EE411_HW14_Soln

# EE411_HW14_Soln - sC R R R R v R R R R v sC R v R R R v R R...

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EE411 Circuit Theory, HW 14 Prof. R. Gharpurey Due Date: May. 11, 2010 Ans:Obviously, there is a zero at ω =2, and 2 poles at ω =20 and ω =100. So the transfer function can be written as + + + 100 1 20 1 2 1 10 ωω ω j j j Ans:(a) nF C C LC 26 . 11 10 15 10 10 2 1 10 15 2 1 3 3 3 = × = × × = π (b)At resonance, we only see the impedance from the resistance. So the current is rms A V 6 20 120 = Ω

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(c) 1 . 47 20 10 10 10 15 2 3 3 = × × × × = = πω R L Q Ans: () RC j LC L j RLC RC j LC RC j L j C j R L j C j R L j Z in ωω ω + + = + + × = + + + × = 2 2 2 1 1 1 1 1 So at resonance, we must have 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 C R LC C R LC RC LC RC RC L RC j L j LC RLC = = = = =
Ans:(a)Applying Thevenin theorem to the source, we have () ωω ω 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ) ( ) ( 2 j j j j j j j j j j j j j s V s V i o + = + + + + = + + + × + = + + + × + = So it resonates at ω =1. And its bandwidth is ω o /Q=3. (b) Applying Thevenin theorem to the source, we have () 3 1 1 1 1 1 1 ) ( ) ( 2 2 j j j j j j j j j j s V s V i o + = + + = + + + × + = So it resonates at ω =1. And its bandwidth is ω o /Q=3.

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Ans:(a)The inputs of the OPAMP is 3 4 4 R R R v s + So we have C R s R R R R C R s R R R v v sC R v R R R sC R R R R v sC R v
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Unformatted text preview: sC R R R R v R R R R v sC R v R R R v R R R R v v s o o s o s s o s s s 2 4 3 2 1 1 3 4 4 2 4 1 3 2 3 4 4 2 2 3 4 4 1 3 3 4 2 3 4 4 1 3 4 4 1 1 1 1 1 1 1 1 + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + × + = ⇒ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + + ⇒ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + = + ⇒ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + = + − (b)High pass filter must have 0 response when s=0. So we must have 4 3 2 1 R R R R = (c) Low pass filter must have 0 response when s= ∞ . So we must have R 4 =0....
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## This note was uploaded on 02/24/2011 for the course EE 411 taught by Professor Lee during the Spring '07 term at University of Texas.

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EE411_HW14_Soln - sC R R R R v R R R R v sC R v R R R v R R...

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