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(a) f (x) = x 2 + 6x + 7 = (x + 7)(x 1). Because f ( 7) = 0 = f (1), f is not one-one. (b) Obviously, f (+1) = f ( 1); f is not one-one. (c) If y = 5x 3 2, then, y + 2 = 5x 3 y+2 = x3 5 y+2 =x 5

Fig. 8-19

0 For the CE ampli er of Fig. 3-10, determine (a) Zin , (b) Zin , and (c) Zo if CC ! 1 but CE cannot be neglected.

Only the principal ($400) is recorded when the note is received, since it represents the amount of the unpaid account. The interest is not due un til the date of collection, 90 days later. At that time, the interest earned will be part of the entry recognizing the receipt of the proceeds from the note: Cash 415

(m2 m1 )2

(a) The small-signal low-frequency equivalent circuit is given in Fig. 8-4. Using (8.11) and (8.13), we have

Substituting (1) into (2) and rearranging give Zin (c) RB shie RE CE hie hfe 1 RE sRE CE RB hie RB hie hfe 1 RE 3

2 2 2

With voltage source vi deactivated (shorted), KVL requires that Ib so that 1 hfe Ib ZE khie hie hfe ZE hie I 0 hie ZE hie b (4)

In this particular case, (3) shows that the input impedance is frequency-dependent, while (5) shows that the output impedance is independent of frequency. In general, however, the output impedance does depend on frequency, through a nite-valued coupling capacitor CC . (See Problem 8.13.)

2m1 m2 = 2 m1 m2 = 1 Conversely, assume that m1 m2 = 1, where m1 and m2 are the slopes of lines L1 and L2 . Then L1 is not parallel to L2 . 2 (Otherwise, by Theorem 4.1, m1 = 1, which contradicts the fact that a square is never negative.) Let L1 intersect L2 at

CHAP. 8]

Interest Income* 15

To examine the combined e ects of coupling and bypass capacitors, let the input coupling capacitor be in nitely large while the output coupling capacitor and the bypass capacitor have practical values in the CE ampli er of Fig. 3-10. For simplicity, assume hre hoe 0 and 0 RB ) Zin . (a) Find the voltage-gain ratio Av s vL =vi . (b) If CE 200 F, CC 10 F, Ri RE 100 ; RC RL 2 k ; hie 1 k ; and hfe 100, determine what parameters control the low-frequency cuto point and whether it is below 100 Hz. (c) Find an expression for the output impedance Zo .

point P (see Fig. 4-13). Let L3 be the line through P and perpendicular to L1 . If m3 is the slope of L3 , then, by what we have just shown, m1 m3 = 1 = m1 m2 or m3 = m2

CC iL + LL _

Fig. 8-20 (a) The small-signal equivalent circuit is given in Fig. 8-20. ZE RE k Then, by KCL, Ie Ib hfe Ib hfe 1 Ib KVL around the input mesh requires that Vi Ri hie Ib ZE Ie Substituting (2) into (3) and solving for Ib yields Ib Vi Ri hie hfe 1 ZE 4 3 2 We rst de ne 1

Current division at the collector node gives IL and Ohm s law and (5) yield V L R L IL RL RC h I RC RL 1=sCC fe b 6 RC h I RC RL 1=sCC fe b 5

Since L2 and L3 pass through P and have the same slope, they must coincide. Since L1 is perpendicular to L3 , L1 is perpendicular to L2 .

Substituting (4) and (1) into (6) and rearranging now lead to the desired voltage-gain ratio: V Av s L Vi hfe RL RC CC s sRE CE 1 hfe 1 RE hie Ri CE RE Ri hie sCC RC RL 1 s 1 hfe 1 RE Ri hie