Homework7_Solutions

Homework7_Solutions - MA 107 HW 7 Modeling And Analysis of...

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MA 107 Modeling And Analysis of Dynamic Systems S06 Professor T-C. Tsao HW 7 1. 9.5 The equation of motion and the transfer function are () 1 I tc t T t s Ts Is c ω =− + = + & I=2,c=4. Thus the steady state output for y(t) =30+5sin3t+2sin5t is 11 1 1 1 30 5s in (3 ) 2cos (5 ) 42 3 4 2 3 4 2 5 4 2 3 4 7.5 0.6935sin(3 0.9828) 0.1857cos(5 1.1903) ss tt t jj j j =⋅ + + + + ⋅+ =+ + 2. 9.9 (c) yss(t)=5 |G(j 0.7)| sin(0.7t+ angle(G(j 0.7))=5*0.1118*sin(0.7t-0.6719) G(j 0.7) = ( 0.6719) 0.7 ( 0.7) 0.0875 0.0696 0.1118 ( 1.4 1)( 3.5 1) j j Gj i e == = ++ 3. 9.35 (See Example 9.4.1 for bandwidth). Also show frequency domain input and out put signal spectrum (line spectrum of Fourier coefficients) and system’s frequency response at the Fourier harmonic frequencies. 4. 9.37 (Use the natural frequency as the bandwidth of 2 nd order systems) Also show frequency domain input and out put signal spectrum and system’s frequency response at the Fourier harmonic frequencies.
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MA 107 Modeling And Analysis of Dynamic Systems S06 Professor T-C. Tsao The bandwidth is somewhat larger than the natural frequency (70 rad/sec) because of the numerator term. From the Bode plot, we identify the bandwidth (-3 dB or 0.707 from the DC gain) to be about 140 rad/sec. So the solution above should have calculated one more term at 40 π rad/sec. The gains and phases may also be read from the Bode plot if the plot gives sufficient resolution.
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This note was uploaded on 09/17/2008 for the course MAE 107 taught by Professor Tsao during the Spring '06 term at UCLA.

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Homework7_Solutions - MA 107 HW 7 Modeling And Analysis of...

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