Ch12s - Chapter 12 Problem Solutions 12.1 (a) Af = A 1 + A...

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Chapter 12 Problem Solutions 12.1 (a) 5 5 55 1 51 0 120 1( 0 ) 1 2 0 ( 1 2 0 ) ( 0) 0 0.008331 f A A A β = + × = = × = (b) 3 3 33 0 120 0 ) 1 2 1 2 0 ) ( 0 0.008133 × = = × = 12.2 (a) 0.15 1 f A A A == + TA = (i) T =∞ (ii) 43 80 dB 10 1.5 10 A A T = = (iii) 15 T = (i) 1 6.667 f A (ii) 6.662 f A = (iii) 6.25 f A = (b) (i) T (ii) 3 2.5 10 T (iii) 25 T = (i) 1 4.00 f A (ii) 3.9984 f A = (iii) 3.846 f A = 12.3 (a) 1 125 1 f A A A ββ =≅ = + 0.0080 = (b)
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[] (125)(0.9975) 124.6875 124.6875 1 (0.008) 124.6875 1 (0.008) A 124.6875 1 0.9975 49,875 f A A A A A A == = + += =− = 12.4 (a) 4 4 3 10 100 1( 1 0 ) 9.9 10 β = + (b) 4 100 ( 0.10) 0.001 0.10% 10 ff f f f dA A dA AA A dA A ⎛⎞ = ⎜⎟ ⎝⎠ = 12.5 4 4 4 3 0.001 ( 0.10) 51 0 500 0 500 0 ) 1.98 10 f f f dA A dA A A A = −= × = × = 12.6 (a) For Fig. P12.6(a) 1 11 1 200 10 12 0 0 0 oo io vv ββ ++ 200 10 40 0 0 0 f A 2 1 1 50 (1 200 )(1 10 ) 1 210 2000 β β =+ + + 2 2000 210 49 0 +− = 1 1 210 44100 4(2000)(49) 2(2000) 0.1126 −± + = = For Fig P12.6(b) 2 2 (200)(10) 40 1 (200)(10) 0.0245 = + = Fig. P12.6(a) (b)
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1 180 180 10 (8.4634)(4.704) 1 (180)(0.1126) 1 (10)(0.1126) 39.81 39.81 40 0.475% 40 f f f f A A A dA A = ⎛⎞ == ⎜⎟ ++ ⎝⎠ = = Fig. P12.6(b) (180)(10) 39.91 1 (180)(10)(0.0245) 39.91 40 0.225% 40 f f f A dA A + = (c) Fig. P12.6(b) is a better feedback circuit. 12.7 (a) ( 10)( 15)( 20) 3000 O OS VV V V εε ε β =− =+ So 3000( ) OO S V + We find 3000 1 3000 O vf S V A V + For 3000 120 0.008 13 0 0 0 vf A = = + (b) Now ( 9)( 13.5)( 18) 2187 O V Then 2187 2187 118.24 1 2187 1 2187(0.008) vf A −− = 120 118.24 % change 100 1.47% change 120 12.8 5 (10 )(4) (50) 8 BB f f k H z = = 12.9 (a) 5 33 (50) (10 )(4) 8 dB dB f k H z = = (b) 5 (10) 40 dB dB f k H z = = 12.10 3 0 (50)(20 10 ) 5 A ×= so 5 0 21 0 A 12.11 Low freq. 0 0 1 f A A A = + 5000 100 0.0098 1 (5000) = = + Freq. response
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12 121 2 2 5000 11 (5000)(0.0098) 1 1 5000 4 9 5000 14 9 5000 50 f ff jj A A A f f jf jf fff f f f jf jf f β ⎛⎞ ++ ⎜⎟ ⎝⎠ == + + = + = ++ + + = Also 0 100 1 f f AB A B A B A A f f f f j j j j f f f f + + +++ So 1 2 100 100 1 50 50 50 A B f f f f j f j f j j j j f f f f f f = + + + Then 1 1 50 50 f f += + and 50 f f = 1 10 f = and 2 2000 f = 1 1 0.002 0.000010 0.002010 50(10) 50(2000) + = + = and 6 111 (50)(10)(2000) 10 B A f f = = Then 6 1 0.002010 10 B B f f 62 3 3 10 1 2.01 10 10 2.01 10 1 0 BB −− + −× + = 36 6 6 34 6 2.01 10 4.0401 10 4(10 )(1) 2(10 ) 2.01 10 2.0025 10 2(10 ) B B f f ×± ×− = × = + sign 3 1.105 10 Hz B f + sign 2 9.05 10 Hz A f
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12.12 (a) Fig. P12.6(a) 1 1 1 11 3 1 200 1 10 200 1 (10)(0.1126) 1 (0.1126) 1 200 (4.704) 1 22.52 940.73 940.73 1 23.52 23.52 1 (23.52) 40 (23. 1 (23.52) f dB f j f A f j f f j f ff jj f jf f ⎡⎤ ⎛⎞ ⎢⎥ ⎜⎟ + ⎝⎠ = + ⎣⎦ + + = ++ == + 52)(100) 2.352 kHz Fig P12.6(b) 1 1 1 3 1 (200)(10) 1 2000 (0.0245)(200)(10) 4 9 1 2000 1 (50)(100) 5 KHz 50 1 (50) f dB f j f A f j f f j f f f j f + + + =⋅ = + (b) Overall feedback wider bandwidth. 12.13 01 2 1 0 0 0 1000 (100) (1) (100)(10) (1)(1) 1000 1 in vA A v A v S vv v N =+ = + 12.14 (a)
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(b) Circuit (b) – less distortion 12.15 (a) Low input R Shunt input Low output R Shunt output Or a Shunt-Shunt circuit (b) High input R Series input High output R Series output Or a series-Series circuit (c) Shunt-Series circuit (d) Series-Shunt circuit 12.16 (a) 45 (max) (1 ) 10(1 10 ) (max) 10 ii i RR T R k =+ ≅Ω
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3 4 10 (min) 10 11 1 0 i i R R k T == ≅ Ω ++ Or (min) 1 i R (b) 44 (max) (1 ) 1(1 10 ) (max) 10 oo o RR T R k =+ = + ≅Ω 4 4 1 (min) 10 1 0 o o R Rk T Or (min) 0.1 o R 12.17 Overall Transconductance Amplifier, o g i i A v = Series output = current signal and Shunt input = current signal. Also, Shunt output = voltage signal and Series input = voltage signal. Two possible solutions are shown.
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Ch12s - Chapter 12 Problem Solutions 12.1 (a) Af = A 1 + A...

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