# exam1_sample - 1 ME360 Solved Examples for Exam#1 Problem 1...

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1 ME360 - Solved Examples for Exam #1 Problem 1. Consider the filter shown in Figure 1. below with R=1000 ohm and C=0.001 F. R C vo vi i Figure 1. (a) Write analytical expressions for and draw sufficiently representative impulse response of the filter, its magnitude spectral density, and phase response. Z R = R, Z C =1 /Cs V o = RI V i =(1 /Cs + R ) I V o = R 1 /Cs + R V i = RCs 1+ RCs V i 1) H ( s )= V o ( s ) V i ( s ) = RCs 1+ RCs 1 RCs +1 1 /RC s +1 /RC plug in R, C H ( s )=1 1 s +1 2) h ( t L 1 { H ( s ) } = δ ( t ) 1 RC ¡ e t/RC ¢ u ( t δ ( t ) e t u ( t ) 3) H ( RCjω 1+ RCjω = 1+ ; | H ( ) | = | ω | 1 2 + ω 2 ,p h a s e = ( ) (1 + )

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2 0 1 2 3 4 5 6 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 Impulse Response of RC Filter Time (sec) h(t) -10 -8 -6 -4 -2 0 2 4 6 8 10 0 0.2 0.4 0.6 0.8 1 magnitude spectral density | H(jw) | -10 -8 -6 -4 -2 0 2 4 6 8 10 -2 -1 0 1 2 phase response w (rad/sec) phase (rad)
3 (b) Classify the filter. Indicate where this filter is used in signal processing applications. Confirm your classification by taking as filter input v i ( t )=4 6sin ¡ 6 πt + π 3 ¢ 4cos ¡ 18 π 6 ¢ , computing the filter output v o ( t ) , and examining the output frequency content. 1) This high-pass filter canbeusedfo r an AC-coupled circuit to filter out DC component . 2) Confirm the classification by v i ( t 6sin(6 + π/ 3) 4cos(18 6) Calculate the magnitudes and phases of the filter frequency response function at ω =6 π, 18 π. H (0) = 0 H ( j 6 π )= j 6 π 1+ j 6 π ; | H ( j 6 π ) | = 6 π 1+(6 π ) 2 1 phase = π 2 tan 1 ( 6 π 1 )=0 . 053( rad ) H ( j 18 π j 18 π 1+ j 18 π ; | H ( j 18 π ) | = 18 π 1+(18 π ) 2 1 phase = π 2 tan 1 ( 18 π 1 . 018( rad ) v o ( t | H (0) | 4 | H ( j 6 π ) | + 3+0 . 053) | H ( j 18 π ) | 4 cos(18 6+0 . 018) = +1 . 1) 4 cos(18 0 . 506) i.e. no dc component and two high frequency components with almost same magnitude. ( c) Choosing three different values of RC, show analytically and graphically how time domain and frequency domain filter characteristics are affected by the RC value. 1) Analytic: let RC = τ for convenience h ( t δ ( t ) 1 RC ¡ e t/RC ¢ u ( t δ ( t ) 1 τ ¡ e t/τ ¢ u ( t ) for τ 1 2 3

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4 0 2 4 6 8 10 12 -2 -1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 Impulse Response of various RC Filters Time (sec) h(t) tau1 tau2 tau3 tau1 < tau2 < tau3 As RC value increases, ( τ increases), 1 τ decreases, impulse response function will become δ ( t ) , which is an all-pass filter.
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## This note was uploaded on 04/18/2010 for the course ME 360 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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exam1_sample - 1 ME360 Solved Examples for Exam#1 Problem 1...

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