hw12_sol - MEEN 364 Fall 2004 Homework Set 12 Due December...

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MEEN 364 Homework Set 12 Fall 2004 November 22, 2004 1 Homework Set 12 – Due December 2, 2004 @ 5:00 PM (1) a) Determine the Gain Margin (GM) and the Phase Margin (PM) of the following transfer function, (do not use MATLAB): () 100 4 Gs ss = + b) For the transfer function given, determine the gain margin and phase margin using MATLAB. Also determine whether the negative-feedback system with a unity controller gain is stable or unstable. Justify your answer. Solution: a) To obtain the phase margin, first we determine the frequency at which the magnitude of G(j ω ) is equal to unity: 100 ()1 4 Gj jj ωω =→ = + 1 22 2 100 1 10000 16 16 = + + 42 16 10000 0 9.6083 /sec rad +− = = The PM is given by adding 180 deg to the phase of G(j ) at that frequency: 1 () 9 0 t a n 4 ⎛⎞ =− ⎜⎟ ⎝⎠ 1 ( ) 180 90 tan 9.6083/ 4 PM G j =+ = PM = 22.6 ° For very small , the phase tends to -90° - 0 = - 90°, and for very high , it tends to -90° - 90° = - 180°. For other frequencies, the phase is higher. Therefore, since the phase curve of the transfer function does not cross the –180 ° line, the gain margin GM of the system is infinity. b) From MATLAB: num=100; den=conv([1 0],[1 4]); sys=tf(num,den); margin(sys);
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MEEN 364 Homework Set 12 Fall 2004 November 22, 2004 2 Bode Diagram Frequency (rad/sec) Phase (deg) Magnitude (dB) -60 -40 -20 0 20 40 Gm = Inf, Pm = 22.603 deg (at 9.6081 rad/sec) 10 0 10 1 10 2 -180 -150 -120 -90 Gain Margin: GM = infinite. Phase Margin: PM = 22.6 deg (at 9.6 rad/sec). Since the Phase Margin (PM) is positive, the negative-feedback system with a unity controller gain is stable. (2) Consider the open loop transfer function () 4 k Gs ss = + a) Determine the gain k such that the phase margin of the Closed-Loop System with Unity Negative Feedback and the above open loop transfer function is 30 ° . (Hint: Consider only positive k ) b) What is the gain margin in this case? ( Note : Do not use MATLAB. Show the details of your work for full credit) Solution: a) The Bode form of the transfer function is:
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MEEN 364 Homework Set 12 Fall 2004 November 22, 2004 3 2 () 4 k Gj j ω = ( 1 ) The phase margin is 30°: 1 4 ( ) 180 tan 180 30 PM G j ⎛⎞ =+ = −− + = ⎜⎟ ⎝⎠ 41 tan 150 4 3 3 −= ° = = With this frequency, the gain G(s) should be 1: 22 2 2 1 16 3072 4 55.42 k k j k ωω =→ = + = = ( 2 ) b) To get the gain margin, we make PM = 0: 1 4 ( ) 180 tan 180 =− = =∞
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hw12_sol - MEEN 364 Fall 2004 Homework Set 12 Due December...

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