sin ,x t Wtt

or sinc ,W Wt

x t

Fig. 3. 42 (b).

With understanding that the value

at t = 0 is obtained as a limit.

As W increases, the frequency-domain reprsentation becomes less

concentrated about = 0, while the time-domain representation X(j)

becomes more concentrated about t = 0.

2. When t = 0, the integral simplifies to W/, i.e.,



Notation for FT pair: 1,FTt

Example 3.28 Inverse FT of An Impulse Spectrum

Find the inverse FT of X(j) = 2().

<Sol.>

1. Inverse FT of X(j):

1

2 12

j tx t e d

2. Notation of FT pair:

1 2FT

Duality between

Example 3.27 and 3.28

Example 3.27 FT of The Unit Impulse

Find the FT of x(t) = (t).

<Sol.> 1. x(t) does not satisfy the Dirichlet conditions, since the discontinuity at the

origin is infinite.

2. FT of x(t): 1j tX j t e dt

8. Properties of Fourier Representation44

8. Properties of Fourier Representation45

8. Properties of Fourier Representation

Periodic Discrete

46

9. Linearity and Symmetry Properties

Linearity property for four Fourier representations

47

;

;

o

o

FT

FS

DTFT j j j

DTFS

z t ax t by t Z j aX j bY j

z t ax t by t Z k aX k bY k

z n ax n by n Z e aX e bY e

z n ax n by n Z k aX k bY k

9. Linearity and Symmetry Properties48

1. Signal z(t):

3 1

2 2z t x t y t

where x(t) and y(t)

2. X[k] and Y[k]:

;2

;2

1 sin 4

1 sin 2

FS

FS

x t X k k k

y t Y k k k

3. FS of z(t):

;2 3 2 sin 2 1 2 sin 2FSz t Z k k k k k


Symmetry Properties: Real and Imaginary Signals

49

Symmetry property for real-valued signal x(t):

1. FT of x(t):

** *( ) ( ) ( )j t j tX j x t e dt x t e dt

(3.37)

2. Since x(t) is real valued, x(t) = x(t). Eq. (3.37) becomes

* ( )j tX j x t e dt

*( ) ( )X j X j (3.38)

X(j) is complex-conjugate symmetric

Re ( ) Re ( ) and Im ( ) Im ( )X j X j X j X j


The conjugate symmetry property for DTFS:

*[ ] [ ]X k X N k

Because the DTFS coefficients are N periodic, and thus

[ ] [ ]X k X N k

The symmetry conditions in all four Fourier representations of real-valued

signals are indicated in Table 3.4.


1. Input signal of LTI system:

cosx t A t

2. Real-valued impulse response of LTI system is denoted by h(t).

/ 2 / 2

j t j tx t A e A e

Euler’s formula

3. Output signal of LTI system:

cos argy t H j A t H j

( arg ( ) ) ( arg ( ) )( ) ( ) ( / 2) ( ) ( / 2)

j t H j j t H jy t H j A e H j A e

Applying Eq. (3.2)

and linearity

Exploiting the symmetry conditions:

( ) ( )H j H j

arg ( ) arg ( )H j H j

Continuous-time case:


Discrete-time case: x[n] = Acos(n )

2. Real-valued impulse response of LTI system is denoted by h[n].

3. Output signal of LTI system:

cos argj jy n H e A n H e

The LTI system modifies the amplitude of the input sinusoid by H(e j)and

the phase by arg{H(e j)}.

x(t) is purely imaginary:

1. x(t) = x(t).

2. Eq.(3.37) becomes


*( ) ( )X j X j (3.39)

Re ( ) Re ( ) and Im ( ) Im ( )X j X j X j X j


Symmetry Properties: Even and Odd Signals

53

1. x(t) is real valued and has even symmetry.

x(t) = x(t) and x ( t) = x(t) x(t) = x( t)

2. Eq.(3.37) becomes


* jX j x e d X j

3. Conclusion:

1) The imaginary part of X(j) = 0: ( ) ( )X j X j

If x(t) is real and even, then X(j) is real.

2) If x(t) is real and odd, then X(j) = X(j) and X(j) is imaginary.

Change of variable = t

10. Convolution Property

Nonperiodic Signals

54

1. Convolution of two nonperiodic continuous-time signals x(t) and h(t):

y t h t x t h x t d

2. FT of x(t ):

( )1

2

j tx t X j e d

3. Substituting this expression into the convolution integral yields

1

2

j ty t H j X j e d

and we identify H(j)X(j) as the FT of y(t).

( ) ( )* ( ) ( ) ( ) ( )FTy t h t x t Y j X j H j (3.40)

1

2

1

2

j t j

j j t

y t h X j e e d d

h e d X j e d


Example

55

2

2

4sinFTx t X j

x t z t z t

1, 1

0, 1

FTt

z t Z jt

10. Convolution Property56

• Convolution of Nonperiodic Signals — Discrete-time case

If DTFT jx n X e DTFT jh n H e and

[ ] [ ]* [ ] ( ) ( ) ( )DTFT j j jy n x n h n Y e X e H e (3.41)

10.2 Filtering57

1. Multiplication in frequency domain Filtering.

3. System Types of filtering:

1) Low-pass filter

2) High-pass filter

3) Band-pass filter

Fig. 3.53 (a), (b), and (c)

The characterization of DT filter is based

on its behavior in the frequency range

< because its frequency

response is 2 -periodic.

4. Realistic filter:

1) Gradual transition band

2) Nonzero gain of stop band

5. Magnitude response of filter: 20log H j 20log jH e or [dB]

2. The terms “filtering” implies that some frequency components of the input

are eliminated while others are passed by the system unchanged.

10.2 Filtering58

10. Convolution Property59

Example 3.33 RC Circuit: Filtering

For the RC circuit depicted in Fig. 3.54, the impulse response for the case where

yC(t) is the output is given by

1 t RC

Ch t e u tRC

Since yR(t) = x(t) yC(t), the impulse response for

the case where yR(t) is the output is given by

1 t RC

Rh t t e u tRC

Figure 3.54 (p. 264)

RC circuit with input x(t) and outputs yc(t) and

yR(t).

1

1CH j

i RC

1

R

i RCH j

i RC


Frequency response of a system

The ratio of the FT or DTFT of the output to the input .

61

For CT system:( )

( )( )

Y jH j

X j

(3.42)

For DT system:( )

( )( )

jj

j

Y eH e

X e

(3.43)

Frequency response


Recover the input of the system from the output

Equalizer

The process of recovering the input from the output is known

as equalization

62

invX j H j Y j and invj j jX e H e Y e

CT case DT case

where inv 1/ ( )H j H j and inv 1/ ( )j jH e H e


Convolution of Periodic Signals

63

Define the periodic convolution of two CT signals x(t) and z(t), each having

period T, as

0

T

y t x t z t x z t d #

where the symbol denotes that integration is performed over a single period

of the signals involved.

2;

( ) ( ) ( ) [ ] [ ] [ ]FS

Ty t x t z t Y k TX k Z k

#


Discrete-Time Convolution of Periodic Sequences

Define the discrete-time convolution of two N-periodic

sequences x[n] and z[n]

64

1

0

N

k

y n x n z n x k z n k

#

2;

[ ] [ ] [ ] [ ] [ ] [ ]DTFS

Ny n x n z n Y k NX k Z k

#

DTFS of y[n]

( ) ( ) ( ) ( )FT

x t z t X j Z j

0;( ) ( ) [ ] [ ]

FSx t z t TX k Z k

#

[ ] [ ] ( ) ( )FT j jx n z n X e Z e

0;[ ] [ ] [ ] [ ]

DTFSx n z n NX k Z k

#

Table 3.5 Convolution Properties

11. Differentiation and Integration

Properties

Differentiation in Time

65

1. A nonperiodic signal x(t) and its FT, X(j), is related by

dejXtx tj

2

1 Differentiating both

sides with respect to

t

dejjXtxdt

d tj

2

1

◆ Differentiation of x(t) in Time-Domain (j) X(j) in Frequency-Domain

( ) ( )FTd

x t j X jdt

00

( ) ( ) ( ) 0d

x t j X jdt

F


Properties

Differentiation in Frequency

66

1. FT of signal x(t):

dtetxjX tj

Differentiating both

sides with respect to

dtetjtxjXd

d tj

2. Differentiation property in frequency domain:

FT djtx t X j

d

◆ Differentiation of X(j) in Frequency-Domain ( jt) x(t) in Time-Domain


Properties

Integration

Integrating nonperiodic signals

67

1. Define dxty

t

( ) ( )

dy t x t

dt (3.51)

2. By differentiation property, we have

1( ) ( )Y j X j

j

(3.52)

Indeterminate at = 0

3. When the average value of x(t) is not zero, then it is possible that y(t) is not

square integrable.

FT of y(t) may not converge!

W can get around this problem by including impulse in the transform, i.e.

1( ) ( ) ( )Y j X j c

j

c can be determined by

the average value of x(t)

C = X(j0)


Properties68

68

Table 3.6 Commonly Used Differentiation and Integration Properties.

( ) ( )FTd

x t j X jdt

0;0

FSdx t jk X k

dt

FT djtx t X j

d

DTFT jdjnx n X e

d

1( ) ( ) ( 0) ( )

tFTx d X j X j

j

12. Time- and Frequency-Shift Properties

Time-Shift Property

69

1. Let z(t) = x(t t0) be a time-shifted version of x(t).

2. FT of z(t):

dtetzjZ tj

dtettx tj

0

3. Change variable by = t t0:

dexjZtj 0

dexe jtj 0 jXe

tj 0

◆ Time-shifting of x(t) by t0 in Time-Domain

(e j t0) X(j) in Frequency-Domain

and( ) ( )Z j X j 0arg{ ( )} arg{ ( )}Z j X j t

12. Time- and Frequency-Shift Properties70

0

0

j tFTx t t e X j

0 0 0;

0

FT jk tx t t e X k

0

0

j nDTFT jx n n e X e

0 0 0;

0

DTFS jk nx n n e X k

12. Time- and Frequency-Shift Properties

Frequency Shift

71

1. Suppose that:

( ) ( )FTx t X j ( ) ( )

FTz t Z jand

2. By the definition of the inverse FT, we have

dejZtz tj

2

1

dejX tj

2

1

Substituting variables = into above Eq. gives

dejXtz tj

2

1

dejXe tjtj

2

1 txe tj

◆ Frequency-shifting of X(j) by in Frequency-Domain [i.e. X(j( ))]

(e t) x(t) in Time-Domain

12. Time- and Frequency-Shift Properties72

FTj te x t X j

0 0 0;

0

jk t FSe x t x k k

[ ]

jDTFTj ne x n X e

0 0 0;

0[ ]jk n FS

e x n X k k

13. Finding Inverse Fourier Transforms

by Using Partial-Fraction Expansions. 73

1. FT X(j) in terms of a ratio of polynomials in j:

jA

jB

ajajaj

bjbjbjX

N

N

N

M

M

01

1

1

01

Assume that M < N. If M N, then we may use long division to express X(j)

in the form

jA

jBjfjX

NM

k

k

k

0

Partial-fraction expansion

is applied to this term

Applying the differentiation property

and the pair to these

terms

( ) 1FTt

2. Let the roots of the denominator

A(j) be dk, k = 1, 2, …, N. These

roots are found by replacing j

with a generic variable v and determining the roots of the polynomial

001

1

1

avavav N

N

N


by Using Partial-Fraction Expansions74

3. For M < N, we may the write

N

k

k

M

k

k

k

dj

jb

jX

1

0

Assuming that all the roots dk, k = 1, 2, …, N, are distinct, we may write

N

k k

k

dj

CjX

1

Ck, k = 1, 2, …, N are determined

by the method of residues

Appendix B

In Example 3.24, we derived the transform pair

1FTdte u t

j d

This pair is valid even if d is

complex, provided that Re{d} < 0.

Assuming that the real part of each dk, k = 1, 2, …, N, is negative, we use

linearity to write

1 1

k

N Nd t FT k

k

k k k

Cx t C e u t X j

j d


by Using Partial-Fraction Expansions

Inverse Discrete-Time Fourier Transform

75

1. Suppose X(e j) is given by a ratio of polynomial in e j, i.e.

11

1

1

01

jNj

N

Nj

N

jMj

Mj

eee

eeeX

Normalized to 1

2. Factor the denominator polynomial as

N

k

j

k

jNj

N

Nj

N edeee1

1

1

1 11

Replace e j with the generic variable v:

01

2

2

1

1

NN

NNN vvvv Find dk, the roots of

this polynomial

3. Partial-fraction expansion:

Assuming that M < N and all the dk are distinct, we may express X(e j)as

N

kj

k

kj

ed

CeX

1 1


by Using Partial-Fraction Expansions

Since

76

1

1

n DTFT

k j

k

d u nd e

The linearity property implies that

N

k

n

kk nudCnx1

Expansions for

repeated roots are

treated in Appendix B.

14. Multiplication Property77

Non-periodic continuous-time signals

1. Non-periodic signals: x(t), z(t), and y(t) = x(t)z(t). Find the FT of y(t).

2. FT of x(t) and z(t):

dvejvXtx jvt

2

1

dejZtz tj

2

1and

dvdejZjvXty tvj

22

1

Change variable: = vInner Part: Z(j) X(j)

Outer Part: FT of y(t)3. FT of y(t):

1 1

2 2

j ty t X jv Z j v dv e d

1( ) ( ) ( ) ( ) ( )* ( )

2

FTy t x t z t Y j X j Z j

(3.56)

Scaled by 1/2


where

dvvjZjvXjZjX *

◆ Multiplication of two signals in Time-Domain

Convolution in Frequency-Domain (1/2)

Non-periodic continuous-time signals

1. Non-periodic DT signals: x[n], z[n], and y[n] = x[n]z[n].

Find the

DTFT of

y[n].2. DTFT of y[n]:

1[ ] [ ] [ ] ( ) ( ) ( )

2

DTFT j j jy n x n z n Y e X e Z e

# (3.57)

◆ Multiplication of two signals in Time-Domain

Convolution in Frequency-Domain (1/2)

where the symbol denotes periodic convolution.

Here, X(e j ) and X(e j ) are 2-periodic, so we evaluate the convolution over

a 2 interval:

jjX j Z j X e Z e d

#


Multiplication property

can

be used to study the

effects

of truncating a time-domain

signal on its frequency-

domain.

Windowing !

Truncate signal x(t) by a

window function w(t) is

represented by

y(t) = x(t)w(t)

Fig. 3. 65 (a)

0 0

Time Interval

T t T


Figure 3.65b (p. 293)

(b) Convolution of the signal

and window FT’s resulting

from truncation in time.

1. FT of y(t):

1

*2

FTy t Y j X j W j

p ; v 2. If w(t) is the rectangular window depicted

in Fig. 3. 65 (b), we have

0sin2

TjW

Smoothing the details

in X(j) and

introducing

oscillation near

discontinuities in X(j).


Example 3.46 Truncating the Impulse Response

The frequency response H(e j) of an ideal discrete-time system is depicted in

Fig. 3. 66 (a). Describe the frequency response of a system whose impulse

response is the ideal system impulse response truncated to the interval M

n M.

<Sol.>

1. The ideal impulse response is the inverse DTFT of H(e j ).

Using the result of Example 3.19, we write

1

sin2

nh n

n

2. Let ht[n] be the truncated impulse response:

,0, otherwise

t

h n n Mh n

[ ] [ ] [ ]th n h n w n

1,

0, otherwise

n Mw n


3. Let [ ] ( )DTFT j

t th n H e

Using the multiplication property in Eq. (3.57), we have

1

2

jj j

tH e H e W e d

4. Since

2/,0

2/,1jeH

sin 2 1 / 2

sin / 2

jM

W e

On the basis of

Example 3.18

and

2/

2/2

1

dFeH j

t

where we have defined

, / 2

0, / 2

j

jjW e

F H e W e

Discussion: refer to

p. 295 in textbook.


Figure 3.66 (p. 294)

The effect of truncating the impulse

response of a discrete-time system.

(a) Frequency response of ideal

system. (b) F() for near zero.

v , u


Figure 3.66 (p. 294)

The effect of truncating the impulse response of a discrete-

time system. (c) F() for slightly greater than /2.(d)

Frequency response of system with truncated impulse

response.

v , u

Ht(ej ) is the area under

F( ) between = /2

and = /2.


1( ) ( ) ( ) ( )

2

FTx t z t X j Z j

;( ) ( ) [ ] [ ]oFS

x t z t X k Z k

1[ ] [ ] ( ) ( )

2

DTFT j jx n z n X e Z e

#

;[ ] [ ] [ ] [ ]oDTFT

x n z n X k Z k

#

Table 3.9 Multiplication Properties of Fourier Representations

15. Scaling Properties86

1. Let z(t) = x(at).

2. FT of z(t) :

( ) ( ) ( )j t j tZ j z t e dt x at e dt

Changing

variable: = at

( / )

( / )

(1/ ) ( ) , 0( )

(1/ ) ( ) , 0

j a

j a

a x e d aZ j

a x e d a

( / )( ) (1/ ) ( ) ,j aZ j a x e d

( ) ( ) (1/ ) ( / ).FTz t x at a X j a

◆ Scaling in Time-Domain Inverse Scaling in Frequency-Domain

Signal expansion or compression!

Refer to Fig. 3.70.

(3.60)


Figure 3.70 (p. 300)

The FT scaling property. The figure assumes that 0 < a < 1.

v


Example 3.48 Scaling a Rectangular Pulse

<Sol.>

Let the rectangular pulse

1, 1( )

0, 1

tx t

t

Use the FT of x(t) and the scaling property to

find the FT of the scaled rectangular pulse

1, 2( )

0, 2

ty t

t

1. Substituting T0 = 1 into the result of Example 3. 25 gives

2( ) sin( )X j

2. Note that y(t) = x(t/2).

2 2( ) 2 ( 2 ) 2 sin(2 ) sin(2 )

2Y j X j

Scaling property and a = 1/2

Fig. 3.71

Substituting T0 = 2 into the result of Example 3.25 can also give the answer!


Figure 3.71

(p. 301)

Application of

the FT scaling

property in Example 3.48.

(a) Original time

signal. (b)

Original FT. (c)

Scaled time signal y(t) =

x(t/2). (d) Scaled

FT Y(j) =

2X(j2).

v

16. Parseval Relationships90

The energy or power in the time-domain representation of a signal is equal to

the energy or power in the frequency-domain representation.

◆ Case for CT nonperiodic signal: x(t)

1. Energy in x(t):

2( )xW x t dt

2. Note that 2( ) ( ) ( )x t x t x t

Express x*(t) in terms of its FT X(j):1

( ) ( )2

j tx t X j e d

1( ) ( )

2

j t

xW x t X j e d dt

The integral inside

the braces is the FT

of x(t).

1( ) ( )

2xW X j X j d

1( ) ( )

2

j t

xW X j x t e dt d

2 21( ) ( )

2x t dt X j d


Representations Parseval Relationships

FT 2 21

( ) ( )2

x t dt X j d

FS 2 2

0

1( ) [ ]

T

kx t dt X k

T

DTFT 22 1

[ ] ( )2

j

nx n X e d

DTFS 1 12 2

0 0

1[ ] [ ]

N N

n kx n X k

N


Example 3.50 Calculating Energy in a Signal

Letsin( )

[ ]Wn

x nn

Use Parseval’s theorem to evaluate

21( )

2

jX e d

22

2 2

sin ( )[ ]

n n

Wnx n

n

<Sol.>

1. Using the DTFT Parseval relationship in Table 3.10, we have

2. Since 1,[ ] ( )

0,

DTFT jW

x n X eW

11 /

2

W

Wd W

Direct calculation in time-

domain is very difficult!

17. Time-Bandwidth Product93

1. Preview: 1,( ) ( ) 2sin( ) /

0,

o FT

o

o

t Tx t X j T

t T

Figure 3.72 (p. 305)

Rectangular pulse illustrating the inverse relationship between the time

and frequency extent of a signal.

Fig. 3.72

2. x(t) has time extent 2T0. X(j) is actually of infinite extent frequency.

v , p

3. The mainlobe of sinc function is fallen in the interval of < /T0, in which

contains the majority of its energy.


4. The product of the time extent T0 and mainlobe width 2/T0 is a constant.

5. Compressing a signal in time leads expansion in the frequency domain and

vice versa.

Signla’s time-bandwidth product !

6. Bandwidth: The extent of the signal’s significant frequency content.

1) A mainlobe bounded by nulls.

Ex. Lowpass filter One half the width of mainlobe.

2) The frequency at which the magnitude spectrum is 1/2 times its peak values.

7. Effective duration of signal x(t):1/ 2

22

2

( )

( )d

t x t dtT

x t dt

(3.63)

8. Effective bandwidth of signal x(t):1/ 2

22

2

( )

( )w

X j dB

X j d

(3.64)


9. The time-bandwidth product for any signal is lower bounded according to

the relationship

1/ 2d wT B (3.65)

We cannot simultaneously decrease the duration and bandwidth of

a signal.

Uncertainty principle !

Example 3.51 Bounding the Bandwidth of a Rectangular Pulse

<Sol.>

Let 1,( )

0,

o

o

t Tx t

t T

Use the uncertainty principle to place a lower

bound on the effective bandwidth of x(t).

1. Td of x(t): 1/ 22o

o

o

o

T

T

d T

T

t dtT

dt


1/ 22

3 1/ 2[(1/(2 ))(1/ 3) ] / 3

o

o o

oo

o

T

T T

d o T dT

T

t dtT T t T

dt

2. The uncertainty principle given by Eq. (3.65) states that

1/(2 )w dB T 3 /(2 )w oB T

18. Duality

Duality of the FT

97

1. FT pair:

1( ) ( )

2

j tx t X j e d

( ) ( ) j tX j x t e dt

and

2. General equation:

1( ) ( )

2

jvy v z e dv

(3.66)

Choose = t and = the Eq. (3.66) implies that

3. Interchange the role of time and frequency by letting = and = t, then

Eq. (3.66) implies that

1( ) ( )

2

j ty t z e d

( ) ( )FTy t z (3.67)

1( ) ( )

2

j ty z t e dt

( ) 2 ( )FTz t y (3.68)

18. Duality98

Example99

Example 3.52 Applying Duality

Find the FT of 1( )

1x t

jt

<Sol.>

1. Note that:1

( ) ( ) ( )1

FTtf t e u t F jj

2. Replacing by t, we obtain

1( )

1F jt

jt

x(t) has been expressed as F(jt).

3. Duality property:

( ) 2 ( ) 2 ( )X j f e u

Duality100

Figure 3.74 (p. 309)

The FT duality property.

v ; p

18. Duality101

1. DTFS Pair for x[n]:

1

0

[ ] [ ] O

Njk n

k

x n X k e

1

0

1[ ] [ ] O

Njk n

n

X k x n eN

and

2. The DTFS duality is stated as follows: if

2;

[ ] [ ]DTFS

Nx n X k

(3.71)

2; 1

[ ] [ ]DTFS

NX n x kN

(3.72)

N = time index; k =

frequency index

18. Duality

DTFT

102

Table 3.11 Duality Properties of Fourier Representations

FT ( ) ( )FTf t F j ( ) 2 ( )FTF jt f

DTFS ; 2 /[ ] [ ]DTFS Nx n X k ; 2 /[ ] (1/ ) [ ]DTFS NX n N x k

FS-DTFT [ ] ( )DTFS jx n X e ;1( ) [ ]FSjtX e x k

( ) [ ] ojk t

k

z t Z k e

( ) [ ]j j n

k

X e x n e

3. Duality relationship between z(t) and X(e j)

Require

T = 2

In the DTFT t in the FS

n In the DTFT k in the FSAssumption: 0 = 1

[ ] ( )DTFS jx n X e

;1( ) [ ]FSjtX e x k

A.1 Wind Instruments

Fipples, pressure-excited reeds, or buzzing lips are

coupled with resonating air column.

A.2 String Instruments

One or more stretched stringed strings are made to vibrate,

typically by bowing, plunking or striking, in conjunction with

resonating boxes or sound boards.

A.3 Percussion Instruments

Include virtually any object that may be struck, typically

consisting of bars, plates, or stretched membrances and

associated resonators

A.4 Cello & Clarinet

A.5 Trumpet & Bass Drum

A.6 Speech

Models

y n a y n k G b x n rk r

r

q

k

p

( ) ( ) ( )

01

0 ( ) z

Pitch period P

Voiced

Unvoiced

V

U

voiced/unvoiced

switch

signal speech

Gain estimate

All-Pole

Filtery(n)

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