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solhwset3 - Solutions to Homework Set 3 - MAE143B, Summer...

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Solutions to Homework Set 3 - MAE143B, Summer II, 2009 1.1 Consider y ( s )= G 1 ( s ) u ( s ) ,G 1 ( s τ 1 s +1 s ( τ 2 s +1) with 0 2 1 then substitution of s = in G 1 ( s ) yields G 1 ( 1+ 1 ω (1 + 1 ω ) = 2 ω τ 2 ω 2 + and with ω> 0th ismakes | G 1 ( ) | = | 1 ω | | || τ 2 ω | = p 1+( τ 1 ω ) 2 ω p τ 2 ω ) 2 G 1 ( (1 + 1 ω ) ( ) (1 + 2 ω )) = tan 1 τ 1 ω π 2 tan 1 τ 2 ω 1.2 The asymptotic behavior of | G 1 ( ) | and G 1 ( ): For 0 ω< 1 1 (low frequency range) follows | G 1 ( ) | = p τ 1 ω ) 2 ω p τ 2 ω ) 2 1 ω 1 = 1 ω G 1 ( )=t a n 1 τ 1 ω π 2 tan 1 τ 2 ω tan 1 0 π 2 tan 1 0= π 2 so plotting log 10 ω versus log 10 | G 1 ( ) |≈ log 10 1 ω = log 10 ω creates an line with slope -1 (or -20dB/dec) for | G 1 ( ) | , while G 1 ( ) ≈− π/ 2radians or -90 degrees. For 1 1 <ω< 1 2 (middle frequency range) follows | G 1 ( ) | = p τ 1 ω ) 2 ω p τ 2 ω ) 2 p ( τ 1 ω ) 2 ω 1 = | τ 1 | ω ω = | τ 1 | G 1 ( a n 1 τ 1 ω π 2 tan 1 τ 2 ω tan 1 ∞− π 2 tan 1 π 2 π 2 =0 so plotting log 10 ω versus log 10 | G 1 ( ) log 10 | τ 1 | creates an line with slope 0 (or 0dB/dec) for | G 1 ( ) | , while G 1 ( ) 0. 2 (high frequency range) follows | G 1 ( ) | = p τ 1 ω ) 2 ω p τ 2 ω ) 2 p ( τ 1 ω ) 2 ω p ( τ 2 ω ) 2 = | τ 1 | ω | τ 2 | ω 2 = | τ 1 | | τ 2 | ω G 1 ( a n 1 τ 1 ω π 2 tan 1 τ 2 ω tan 1 π 2 tan 1 = π 2 π 2 π 2 = π 2 so plotting log 10 ω versus log 10 | G 1 ( ) log 10 | τ 1 | creates an line with slope -1 (or 0dB/dec) for | G 1 ( ) | , while G 1 ( ) 2rad ians or -90 degrees. 1
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Amplitude and Phase Bode plot Amplitude 1/ τ 1 1/ τ 2 | τ 1 | Freq. [rad/s] actual Bode plot asymptotes -100 -80 -60 -40 -20 0 20 Phase 1/ τ 1 1/ τ 2 Freq. [rad/s] Figure 1: Amplitude (top) and Phase (bottom) Bode plot of G 1 ( s ) with asymptotes. 1.3 Using the results from question 1.2, Figure 1 can be created. 1.4 Compared to the previously de±ned G 1 ( s ), consider now y ( s )= G 2 ( s ) u ( s ) ,G 2 ( s τ 1 s +1 s ( τ 2 s +1) where τ 1 < 0 and still 0 2 < | τ 1 | It can be seen from the answers for | G 1 ( ) | that | G 2 ( ) | = | G 1 ( ) | as either τ 2 1 > 0or | τ 1 | > 0appearsin | G 1 ( ) | , making the sign of τ 1 irrelevant. The phase does change, as G 2 ( (1 + 1 ω ) ( ) (1 + 2 ω )) = tan 1 | τ 1 | ω π 2 tan 1 τ 2 ω Going over the asymptotic behavior of G 2 ( ) for the same low, middle and high frequency ranges evaluated under question 1.2 now yields For 0 ω< 1 1 (low frequency range) follows G 2 ( tan 1 | τ 1 | ω π 2 tan 1 τ 2 ω ≈− tan 1 0 π 2 tan 1 0= π 2 so G 2 ( ) remains at π/ 2 radians or -90 degrees at the low frequency range. 2
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For 1 / | τ 1 | <ω< 1 2 (middle frequency range) follows G 2 ( )= tan 1 | τ 1 | ω π 2 tan 1 τ 2 ω ≈− tan 1 ∞− π 2 tan 1 0= π 2 π 2 = π so G 2 ( ) becomes π radians or 180 degrees, compared to the angle of 0 for G 1 ( ) for the same middle frequency range.
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This note was uploaded on 05/27/2010 for the course MAE MAT143B taught by Professor Linearcontrol during the Spring '10 term at UCSD.

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solhwset3 - Solutions to Homework Set 3 - MAE143B, Summer...

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