HW5S - EE202 HW#5 Solutions 12.2 600 V1 20 mH 300 V2 V1...

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EE202 HW#5 Solutions 12.2 2 1 2 1 2 1 // ( ) ( ) 600 // z z v s v s z z = + ° , 2 5 1 6 1 ( ) 3 10000 18 1.8 10 v s s T s v s s = = = + + ! (a) T(s =0) = 0 (dc gain) T(s = )= 1/3 (infinite energy gain) = -9.5 dB Cutoff frequency ω c = 10000, f c = ω / 2 π = 1.6 KHz, so it is high-pass. (b) (c) ω = 0.5 ω c = 5000 2 2 1 5000 ( ) 0.149 3 5000 10000 T j ! = = + ω = ω c = 10000 2 2 1 10000 ( ) 0.2357 3 10000 10000 T j ! = = + ω = 2 ω c = 20000 2 2 1 20000 ( ) 0.2981 3 10000 20000 T j ! = = + 12.3 Find: T v (s) = V 2 (s)/V 1 (s) (a) DC gain, INF, Cutoff frequency, type of response (b) Sketch Straight Line (c) What Element would you change to increase the passband gain to 10? (a) ω c = 200 rad/s, ω ( ) = 0, ω (0) = 4 = 12 dB, Low pass transfer-function. 1 1 0 1/ x x V V V sc R ! ! + = , 1 1 1 ( 1/ ) x V V sc R R ! + = , 1 1 1 x V V R sc = + 2 2 3 0 x x V V V R R ! ! + = , 1 1 2 2 1 3 1 3 0 ( 1) ( 1) V V V R R sc R R sc R ! ! + = + + 300 Ω 20 mH 600 Ω + V 2 - + V 1 - Z 1 =0.02S + V 2 - + V 1 - Z 2 =300 ω c 20 dB/dec -9.5 dB 20 log(T) 30k Ω 20 k Ω + V 2 (t) - 250 nF 10 k Ω + V 1 (t) -
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800 ( ) 200 v T s s = + (b) (c) R3 = 90 k Ω 12.6 FIND T v (s) = 2 1 ( ) ( ) v s v s 1 0 1/ x x v v v sc R ! ! = , 2 0 x x v v v R R ! ! + = 2 ( ) 1/ v s T s s RC = + , RC = (5000)2 π = 31416 rad/s C = 0.0001 F, R = 32 Ω 12.11 DC gain of -6 dB, f = 500 Hz ω = 2 π f = 3.146 × 10 3 rad/s Dc gain = -6 dB => |T(j ω )| = 1/ 2 2 1 1 2 R R = ; 3 2 1 3.146 10 R C = ! rad/s If C = 1 μ F, R 2 = 320 Ω , thus R 1 = 640 Ω 12.12 1 st order high pass filter T( ω ) = 5, ω c = 1.5 KHz = 9424.78 rad/s , 2 1 1 ( / ) ( ) 1/ R R s T s s R C ! = + , ω c = 1/R 1 C, IF C = 1nF, R
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