quiz 1 solutions

# H j 10 90 tan 1 1 45 based on this information we can

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H ( j 10) = 90 - tan 1 (1) = 45 Based on this information, we can sketch the plots for the magnitude and the phase. The plots below show the magnitude and phase response for 0 ω 30 rad/s. 0 5 10 15 20 25 30 0.2 0.4 0.6 0.8 1.0 0 5 10 15 20 25 30 0 45 90 Dr. Vahe Caliskan 1 of 4 Handed out: January 31, 2013

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ECE 412 Introduction to Filter Synthesis Quiz #1 Solutions University of Illinois at Chicago Spring 2013 Problem 2 (2 points) The magnitude response | H ( ) | and the phase response H ( ) for a filter is given below. 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 2 3 4 5 6 7 8 9 10 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 -60 -50 -40 -30 -20 -10 0 ω (rad/s) ω (rad/s) | H ( ) | H ( ) (degrees) (a) If a voltage v in ( t ) = 2 sin( t + 30 ) is applied as an input, estimate the output voltage v out ( t ). (b) If a voltage v in ( t ) = 2 cos ( 7 2 t - 45 ) is applied as an input, estimate the output voltage v out ( t ). Solution An input sinusoid with amplitude A , angle θ a at frequency ω a can be written as v in ( t ) = A sin( ω a t + θ a ). For this input, the output of the filter will be v out ( t ) = A | H ( a ) | sin( ω a t + θ a + H ( a )) where | H ( a ) | and H ( a ) are the magnitude and phase of the transfer function evaluated at ω = ω a which can be read from the graphs above. (a) For v in ( t ) = 2 sin( t + 30 ) the input frequency is 1 rad/s. From the frequency response plot we see that | H ( j 1) | ≈ 7 and H ( j 1) ≈ - 40 . As a result
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