solmidterm05_mae143b

solmidterm05_mae143b - Name Student Midterm MAE143B Fall...

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Unformatted text preview: Name: Student #: Midterm - MAE143B, Fall 2005 November 15, 2005 11:00am-11:50pm, HSS1330 Solutions open-book and open-notes midterm exam use the available space to derive your results, attach extra paper if necessary use of any electronic equipment (calculator, phone, PDA) not allowed during exam Consider a linear dynamic system G characterized by the transfer function model y (s) = G(s)u(s), G(s) = 1−s √ s 2 + 2s + 1 (1) 1. Show that the system is stable and has a right half plane zero. Derive the value of the undamped natural frequency and the damping ratio of this system. [15pt] The poles are found by solving s2 + √ 2s + 1 = 0, 1 1 and by using the quadratic equation the roots can be shown to be located at s = − √2 ± √2 j . Therefore, the system is stable since the real part of the poles is negative. The dampening ratio β and the undamped natural frequency ωn can be found via the following comparison: 2 ωn = 1 √ 2βωn = 2. This gives β = 1 √. 2 The zero is found by solving 1 − s = 0, and therefore a zeros is located are the point s = 1, which is in the RHP. 2. Compute the impulse response y (t) of this dynamcal system. [15pt] 1−s √ s2 + 2s + 1 1 √ √ 2 2 = 1 (s + √2 )2 + y (s) = 1 2 − √ 2s (s + 1 √ 2 1 √ )2 2 + 1 2 From the Table of Laplace Transforms in Franklin, Powell, Emami-Naeini entry 20, we get √ √ d −√ t 1 1 1 e 2 sin( √ t) sin( √ t) − 2 dt 2 2 √ −√ t 1 1 1 1 1 1 −√ t −√ t 2e 2 sin( √ t) + e 2 sin( √ t) − e 2 cos( √ t) = 2 2 2 √ 1 1 1 −√ t = e 2 (( 2 + 1) sin( √ t) − cos( √ t)) 2 2 y (t) = 1 −√ t 2e 2 1 3. As G(s) is stable, the output y (t) to a sinusoidal input signal u(t) = cos ωt is given by y (t) = M (ω ) cos(ωt + φ(ω )) (2) for large values of t. Examine the approximate (asymptotic) behavior of M (ω ) and φ(ω ) for the low (ω < 1 rad/s), and high (ω > 1 rad/s) frequency range. Comment on the slope of log10 M (ω ) when plotted against log10 ω over these frequency ranges. [20pt] √ 1 + ω2 1 − jω √ M (ω ) = = −ω 2 + 2jω + 1 (1 − ω 2 )2 + 2ω 2 √ √ 1 + ω2 ω2 ω 1 lim M (ω ) = lim = lim = lim 2 = lim = 0. ω →∞ ω →∞ ω →∞ ω (1 − ω 2 )2 + 2ω 2 ω→∞ (ω 2 )2 ω→∞ ω Taking log10 of M (ω ) at very high frequencies gives 1 log10 (M (ω )) = log10 ( ) = log10 1 − log10 ω. ω From this equation it can be seen that M (ω ) will have slope of −1 in log-scale at high √ √ frequencies. 1 + ω2 1+0 = lim =1 lim M (ω ) = lim 2 )2 + 2ω 2 ω →0 ω →0 ω →0 (1 − ω (1 − 0)2 Since M (ω ) approaches a constant value at low frequencies, the slope will be zero. √ 1 − jω 2ω −ω √ ) φ(ω ) = ∠ ) − arctan( = arctan( 2+ 1 1 − ω2 −ω 2jω + 1 √ √ −ω 2ω 2 lim φ(ω ) = lim arctan( ) = −90 − lim arctan( ) − arctan( ) = −270 ω →∞ ω →∞ ω →∞ 1 1 − ω2 −ω √ −ω 2ω ) = arctan(−0) − arctan(0) = 0 − 0 = 0 lim φ(ω ) = lim arctan( ) − arctan( ω →0 ω →0 1 1 − ω2 4. In the two figures below, sketch the amplitude and phase Bode plot of the transfer function model G(s) in (1). Use solid lines to sketch the actual Bode plots, use dashed lines to indicate the asymptotic behavior of the Bode plots. [10pt] 2 10 90 0 1 φ(ω ) [deg] M (ω ) 10 0 10 −90 −180 −1 10 −270 −2 10 −2 10 −1 10 0 10 1 10 −360 −2 10 2 10 ω rad/s −1 10 0 10 ω rad/s 2 1 10 2 10 Now consider the standard 2nd order system characterized by the transfer function model 2 ωn y (s) = G(s)u(s), G(s) = 2 2 s + 2βωn s + ωn (3) controlled by a PD-controller C (s) = Kp + Kd s using negative feedback: u(s) = r(s) − C (s)y (s). 5. Show that the closed-loop transfer function T (s) from r(s) to y (s) can be written as the standard 2nd order system 2 ωn with ωn = ωn ¯ ¯¯ s2 + 2β ωn s + ωn ¯2 ¯ β + ωn Kd /2 . [15pt] 1 + Kp and β = 1 + Kp 2 ωn 2 G(s) s2 + 2βωn s + ωn =2 Tyr (s) = 2 2 1 + G(s)C (s) s + 2βωn s + ωn ωn +2 2 2 Kp + Kd s s2 + 2βωn s + ωn s + 2βωn s + ωn 2 ωn =2 2 2 s + (2βωn + Kd ωn )s + ωn (1 + Kp ) T (s) = This gives the following relationships: 2 ωn = ωn (1 + Kp ) ¯2 2 ¯¯ 2β ωn = 2βωn + Kd ωn Solving gives ωn = ωn ¯ 2 2 ¯ 2βωn + Kd ωn = 2βωn + Kd ωn = (1 + Kp ) and β = 2¯ n ω 2ωn (1 + Kp ) β+ Kd 2ωn (1 + K p ) . 6. In case the uncontrolled system G(s) in (3) has no damping and a resonance frequency of 1 rad/s, compute the numerical values Kp and Kd of the PD-controller to triple the undamped resonance frequency and bring the damping ratio up to 1/3. [10pt] No damping means β = 0 and with the given formulae of question 5 and the desire of tripling the resonance frequency ωn we see ωn = ωn 1 + Kp = 3ωn ⇒ Kp = 9 − 1 = 8 ¯ √ β + ωn Kd /2 Kd /2 1 3 ¯ β= = = √ ⇒ Kd = 2 √ = 2 3 1 + Kp 1 + Kp 3 3 7. Encircle the word ‘increase’ or ‘decrease’ to make the following statements true. [5pt] Compared to the uncontrolled system G(s) in (3), application of a PD controller will • • • • • increase/ decrease the settling time increase /decrease the damping increase/ decrease rise time increase/ decrease peak time increase/ decrease overshoot end of midterm 3 ...
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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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