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Unformatted text preview: Problem 1 1.1 Gain Margin A PI Controller with zero at- 1 (like the assignment told us) is given by C ( s ) = k p s + 1 s , (1) so that our open loop looks like that: T ol ( s ) = P ( s ) C ( s ) = k p 50 ( s + 1)( s + 5)( s + 12) s + 1 s = k 1 s ( s + 5)( s + 12) (2) (I absorbed the 50 into k p , so k = 50 k p ). Youll find in the course of your (hopefully long) control engineering careers that using the zero of the PI Controller to compensate the slowest pole of the Plant is a quite common thing to do. But youll also see in Problem 3 that it can be totally wrong to do that. It should be clear by now that the closed-loop transfer function is then given by T cl ( s ) = T ol ( s ) 1 + T ol ( s ) = Numerator( T ol ( s )) Numerator( T ol ( s )) + Denominator( T ol ( s )) (3) and thus whether or not we have a stable closed loop hinges only on the closed-loop denominator Numerator( T ol ( s )) + Denominator( T ol ( s )) = k + s ( s + 5)( s + 12) = s 3 + 17 s 2 + 60 s + k. (4) Using the fact given to you on the sheet ( s 3 + as 2 + bs + c is stable a,b,c > 0 and ab > c ) 17 60 > k k p < 17 60 50 = 20 . 4 . (5) 3 Points 1.2 Marginal stability in the Bode Plot The result of this Matlab code is shown in Figure 1 P = zpk(,-[1 5 12],50); C = tf([1 1],[1 0]);C = tf([1 1],[1 0]); bode(series(20.4*C,P),k); grid on; % --- this is only to make the lines thicker h=findobj(gcf,type,line,visible,on); set(h,LineWidth,2); % --- so they are visible when printed. The slope at the crossover frequency is -40dB/Decade, corresponding to a phase of- 180 . Since this is a marginally stable system, we can conclude that -40dB/Decade is more than gentle. 2 Points for the Plot, 2 more if you comment on slope and phase at crossover 1-60-40-20 20 40 60 Magnitude (dB) 10-1 10 10 1 10 2-270-225-180-135-90 Phase (deg) Bode Diagram Frequency (rad/sec) Figure 1: Bode Plot of the system at the margin of stability. Note that the crossover frequency coincides with the frequency of 6 H ( j ) =- 180 . 1.3 More Bode Plots Figure 2 shows the requested bode plots. Since the phase of a positive constant is 0 , the phase of the bode plot remains the same, no matter what the gain is. The magnitude however is raised or lowered. If the gain is 10 times the maximum gain, the magnitude is raised, so the crossover frequency moves into the region where the slope is almost -60dB/Decade and the phase is around -225 , so well below -180 . Conversely, the smaller gain lowers the magnitude and moves the crossover frequency into the area with slope around -20dB/Decade, phase around -115 . So it appears as if a phase of less than -180 at crossover renders the system unstable, and a phase of more than -180 at crossover indicates a stable system. 2 for the Bode Plot, 4 more for good explanations/comments 1.4 Lead Controller The requested Lead Controller has transfer function L ( s ) = 4 s +5 s +20 . Convince yourselves that it fulfills the....
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- Spring '10