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(7.10) (7.11) (d) .2 6 .3 9 (e) x 3 when x 8 4 6. (7.11) 1 4 (7.11) 4:5y and 2 :15 2 4 :5y 14 :3y (7.12) 1 3 (c) 2x 8 y y 10 1 1 2

be the unit sample response of a linear shift-invariant system. If the input to this system is a unit step,

7.15. Which of the following is not a proportion 4 24 3 7 25 10 (a) (b) (c) 3 18 3 12 45 18

x(n) =

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The quotient rule allows us to extend Rule 4 of 12: RULE 7. Dx (x k ) = kx k 1 for any integer k (positive, zero, or negative). For a proof, see Problem 13.3. EXAMPLES

find limn,,

7.16. From each of the following, form a new proportion whose first term is x. Then find x. 9 x 20 5 2 3 1 a x 5 11 (a) (b) x (c) x (d) (e) x 2 4 b 5 10 20 7.17. Find x in each of these pairs of proportions: (a) a :b (b) 5:7 c : x and a: b x:42 and 5 :7 c: d 35:42 (c) 2:3x (d) 7: 5x

y ( n ) where y ( n ) = h ( n ) x ( n ) . y(n) = h ( n ) * x ( n ) =

14 :3y and 7 :18

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h(k)x(n - k )

7.18. Find x in each of the following proportions: x y x y x 7 7 (a) (b) 8 4 3 6

if x ( n ) is a unit step,

(a) (b) Dx Dx 1 x 1 x2 1 1 = Dx (x 1 ) = ( 1)x 2 = ( 1) 2 = 2 x x 2 = Dx (x 2 ) = 2x 3 = 3 x

7.19. Find x in each part of Fig. 7-45.

lim y(n) =

h(k)

(7.13)

k=-03

Fig. 7-45

13.1 Prove Rule 3, the product rule: If f and g are differentiable at x, then Dx (f (x) g(x)) = f (x) Dx g(x) + g(x) Dx f (x)

SIGNALS AND SYSTEMS Because u(k) is zero fork < 0, and u(n - k) is zero fork > n, this sum may be rewritten as follows:

(7.13)

- 90 + 90(0.9)"]u(n)

Fig. 7-46

y(n) = x ( n ) * 0 ) when h(n) = (;)"u(n)

7.21. Find x in each part of Fig. 7-47.

CHAP. 13]

x ( n ) = ( i ) " [ u ( n ) u(n - 101)l -

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To evaluate this sum, which depends on n, we consider three cases. First, for n c 0, the sum is equal to zero because u(n - k ) = 0 for 0 5 k 5 100. Therefore,

(7.14)

Second, note that for 0 5 n 5 100, the step u(n - k) is only equal to 1 fork 5 n. Therefore,

Fig. 7-47

I (f)" -1(-f1)

By simple algebra, f (x + h)g(x + h) f (x)g(x) = f (x + h)[g(x + h) g(x)] + g(x)[f (x + h) f (x)] Hence, Dx (f (x) g(x)) = lim = lim f (x + h)g(x + h) f (x)g(x) h f (x + h)[g(x + h) g(x)] + g(x)[f (x + h) f (x) h f (x + h)[g(x + h) g(x)] g(x)[f (x + h) f (x)] + lim h h h 0 g(x + h) g(x) f (x + h) f (x) + lim g(x) lim h h h 0 h 0 h 0

7.22. Find x in each part of Fig. 7-48.

= 3(;)"[1 - (f)"+']

(7.15)

Finally, for n 2 100, note that u ( n - k ) is equal to I for all k in the range 0 5 k 5 100. Therefore,

=(f)"

Fig. 7-48

= f (x) Dx g(x) + g(x) Dx f (x) In the last step, lim f (x + h) = f (x) follows from the fact that f is continuous at x, which is a consequence of Theorem 13.1.

1 - ($O'

(7.16)

(f) ]

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