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

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Unformatted text preview: xercise E12.1 For a system specified by the equation y [k y[k] = 1.983 cos ( fk - 0.2 - 0.916) = 1.983cos ( fk - 1.116) (12.11) = k T) results in a discrete-time f[k] = cos 1500kT For T (12.12) = 0.001, t he input is f[k] In this case, n = 1.5. = cos (1.5k) (12.13) + 1J - 0.5y[kJ = f[kJ find the amplitude and phase response. Find the system response to sinusoidal input cos ( 1000t- t) sampled every T = 0 .5 rns. Answer: LH[dflJ = - tan- 1 [~l c os f l-O.5 IH[dflJI = , (1.25 1 c os I i y[kJ = 1.639 cos (0.5k - Figure 12.2 shows the input J[k] a nd t he corresponding system response. ( c) A sinusoid cos 1500t sampled every T seconds (t sinusoid 721 F requency R esponse o f D iscrete-Time S ystems f - 0.904) = 1.639 cos (0.5k - 1.951) \l ;::,. E xercise E12.2 Show that for an ideal delay (H[zJ = l iz)' the amplitude response IH[ejflJl = 1, and the phase response LH[ejflJ = -no Thus, a pure time-delay does not affect the amplitude gain of sinusoidal input, but it causes a phase shift (delay) of n radians in a discrete sinusoid of frequency n. Thus, in the case of an ideal delay, we see that the phase shift a t the output is proportional to the frequency of the input sinusoid (linear phase shift). \l According t o Eqs. (12.8a) and (12.8b) T he Periodic N ature o f t he Frequency Response IH[e j 1.5J1 = = 0.809 (12.14) 0.8 sin (1.5) ] = _ 1 -0.8cos(1.5) 0 .702rad (12.15) 1 ) 1.64 - 1.6 cos (1.5) LH[ejL5] = _ t an - 1 [ These values also could be read directly from Fig. 12.1 corresponding to Therefore y[k] o = 0.809 cos (1.5k - n= 0.702) • C omputer E xample C 12.1 Using MATLAB, find the frequency response of the system in Example 12.1. 1.5. F igure 12.1 shows t hat for t he s ystem i n E xample 12.1, t he f requency r esponse n w ith p eriod 211". T his f act is n ot a coincidence. U nlike t he f requency r esponse o f a c ontinuous-time s ystem, t he f requency response o f e very L TID s ystem is a p eriodic f unction o f n w ith p eriod 211". T his f act follows from t he v ery s tructure o f t he f requency response H [e jnk ]. I ts a rgument eink is a p eriodic f unction o f n w ith p eriod 211". T his f act will a utomatically r ender H [e ink ] p eriodic. T here is a physical reason for t his p eriodicity a nd t he p eriodicity o f H [e ink ] s hould n ot c ome a s a s urprise. We know t hat d iscrete-time s inusoids separated b y v alues o f n i n i ntegral m ultiples o f 211" a re i dentical. Therefore, t he s ystem r esponse t o s uch s inusoids (or exponentials) is also identical. T hus for discrete-time systems, we n eed t o p lot t he f requency response o nly o ver t he f requency r ange f rom -11" t o 11" ( or f rom 0 t o 211"). I n a r eal s ense, d iscrete-time s inusoids of frequencies o utside t he f undamental r ange o f frequencies d o n ot e xist ( although t hey e xist i n a t echnical sense). H [e in ] is a p eriodic f unction o f 722 12 Frequency Response a nd Digital Filters 12.2 Frequency...
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