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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)km I 2 3 4 g [ km] 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)km e [ k]
g [ km] 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.74) 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 imeDomain A nalysis o f D iscreteTime S ystems
E~ercise 9.4 S ystem r esponse t o E xternal I nput: T he Z eroState 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 gtape 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.
 Spring '13
 Bayliss
 Signal Processing, The Land

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