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Unformatted text preview: Problem Set #1 16.30: Estimation and Control of Aerospace Systems Posted: Feb 14, Due: Feb 21 Problem 1 Analyze the stability of LTI systems described by the following transfer functions, when enclosed in a unity feedback loop, using (i) the root locus method, (ii) the Nyquist plot, and (iii) the Bode plot. 1. G ( s ) = 10( s + 1) ( s + 5)( s + 0 . 2) 2. G ( s ) = 10( s + 1) s ( s + 5)( s + 0 . 2) 3. G ( s ) = 10( s + 1) s ( s + 5)( s . 2) 4. G ( s ) = 10( s 1) s ( s + 5)( s + 0 . 2) Solution: (The following rootlocus and Nyquist plot sketches are not drawn to scale. In the root locus, the dashed loci correspond to negative controller gains; in the Nyquist plot, the dashed line correspond to negative frequencies and to the parts of the plot at infinity. ) Part 1: G ( s ) = 10( s +1) ( s +5)( s +0 . 2) This is a simple system with two stable real poles, and one minimumphase zero. Notice that the root locus gain, the standard (Bode) gain, and the DCgain are all equal to 10. Analyzing, e.g., the Nyquist plot, it can be recognized that the system is stable for all positive controller gains, and for small, negative gains, not exceeding (in magnitude) 1 / 10 . Summarizing, the stabilizing controller gains are such that k > 1 / 10 . The phase margin is greater than 90 degrees for all positive gains. Notice that it is hard to draw conclusions on the closedloop stability for negative controller gain through the Bode plot. Re Im Re Im0.215 10 stable 1 unstable pole stable  G ( j )  10 0.2 5 0.2 5 G ( j )90 2 Part 2: G ( s ) = 10( s +1) s ( s +5)( s +0 . 2) Now we add an integrator. Again, the system is stable for all positive controller gains. Now, however, the phase margin can get arbitrarily low, as the controller gain is increased (e.g., the phase on the Bode plot never drops below 180 degrees, but gets arbitrarily close). In addition, the closedloop system is unstable for all negative controller gains. Summarizing, the closedloop system is stable for k > . Re Im Re Im0.215 10 stable 1 unstable pole  G ( j )  10 0.2 5 0.2 5 G ( j )90 3 Part 3: G ( s ) = 10( s +1) s ( s +5)( s . 2) Now we made one of the poles unstable. Notice the fact that now the Bode gain is negative. From the analysis of the Nyquist plot now you see that the closedloop system will be stable for controller...
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This note was uploaded on 03/17/2008 for the course COURSE 16.30 taught by Professor Frazzolli during the Spring '07 term at MIT.
 Spring '07
 Frazzolli

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