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

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Unformatted text preview: 3)k ( 0.8)k:' - (0.3)k+l u[k] (0.3) (0.8 - 0.3) = 2 [(0.8)k+l - (0.3)k+'] u[k] f:" • E xercise E 9.6 Show that (O.8)ku[k} * u[k} = 5[1 - (O.8)k+l}u[k} \ l Convolution Sum from a Table J ust a s in t he c ontinuous-time case, we have p repared a t able o f convolution s ums ( Table 9.1) from which convolution sums m ay b e d etermined d irectly for a v ariety o f signal pairs. For example, t he c onvolution in E xample 9.6 c an b e r ead directly from t his t able ( Pair 4) a s We shall evaluate t he c onvolution s um first by a n a nalytical m ethod a nd l ater w ith g raphical aid. • (9.52) (0.8)kU[k] * ( 0.3/u[k] = (0.8)k+l - (0.3)k+l u[k] 0.8 - 0.3 = 2[(0.8)k+l - (0.3)k+l]u[k] We shall d emonstrate t he u se of t he c onvolution table in t he following example. E xample 9 .6 Determine e[k] = f [k] * g[k] for • f [k] = (0.8)ku[k] and g[k] = (0.3)k u [k] (9.49) E xample 9 .7 Find the (zero-state) response y[k] of an LTID system described by the equation We have y[k + 2 ]- 0.6y[k + 1 ]- 0.16y[k] e[k] = L N ote t hat and g[k - m] = (0.3)k- mu[k - m] 5J[k + 2] if the input J[k] = 4 - u[k]. T he unit impulse response of this system is obtained in Example 9.5. f[m]g[k - m] m =-oo J[m] = (0.8)mu[m] = k (9.50) B oth J[k] and g[k] are causal. Therefore, [see Eq. (9.48)] t The width of a signal is one less than the number of its elements (length). For instance, the signal in Fig. 9.3h has 6 elements, but has a width of only 5. Thus, if /I[k} and h[k} have widths of W , and W2, respectively, then the width of c[k} is W , + W2. h[k] = [( _0.2)k + 4(0.8)k]u[k] (9.53) Therefore y[k] =J[k] * h[k] = (4)-ku [k] * [( -0.2)ku[k] + 4(0.8)k U[kJ] = (4)-ku [k] * (-0.2)ku[k] + (4)-ku[k] * 4(0.8)kU[k] (9.54) 9.4 S ystem r esponse t o E xternal I nput: T he Z ero-State R esponse 591 N ote t hat '" 8 .., ;" .., e'n ( 4)-ku[k] 0) = Wk u [k] = (O.25)ku[k] T herefore 0) 8 y [k] <<a 11 ~ .... .0, ., = (O.25)ku[k] * ( -O.2)ku[k] + 4(O.25)ku[k] * (O.8)ku[k] We use P air 4 ( Table 9.1) t o find t he a bove convolution sums. c-. r 0) .... u '" is .~ '" £. c r-. ~ 1\ < £: 'H-. ~ _ [ (O.25)k+l - (_O.2)k+l y [k] 0.25 _ ( -0.2) -:s: ~ ~ ~ ; :l 8 0 ,..... 8 + 0, ~ ~ 1 ., c r-. en 0:> '" 0 ~ 'H-. ~ * ~ 8 ;:! I:: .S ~ ;:! '0 > I:: 0 0 ..:: ~ ~ .... .,; ~ * r.1 ..:: ~ ~ ~ ~ 1 ::5.. ...... ir l lr I,..... ,..... ~ r ~Ik + :; ; :l , ...----, ""c-. ; :l ~ ~ ;; ~ : ,..... + ~ ""c-. c-. rr , 1 + r ""~ ~ ~ ~ ~ ;:l ; :l ~Ik r ""~ r 1 ..., ~ :I~ + ~ ;; --;::: , ~ 1 """ ~ + .... ~ ,..... ""~ r ""c-. , ...----, ""~ r c-. S!-.. c-. r < ::1 r ~ ,..... ~ ~c-. ...,~ ~ ;; ".".t:. ,..... ~ ..., + -£ ,.....,'" ~ +r , ,..... ,..... 1 "!":. r ~ 1 <:!> ..:: ~ ...... ~ ; :l ~ ;:l ..!.. ~ ""c-. ""c-. r r 0 '0" u We c an e xpress y [k] a s r ---., () £. ~ ", " "c-. r N y [k] ~I"- . .5 "'- " = [ -1.26(4)-k + 0.444(-0.2)k ;:l r ~ ..., ; :l ~ ;; ""r ~ ;.., .; .i::~ ~ ~ -= ,.....I~ ~ i ). =. ~ II ~ ; :l ""c-. r E...
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