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8.22 Show that if two linear networks are connected in cascade to form a new network such that T j! T1 j! T2 j! , then the composite Bode plot is obtained by adding the individual amplitude ratios Mdb1 and Mdb2 and phase angles ( 1 and 2 ) associated with T1 j! and T2 j! at each frequency. Show that (8.3) follows from the evaluation of k1 and k2 of (8.2).

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CHAPTER 12: Negotiable Instruments Solution: (a) Nov. 1 Notes Receivable Accounts Receivable Jan. 30 Cash Interest Income Notes Receivable (b) Dec. 31 Interest Receivable Interest Income 20,000 20,000 20,725 725 20,000 483.33 483.33 *

CHAP. 8]

An ampli er has a Laplace-domain transfer function (voltage-gain ratio) given by Av s Vo Ks Vi s 100 s 105

[letting u = sin 2x]

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(a) If an asymptotic Bode plot of Av j! is made, over what values of frequency (in the midfrequency range) is the gain constant in amplitude (b) Find the midfrequency gain in decibels if Ks 108 . (c) Within 2 percent accuracy, over what range of frequencies is the exact gain constant Ans: a 100 ! 105 rad=s; b MdbM 60 db; c Mdb ! 58:8 db for 500 ! 5 104 rad/s 8.25 In Problem 8.19, the gain of the FET ampli er does not depend on the Miller capacitance CM ; however, the situation changes if the source resistance is nonzero. (a) Add a source resistance Ri to Fig. 8-26, and nd an expression for the voltage-gain ratio. (b) Evaluate the gain for Ri 0 and for Ri 100 if Cgs 3 pF, Cds 1 pF; Cgd 2:7 pF; rds 50 k ; gm 0:016 S; RL RD 2 k ; RG 1 M ; and f 50 MHz. Ans: a b 8.26 Av s gm RG = Ri RG s Cds Cdg gds GD GL s RG kRi Cgs CM 1

For Ri 0; Av 10:348j131:538; for Ri 100; Av 3:49j61:268

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(c) From Problem 34.6(a), we know how to integrate the rst power of tan x, tan x dx = ln |sec x| + C Higher powers are handled by means of a reduction formula. We have, for n = 2, 3, , tann x dx = = = = Similarly, from sec x dx = ln |sec x + tan x| + C and the reduction formula of Problem 38.13(a), we can integrate all powers of sec x. (d) Antiderivatives of the forms sin Ax cos Bx dx, sin Ax sin Bx dx, cos Ax cos Bx dx can be computed by using the identities tann 2 x(tan2 x) dx = tann 2 x sec2 x dx un 2 du tann 1 x n 1 tann 2 x(sec2 x 1) dx tann 2 x dx [let u = tan x] (39.1)

Consider the high-pass lter circuit of Fig. 8-15(a). (a) Show that as ! becomes large, the amplitude ratio Mdb actually approaches 20 log RL = RL RS as indicated in Fig. 8-15(b). (b) Show that jM 2 j!L j, where !L 1=C RL RS , has the value 1 jM 2 j1 j 1 RL = RL RS 2 . 2 2 In the high-pass lter circuit of Fig. 8-15(a), the source impedance RS 5 k . If the circuit is to have a high-frequency gain of 0.75 and a break or cuto frequency of 100 rad/s, size RL and C. Ans: RL 15 k ; C 0:5 F In the circuit of Fig. 3-20, replace VS with a sinusoidal source to give the small-signal circuit of Fig. 8-6. (a) If the impedance of the coupling capacitor is not negligible, nd the current-gain ratio Ai s IL =IS . (b) Determine the low-frequency cuto point. Ans: a Ai hfe RB = RB hie ; b the gain is independent of frequency down to f 0 Show that the RC network of Fig. 8-29 is a high-pass lter. Ans: Determine its low-frequency cuto point. !L R1 R2 R3 R1 R2 R3 C3

*$20,000 times 14.5% times 1/6 (November and December)

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