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

# 99 w ith i kj 0 t hat is 912a qejyokj 0 or 912b or

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Unformatted text preview: e Response h[kJ () = t an- 1 (-0.172) = - 0.17 rad. C onsider a n n th-order s ystem s pecified b y t he e quation S ubstituting () = - 0.17 r adian in eeos () = 2.308 yields e = 2.34 a nd yo[kJ = 2.34(0.9)k cos (~k - 0.17) U nit I mpulse R esponse ( En + a n_IE n - (bnEn k:::: 0 O bserve t hat here we have used radian u nit for b oth fJ a nd (). We also could have used t he degree unit, although i t is n ot recommended. T he i mportant c onsideration is to be consistent a nd t o use t he s ame units for b oth fJ a nd (). • E xercise E 9.2 Find and sketch the zero-input response for the systems described by the following equations: ( a) y[k + 1 ]- O.8y[k] = 3 f[k + I] ( b) y[k + 1] + O.8y[k] = 3 f[k + 1]. In each case the initial condition is Y[-I] = 10. Verify the solutions by computing the first three terms using the iterative method. Answer: (a)8(O.8)k ( b)-8(-O.8)k. V 1 + . .. + a lE + a o)y[k] = + b n_IE n - 1 + ... + b lE + b o)f[k] ( 9.27a) or Q [E]y[k] = P [E]f[k] ( 9.27b) T he u nit i mpulse r esponse h [k] is t he s olution o f t his e quation for t he i nput 6[k] w ith a ll t he i nitial c onditions z ero; t hat is, b. b. E xercise E 9.3 Find the zero-input response of a system described by the equation y[k] + O.3y[k - 1] - O.ly[k - 2] = f [k] + 2 f[k - 1] The initial conditions are y o[-l] = 1 and Yo[-2] = 33. Verify the solution by computing the first three terms iteratively. Answer: York] = (O.2)k + 2( -O.5)k V b. E xercise E 9.4 Find the zero-input response of a system described by the equation y[k] s ubject t o i nitial c onditions h [-l] = h [-2] = . .. = h [-n] = 0 ( 9.29) E quation ( 9.28) c an b e s olved t o d etermine h [k] i teratively o r i n a c losed form. T he f ollowing e xample d emonstrates t he i terative s olution. . Example 9 .4: I terative D etermination o f h [kj Find h[kJ, t he u nit impulse response o f a s ystem described by t he e quation y[kJ - 0.6y[k - IJ - 0.16y[k - 2J = 5f[kJ (9.30) To determine t he u nit impulse response, we m ust l et t he i nput f[kJ = 6[kJ a nd t he o utput y[kJ = h[kJ in t he above equation. T he r esulting equation is (9.31) s ubject t o zero initial s tate; t hat is, h [-IJ = h [-2J = O. S etting k = 0 i n t his equation yields h[OJ - 0.6(0) - 0.16(0) = 5(1) =} h[OJ = 5 Next, setting k = 1 i n Eq. (9.31) a nd using h[OJ = 5, we o btain <::) C omputer E xample C 9.2 F ind a nd sketch t he z ero-input response for t he s ystem described by + 0.81)y[kJ u sing t he i nitial conditions y[-IJ=2, a nd y[-2J=I. y =[l 2J; y =y'; k =-2:16; k =k'; f or m =1:length(k)-2 Y = 1.56*y(m+l)-O.81 * y(m); Y zi=[Yzi;Yj; e nd s tem(k,Yzi) 0 ( 9.28) h[kJ - 0.6h[k - IJ - 0.16h[k - 2J = 56[kJ + 4y[k - 2J = 2f[kJ T he initial conditions are yo[-IJ = - 2~ and yo[-2J = ~. Verify the solution by computing t he first three terms iteratively. Answer:yo[k] = (2)kcos(~k -~) V (E2 - I.56E Q [E]h[k] = P [E]6[k] = (E + 3)f[kJ h[IJ - 0.6(5) - 0.16(0) = 5(0) =} h[lJ = 3 • C ontinuing t his way, we c an d etermine a ny n umber o f t erms o f h [k]. U nfortunately, s uch a s olutio...
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