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Selecting congruent triangles From each set of three triangles in Fig. 3-6 select the congruent triangles and state the congruency principle that is involved.

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

EXAMPLE 9.4 Given V1 25:0 V1 =V2 143:138 and V2 11:2 26:578, nd the ratio V1 =V2 and the sum V1 V2 .

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(a) nI (b) nII (c) nI nII, by SAS. In nIII, the right angle is not between 3 and 4. nIII, by ASA. In nI, side 10 is not between 70 and 30 . nII nIII by SSS.

25:0 143:138 2:23 116:568 1:00 j1:99 11:2 26:578 V1 V2 20:0 j15:0 10:0 j5:0 10:0 j20:0 23:36

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

Fig. 3-6

116:578

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)

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Determining the reason for congruency of triangles In each part of Fig. 3-7, nI can be proved congruent of nII. Make a diagram showing the equal parts of both triangles and state the congruency principle that is involved.

A sinusoidal voltage or current applied to a passive RLC circuit produces a sinusoidal response. With time functions, such as v t and i t , the circuit is said to be in the time domain, Fig. 9-7(a); and when the circuit is analyzed using phasors, it is said to be in the frequency domain, Fig. 9-7(b). The voltage and current may be written, respectively, v t V cos !t Re Ve j!t i t I cos !t Re Ie

Fig. 3-7

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.)

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The ratio of phasor voltage V to phasor current I is de ned as impedance Z, that is, Z V=I. The reciprocal of impedance is called admittance Y, so that Y 1=Z (S), where 1 S 1 1 1 mho. Y and Z are complex numbers.

Fig. 3-8

Fig. 9-7

CHAP. 8]

Finding parts needed to prove triangles congruent State the additional parts needed to prove nI nII in the given figure by the given congruency principle.

When impedance is written in Cartesian form the real part is the resistance R and the imaginary part is the reactance X. The sign on the imaginary part may be positive or negative: When positive, X is called the inductive reactance, and when negative, X is called the capacitive reactance. When the admittance is written in Cartesian form, the real part is admittance G and the imaginary part is susceptance B. A positive sign on the susceptance indicates a capacitive susceptance, and a negative sign indicates an inductive susceptance. Thus, Z R jXL Y G jBL and and Z R jXC Y G jBC

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 .

Fig. 3-9

The relationships between these terms follow from Z 1=Y. Then, R G B2 R G 2 R X2 G2 and and X B B2 X B 2 R X2 G2

CC iL + LL _

CHAP. 9]

(a) In Fig. 3-9(a) by SSS. (b) In Fig. 3-9(a) by SAS. (c) In Fig. 3-9(b) by ASA. (d) In Fig. 3-9(c) by ASA. (e) In Fig. 3-9(c) by SAS.