ELEG3503 Introduction to Digital Signal Processing

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1 ELEG3503 Itroductio to Digital Sigal Processig 1 Itroductio 2 Basics of Sigals ad Systems 3 Fourier aalysis 4 Samplig 5 Liear time-ivariat (LTI) systems 6 z-trasform 7 System Aalysis 8 System Realizatio 9 Filter Desig

2 5 Liear time-ivariat (LTI) systems (2) 5.1 Itroductio 5.2 Impulse Respose 5.3 Covolutio 5.4 Eigefuctio of a DT aperiodic LTI system 5.5 Four types of Fourier Trasforms & eigefuctios 5.6 Examples: Low-pass & High-pass Filterig

3 5.1 Itroductio (3) Cosider a DT Liear system ad a iput x[]. x[] ca be expressed as a liear combiatio of shifted samples. For example, x[] 1 [] + 2 [-1] + 1 [-2]. x[0] x[1] x[2] I geeral, x[].+ x[-1][+1] + x[0][] + x[1][-1]+ x[2][-2] + k x[k] [-k] 3

4 ote: [] 1 for 0 0 otherwise 5.2 Impulse Respose (4) [] h[] 5.1 Itroductio 5.2 Impulse Respose 5.3 Covolutio 5.4 Eigefuctio types of FT & Eigefuctios 5.6 Examples Whe x[] [], the the output is called impulse respose which is ofte represeted by h[]. For example, h[] 5 [] + 1 [-1]. h[0] h[1] [] DT System h[] If we kow the impulse respose of a system, the we also kow the output y[] for ay iput x[], its frequecy respose, trasiet respose, steady-state respose, if it is stable or ot stable, causal or o-causal. 4

5 The output y[] is equal to the covolutio sum of the impulse respose h[] ad the iput x[], i.e. k 5.3 Covolutio (5) y[] x[] h[] x [ k] h[ k] h[] x[] h [ k] x[ k] k What is covolutio? We shall explai usig a example. Cosider a DT system of impulse respose h[] which has a iput x[] x[0] [] + x[1] [-1] + x[2] [-2] 5.1 Itroductio 5.2 Impulse Respose 5.3 Covolutio 5.4 Eigefuctio types of FT & Eigefuctios 5.6 Examples h[] y[]? 5

6 cotiued. Cosider a DT Liear system of impulse respose h[] ad iput x[] x[0] [] + x[1] [-1] + x[2] [-2] [] h[] h[0] [] h[1][-1] x[] The DT system is liear so that y[] y 1 [] + y 2 [] + y 3 [] y[] h[] x[0] [] y 1 [] x[0]h[0] [] + + x[1] [-1] x[2] [-2] The DT system is time-ivariat so that y 2 [] x[1]h[0] [-1] ad + x[1]h[1] [-2] y 3 [] x[2]h[0] [-2] + x[2]h[1] [-3] + y 2 [] + y 3 [] x[0]h[1] [-1] x[1]h[0] [-1] x[1]h[1] [-2] x[2]h[1] [-3] x[2]h[0] [-2] 6

7 cotiued. Cosider a DT Liear system of impulse respose h[] ad iput x[] x[0] [] + x[1] [-1] + x[2] [-2] x[] 0 x[0] 1 x[1] 2 x[2] 3 0 [] 4 0 h[] h[0] [] h[1][-1] explaatio x[] has 3 o-zero elemets. h[] h[] has 2 o-zero elemets h[0] ad h[1]. h[0] x[0] h[1] x[0] x[0] geerates x[0]h[0] at 0 & x[0]h[1] at 1 h[0] x[1] h[1]x[1] x[1] geerates x[1]h[0] at 1 & x[1]h[1] at 2 k0 k1 h[0] x[2] h[1] x[2] y[0] y[1] y[2] y[3] h[0] x[0-0] + h[1] x[0-1] h[0] x[1-0] + h[1] x[1-1] h[0] x[2-0] + h[1] x[2-1] h[0] x[3-0] + h[1] x[3-1] x[2] geerates x[2]h[0] at 2 & x[2]h[1] at 3 y[] x[] h[] k h [ k] x[ k] k0 k1 k2 y[0] h[0-0] x[0] + h[0-1] x[1] + h[0-2] x[2] y[1] h[1-0] x[0] + h[1-1] x[1] + h[1-2] x[2] y[2] h[2-0] x[0] + h[2-1] x[1] + h[2-2] x[2] y[3] h[3-0] x[0] + h[3-1] x[1] + h[3-2] x[2] k x [ k] h[ k] 7

8 If x[ ] F ( DT Covolutio ) (8) j j X ( e ); h[ ] F H( e ); the x[] h[] covolutio i time domai k x [ k] h[ k] F X(e j ) H(e j ) multiplicatio i frequecy domai h[] x[] h[] impulse respose F X(e j ) x frequecy respose F H(e j ) y[] F Y(e j )

9 5.4 Eigefuctio of a DT aperiodic LTI system (9) If the iput to a DT aperiodic LTI system is a complex expoetial iput e j, the the output is always the same complex expoetial with oly a chage i amplitude ad phase equal to H(e j ), i.e. e j h[] or H(e j ) H(e j ) e j where H(e j ) is the frequecy respose of the system. Also, H(e j ) is equal to the DTFT of its impulse respose h[]. A sigal for which the system output is just a costat times this iput sigal is called a eigefuctio of the system (e j i this case). The costat is called the eigevalue (frequecy respose H(e j ) i this case). The complex expoetial e j is a eigefuctio of ay DT aperiodic LTI system.

10 Prove: e j is a eigefuctio of ay DT aperiodic LTI system Pf: Cosider a DT LTI system with impulse respose h[] ad frequecy respose H(e j ). The iput is e j. We eed to prove that the output y[] is always H(e j ) e j for ay H(e j ). x[] e j h[] or H(e j ) y[] H(e j ) e j y[] k h [ k] x[ k] h[ k] e e k j k h[ k] e j( k ) j k ote: The DTFT of x[] X j e x e j e H ( e j j ) Therefore, e j is a eigefuctio of ay DT aperiodic LTI system. 10

11 What is e j? e j cos + j si ad so e j + e -j 2 cos e j H(e j ) H(e j ) e j + + H(e -j ) e -j e -j 2 cos 2 R cos[ + ] where H(e j ) R e j H(e j ) R e -j & Ay sigal x[] ca be recostructed from the cotiuous sum of all the complex expoetial fuctios e j weighted by its spectrum X(e j ) x 1 j X e 2 e j d X(e j ) e j + X(e -j ) e -j H(e j ) H(e j ) X(e j ) e j + H(e -j ) X(e -j ) e -j where X(e j ) Pe j & X(e j ) Pe -j 2 P cos [ + ] 2 R P cos[ + + ] 11

12 5.5 Four types of Fourier Trasforms & eigefuctios (12) 4 type of sigals aperiodic periodic cotiuous-time x(t) x p (t) of period T o discrete-time x[] x p [] of period N The 4 recostructio equatios i Fourier aalysis aperiodic periodic cotiuous-time CTFT j t x t X e jt d FS 1 2 jk0 x t a e p k t e j k(2/t ) t 0 k period T 0 2/ 0 1 X e j discrete-time DTFT e j DFT x 2 e d x p period N 1 1 N N k 0 X p k e 2 jk( ) N jk(2/n) The 4 Fourier aalysis techiques ad eigefuctios aperiodic e jt cotiuous-time CTFT Fourier series discrete-time DTFT DFT periodic e j e jk(2/t 0 ) t e jk(2/n )

13 Steady state respose aalysis Trasform Sigal characteristics Eigefuctio Fourier trasform cotiuous time aperiodic e j t Fourier series cotiuous time periodic e j (2k/T o ) t Discrete-time FT discrete time aperiodic e j DFT discrete time periodic e j (2k/N) where T 0 is the period. where N is the period. Steady state & trasiet respose aalysis Trasform Sigal characteristics Eigefuctio Laplace trasform cotiuous time aperiodic e t e j t or e ( + j) t z-trasform discrete time aperiodic e e j or (r e j ) or e s t or z go to slide 10 13

14 5.6 Examples: Low-pass & High-pass Filterig (14) Low pass filter for CT Sigal x(t) F X() X() h(t) impulse respose F x frequecy respose H() H() H() y(t) F Y() Y()

15 Low pass filter for DT Sigal x[] F X(e j ) X(e j ) high frequecy impulse respose h[] F x frequecy respose H(e j ) H(e j ) y[] F Y(e j ) Y(e j ) A low pass filter takes i several iput samples ad the computes a weigthed average of these samples as the output. e.g. >hrow[ ] >freqz(hrow) 15

16 ximread('c:\lea.tif'); imshow(x) hrow is a lowpass filter for row hcol is a lowpass filter for colum hrow[ ]; hcol[0.1; 0.2; 0.4; 0.2; 0.1]; lowpass filterig yrowimfilter(x,hrow); imshow(yrow) ycolimfilter(x,hcol); imshow(ycol) 16

17 ximread('c:\lea.tif'); imshow(x) hrow is a lowpass filter for row hcol is a lowpass filter for colum hrow[ ]; hcol[0.1; 0.2; 0.4; 0.2; 0.1]; lowpass filterig yrowimfilter(x,hrow); imshow(yrow) ycolimfilter(x,hcol); imshow(ycol) 17

18 ximread('c:\lea.tif'); imshow(x) hrow is a lowpass filter for row hcol is a lowpass filter for colum hrow[ ]; hcol[0.1; 0.2; 0.4; 0.2; 0.1]; lowpass filterig yrowimfilter(x,hrow); imshow(yrow) ycolimfilter(x,hcol); imshow(ycol) 18

19 High pass filter for CT Sigal X() x(t) F X() h(t) y(t) impulse respose F F x H() frequecy respose Y() H() Y() 19

20 High pass filter for DT Sigal high X(e j frequecy ) low frequecy x[] F X(e j ) impulse respose h[] F x frequecy respose H(e j ) H(e j ) y[] F Y(e j ) Y(e j ) A high pass filter takes i several iput samples ad the computes a weigthed differece of these samples as the output. e.g. >hrow[1/10-2/10 12/10-2/10 1/10] ; >freqz(hrow) 20

21 ximread('c:\lea.tif'); imshow(x) hrow is a highpass filter for row hcol is a highpass filter for colum hrow[ ]; hcol[0.1; -0.2; 1.2; -0.2; 0.1]; highpass filterig yrowimfilter(x,hrow); imshow(yrow) ycolimfilter(x,hcol); imshow(ycol) 21

22 ximread('c:\lea.tif'); imshow(x) hrow is a highpass filter for row hcol is a highpass filter for colum hrow[ ]; hcol[0.1; -0.2; 1.2; -0.2; 0.1]; highpass filterig yrowimfilter(x,hrow); imshow(yrow) ycolimfilter(x,hcol); imshow(ycol) 22

23 Ideal Low pass filter for CT Sigal x(t) X() F X() impulse respose h(t) F x frequecy respose H() H() y(t) F Y() Y() 23

24 Ideal Low pass filter for DT Sigal x[] F X(e j ) X(e j ) h[] y[] impulse respose F F x H(e j ) frequecy respose Y(e j ) H(e j ) Y(e j ) 24

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