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Admittance Diagram Figure 9-9, an admittance diagram, is analogous to Fig. 9-8 for impedance. Shown are an admittance Y1 having capacitive susceptance and an admittance Y2 having inductive susceptance, together with their vector sum, Y1 Y2 , which is the admittance of a parallel combination of Y1 and Y2 .

Since there is at least a decade of frequency (in which the gain can attenuate from its midfrequency value) between !3 and the other (lower) break frequencies, !3 must be the low-frequency cuto !L . Then fL (c) !3 509:1 81:02 Hz < 100 Hz 2 2

(c) Since nI nII, by ASA, corresponding sides are congruent. Then 2x 3y 8 and x 2y. Substituting 2y for x in the first of these equations, we obtain 2(2y) 3y + 8 or y 8. Then x 2y 16.

Fig. 9-9

PROOF:

As in Problem 8.12, Ib 0 if vi is deactivated; a driving-point source replacing RL would then see a frequency-dependent output impedance given by Zo Zdp RC 1 sCC 8

In view of the analogy between impedance in the frequency domain and resistance in the time domain, Sections 3.6 and 3.7 imply the following results. (1) Impedances in series divide the total voltage in the ratio of the impedances: Vr Zr Vs Zs See Fig. 9-10. or Vr Zr V Zeq T

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Statements 1. BF ' AC 2. /5 /6 3. BF > BF 4. BF ' DE 5. /1 is the complement of /3. /2 is the complement of /4. 6. /3 /4 7. /1 /2 8. nI nII 9. AF > FC 1. Given

Fig. 9-10

Assume that the coupling capacitors in the CS MOSFET ampli er of Fig. 4-25 are identical. Determine the voltage-gain ratio (a) for any frequency and (b) for midfrequency operation.

2. $ form rt. ; rt. are 3. Reflexive property 4. Given 5. Adjacent angles are complementary if exterior sides are to each other. 6. Given 7. Complements of are 8. ASA s 9. Corresponding parts of congruent n are

Fig. 9-11

(a) The equivalent circuit is drawn in Fig. 8-21. By voltage division, where RG R1 kR2 R1 R2 R1 R2 1 RG sRG CC Vgs V V RG 1=sCC i sRG CC 1 1 Current division at the drain node yields IL from which Vo RL IL sgm RD RL rds CC = RD rds V sCC RD rds = RD rds RL 1 gs 3 sCC RD rds = RD rds gm Vgs RD krds g V RD krds 1=sCC RL m gs sCC RD rds = RD rds RL 1 2

Proving a congruency problem stated in words Prove that if the opposite sides of a quadrilateral are equal and a diagonal is drawn, equal angles are formed between the diagonal and the sides.

(2) Impedances in parallel (admittances in series) divide the total current in the inverse ratio of the impedances (direct ratio of the admittances): Ir Zs Yr Is Zr Ys See Fig. 9-11. or Ir Zeq Y IT r IT Zr Yeq

1 2T

If the opposite sides of a quadrilateral are congruent and a diagonal is drawn, congruent angles are formed between the diagonal and the sides. Quadrilateral ABCD AB > CD, BC > AD AC is a diagonal. To Prove: /1 /4, /2 /3 Plan: Prove nI nII

Consider the frequency-domain network of Fig. 9-12. Applying KVL, as in Section 4.3, or simply by inspection, we nd the matrix equation

PROOF:

CHAP. 9]

3 2T

for the unknown mesh currents I1 ; I2 ; I3 . Here, Z11 ZA ZB , the self-impedance of mesh 1, is the sum of all impedances through which I1 passes. Similarly, Z22 ZB ZC ZD and Z33 ZD ZE are the self-impedances of meshes 2 and 3.

Given:

Fig. 1-22 (a) The integral in (1.20) is simply the area under the f t curve for one period. average current as 1 T T 4 1 2:5 A T 2 2 We can, then, nd the

Fig. 9-12

PRINCIPLE 1:

I0

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