HW5S - EE202 HW#5 Solutions 12.2 2 1 2 1 2 1 // ( ) ( ) 600...

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Unformatted text preview: 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 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 x x V V V R R ! ! + = , 1 1 2 2 1 3 1 3 ( 1) ( 1) V V V R R sc R R sc R !...
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This note was uploaded on 05/24/2008 for the course EE 202L taught by Professor Katsouleas during the Spring '08 term at USC.

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HW5S - EE202 HW#5 Solutions 12.2 2 1 2 1 2 1 // ( ) ( ) 600...

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