signal and system1 dept. information and communication eng. 1 signal and system ( chapter 7:...

1 Dept. Information and Communication Eng. 1

Signal and System( Chapter 7: Spectral Analysis I )

for Continuous-Time Signals

Prof. Kwang-Chun [email protected]: 02-760-4253 Fax:02-760-4435


Fourier Analysis


We desire a measure of the frequencies presented in a wave. This will lead to a definition of the term, the spectrum

Plane waves have only one frequency, .

This light wave has many frequencies. And the frequency increases in time (from red to blue).

It will be nice if our measure also tells us when each frequency occurs !

What do we hope to achieve with the Fourier analysis?

Ligh

t el

ectr

ic f

ield

Time


A complete sine wave in the time domain can be represented by one single spike in the frequency domain.

The frequency domain is more compact and useful when we are dealing with more than one sine wave. For example, figure shows three sine waves, each with different

amplitude and frequency. All can be represented by three spikes in the frequency domain.




( Time vs. Frequency Relationship of signal )

Is Frequency Domain Better ?

In time domain, all frequency components of signals are summed together

In frequency domain, complex signals are separated into their frequency components


Fourier Family


Fourier Series

Fourier series formula: In 1807, assert that a periodic signal f(t) can be

written as an infinite sum of mutual orthogonal signals

That is,

where the mutual orthogonal signals satisfy the orthogonal condition

( ) ( )m

m mm

f t c t


One of the most popular of orthogonal signals is exponential (or trigonometric) signal like

Thus, we have

where the fundamental frequency is

, which is called exponential Fourier series

0

for ( ) ( )

0 for

T

m n

T m nt t dt

m n

Similar to the orthogonality between unit vectors

1 for 0 for x y

x yx y

ì =ïï=íï ¹ïîa a

ojm te

( ) om

jm tm

mf t c e

2

o T

Fourier Series


Now, let’s determine the coefficient cm of Fourier seriesMultiplying into both sides and integrating over

one period, it yields

Trigonometric form of Fourier series:Rearranging the complex (exponential) form, it is

ojn te

( )

0 0

( ) o oT T

jn t j m n tm n

m

f t e dt c e dt c T

From orthogonal condition, it is T

0

1 ( ) oT

jn tnc f t e dtT

0

1 ( ) oT

jm tmc f t e dtT

Setting n to m Note: m mc c

Fourier Series


Since cm is complex value, that is, , f (t) becomes

Thus,

where

01

( ) o ojm t jm tm mm

f t C c e c e

mj

m mc c e

( ) ( )0

1

01

( )

2 cos cos 2 sin sin

o m o mj m t j m tm m

m

m m o m m om

f t C c e c e

C c m t c m t

=am =bm=a0

01

( ) cos sinm o m om

f t a a m t b m t

0 0

2 2( )cos( ) , ( )sin( )T T

m o m oa f t m t dt b f t m t dtT T

Fourier Series


Example 7.1:Find Fourier series for the periodic waveform

Solution:Evaluate the coefficient integration

0.5 2 2

0.523

1 1 1 sin3 3 3

o

o

o

o

m mj jjm t

mo

e e mc e dtjm m

MatLab script:cm=int('(1/3)*exp(-i*m*(2*pi/3)*t)',-0.5,0.5);simple(cm)

Fourier Series


Thus, the Fourier series is

1( ) sin3

o ojm t jm tm

m m

mg t c e em

t-3 -2 -1 0 1 2 3

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

( )pg t

5m 20m

(Partial sums of Fourier series for m)

Fourier Series


MatLab script of Fourier series:clear;t = -3:0.01:3; omega_0 = (2*pi)/3;gp_L = 0; gp_H = 0;for m = -5:5

x = m/3;gp_L = gp_L + (1/3)*sinc(x)*exp(i*m*omega_0*t);

end;for m = -20:20

x = m/3;gp_H = gp_H + (1/3)*sinc(x)*exp(i*m*omega_0*t);

end;plot(t,real(gp_L),t,real(gp_H)); grid;xlabel('t'); ylabel('g_p(t)');title('Fourier series approximation to g(t)');legend('m=5','m=20')

Fourier Series


Then, the spectrum of is a plot of its vs. ( )g t mc om

(Amplitude spectrumof trigonometric form)

Since 0,2 cos 2

2 sin 0

m

m m m m

m m m

a c c

b c

For EvenSignals

(Amplitude spectrum of exponential form)

1 sin3

Fourier Series


Key Points:Determine a signal’s spectral power distribution

Find the percentage of total power in each frequency component

Define the average power of function as

Expressing in the form of Fourier series gives

( )f t

2

0

1 ( )T

avP f t dtT

i(t)

R=1

Instantaneous Power:2 2( ) ( )P i t R i t

Average Power: 20

1 ( )T

avP i t dtT

Parseval’s Theorem


That the signal power can be determined in either the time or the frequency domain is called Parseval’s theorem

0 0

2

1 1( ) ( )o oT T

jm t jm tav m m

m m m

P f t c e dt c f t e dtT T

c c c

22

0

1 ( )T

av mP f t dt cT

Time domain Frequency domain



Thus, the power spectrum of in previous example is

Find the percentage of total power to 4th-harmonic power

Solution:

( )g t



The average power is

And the power of 4th-harmonic is

Thus, the percentage is

Note:If P0=80% and P1=19%, then the signal can be

expressed approximately as

0.5 3

0 2.5

1 11 13 3av

P dt dt

41 4sin 0.07

4 3c p

p= - 24 42P c

4100 3%av

PP

( )1 1( ) sin3 3

o oj t j tg t e ew wpp

-+ +



A method to express the spectrum of aperiodic (non-periodic) signals

Suppose that a signal is aperiodic signalThen, we can construct a new periodic signal ,

dependent on . That is,

( )f t

( )f t( )Tf t

lim ( ) ( )TT f t f t

Fourier Transform

( )Tf t

( )f t

T T

/ 2T/ 2T


The exponential Fourier series is given by

Since integrating over (-T/2,T/2) is the same as integrating over , we have

Now, let’s define asThen, it becomes

/ 2

/ 2

1( ) , ( )o oT

jm t jm tT m m T

m T

f t c e c f t e dtT

2

o T

( )Tf t( )f t ,

1 ( ) ojm tmc f t e dtT

where

0( )F m 00( ) ( )jm tF m f t e dt

00 00 0

lim ( ) lim ( ) lim ( )T

F m F m F m

Because is infinitesimal

( ) ( ) j tF f t e dt

0

(Fourier Transform)

Fourier Transform


Similarly, since the Fourier coefficient is written as

, the periodic signal then becomes

Consequently, we have

Example 7.2:Determine the Fourier transform of

0( )m

F mcT

0 0 0( ) ( )( )2

o ojm t jm tT

m m

F m F mf t e eT

0 0 0

( )lim ( ) lim ( ) lim2

jm tT TT m

F mf t f t e

1( ) ( )

2j tf t F e d

(Inverse Fourier Transform)

0, 0( )

, 0tt

f te t

Fourier Transform


(Spectra resulting from 0.5-sec rectangular pulse)Fourier series are seen to be samples of Fourier transform taken at frequencies m/T

Fourier Transform


Solution:

(1 ) (1 )

00 0

1 1( )1 1

t j t j t j tF e e dt e dt ej j

2

( )1

1

F

1

( )

tan1

F

-4 -3 -2 -1 0 1 2 3 40

0.2

0.4

0.6

0.8

1

-4 -3 -2 -1 0 1 2 3 4-2

-1

0

1

2

Fourier Transform


Properties of Fourier transformLinearity:

Duality:If is a Fourier transform pair, then

Time and Frequency Scaling:For any positive number a,

Convolution in Time-Domain:If f(t) and h(t) are Fourier transformable, then

1 1 2 2 1 1 2 2( ) ( ) ( ) ( )a f t a f t a F a F

F( ) ( )f t F

( ) 2 ( )F t f

1( )f at Fa a

( ) ( ) ( )h t f d H F

F

F

F

F

Fourier Transform


Differentiation:If f(t) has an n-th derivative, then

Time and Frequency Shift:

Modulation:If a signal f(t) is multiplied by a cosine or a sine, we

have

( ) nnf t j F

0 00 0( ) , ( )j t j tf t t F e f t e F

0 0 0

0 0 0

1( ) cos( ) ,2

( )sin( )2

f t t F F

jf t t F F

F

F F

F

F

Fourier Transform


Example 7.3:Determine the Fourier transform of

Solution:Using duality,

Applying modulation property, we obtain

Similarly,

0 0( ) cos( ) or sin( )f t t t

1 2 ( )

0 0 0

0 0

11 cos( ) 22

t

0 0 01 sin( )t j

F

F

F

Fourier Transform


Key Points: Improve partial sums of the Fourier series

Identify several window functions

Recall the partial sums of rectangular pulseOne main purpose of Fourier series is to provide an

equation to represent a particular shape

To make it practical to use, the series is limited to a finite number of terms (partial sums)

When a partial sum is used, the waveform shows rippling and Gibb’s phenomenon

Windows


t-3 -2 -1 0 1 2 3

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

( )pg t

5m

20m

Gibb’s Phenomenon

How to disappear?

(As the partial sum becomes the infinite sum)

Windows


Then, the waveform can be improved by a specific windowing process

Windowing process:A process of multiplying an infinite length sequence by a window function

Windows


-0.2

0

0.2

0.4

0.6

0.8

1

1.2

-3 -2 -1 0 1 2 3

Rectangular window ( 20 20)m

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

-3 -2 -1 0 1 2 3

Hamming window

( Partial Fourier series of a rectangular pulse is demonstrated with rectangular and hamming windows )

Windows


Rectangular window:They all multiply the d-c term by unity

The d-c term only set the level of waveform, but does not affect its shape

Other windows like triangular, hanning and hamming windows (see Table in the next slide):Reduce the series coefficients of each successively higher

harmonic to zero by decreasing percentages

The spectra of significant harmonics are so weighted that the partial sums of Fourier series enhance

Windows


Example 7.4:Fourier series of a discontinuous function is

The series is to be approximated by a five-term partial sums

3 2 2 3( ) 0.6 2 2 2 0.6j t j t jt jt j t j tf t e e e e e e 2c 1c 0c 1c 2c

( Common Window Functions )

Windows


Modify the partial sums with a triangular window to reduce rippling

Solution:Since the triangular window gives for M=2 (five terms)

the triangular windowed five-term partial sum is

1 2 2 1( ) 1 13 3 3 3 3m

w m

2w 1w 0w 1w 2w

2 21 2 2 1( ) 2 2 1 23 3 3 3

j t jt jt j twf t e e e e

2 2w c 1 1w c 0 0w c 1 1w c 2 2w c

Windows


0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

-10 -8 -6 -4 -2 0 2 4 6 8 10

f(t)

fw(t)

( Partial Fourier series of a discontinuous function demonstrated with and without windows )

Windows


Transfer function 1/(s+2) and input e -t u(t)

Either way takes about the same amount of work

By Fourier Transform By Laplace Transform1 1Transform the input

1 11 1Transform

2 2

1 1 1 1Calculate output1 2 1 2

1 11 11 21 2

Invers

s j

F F sj s

h(t) H H s H sj s

Y H F Y s H s F s

j j s s

s sj j

2 2e Transform t t t ty t e e u t y t e e u t

Comparison between Transforms


By Fourier Transform By Laplace Transform1 1Transform the input

1 1Transform 2 2

1 12

1 1Calculate output 12 2

1 1 1 12 2 2 2

s j

F F sj s

h(t) H H s H sj s

Y H F

Y s H s F sj j

s sj j j

j j

2 2

21 1 12 2

1 1Inverse Transform 1 1 2 2

t t

s s

y t e u t y t e u t

Transfer function 1/(s+2) and input u(t)

Laplace is Easier !



Frequency response for periodic signals


00 0 0

0 0 0cos( )t F

0 2 20

cos ( ) st u ts

L

0

1( ) sin3

ojm t

m

mg t em

(Fourier series)

Fourier is more Clear !

0.5 0.5 0.5

3 30.5

1( )1 1

s sst

s s

e eg t e dte s e


Problem 7.1: Compute the Fourier series for the odd function

Problem 7.2: Consider the computer clock signal with a pulse rate of 8

million pulses per second ( ) and amplitude of 4 V and a pulse width of Find the Fourier series in trigonometric form

Homework Assignments

( ) ,2 2

f x Ax x

8 MHzcf0.05 sec


Problem 7.3: Compute the Fourier series for the function

Hint: Expand in terms of exponentials

Problem 7.4: Compute the Fourier transform of the triangular pulse Using MatLab, plot the spectrum

5( ) siny t t



Problem 7.5: Compute the Fourier transform of the double gate function Using MatLab, plot the spectrum at and

Problem 7.6: Plot the Fourier transform for positive frequencies for the

pulse of width centered at the originOn the same plot, compare the transforms for

2 secT 5 secD

4,8,16 sec

On the same plot, compare the transforms for


signal and system1 dept. information and communication eng. 1 signal and system ( chapter 7:...

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