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Convolutions 2.23. Let X and Y be random variables having joint density function f (x, y). Prove that the density function of U X Y is

The pole-zero plot in Fig. 8-39 shows a pole at s 0 and zeros at s 50 j50. method to evaluate the transfer function at the test point j100. Ans: H j100 223:6 26:578

X Y, V X, where we have arbitrarily added the second equation. Corresponding to these we have y, v x or x v, y u v. The Jacobian of the transformation is given by 'x 'u 4 'y 'u 'x 'v 4 'y 'v 0 1 1 2 1

Fig. 8-38

Thus by Theorem 2-4, page 42, the joint density function of U and V is g(u, v) f (v, u v)

g(u)

Fig. 8-39

Method 2 The distribution function of U by x y u, i.e.,

From experience, I have found that if N and est N are not within about 30 percent of each other, it may not be reasonable to apply any of the other software science measures.

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and L 0:1 H in the other. First, apply an excitation, Ii s , and obtain the natural frequency from the denominator of the network function. Try di erent locations for applying the current source. Second, insert a voltage source, Vi s , and obtain the natural frequency. Ans: 300 Np/s in all cases

G(u)

G(u)

Obtain the natural frequencies and the complete current, i in if . Ans: 8:5 Np/s, 23:5 Np/s; i 2:25e 8:5t 0:25e 23:5t 2:5 A

Fig. 2-16

Fig. 8-40

g(u)

2.24. Work Problem 2.23 if X and Y are independent random variables having density functions f1(x), f2( y), respectively.

, L 2 H, and C 0:25 F. Simultaneously apply magnitude and frequency scaling, with Km 2000 and Kf 104 . What are the scaled element values Ans: 2000

Halstead thought of volume as a 3D measure, when it is really related to the number of bits it would take to encode the program being measured.4 In other words: V N log2 1 2

0 0 Y.

; 0:4 H; 12:5 mF At a certain frequency !1 , a voltage V1 25 08 V applied to a passive network results in a current I1 3:85 308 (A). The network elements are magnitude-scaled with Km 10. Obtain the current which results from a second voltage source, V2 10 458 V, replacing the rst, if the second source frequency is !2 103 !1 . Ans: 0:154 158 A In the circuit of Fig. 8-41 let R1 C1 R2 C2 10 3 . Find v2 for t > 0 if: (a) v1 cos 1000t u t , (b) v1 sin 1000t u t . Ans: a v2 sin 1000t ; b v2 1 cos 1000t In the circuit of Fig. 8-42 assume R 2 k

f2 (y)

3e 0

, C 10 nF, and R2 R1 and v1 cos !t. following frequencies: (a) !0 5 104 rad/s, (b) !1 105 rad/s. Ans: a v2 2 sin !0 t; b v2 0:555 cos !1 t 146:38 Find v2 for the

By Problem 2.24 the required density function is the convolution of f1 and f2 and is given by g(u) In the integrand f1 vanishes when v

Fig. 8-41

g(u)

Calculate V for the code of Example 5.10. V 26 log2 13 26 3:7 96:2

e3u)

Fig. 8-42

0 and g(u)

Noninverting integrators. In the circuits of Fig. 8-43(a) and 8-43(b) nd the relationship between v2 and v1 . Ans: a v1 RC=2 dv2 =dt; b v1 2RCdv2 =dt

Check:

f2 * f1 (Property 1, page 43).

In the circuit of Fig. 8-44 nd the relationship between v2 and v1 . v2 R2 v1 = R1 R2 . Ans: R1 R2 C1 C2 dv2 dv R1 R2 v2 R1 R2 C1 1 R2 v1 dt dt

1 3

CHAP. 8]

The volume gives the number of bits necessary to encode that many di erent values. This number is hard to interpret.

w) f2(w)( dw)

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