q12 - at which C F starts to become a short circuit. What...

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ECE 3050 Analog Electronics Quiz 12 April 8, 2009 Professor Leach Name Instructions. No calculators are allowed on this quiz. Print your name in the space above. Honor Code: I have neither given nor received help on this quiz. Initials 1 of 2. Given the transfer function T ( s )=40 ( s/ 300) 2 ( s/ 300) 2 + b ( s/ 300) + 1 (a) For the case b =2 , factor the denominator of the transfer function. Sketch and label the Bode magnitude plot, both the straight-line approximation and the smooth curve, on the log-log scales below. (b) On the same graph, sketch and label the plot for the case b =0 . 1 . Why is it that you do not factor the denominator for this case? Answer: Because the roots are complex. 1000 0.01 T a s n T b s n 100000 1 ω n 1 10 100 1 10 3 1 10 4 1 10 5 0.01 0.1 1 10 100 1 10 3 2o f2 . (a)Atdc(zerofrequency) ,whatis V o /V i ? Answer: R F / ( R 1 + R 2 ) (b) As ω is increased, it is given that
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Unformatted text preview: at which C F starts to become a short circuit. What is V o /V i in the band between these two frequencies? Answer: − R F /R 2 (c) For ω → ∞ , what is V o /V i ? Answer: (d) Use the information from parts (a) through (c) to sketch the general form of the straight-line approximation to the Bode magnitude plot. (e) Solve for the transfer function for V o /V i and use it to solve for the equations for the break frequencies 1 in the Bode plot of part (d). V o V i = − Z F Z 1 = − R F 1 + R F C F s ( R 1 + R 2 ) 1 + R 1 C 1 s 1 + R 1 k R 2 C 1 s = − R F R 1 + R 2 1 + R 1 k R 2 C 1 s (1 + R 1 C 1 s ) (1 + R F C F s ) Zero at ω = 1 / ( R 1 k R 2 C 1 ) . Poles at ω = 1 /R 1 C 1 and ω = 1 /R F C F 2...
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This note was uploaded on 05/26/2011 for the course ECE 3050 taught by Professor Hollis during the Summer '08 term at Georgia Tech.

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q12 - at which C F starts to become a short circuit. What...

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