Signal Processing and Linear Systems-B.P.Lathi copy

# b w rite t he difference e quation r elating t he o

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Unformatted text preview: g i nterval T , t he r esulting s ignal i s a d iscrete-time s inusoid c os w kT. T herefore, a ll t he r esults d eveloped i n t his s ection a pply i f w e s ubstitute w T f or n: 5 n = wT • E xample 1 2.1 F or a s ystem specified by t he e quation y[k + 1 ]- 0.8y[k] = f [k + 1] o f ind t he s ystem r esponse t o t he i nput ( a) l k = 1 ( b) cos [ifk - 0.2] Q- 31t 21t It ( c) a s ampled sinusoid cos 1500t w ith s ampling interval T = 0.001. T he s ystem e quation c an b e e xpressed as ( E - 0.8)y[k] = E f[k] Q- T herefore, t he t ransfer function o f t he s ystem is H[z] = z z _ 0.8 = 1 1 _ 0 .8z 1 T he f requency response is Fig. 1 2.1 1 H [eif!] = F requency response o f a n L TID s ystem i n E xample 12.1. (12.7) 1 - 0.8e- i f! F igure 12.1 shows p lots o f a mplitude a nd p hase r esponse a s f unctions o f fl. W e now c ompute t he a mplitude a nd t he p hase r esponse for t he v arious inputs: 1 1 - 0.8(cos f l- j s in f l) ( a) f (k]=l k =1 S ince 1k = ( eif!)k w ith f l = 0, t he a mplitude r esponse is H [ei°]. F rom E q. (12.8a) 1 ( 1- 0.8 cos fl) + j O.8sin fl we o btain T herefore I H[e i f!lI = 1 ) (1 - 0.8 cos f l)2 ~ H[eJ ] + (0.8 s in fl)2 1 = ) 1.64 - 1.6 cos (0) (12.8a) T herefore a nd v '1.64 - 1.6 cos f l a nd L H[ e if!] = - tan -1 [ 0 .8 s in f l ] 1 _ 0 .8 cos f l (12.8b) T hese values also c an b e r ead d irectly from Figs. 12.1a a nd 12.1b, respectively, corresponding t o f l = O. T herefore, t he s ystem r esponse t o i nput 1 is T he a mplitude r esponse IH[eif!]1 c an also be o btained b y observing t hat IHI2 = H H *. T herefore y[k] = 5 (lk) IH[eif!]12 = H [eif!]H*[eif!] ( b) f [k] = H [eif!]H[e-if!] H ere fl = (12.9) = cos [ifk - 0.2] if. According t o Eqs. I H[e i "/611 = F rom E q. (12.7) i t follows t hat w hich yields t he r esult found earlier in Eq. (12.8a). (12.10) =5 (12.8) 1 " ) 1.64 - 1.6 cos '6 L H[ei,,/6]=-tan-1 ( 1.64 - 1.6 cos f l = -1- = 5 = 5LO v'Q.04 = 1.983 0 .8sin if ,,] = -0.916rad. 1 - 0.8 cos '6 T hese values also c an b e r ead d irectly from Figs. 12.1a a nd 12.1b, respectively, corresponding t o f l = if. T herefore 720 12 Frequency R esponse a nd D igital F ilters 2 ~Y[k) J [k) \ ~ -6 6 o 11 18 12 24 Ii 12.1 n um=[1 0); d en=[1 - 0.8); W =-pi:pi/100:pi; H =freqz(nurn,den, W ); m ag=abs(H); p hase=180/pi*unwrap(angle(H»; s ubplot(2,1,1); p lot(W,mag);grid; s ubplot(2,1,2); p lot(W,phase);grid 0 C omment F ig. 1 2.2 Sinusoidal input and t he corresponding o utput of a n LTID system in Example 12.1. F igure 12.1 shows p lots o f a mplitude a nd p hase r esponse as functions o f n. T hese p lots a s well a s E qs. (12.8) i ndicate t hat t he f requency response o f a d iscretetime s ystem is a c ontinuous ( rather t han d iscrete) function o f f requency n. T here is n o c ontradiction h ere. T his b ehavior is merely a n i ndication o f t he f act t hat t he f requency v ariable n is c ontinuous ( takes o n a ll possible values) a nd t herefore t he s ystem r esponse e xists a t e very v alue of n. ;::,. E...
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## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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