lec07_print

# lec07_print - MAE143b Linear Control Theory and...

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MAE143b Linear Control - Theory and Applications Lecture 7, Tuesday Aug. 25, 2:00-4:50pm Prof. R.A. de Callafon [email protected] CONTENTS OF THIS LECTURE Midterm from 2pm-3:30pm Frequency response analysis ( sec. 6.1 ) – recap frequency response & Bode plots – examples of 1st order lead/lag compensators – Bode plot of integrator/di±erentiator – Bode plot of standard 2nd order system – Bode plots of higher order systems – Example: lab demo MAE143b, UCSD, Summer II 2009, R.A. de Callafon – Lecture 7, Page 1 Dynamic Response - recap frequency response General formulation – after transient response has decayed, the output response y ( t ) on a sinusoidal input signal u ( t )=cos ωt is going to be y ( t )= M ( ω )cos( + φ ( ω )) where M ( ω | G ( ) | = | G ( s ) | s = and φ ( ω )=tan 1 Im { G ( ) } Re { G ( ) } = 6 G ( ) M ( ω )isthe magnitude of the TF G ( s ) evaluated at s = φ ( ω ) is called the phase of the TF G ( s ) evaluated at s = MAE143b, UCSD, Summer II 2009, R.A. de Callafon – Lecture 7, Page 2

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Dynamic Response - recap Bode plots Basically, for a SISO dynamical system, the transfer function G ( s ) is a scalar complex function of the complex (Laplace) vari- able s . When evaluating the transfer function G ( s ) along s = jω, ω > 0 (positive imaginary axis in s-plane) you evaluate again a complex function along the complex variable with ω> 0. Plotting the amplitude M ( ω ):= | G ( ) | of the complex function and the angle φ ( ω )= 6 G ( ) gives a graphical interpretation of your transfer function G ( s ) ! Amplitude Bode Plot : Plotting of amplitude M ( ω ) as a func- tion the frequency ω using a logarithmic scale for both ω (x-axis) and M ( ω )(y-ax is) Pha seBodeP lo t : Plotting of phase φ ( ω ) as a function the frequency ω using a logarithmic scale for ω (x-axis) and a linear scale for φ ( ω MAE143b, UCSD, Summer II 2009, R.A. de Callafon – Lecture 7, Page 3 Frequency Response - Frst order lead system Recap - the Amplitude Bode plot of the lead compensator G ( s K τ 1 s +1 τ 2 s 1 2 > 0 ±or small frequencies ( ω 0): log 10 | G ( ) | =log 10 | K | q 1+ τ 2 1 ω 2 q τ 2 2 ω 2 10 | K | ±or middle frequencies ( 1 τ 1 <ω< 1 τ 2 ) log 10 | G ( ) | =l o g 10 K q 1+ τ 2 1 ω 2 q 1+ τ 2 2 ω 2 10 K q τ 2 1 ω 2 1 o g 10 Kτω 10 ω +(log 10 K +log 10 τ 1 ) ±or high frequencies ( 1 τ 2 ) log 10 | G ( ) | 10 K q τ 2 1 ω 2 q τ 2 2 ω 2 10 K q τ 2 1 ω 2 q τ 2 2 ω 2 10 K · τ 1 τ 2 MAE143b, UCSD, Summer II 2009, R.A. de Callafon – Lecture 7, Page 4
Frequency Response - frst order lead system Recap - With G ( )= K 1+ jωτ 1 1+ 1 we have 6 G ( 6 (1 + 1 ) 6 (1 + 2 )=tan 1 ωτ 1 tan 1 2 For small ±requencies ( ω 0) we have 6 G ( )=0 For middle ±requencies ( 1 τ 1 <ω< 1 τ 2 )wehave 6 G ( )=+ π/ 2 For high ±requencies ( ω> 1 τ 2 6 G ( This is also called a lead compensator : it provides phase ad- vancement in the middle ±requency range ( 1 τ 1 1 τ 2 ) MAE143b, UCSD, Summer II 2009, R.A. de Calla±on – Lecture 7, Page 5 Frequency Response - frst order lead system So the lead compensator G ( s K τ 1 s +1 τ 2 s 1 2 > 0 will have a Bode plot with the ±ollowing shape 1 1 1 2 K K · τ 1 τ 2 Amplitude Bode plot 1 1 1 2 0deg 90 deg Phase Bode plot

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lec07_print - MAE143b Linear Control Theory and...

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