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

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

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Unformatted text preview: ] g [ -m] (e) (d) 0 g [ -m] I 2 4 3 4 m- 3 2 o m_ .--f[m] :" .......... / (e) e[k] = f [k] * g[k] --_ .. - o where f [k] and g[k] are depicted in Figs. 9.3a and 9.3b, respectively. We are given f [k] = (0.8)k and g[k] = (0.3)k Therefore f[m] = (0.8)m and g[k - m] = ( 0.3)k-m I 2 3 4 g [ k-m] m- k> 0 ~! Figure 9.3f shows the general situation for k 2: O. T he two functions f [m] and g[k - m] overlap over the interval 0 :S m :S k. Therefore (f) k e[k] = L f[m]g[k - m] k m- m =O k = L (O.8)m(0.3)k-m e [ k] g [ k-m] m =Q ..... f [m] k <0 (h) k = (0.3)k L m =O ( 0.8)m 0.3 k 2:0 (see Sec. B.7-4) For k < 0, there is no overlap between f[m] and g[k - mJ, as shown in Fig. 9.3g so t hat e[k] = 0 k <O e[k] = 2 [(0.8)k+l - (0.3)k+l] u[k] o ( g) m- o 2 r 345 k- Fig. 9 .3 Graphical understanding of convolution of f [k] and g[k] for Example 9.7. a nd which agrees with the earlier result in Eq. (9.52). • k (9.56) 594 t::. 9 T ime-Domain A nalysis o f D iscrete-Time S ystems E~ercise 9.4 S ystem r esponse t o E xternal I nput: T he Z ero-State R esponse 595 E 9.10 F ind (O.S)ku[k] * u[k] graphically and sketch the result. Answer: 5(1 - (O.S)k+l )u[k] \ l J [k] An A lternative Form o f Graphical Procedure: T he Sliding Tape Method 5 T his a lgorithm is convenient w hen t he s equences I [k] a nd g[k] a re s hort o r w hen t hey a re a vailable o nly i n g raphical f orm. T he a lgorithm is b asically t he s ame a s t he g rapb.ical p rocedure i n F ig. 9.3. T he o nly difference is t hat i nstead o f p resenting t he d ata a s g raphical p lots, we display i t as a s equence o f n umbers o n t apes. O therwise t he p rocedure is t he s ame, a s will b ecome c lear i n t he following e xample . • ( a) * , [':1 e[OJ [[[[[ k- 5 E~ample 9 .9 Using t he sliding t ape method, convolve t he two sequences l[kJ andg[kJ depicted in Fig. 9.4a and 9.4b, respectively. In this procedure we write t he sequences l[kJandg[kJ in t he s lots of two tapes: I t ape a nd 9 t ape (Fig. 9.4c). Now leave t he I t ape s tationary (to correspond t o l[mJ). T he g[-mJ t ape is o btained by time inverting t he g[mJ t ape a bout the origin (k = 0) so t hat t he slots corresponding t o f(OJ a nd g[OJ remain aligned (Fig. 9.4d). We now shift the inverted t ape by k slots, multiply values on two tapes in adjacent slots, a nd a dd all t he p roducts t o find e[kJ. Figures 9.4d, e, f, g, h, i, a nd j show t he cases for k = 0 , 1 ,2,3,4,5, a nd 6, respectively. For t he case of k = 0, for example (Fig. 9.4d) (b) k- m=O .j, Jlml f tape c[O] = 0 -+ g tape -+ ( d) c [1] = I rotate the g-tape about the verticle axis as showen in (d) =0 x 1=0 ( e) F or k = 1 (Fig. 9.4e) c[lJ c [2] = 3 = (0 x 1) + (1 x 1) = 1 Similarly, + (1 x 1) + (2 x 1) = 3 e[3J = (0 x 1) + (1 x 1) + (2 x 1) + (3 x 1) = 6 e[4J = (0 x 1) + (1 x 1) + (2 x 1) + (3 x 1) + (4 x 1) = 10 e[5J = (0 x l) + (1 x 1) + (2 x l) + (3 x l) + (4 x l) + (5 x l) = e[6J = (0 x 1) + (1 x 1) + (2 x 1) + (3 x 1) + (4 x 1) + (5 x 1) = ( f) e[2J = (0 x 1) c [...
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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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