1 fourier representation of signals and lti systems. chapter 3 school of computer and communication...

1

Fourier Fourier Representation Representation of Signals and of Signals and LTI Systems.LTI Systems.

CHAPTER CHAPTER 33

School of Computer and Communication School of Computer and Communication Engineering, UniMAPEngineering, UniMAP

Hasliza A Rahim @ SamsuddinHasliza A Rahim @ Samsuddin

EKT 230 EKT 230

2

3.1 Introduction. 3.1 Introduction.

3.2 Complex Sinusoids and Frequency Response of LTI 3.2 Complex Sinusoids and Frequency Response of LTI Systems. Systems.

3.3 3.3 Fourier Representation of Four Classes of Signals.Fourier Representation of Four Classes of Signals.

3.4 Discrete Time Periodic Signals: Discrete Time 3.4 Discrete Time Periodic Signals: Discrete Time Fourier Series.Fourier Series.

3.53.5 Continuous-Time Periodic Signals: Fourier Series. Continuous-Time Periodic Signals: Fourier Series.

3.6 Discrete-Time Non Periodic Signals: Discrete-Time 3.6 Discrete-Time Non Periodic Signals: Discrete-Time Fourier Transform.Fourier Transform.

3.7 Continuous-Time Non Periodic Signals: Fourier 3.7 Continuous-Time Non Periodic Signals: Fourier Transform.Transform.

3.8 Properties of Fourier Representation.3.8 Properties of Fourier Representation.

3.9 Linearity and Symmetry Properties.3.9 Linearity and Symmetry Properties.

3.10 Convolution Properties.3.10 Convolution Properties.

3.0 Fourier 3.0 Fourier Representation of Signals Representation of Signals and LTI Systems.and LTI Systems.

3

Signals are represented as superposition's of complex sinusoids which leads to a useful expression for the system output and provide a characterization of signals and systems.

Example in music, the orchestra is a superposition of sounds generated by different equipment having different frequency range such as string, base, violin and ect. The same example applied the choir team.

Study of signals and systems using sinusoidal Study of signals and systems using sinusoidal representation is termed as representation is termed as Fourier AnalysisFourier Analysis introduced by Joseph Fourier (1768-1830).introduced by Joseph Fourier (1768-1830).

There are four distinct Fourier representations, each applicable to different class of signals.

3.1 Introduction.3.1 Introduction.

4

3.2 Complex Sinusoidal 3.2 Complex Sinusoidal and Frequency Response and Frequency Response of LTI System.of LTI System. The response of an LTI system to a sinusoidal

input that lead to the characterization of the system behavior that is called frequency response of the system.

An impulse response h[n] and the complex sinusoidal input x[n]=ejn.

5

kj

k

j ekheH

njj

kj

k

nj

nj

knj

k

k

eeH

ekheny

outefactor

ekhny

knxkhny

Discrete-Time (DT)Discrete-Time (DT)

Derivation:

Frequency Response:

Cont’d…Cont’d…

6

dehjH tj

tj

jtj

tj

ejH

dehe

dehty

Continuous-Time (CT)Continuous-Time (CT)Derivation:

Frequency Response:

Cont’d…Cont’d…

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.

7

Example 3.3Example 3.3: : Frequency Response.Frequency Response.The impulse response of a system is given asThe impulse response of a system is given as

Find the expression for the Find the expression for the frequency responsefrequency response, and , and plot the magnitude and phase response.plot the magnitude and phase response.Solution:Solution:Step 1Step 1:: Find frequency response, Find frequency response, HH((jj).). Substitute h(t) into equation below,

tueRC

th RC

t

1

RCj

RC

RCj

RC

e

RCj

RC

deRC

deueRC

jH

RCj

RCj

jRC

1

1

101

11

1

11

1

1

0

1

0

1

dehjH tj

8

2

2 1

1

||

RC

RCjH

Step 2Step 2:: From frequency response, get the From frequency response, get the magnitude & phase response.magnitude & phase response.The magnitude response is,

Cont’d…Cont’d…

9

Figure 3.1: Frequency response of the Figure 3.1: Frequency response of the RCRC circuit (a) circuit (a) Magnitude response.Magnitude response.

(b) Phase response.(b) Phase response.

The phase response is arg{H(j)}= -arctan(RC)

.

Cont’d…Cont’d…

10

3.3 Fourier 3.3 Fourier Representation of Four Representation of Four Class of Signals.Class of Signals. There are four distinct Fourier representationsfour distinct Fourier representations, class of

signals;(i) Periodic Signals.(i) Periodic Signals.(ii) Nonperiodic Signals.(ii) Nonperiodic Signals.(iii) Discrete-Time Periodic Signals.(iii) Discrete-Time Periodic Signals.(iv) Continuous-Time Periodic Signals.(iv) Continuous-Time Periodic Signals.

Fourier Series (FS) applies to continuous-time periodic signals. Discrete Time Fourier Series (DTFS) applies to discrete-time

periodic signals. Nonperiodic signals have Fourier Transform (FT)

representation. Discrete Time Fourier Transform (DTFT) applies to a signal

that is discrete in time and non-periodic.

11

Time Property Periodic Nonperiodic

Continuous (t)

Fourier Series (FS)

[Chapter 3.5]

Fourier Transform (FT) [Chapter 3.7]

Discrete (n)

Discrete Time Fourier Series

(DTFS)[Chapter 3.4]

Discrete Time Fourier Transform

(DTFT)[Chapter 3.6]

Table 3.1: Relation between Time Properties of a Signal an the Table 3.1: Relation between Time Properties of a Signal an the Appropriate Fourier Representation.Appropriate Fourier Representation.

Cont’d…Cont’d…

12

Cont’d…Cont’d…

(1) Discrete (1) Discrete Time Time

Periodic Periodic SignalSignal

(1) Discrete (1) Discrete Time Time

Periodic Periodic SignalSignal

(2) (2) Continuous Continuous

Time Time Periodic Periodic SignalSignal

(2) (2) Continuous Continuous

Time Time Periodic Periodic SignalSignal

(3) (3) Continuous Continuous Time Non Time Non Periodic Periodic SignalSignal

(3) (3) Continuous Continuous Time Non Time Non Periodic Periodic SignalSignal

(4) Discrete (4) Discrete Time Non Time Non Periodic Periodic SignalSignal

(4) Discrete (4) Discrete Time Non Time Non Periodic Periodic SignalSignal

Fourier Fourier RepresentRepresent

ationation

Fourier Fourier RepresentRepresent

ationation

13

3.3.1 3.3.1 Periodic SignalsPeriodic Signals: : Fourier Series Fourier Series RepresentationRepresentation If x[n] is a discrete-time signaldiscrete-time signal with fundamental

period NN then x[n] of DTFS is represents as,

where oo= 2= 2/N/N is the fundamental frequency of x[n].

The frequency of the kth sinusoid in the superposition is ko

njk

k

oekAnx ˆ

14

If x(t) is a continuous-time signalcontinuous-time signal with fundamental period T, x(t) of FS is represents as

where = 2= 2/T/T is the fundamental frequency of x(t). The frequency of the kth sinusoid is k and each sinusoid has a common period T.

A sinusoid whose frequency is an integer multiple of a fundamental frequency is said to be a harmonic of a sinusoid at the fundamental frequency. For example ejk

t is the kth harmonic of

ejt.

The variable k indexes the frequency of the sinusoids, A[k] is the function of frequency.

tjk

k

oekAtx ˆ

Cont’d…Cont’d…

15

3.3.2 3.3.2 Nonperiodic SignalsNonperiodic Signals: :

Fourier-Transform Fourier-Transform Representation.Representation. The Fourier Transform representations employ

complex sinusoids having a continuum of frequencies.

The signal is represented as a weighted integral of complex sinusoids where the variable of integration is the sinusoid’s frequency.

Continuous-time sinusoidsContinuous-time sinusoids are used to represent continuous signal in FT.

where X(j)/(2) is the “weight” or coefficient applied to a sinusoid of frequency in the FT representation

dejXtx tj

2

1ˆ

16

Discrete-time sinusoidsDiscrete-time sinusoids are used to represent discrete time signal in DTFT.

It is unique only a 2 interval of frequency. The sinusoidal frequencies are within 2 interval.

deeXnx njj

2

1ˆ

Cont’d…Cont’d…

17

3.4 Discrete Time Periodic 3.4 Discrete Time Periodic Signals: Signals: Discrete Time Discrete Time Fourier Series.Fourier Series. The Discrete Time Fourier Series representation;

where x[n] is a periodic signal with period NN and fundamental frequency =2=2/N/N. XX[[kk]] is the DTFS coefficient of signal x[n].

The relationship of the above equation,

njkN

k

oekXnx

1

0

njkN

n

oenxN

kX

1

0

1

kXnx oDTFS ;

18

Given N value of x[n] we can find X[k]. Given N value of X[k] we can find x[n] vise-versa. The X[k] is the frequency-domain

representation of x[n]. DTFS is the only Fourier representation that can

be numerically evaluated using computer, e.g. used in numerical signal analysis.

Cont’d…Cont’d…

19

Example 3.1:Example 3.1: Determining DTFS Determining DTFS Coefficients.Coefficients.Find the Find the frequency-domainfrequency-domain representation of the representation of the signal in Figure 3.2 below.signal in Figure 3.2 below.

Figure 3.2: Time Domain Signal.Figure 3.2: Time Domain Signal.

Solution:Solution:Step 1Step 1: Determine N and : Determine N and ..

The signal has period N=5, so =2/5.

Also the signal has odd symmetry, so we sum over n = -2 to n = 2 from equation

20

Step 2Step 2: Solve for the frequency-domain, : Solve for the frequency-domain, XX[[kk].].From step 1, we found the fundamental frequency, N

=5, and we sum over n = -2 to n = 2 .

5/45/205/25/4

5/22

2

1

0

210125

1

5

1

1

jkjkjjkjk

njk

n

njkN

n

exexexexex

enxkX

enxN

kX o

Cont’d…Cont’d…

21

From the value of x{n} we get,

Step 3Step 3:: Plot the magnitude and phase of DTFS. Plot the magnitude and phase of DTFS.From the equation, one period of the DTFS

coefficient X[k], k=-2 to k=2, in the rectangular and polar coordinate as

5/2sin15

1

2

1

2

11

5

1 5/25/2

kj

eekX jkjk

760.0

531.0

276.05

5/2sin

5

11

232.05

5/4sin

5

12

j

j

ejX

ejX

Cont’d…Cont’d…

22

The above figure shows the magnitude and phase of X[k] as a function of frequency index k.

Figure 3.3: Magnitude and phase of the DTFS coefficients for Figure 3.3: Magnitude and phase of the DTFS coefficients for the signal in Fig. 3.2.the signal in Fig. 3.2.

531.0

760.0

0

232.05

5/4sin

5

12

276.05

5/2sin

5

11

2.05

10

j

j

j

ejX

ejX

eX

Cont’d…Cont’d…

23

Just to Compare, for different range of Just to Compare, for different range of nn..Calculate X[k] using n=0 to n=4 for the limit of the sum.

This expression is different from which we obtain from using n=-2 to n=2. Note that,

n=-2 to n=2 and n= 0 to n=4, yield equivalent expression for the DTFS coefficient.

.

5/85/2

5/85/65/45/20

2

1

2

11

5

1

432105

1

jkjk

jkjkjkjj

eekX

eeeeexkX

5/2

5/225/8

jk

jkjkjk

e

eee

Cont’d…Cont’d…

24

3.5 Continuous-Time 3.5 Continuous-Time Periodic Signals: Periodic Signals: Fourier Fourier Series.Series.Continues time periodic signals are represented by

Fourier series (FS). The fundamental period is T and fundamental frequency o =2/T.

X[k] is the FS coefficient of signal x(t).

Below is the relationship of the above equation,

k

tjk oekXtx )(

dtetxT

kX tjkT

o0

1)(

)(; kXtx oFS

25

The FS coefficient X(k) is known as frequency-domain representation of x(t).

Cont’d…Cont’d…

26

Example 3.2:Example 3.2: Direct Calculation of FS Direct Calculation of FS Coefficients.Coefficients.

Determine the FS coefficients for the signal Determine the FS coefficients for the signal xx((tt) ) depicted in Figure 3.4.depicted in Figure 3.4.

Solution:Solution: Figure 3.4: Time Domain Signal.Figure 3.4: Time Domain Signal.

Step 1Step 1: Determine T and : Determine T and 00..

The period of x(t) is T=2, so 0=2/2 = On the interval 0<=t<=2, one period of x(t) is expressed as x(t)=e-2t, so it yields

Step 2Step 2: Solve for : Solve for XX[[kk].]. dtekX

dteekX

tjk

tjkt

2

0

2

2

0

2

2

1)(

2

1)(

dtetxT

kX ojkT

0

1)(

27

Step 3Step 3: Plot the magnitude and phase spectrum.: Plot the magnitude and phase spectrum.

- From the above equation for example k=0, X(k)= (1-e-4)/4 = 0.245.

- Since e-jk2=1 Figure 3.5 shows the magnitude spectrum|X(k)| and the phase spectrum arg{X(k)}.

24

1)(

124

1)(

)2(2

1)(

4

24

2

0

2

jk

ekX

eejk

kX

ejk

kX

jk

tjk

Cont’d…Cont’d…

28

Figure 3.5: Magnitude and Phase Spectra.Figure 3.5: Magnitude and Phase Spectra..

Cont’d…Cont’d…

29

3.6 Discrete-Time Non 3.6 Discrete-Time Non Periodic Signals: Periodic Signals: Discrete-Discrete-Time Fourier Transform.Time Fourier Transform. The DTFT representation of time domain signal,

X[k] is the DTFT of the signal x[n].

Below is the relationship of the above equation,

deeXnx njj

21

njj enxeX

jDTFT eXnx

30

The FT, X[j], is known as frequency-domain representation of the signal x(t).

The above equation is the inverse FT, where it map the frequency domain representation X[j] back to time domain.

deeXnx njj

21

Cont’d…Cont’d…

31

3.7 Continuous-Time 3.7 Continuous-Time Nonperiodic Signals: Nonperiodic Signals: Fourier Transform.Fourier Transform. The Fourier transform (FT) is used to represent

a continuous time nonperiodic signal as a superposition of complex sinusoids.

FT representation,

where,

Below is the relationship of the above equation,

dejXtx tj

2

1

tjetxjX

jXtx FT

32

Example 3.4:Example 3.4: FT of a Real Decaying FT of a Real Decaying Exponential.Exponential.

Find the Fourier Transform (FT) of Find the Fourier Transform (FT) of xx((tt) =e) =e-at-at uu((tt).).

Solution:Solution:The FT does not converge for a<=0, since x(t) is not

absolutely integrable, that is

ja

eja

dte

dtetuejX

haveweafor

adte

tja

tja

tjat

at

1

1

)(

,0

0,

0

)(

0

)(

0

0

33

Converting to polar form, we find that the magnitude and phase of X(jw) are respectively given by

This is shown in Figure 3.6 (b) and (c).

ajX

and

ajX

arctanarg

1

2

122

Cont’d…Cont’d…

34Figure 3.6: (a) Real time-domain exponential signal. (b) Figure 3.6: (a) Real time-domain exponential signal. (b)

Magnitude spectrum. Magnitude spectrum. (c) Phase spectrum.(c) Phase spectrum.

.

Cont’d…Cont’d…

35

3.8 Properties of Fourier 3.8 Properties of Fourier Representations.Representations. The four Fourier representationsfour Fourier representations discussed in

this chapter are summarized in Table 3.2. Attached Appendix C is the comprehensive table

of all properties.

Table 3.2 : Fourier Representation.

36

Table 3.2: The Four Fourier Representations.

t

periodhastx

dtetxT

kX

ekXtx

SeriesFourier

tjkT

tjk

k

2

1

0

0

0

0

2

2

1

periodhaseX

enxeX

deeXnx

TransformFourierTimeDiscrete

j

njj

njj

dtetxjX

dwejXtx

TransformFourier

tj

tj

2

1

Time Domai

n

Periodic(t,n)

Non periodic(t,n)

CONTINUOUS

(t)

NONPERIODIC

(k,w)

DISCRETE

(t)

PERIODIC

(k,)

Discrete (k) Continuous () Freq. Doma

in

2

2

1

periodhaseX

enxeX

deeXnx

TransformFourierTimeDiscrete

j

njj

njj

dtetxjX

dwejXtx

TransformFourier

tj

tj

2

1

n

NperiodhavekXandnx

enxN

kX

ekXnx

SeriesFourierTimeDiscrete

njkN

n

njkN

k

2

1

0

1

0

1

0

0

0

Cont’d…Cont’d…

37

3.9 Linearity and 3.9 Linearity and Symmetry Properties.Symmetry Properties. All four Fourier representations involve linear

operations.

The linearity property is used to find Fourier representations of signals that are constructed as sums of signals whose representation are already known.

kbYkaXkZnbynaxnz

ebYeaXeZnbynaxnz

kbYkaXkZtbytaxtz

jbYjaXjZtbytaxtz

o

o

DTFS

jjjDTFT

FS

FT

;

;

38

Example 3.5:Example 3.5: Linearity in the Fourier Linearity in the Fourier Series.Series.Given Given zz((tt) which is the periodic signal. Use the ) which is the periodic signal. Use the linearity property to determine the FS coefficients linearity property to determine the FS coefficients ZZ[[kk]]..

Solution:Solution:From Example 3.31(Text) we have

The linearity property implies that

.

tytxtz2

1

2

3

22

1

42

32;

kSin

k

kSin

kkZtz FS

2

1

4

1

2;

2;

kSin

kkYty

kSin

kkXtx

FS

FS

39

3.9.1 Symmetry 3.9.1 Symmetry Properties: Properties: Real and Imaginary Real and Imaginary Signal.Signal.

Table 3.3: Symmetry Properties for Fourier Representation ofTable 3.3: Symmetry Properties for Fourier Representation of

Real-and Imaginary–Valued Time Signals.Real-and Imaginary–Valued Time Signals.

Representation

Real-Valued Time Signal

Imaginary-Valued Time

Signal

FT X*(j) = X(-j) X*(j) = -X(-j)

FS X*[k] = X[-k] X*[k] = -X[-k]

DTFT X*(ej) = X(e-j) X*(ej) = -X(e-j)

DTFS X*[k] = X[-k] X*[k] = -X[-k]

40

Change of variable = -t.

The only way that the condition X*(j) =X(j) holds is for the imaginary part of X(j) to be zero.

3.9.2 Symmetry 3.9.2 Symmetry Properties: Properties: Even and Odd Signal.Even and Odd Signal.

jX

dexjX j

*

dtetxjX tj )(*

41

If time signal is real and even, then the frequency-domain representation of it also real.

If time signal is real and odd, then the frequency-domain representation is imaginary.

Cont’d…Cont’d…

42

The convolution of the signals in the time domain transforms to multiplication of their respective Fourier representations in the frequency domain.

The convolution property is a consequences of complex sinusoids being eigen functions of LTI system.

3.10.1 Convolution of Non periodic Signals.3.10.1 Convolution of Non periodic Signals.

3.10.2 Filtering.3.10.2 Filtering.

3.10.3 Convolution of Periodic Signals.3.10.3 Convolution of Periodic Signals.

3.10 Convolution 3.10 Convolution Property. Property.

43

3.10.1 Convolution of Non 3.10.1 Convolution of Non Periodic Signals.Periodic Signals. Convolution of two non periodic continuous-time

signals x(t) and h(t) is defines as

express x(t-) in term of FT:

.

*

dtxh

txthty

dejXtx tj

2

1

44

Substitute into convolution integral yields,

dejXdeh

ddeejXhty

tjj

jtj

2

1

2

1

dejXjHty tj

2

1

Cont’d…Cont’d…

45

Convolution of h(t) and x(t) in the time domain corresponds to multiplicationmultiplication of Fourier transforms, H(j) and X(j) in the frequency frequency domaindomain that is;

Discrete-timeDiscrete-time nonperiodic signals, if

.* jHjXjYtxthtyFT

.*

jjjDTFT

jDTFT

jDTFT

eHeXeYnxnhny

theneHnh

andeXnx

Cont’d…Cont’d…

46

Example 3.6:Example 3.6: Convolution in Frequency Convolution in Frequency Domain.Domain.Let Let xx((tt)= (1/()= (1/(tt))sin())sin(tt) be the input of the system with ) be the input of the system with impulse response impulse response hh((tt)= (1/()= (1/(tt))sin(2))sin(2tt). Find the output ). Find the output y(t)=y(t)=xx((tt)*)*hh((tt))

Solution:Solution: From Example 3.26 (text, Simon) we have,

Since

We conclude that.

,0

,1

*

jY

thatfollowsit

jHjXjYthtxty FT

2

2

,0

,1

,0

,1

jHth

jXtx

FT

FT

.sin1

tt

ty