eleg 3124 systems and signals ch. 4 fourier transform...find the fourier transform of x(t) cos(z 0...

Department of Electrical EngineeringUniversity of Arkansas

ELEG 3124 SYSTEMS AND SIGNALS

Ch. 4 Fourier Transform

Dr. Jingxian Wu

[email protected]

2

OUTLINE

• Introduction

• Fourier Transform

• Properties of Fourier Transform

• Applications of Fourier Transform

3

INTRODUCTION: MOTIVATION

• Motivation:

– Fourier series: periodic signals can be decomposed as the

summation of orthogonal complex exponential signals

tjnctxn

n 0exp)( +

−=

=

• each harmonic contains a unique frequency: n/T

=T

n dttjntxT

c0

0exp)(1

How about aperiodic signals ?

( )=T• time domain ➔ frequency domain

t

x(t)

Time domain Frequency domain

4

INTRODUCTION: TRANSFER FUNCTION

• System transfer function

• System with periodic inputs

tje

)(th)( He tj

+

−= dttjthH exp)()(

tjne 0

)(th)( 0

0 nHe

tjn

tjn

n

nec 0+

−= )(th)( 0

0 nHec

tjn

n

n+

−=

)(tx)(th

)( 00

nHectjn

n

n+

−=

5

OUTLINE

• Introduction




6

FOURIER TRANSFORM

• Inverse Fourier Transform


– given x(t), we can find its Fourier transform

– given , we can find the time domain signal x(t)

– signal is decomposed into the “weighted summation” of complex exponential functions. (integration is the extreme case of summation)

+

−

−= dtetxX tj )()(

+

−=

deXtx tj)(2

1)(

)(X

)(X

➔)(tx )(X

7

FOURIER TRANSFORM

• Example

– Find the Fourier transform of )/()( trecttx =

t

x(t)

t

x(t)

8

FOURIER TRANSFORM

• Example

– Find the Fourier transform of |)|exp()( tatx −= 0a

9

FOURIER TRANSFORM

• Example

– Find the Fourier transform of )()exp()( tuattx −= 0a

10

FOURIER TRANSFORM

• Example

– Find the Fourier transform of )()( attx −=

11

FOURIER TRANSFORM: TABLE

12

FOURIER TRANSFORM

+

−dttx |)(|

)()exp()( tuttx =

• Example

–

• The existence of Fourier transform

– Not all signals have Fourier transform

– If a signal have Fourier transform, it must satisfy the following two

conditions

• 1. x(t) is absolutely integrable

• 2. x(t) is well behaved

– The signal has finite number of discontinuities, minima,

and maxima within any finite interval of time.

13

OUTLINE

• Introduction




14

PROPERTIES: LINEARITY

• Linearity

– If

– then

)()( 11 Xtx )()( 22 Xtx

)()()()( 2121 bXaXtbxtax ++

• Example

– Find the Fourier transform of )(4)()2exp(3)/(2)( ttuttrecttx +−+=

15

PROPERTY: TIME-SHIFT

• Time shift

– If

– Then

)()( Xtx

]exp[)()( 00 tjXttx −−

• Time shift

– If

– Then

• Review: complex number

jbacjcecc j +=+== )sin(||)cos(||||

cos|| ca = sin|| cb =

22|| bac += )/tan( aba=

phase shift

time shift in time domain ➔ frequency shift in frequency domain

– Phase shift of a complex number c by : 0 )(exp||)exp( 00 += jcjc

16

PROPERTY: TIME SHIFT

• Example:

– Find the Fourier transform of 2)( −= trecttx

17

PROPERTY: TIME SCALING

• Time scaling

– If

– Then

• Example

– Let , find the Fourier transform of

)()( Xtx

aX

aatx

||

1)(

( ) 2/1)( −= rectX )42( +− tx

18

PROPERTY: SYMMETRY

• Symmetry

– If , and is a real-valued time signal

– Then

)()( Xtx )(tx

)()( * XX =−

19

PROPERTY: DIFFERENTIATION

• Differentiation

– If

– Then

)()( Xtx

)()(

Xjdt

tdx ( ) )(

)( Xj

dt

txd n

n

n

• Example

– Let , find the Fourier transform of ( ) 2/1)( −= rectXdt

tdx )(

20

PROPERTY: DIFFERENTIATION

• Example

– Find the Fourier transform of

(Hint: )

)sgn()( ttx =

)()sgn(2

1tt

dt

d=

21

PROPERTY: CONVOLUTION

• Convolution

– If ,

– Then

)()( Xtx )()( Hth

)()()()( HXthtx

)(tx

)(th)()( thtx )(X

)(H)()( HX

time domain frequency domain

22

PROPERTY: CONVOLUTION

• Example

– An LTI system has impulse response

If the input is

Find the output

( ) )(exp)( tuatth −=

( ) )()exp()()(exp)()( tuctactubtbatx −−+−−=

)0,0,0( cba

23

PROPERTY: MULTIPLICATION

• Multiplication

– If ,

– Then

)()( Xtx )()( Mtm

)()(2

1)()(

MXtmtx

24

PROPERTY: DUALITY

• Duality

– If

– Then

)()( Gtg

)(2)( − gtG

25

PROPERTY: DUALITY

• Example


(recall: )

=

2)(

tSath

2sinc )/(rect t

26

PROPERTY: DUALITY

• Example

– Find the Fourier transform of 1)( =tx

tjetx 0)(

=– Find the Fourier transform of

27

PROPERTY: SUMMARY

28

PROPERTY: EXAMPLES

• Examples

– 1. Find the Fourier transform of )cos()( 0ttx =

– 2. Find the Fourier transform of )()( tutx =

1)sgn(2

1)( += ttu

jt

2)sgn(

29

PROPERTY: EXAMPLES

• Examples

– 3. A LTI system with impulse response

Find the output when input is

)(exp)( tuatth −=

)()( tutx =

– 4. If , find the Fourier transform of

(Hint: )

)()( Xtx −

t

dx )(

)()()( tutxdxt

= −

30

PROPERTY: EXAMPLES

• Example

– 5. (Modulation) If ,

Find the Fourier transform of

)()( Xtx )cos()( 0ttm =

)()( tmtx

– 6. If , find x(t)

ja

X+

=1

)(

31

PROPERTY: DIFFERENTIATION IN FREQ. DOMAIN

• Differentiation in frequency domain

– If:

– Then:

)()( Xtx

n

nn

d

Xdtxjt

)()()( =−

PROPERTY: DIFFERENTIATION IN FREQ. DOMAIN

32

),()exp( tuatt − 0a

• Example


33

PROPERTY: FREQUENCY SHIFT

• Frequency shift

– If:

– Then:

)()( Xtx

)()exp()( 00 − Xtjtx

• Example

– If , find the Fourier transform ( ) 2/1)( −= rectX )2exp()( tjtx −

34

PROPERTY: PARSAVAL’S THEOREM

• Review: signal energy

+

−= dttxE 2|)(|

• Parsaval’s theorem

+

−

+

−=

dXdttx 22 |)(|

2

1|)(|

35

PROPERTY: PARSAVAL’S THEOREM

• Example:

– Find the energy of the signal )()2exp()( tuttx −=

36

PROPERTY: PERIODIC SIGNAL

• Fourier transform of periodic signal

– Periodic signal can be written as Fourier series

tjnctxn

n 0exp)( +

−=

=

– Perform Fourier transform on both sides

)(2)( 0 ncXn

n −= +

−=

37

OUTLINE

• Introduction




38

APPLICATIONS: FILTERING

• Filtering

– Filtering is the process by which the essential and useful part of a

signal is separated from undesirable components.

• Passing a signal through a filter (system).

• At the output of the filter, some undesired part of the signal

(e.g. noise) is removed.

– Based on the convolution property, we can design filter that only

allow signal within a certain frequency range to pass through.

)(tx

)(th)()( thtx )(X

)(H)()( HX

time domain frequency domain

filter filter

39


• Classifications of filters

Low pass filter

Band pass filterBand stop (Notch) filter

PassbandStop

band PassbandStop

band

High pass filter

Passband Stop

band

Stop

band

Stop

bandPassband Passband

40

APPLICATION: FILTERING

• A filtering example

– A demo of a notch filter

)(X

)(H

)()( HX

Corrupted sound Filter Filtered sound

41


• Example

– Find out the frequency response of the RC circuit

– What kind of filters it is?

RC circuit

42

APPLICATION: SAMPLING THEOREM

• Sampling theorem: time domain

– Sampling: convert the continuous-time signal to discrete-time signal.

+

−=

−=n

nTttp )()(

sampling period

)()()( tptxtxs =

)(tx

Sampled signal

43


• Sampling theorem: frequency domain

– Fourier transform of the impulse train

• impulse train is periodic

+

−=

+

−=

=−=n

tjn

sn

sse

TnTttp

11

)()(

• Find Fourier transform on both sides

+

−=

−=n

s

s

nT

P )(2

)(

• Time domain multiplication ➔ Frequency domain convolution

)()(2

1)()(

PXtptx

+

−=

−n

s

s

nXT

tptx )(1

)()(

s

sT

2=

Fourier series

44


• Sampling theorem: frequency domain

– Sampling in time domain ➔ Repetition in frequency domain

Time domain Frequency domain

45


• Sampling theorem

– If the sampling rate is twice of the bandwidth, then the original

signal can be perfectly reconstructed from the samples.

Bs 2

Bs 2

Bs 2=

Bs 2

Frequency domain

46

APPLICATION: AMPLITUDE MODULATION

• What is modulation?

– The process by which some characteristic of a carrier wave is

varied in accordance with an information-bearing signal

modulationInformation

bearing signal

Carrier wave

Modulated signal

• Three signals:

– Information bearing signal (modulating signal)

• Usually at low frequency (baseband)

• E.g. speech signal: 20Hz – 20KHz

– Carrier wave

• Usually a high frequency sinusoidal (passband)

• E.g. AM radio station (1050KHz) FM radio station

(100.1MHz), 2.4GHz, etc.

– Modulated signal: passband signal

47


• Amplitude Modulation (AM)

)2cos()()( tftmAts cc =

– A direct product between message signal and carrier signal

Mixer

Local

Oscillator

)(tm

)2cos( tfA cc

)(ts

Amplitude modulation

48


• Amplitude Modulation (AM)

)()(2

)( ccc ffMffM

AfS ++−=

Amplitude modulation

eleg 3124 systems and signals ch. 4 fourier transform...find the fourier transform of x(t) cos(z 0...

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