Ch15s - Chapter 15 Problem Solutions 15.1 (a) For example:...

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Chapter 15 Problem Solutions 15.1 (a) For example: Low-Frequency: 22 11 0 9 ⎛⎞ =+ = = ⎜⎟ ⎝⎠ o i v RR vR R Corner Frequency: 35 1 51 0 3 . 1 81 0 2 π == × fR C RC (b) For Example: 2 2 2 1 1 1 o i R jC v R R j R C ω + So, set 2 1 15 R R = For example, 12 10 , 150 Rk R k =Ω= Ω 5 2 3 3 1.06 10 ( 1 5 1 0 ) dB RC f ππ = × × Then 70.7 = Cp F 15.2 (a) 3 0.447 7 1( 2 ) 1 = =− + + vv dB A Ad B f f
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(b) 24 3 11 0.2425 12.3 1( 2 ) 1 == = =− + ⎛⎞ + ⎜⎟ ⎝⎠ vv dB A Ad B f f (c) 26 3 0.1240 18.1 2 ) 1 = + + dB A B f f 15.3 (a) Figure 15.6 6 3 12 4 3 1 2 7.958 10 2 2 (20 10 ) Let 10 K 795.8 pF So 0.707 562.6 pF 1.414 1.125 nF f RC RC f RRR C CC π = × = = (b) (i) 4 1 0.777 18 1 20 T + (ii) 4 1 0.707 20 1 20 T + (iii) 4 1 0.637 22 1 20 T + 15.4 Use Figure 15.10(b) 3 1 2 dB f RC = or 6 3 1 3.18 10 2( 5 01 0) RC × × For example, let 100 pF C = Then 31.8 k R 1 2 3 And 8.97 k 22.8 k 157 k R R R From Equation (15.26) 6 3 1 1 dB T f f = +
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We find f kHz | T | 30 0.211 35 0.324 40 0.456 45 0.589 15.5 From Equation (15.7). 12 4 1 2 3 () YY Ts Y Y Y Y = ++ + For a high-pass filter, let , YY s C == 3 3 1 , Y R = and 4 4 1 Y R = Then 22 43 11 1 = ⎛⎞ + ⎜⎟ ⎝⎠ = sC s C sC sC RR sR C sR C Define 33 rR C = and 44 C = 1 s rs r = Set s j ω = 2 4 34 1/2 2 2 4 1 1 1 1 14 () 1 ωω = = −− = ⎧⎫ ⎪⎪ =− + ⎨⎬ ⎩⎭ Tj jr jj rr j r r For a maximally flat filter, we want 0 dT d →∞ = Taking the derivative, we find 3/2 2 2 2 3 3 2 4 4 4 1 2 4 ( 2 ) 1 2 dT j dr r r r r r r r + × +
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or 32 3 2 34 4 2 4 () 0 41 8 1 1 2 1 dT j d rr r r ω ωω →∞ = ⎡⎤ ⎛⎞ =− ⎢⎥ ⎜⎟ ⎝⎠ ⎣⎦ Then 22 4 11 2 10 r −− = So that 12 2 = = Then the transfer function can be written as: 1/2 2 33 222 333 3 14 () 1 (2 ) (4 ) 111 1 4( ) 1 1 4( ) ωωω ⎧⎫ ⎪⎪ + ⎨⎬ ⎩⎭ + + =+ Tj rrr r 3 dB frequency 3 21 r = or rR C == Define 1 RC = So that 3 2 = R R We had 2 = or 344 3 2( ) 2 RC R R = = So that 4 2 =⋅ RR 15.6 From Equation (15.25) 2 3 3 1 25 dB 0.0562 1 20 2 10 = + N dB dB TT f f f f So
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2 22 1 0.0562 1( 2 ) 1 (2) 316.6 (2) 315.6 2 ln (2) ln (315.6) 4.15 5 A 5-pole filter N NN N = + += = ⋅= = = 15.7 2 3 1 1 N dB T f f = ⎛⎞ + ⎜⎟ ⎝⎠ At 12 kHz, f = 0.9 T = 33 2 2 3 11 0.9 12 12 1 10 . 2 3 4 6 (0.9) == ++ =− = dB dB N dB f ff f Also 2 3 2 2 3 2 2 3 4 2 3 24 1 0.01 14 1 14 1 1 9999 (0.01) 14 14 9999 4.262 10 12 0.2346 12 (1.16667) 4.262 10 35 = + = === × = N dB N dB N N dB N dB N f f f f N Then 2 3 2 3 1 0.014286 2 3 1 0.9 12 1 12 0.2346 12 (0.2346) (0.2346) 0.9795 N dB N dB N dB f f f = + = = So 3 12.25 kHz = dB f 15.8
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2 3 1 1 N dB T f f = ⎛⎞ + ⎜⎟ ⎝⎠ (a) For 3 N = 6 1 0.2841 1( 1 . 5 ) T == + (b) For 5 N = 10 1 0.1306 1 . 5 ) T + (c) For 7 N = 14 1 0.05843 1 . 5 ) T + 15.9 Consider For low-frequency: 23 123 o i vR R R R + = ++ For high-frequency: 2 12 o i v R R = + So we need 2 25 RR R RRR + = + Let 50 k += Ω and 21 1.5 48.5 Rk R k Then
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3 3 3 1.5 1.5 25 144 50 50 + ⎛⎞ = ⎜⎟ + ⎝⎠ R Rk R Connect the output of this circuit to a non-inverting op-amp circuit. At low-frequency: 23 1 123 1.5 144 0.75 48.5 1.5 144 oi i i RR vv v v RRR + + =⋅ = = ++ + + Need to have 25. o v = 55 5 1 44 4 25 1 1 (0.75) 32.3 ⎛⎞⎛⎞ ==+ ⋅=+ = ⎜⎟⎜⎟ ⎝⎠⎝⎠ oo i R v R To check at high-frequency.
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This homework help was uploaded on 02/10/2008 for the course ECE 138 taught by Professor N/a during the Spring '08 term at CSU Fresno.

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Ch15s - Chapter 15 Problem Solutions 15.1 (a) For example:...

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