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EX6_5.CIR vi 1 0 SIN(0V 250mV 10kHz) Ri 1 2 100ohm CC1 2 3 1000uF CC2 4 7 1000uF R1 3 0 6kohm R2 3 6 50kohm RC 6 4 1kohm RE 5 0 100ohm RL 7 0 1kohm VCC 6 0 15V Q 4 3 5 QNPNG .MODEL QNPNG NPN(Is=10fA Ikf=150mA Isc=10fA Bf=150 + Br=3 Rb=1ohm Rc=1ohm Va=75V Cjc=10pF Cje=15pF) .TRAN 1us 0.1ms .PROBE .END

(d) cos ( 4x)

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After executing hEx6_5.CIRi, the plots of Fig. 6-8 can be generated by use of the Probe feature of PSpice. resulting h-parameter value is indicated on each of the four plots of Fig. 6-8.

CHAPTER 8: Summarizing and Reporting via the Worksheet Solution:

CHAP. 6]

(c) sin x = 1 (d) cos x = 1 2 3 1 (g) sin x = (h) cos x = (e) sin x = 1 (f ) cos x = 2 2 2 [Hint: By Theorem 26.1, only two of parts (a), (b), and (d) need be solved.]

Fig. 6-8 Example 6.6. For the ampli er of Example 6.5, use SPICE methods to determine (a) the input impedance Zin , (b) the current gain Ai , and (c) the voltage gain Av . Netlist code that describes the ampli er circuit follows:

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Ex6_6.CIR vi 1 0 AC 1V Ri 1 2 100ohm CC1 2 3 1000uF CC2 4 7 1000uF R1 3 0 6kohm R2 3 6 50kohm RC 6 4 1kohm RE 5 0 100ohm RL 7 0 1kohm VCC 6 0 15V Q 4 3 5 QNPNG .MODEL QNPNG NPN(Is=10fA Ikf=150mA Isc=10fA Bf=150 + Br=3 Rb=1ohm Rc=1ohm Va=75V Cjc=10pF Cje=15pF) .AC LIN 1 100Hz 100Hz .PRINT AC Vm(1) Vp(1) Vm(7) Vp(7) .PRINT AC Im(Ri) Ip(Ri) Im(RL) IP(RL) .END

27.11 Find the derivatives of the following functions: (a) 4 sin3 x sin x (e) x (i) cos (2x 2 3) (b) sin x + 2 cos x 1 cos x (f ) x2 3 (5x + 4) (j) sin (c) x sin x (g) 5 sin 3x cos x (k) cos 2x (d) x 2 cos 2x (h) cos2 2x (l) sin3 (sin2 x)

(a) Execute hEx6_6.CIRi and poll the output le to nd the values of input voltage and current. V 1 1 Zin 4:056 k I vi 2:465 10 4

Thus,

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27.12 Calculate the following limits: sin x cos x 1 (b) lim (a) lim 3x 2x x 0 x 0 3x sin 4x sin (f ) lim (e) lim x 0 5x x 0 4x 3 1 x sin 4x sin2 x (j) lim x cos 2 (i) lim x 0 x 0 x2 x 27.13 Sketch the graphs of the following functions: (a) f (x) = sin2 x (d) h(x) = cos x cos2 x (g) f (x) = sin (x 1) (b) g(x) = sin x + cos x (e) g(x) = |sin x|

(a) B, Dr. (b) B, Dr. (c) B, Cr. (d) B, Cr. (e) B, Dr. (f ) I, Cr. (g) I, Dr. (h) B, Cr. Solved Problem 8.2 Joe Hurt owns and operates Rent-a-Wreck Com pany, a used car rental business. The following is a trial balance before the end of the month adjustments. Dr. 1,940 1,575 1,740 2,900 16,500 Cr.

(b) The output le contains the magnitudes and phase angles of the input and output voltages. Vm 7 4:649 4:649 Av Vm 1 1 The negative sign accounts for the 1808 phase shift [see Vp(7)] of V(7) with respect to V(1). (c)

Hence,

sin 2x x 0 sin 3x sin x 2 (g) lim x x 0 sin2 (x 4) (k) lim x 4 (x 4)2

The output le values of Ip(Ri) and Ip(RL) show the two signals to be 1808 out of phase. The current gain is found as Ai Im RL 4:649 10 3 18:86 Im Ri 2:465 10 4

6.1 For the CB ampli er of Fig. 3-23, nd the voltage-gain ratio Av vL =vS using the tee-equivalent small-signal circuit of Fig. 6-3.

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