HW#3_solution

HW#3_solution - HW3 1. Plausible approximate values f(Hz) 0...

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HW3 1. Plausible approximate values If the frequency response of this amplifier is 4 100 () 1( ) 10 Tj j ω = + , then 14 4 100 | | , () t a n ( / 1 0 ) 1 ( /10 ) ωω = ∠= + . Thus 2 1 51 61 100 | ( 1000) | 20log( ) 39.96 1 (1000/10000) ( 1000) tan (1000/10000) 5.7 ( 10 ) tan (10) 84.3 ( 10 ) tan (100) 89.4 T j dB = = + = = −° = = = = f(Hz) |T|(dB) T ∠° 0 40 0 100 40 0 1000 40 0 4 10 37 -45 5 10 20 -90 6 10 0 -90 0 20 40 60 1 10 100 1000 10000 100000 1000000 |T|dB
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2. f(Hz) 0 10 2 10 3 4 5 6 7 8 |T|(dB) 0 20 37 40 40 40 37 20 0 3. 2s 1 1/s 1/s 1 + - () i Vs + - o 1||(1/s) 1||(1/s) 1/s 1||(1/s) 1||(1/s) 1/s i 1 1 i s + + - o Circuit (a) Circuit (b) Circuit (c) At Circuit (c), 32 2 1|| (1/ ) 1 1||(1/ ) 1/ 1||(1/ ) 1 0.5 0.5 2 2 1 ( 1)( 1) oi o s ss s s s s = ++ + = = + + + + 2 0.5 ( ) ( o i s s s = + (1) Poles are given by : 2 ( 0 13 1 22 s Poles are s and s j + ++ = = = −± 0 10 20 30 40 50 1 100 10000 1000000 100000000 |T|dB
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4. 22 0 ( ) ( 0, 0) Ks Hs K B s Bs ω = >> ++ which is the general network function of a second-order bandpass filter having center frequency 0 , bandwidth B and 0 | ( )| K Hj B = . 12 21 2 0 0 0 2 10 / , 1000 / 990 10000 100 | ( ) | 1, 990 990 990 ( ) (1) 990 10000 cc rad s rad s B K If H j K s ss ωω = = = −= = = ⇒= = = ∴= 0 V s V R sL 1/ sC 2 2 1 () ||1/ 1 ( ) 1 ( ) 1 (1/ ) (2) (1/ ) s o Vs sL sC sL sC V s R sL sC R sL sC sC R sL sC s RLC sC R RC s s RC s LC + = = = + + + = = + + = By Eq. (1) and Eq. (2) 11 990, 10000 1 1 1.01 990 1000 1 0.99 1.01 10000 RC LC Let R K CF LH µ = = = = = × = = ×
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5. 12 21 2 0 0 2 100 / , 10000 / 9900 1000000 1000 ( ) (1) 9900 1000000 cc rad s rad s B Ks Hs ss ωω ω = = = −= = = ⇒=
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This note was uploaded on 10/16/2010 for the course EE 201 taught by Professor Kruempel during the Spring '06 term at Iowa State.

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HW#3_solution - HW3 1. Plausible approximate values f(Hz) 0...

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