chapter 3: fourier representation of signals and lti...

Chapter 3: Fourier Representation of Signals

and LTI Systems

Chih-Wei Liu

Outline Introduction Complex Sinusoids and Frequency Response Fourier Representations for Four Classes of Signals Discrete-time Periodic Signals Fourier Series Continuous-time Periodic Signals Discrete-time Nonperiodic Signals Fourier Transform Continuous-time Nonperiodic Signals Properties of Fourier representations Linearity and Symmetry Properties

Convolution Property

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Outline

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Differentiation and Integration Properties

Time- and Frequency-Shift Properties

Finding Inverse Fourier Transforms

Multiplication Property

Scaling Properties

Parseval Relationships

Time-Bandwidth Product

Duality

Introduction

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In this chapter, we represent a signal as a weighted superposition of complex sinusoids. AKA Fourier analysis The weight associated with a sinusoid of a given frequency

represents the contribution of that sinusoid to the overall signal.

Four distinct Fourier representations:

)(

)(

)(

)()()()(

jHe

dehe

deh

thtxty

tj

jtj

tj

Frequency Response of LTI System

5

The response of the LTI system to a sinusoidal input ejt : H{x(t)=ejt}= ejt

H(j)

For discrete-time case, the response of the LTI system to a sinusoidal input ejn is H{x[n]=ejn}= ejn H(ej)

LTI System h(t)

tjetx )(

( ) ( ) jH j h e d

constantDependent on , but independent on t

k

kjj ekheH ][)(

)(

][

][

][][][

)(

jnjk

kjnj

k

knj

eHe

ekhe

ekh

nhnxny

Dependent on , but independent on n

Frequency Response of LTI System

6

Frequency response of a continuous-time LTI system

Frequency response of the LTI system can also be represented by

Magnitude response Phase response

( ) ( ) jH j h e d

k

kjj ekheH ][)(

LTI Systemx(t) or x[n] y(t) or y[n]

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

)( jH)}(arg{ jH

Example 3.1 RC Circuit System

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The impulse response of the RC circuit system is derived in Example 1.21 as

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

Find an expression for the frequency response, and plot the magnitude and phase response.

<Sol.>

de

RCRC

j

0

11

1

0

1 11

jRCe

RC jRC

RC

j

RC1

1

Frequency response:

deueRC

jH jRC

1

Magnitude response:

2

2 1

1

RC

RCjH

Phase response:

RCjH arctanarg

H j

Low-pass filter

Another Meaning for Frequency Response

8

The eigenfunction of the LTI system (t):

The eigen-representation of the LTI system

By representing arbitrary signals as weighted superposition of eigenfunctionejt, then

LTI System h(t)

M

k

tjkk

kejHatxHty1

)()}({)(

)()( tty

eigenvalueIndependent of t

( ) j tH j e H

j te

Hj ne j j nH e e

][][ nnx ][][ nny

M

k

tjk

keatx1

)()( ttx

the weights describe the signal as a function of frequency. (frequency-domain representation)

Multiplication in frequency domain,c.f. convolution in time-domain

Fourier Analysis

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Non-periodic signals have (continuous) Fourier transform representations, while periodic signals have (discrete) Fourier series representations.

Why Fourier series representations for Periodic signals Periodic signal can be considered as a weighted superposition of

(periodic) complex sinusoids (using periodic signals to construct a periodic signal)

Recall that the periodic signal has a (fundamental) period, this implies that the period (or frequency) of each component sinusoid must be an integer multiple of the signal’s fundamental period (or frequency) in frequency-domain analysis, the weighted complex sinusoids look like a discrete series of weighted frequency impulse Fourier series representation

Question: Can any a periodic signal be represented or constructed by a weighted superposition of complex sinusoids?

k

tjkekAtx 0][)(ˆ

k

njkekAnx 0][][ˆ

Approximated Periodic Signals

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Suppose the signal is approximated to a discrete-time periodic signal x[n] with fundamental period N, where 0 = 2/N.

Since , there are only Ndistinct sinusoids of the form : e.g. k=0, 1, …, N-1

Accordingly, we may rewrite the signal as

For continuous-time case, we then have , where 0 = 2/T is the fundamental frequency of periodic signal x(t)

Although is periodic, is distinct for distinct k0

Hence, an infinite number of distinct terms, i.e.

njknjknjnjknjNnNkj eeeeee 00000 2)(

njke 0

1

0

0][][ˆN

k

njkekAnx

tjke 0 tjke 0

k

tjkekAtx 0][)(ˆ

DTFS

FS

Approximation Error

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Mean-square error (MSE) performance:

We seek the weights or coefficients A[k] such that the MSE is minimum

The DTFS and FS coefficients (Fourier analysis) achieve the minimum MSE (MMSE) performance.

1 2

0

1 [ ] [ ]N

n

MSE x n x n dtN

2

0

1 ( ) ( )T

MSE x t x t dtT

Frequencies separated by an integer multiple of 2 are identical

Fourier Analysis

12

Why Fourier transform representations for Non-periodic signals Using periodic sinusoids (the same approach) to construct a non-

periodic signal, there are no restrictions on the period (or frequency) of the component sinusoids there are generally having a continuum of frequencies in frequency-domain analysis Fourier transform representation

Fourier transform: Continuous-time case

Discrete-time case

deeXnx njj

)(

21][ˆ

deeXtx tjj

)(

21)(ˆ

FT

DTFT

1

0

0][][ˆN

k

njkekAnxDTFS

k

tjkekAtx 0][)(ˆ

FS

Outline Introduction Complex Sinusoids and Frequency Response Fourier Representations for Four Classes of Signals Discrete-time Periodic Signals Continuous-time Periodic Signals Discrete-time Nonperiodic Signals Fourier Transform Continuous-time Nonperiodic Signals Properties of Fourier representations Linearity and Symmetry Properties

Convolution Property

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Discrete-Time Fourier Series (DTFS)

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The DTFS-pair of a periodic signal x[n] with fundamental period N and fundamental frequency 0=2/N is

The DTFS coefficients X[k] are called the frequency-domain representation for x[n]

The value k determines the frequency of the sinusoid associated with X[k] The DTFS is exact. (Any periodic discrete-time signal can be described in terms

of DTFS coefficients exactly) The DTFS is the only one of Fourier analysis that can be evaluated and

manipulated in computer for a finite set of N numbers.

0

1

0

[ ] [ ]N

jk n

k

x n X k e

0

1

0

1[ ] [ ]N

jk n

n

X k x n eN

0;DTFSx n X k

Example 3.2 DTFS Coefficients

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Find the frequency domain representation of the signal depicted in Fig. 3.5.

<Sol.>1. Period: N = 5 o = 2/5

2. Odd symmetry We choose n = 2 to n = 23. Fourier coefficient:

2

2 /5

2

15

jk n

n

X k x n e

5/45/205/25/4 2101251 jkjkjjkjk exexexexex

2 /5 2 /51 1 1[ ] {1 }5 2 21 {1 sin( 2 / 5)}5

jk jkX k e e

j k

Example 3.2 (conti.)

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If we calculate X[k] using n = 0 to n = 4:

5/85/65/45/20 4321051 jkjkjkjkj exexexexexkX

5/85/2

21

211

51 jkjk ee 8 / 5 2 2 / 5 2 / 5jk jk jk jke e e e since

2 /5 2 /51 1 1[ ] {1 }5 2 21 {1 sin( 2 / 5)}5

jk jkX k e e

j k

The same expression for the DTFS coefficients !!!

Example 3.2 (conti.)

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[ ] Magnitude spectrum of [ ]X k x n[ ]X k Even function

arg [ ] Phase spectrum of [ ]X k x n arg [ ]X k

Odd function

Example 3.3 Computation by Inspection

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Determine the DTFS coefficients of x[n] = cos (n/3 + ), using the method of inspection.

<Sol.>1. Period: N = 6 o = 2/6 = /32. Using Euler’s formula, x[n] can be expressed as

( ) ( )3 3

3 31 1[ ]2 2 2

j n j nj n j nj je ex n e e e e

(3.13)

3. Compare Eq. (3.13) with the DTFS of Eq. (3.10) with o = /3, written by summing from k = 2 to k = 3:

3

/ 3

22 / 3 / 3 / 3 2 / 3

[ ]

[ 2] [ 1] [0] [1] [2] [3]

jk n

kj n j n j n j n j n

x n X k e

X e X e X X e X e X e

;

3

/ 2, 1/ 2, 1

0, otherwise on 2 3

j

DTFS j

e kx n X k e k

k

Example 3.4

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Find the DTFS coefficients of the N-periodic impulse train .

<Sol.>

1

2 /

0

1 1Njkn N

n

X k n eN N

1. Period: N. 2. By (3.11), we have

l

x n n lN

Example 3.6

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Find the DTFS coefficients for the N-periodic square wave given by

<Sol.>1. Period = N, hence o = 2/N2. It is convenient to evaluate DTFS coefficients over the interval n = M to n = NM1.

MNnM

MnMnx

,0,1

3. For k = 0, N, 2N, …, we have 1o ojk jke e

,2 , ,0 ,1211][ NNkN

MN

kXM

Mn

M

Mn

njkMN

Mn

njk eN

enxN

kX 001][1][

1

For k 0, N, 2N, …, we have

,2 , ,0 ,1

11][0

000

)12(

NNke

eN

eeN

kX jk

MjkMjkM

Mn

njk

Example 3.6 (conti.)

21

0 0

0

(2 1)1[ ] , 0, , 2 , ......1

jk M jk M M

jk

e eX k k N NN 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 11

jk M jk M jk M jk M

jk jk jk jk

e e e eX kN e e N e e

,

2/sin2/12sin1

0

0

k

MkN

kX

0, , 2 ,k N N

The numerator and denominator of above Eq. are divided by 2j

The DTFS coefficients for the square wave, assuming a period N = 50: (a) M = 4. (b) M = 12.

Even symmetry

Symmetry Property of DTFS Coefficients

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If X[k] = X[-k], it is instructive to consider the contribution of each term in

of period N

Assume that N is even, so that N/2 is integer. o = 2/N Rewrite the DTFS coefficients by letting k range from N/2 +1 to N/2, i.e.

Define new set of coefficients

0

1

0

[ ] [ ]N

jk n

k

x n X k e

2/

12/

0

N

Nk

njkekXnx

/ 2 1

01

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

, 0, / 22 , 1, 2, , / 2 1X k k N

B kX k k N

/ 2

00

[ ] [ ]cos( )N

k

x n B k k n

A similar expression may be derived for N odd.

0

0[ ] [ ]cos( )

J

Jk

x n B k k n

Example 3.7

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The contribution of each term in DTFS series to the square wave may be illustrated by defining the partial-sum approximation to x[n] as

where J N/2. This approximation contains the first 2J + 1 terms centered on k = 0 in the square wave above. Assume a square wave has period N = 50 and M = 12. Evaluate one period of the Jth term and the 2J + 1 term approximation for J = 1, 3, 5, 23, and 25<Sol.>

J = 1

J = 3

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J = 5

J = 23

J = 25

The coefficients B[k] associated with values of k near zero represent the low-frequency or slowly varying features in the signal, while the coefficients associated with values of k near N/2 represent the high frequency or rapidly varying features in the signal.

Fourier Series (FS)

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The DT-pair of a periodic signal x(t) with fundamental period T and fundamental frequency 0=2/T is

The FS coefficients X[k] are called the frequency-domain representation for x(t) The value k determines the frequency of the sinusoid associated with X[k] The infinite series in x(t) is not guaranteed to converge for all possible signals.

Suppose we define

0;FSx t X k

0( ) [ ] jk t

k

x t X k e

0

0

1[ ] ( )T jk tX k x t e dt

T

k

tjkekXtx 0][)(ˆ )( ? toapproach tx If x(t) is square integrable, then

T

dttxtxT 0

2

01

a zero power in their differences.

take one period of x(t)

Remarks

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A zero MSE does not imply that the two signals are equal pointwise.

Dirichlet’s conditions:1. x(t) is bounded

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

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

Pointwise convergence of and x(t) is guaranteed at all t except those corresponding to discontinuities satisfying Dirichlet’sconditions.

If x(t) satisfies Dirichlet’s conditions and is not continuous, then converges to the midpoint of the left ad right limits of x(t) at each discontinuity.

x̂ t

x̂ t

2 42 4 2

0

1 1 112 2 4 2 4 2

jk t jk eX k e e ejk jk jk

Example 3.9

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Determine the FS coefficients for the signal x(t).

1. The period of x(t) is T = 2, so o=2/2 =.2. Take one period of x(t): x(t) = e2t, 0 t 2. Then

2 2 22

0 0

1 12 2

jk tt jk tX k e e dt e dt

<Sol.>

The Magnitude of X[k] the magnitude spectrum of x(t)

The phase of X[k] the phase spectrum of x(t)

Example 3.10

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Determine the FS coefficients for the signal x(t) defined by 4l

x t t l

<Sol.>1. Fundamental period of x(t) is T = 4, each period contains an impulse.

2. By integrating over a period that is symmetric about the origin, 2 < t 2, to obtain X[k]:

2 / 2

2

1 14 4

jk tX k t e dt

3. The magnitude spectrum is constant and the phase spectrum is zero.

Example 3.11 Computation by Inspection

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Determine the FS representation of the signal 4/2/cos3 ttx<Sol.>1. Fundamental frequency of x(t) is o= 2/4= /2, so T = 4.

/ 2( ) [ ] jk t

k

x t X k e

/ 2 / 4 / 2 / 4

/ 4 / 2 / 4 / 23 332 2 2

j t j tj j t j j te ex t e e e e

2. Rewrite the x(t) as

3. Compare with / 4

/ 4

3 , 123[ ] , 120, otherwise

j

j

e k

X k e k

/4

/4

Example 3.12 Inverse FS

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Find the (time-domain) signal x(t) corresponding to the FS coefficientsAssume that the fundamental period is T=2.

20/2/1 jkk ekX

<Sol.>1. Fundamental frequency: o= 2/T= . Then

/ 20 / 20

0 11/ 2 1/ 2k kjk jk t jk jk t

k kx t e e e e

0( ) [ ] jk t

k

x t X k e

1

20/

0

20/ 2/12/1l

tjljll

k

tjkjkk eeee

12/11

12/11

120/20/

tjtj ee

tx

Example 3.13

31

Determine the FS representation of the square wave:

<Sol.>1. The period is T, so the fundamental frequency o= 2/T. 2. We consider the interval T/2 t T/2 to obtain the FS coefficients. Then

00 0

0

0

0

0

0 0 0 0

/ 2

/ 2

0

0

0 0

0

1 1

1 , 0

2 , 02

2sin, 0

T Tjk t jk t

T T

Tjk t

T

jk T jk T

X k x t e dt e dtT T

e kTjk

e e kTk j

k Tk

Tk

(2) For k = 0, we have

0

0

0210T

T TT

dtT

X

(1) For k 0, we have

By means of L’Hôpital’s rule

TT

TkTk

k

0

0

00

0

2sin2lim

0

00sin2

TkTk

kX

Example 3.13 (conti.)

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Figure 3.22 The FS coefficients, X[k], –50 k 50, for three square waves. (a) To/T = 1/4 . (b) To/T = 1/16. (c) To/T = 1/64.

Sinc Function

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Maximum of sinc(u) is unity at u = 0, the zero crossing occur at integer values of u, and the amplitude dies off as 1/u.

The portion of sinc(u) between the zero crossings at u = 1 is known as the mainlobe of the sinc function.

The smaller ripples outside the mainlobe are termed sidelobes

sin( )sinc( ) uuu

More on the FS Pairs

34

The original FS pairs are described in exponential form: Let’s consider the trigonometric form

For real-valued signal x(t): , and

0( ) [ ] jk t

kx t X k e

1

][arg][arg

1

)][][(]0[

)][][(]0[][)(

00

000

k

kXjtjkkXjtjk

k

tjktjk

k

tjk

ekXekXX

ekXekXXekXtx

][][ kXkX ][arg][arg kXkX

100

01

1

)][arg()][arg(

)sin()][sin(arg)cos()][cos(arg][2]0[

)][argcos(][2]0[

)(][]0[)( 00

k

k

k

kXtkjkXtkj

tkkXtkkXkXX

kXtkkXX

eekXXtx

0 01

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

x t B B k k t A k k t

Rewrite the signal as we have

Trigonometric FS Pair for Real Signals

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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

0 01

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

x t B B k k t A k k t

where

],0[]0[ XB ]}[Re{2

)][cos(arg][2][kX

kXkXkB

]}[Im{2

)][sin(arg][2][kX

kXkXkA

Or, if we use trigonometric FS representation for a real-valued periodic signal x(t) with period T, then

Example 3.15

36

Let us find the FS representation for the output y(t) of the RC circuit in response to the square-wave input depicted in Fig. 3.21, assuming that To/T = ¼, T = 1 s, and RC = 0.1 s.

1. If the input to an LTI system is expressed as a weighted sum of sinusoids (eigenfunctions), then the output is also a weighted sum of sinusoids.

2. Input:

k

tjkekXtx 0

3. Output:

k

tjkekXjkHty 00

0;0

FSy t Y k H jk X k

4. Frequency response of the RC circuit:

RCj

RCjH/1

/1

<Sol.>

0

00sin2

TkTk

kX

5. Substituting for H(jko) with RC = 0.1 s and o = 2, and To/T = ¼

kk

kjkY 2/sin

10210