Design a Compensator with and without system bandwidth constraints GH ( s ) is a
ID: 1921887 • Letter: D
Question
Design a Compensator with and without system bandwidth constraints GH ( s ) is a minimum - phase system which has the Bode plot shown below. It is desired to have a phase margin of 45 degree s. There is no requirement on bandwidth or speed of response. The steady - state errors are satisfactory but must not increase. Design a compensator to achieve the desired phase margin without increasing the steady - state errors. Draw the Bode plot for the resulting closed loop system. Confirm that the specification are met. GH ( s ) is a minimum - phase system which has the Bode plot shown below. It is desired to increase the phase margin by 30 degree s beyond its present value and also to slightly increase the closed - loop system bandwidth. Design a compensator for this purpose. Draw the Bode plot for the resulting closed - loop Confirm that the specification are met.Explanation / Answer
In the question itself it is stated that there is no requirement on the bandwidth and speed of response but the steady state response will not increase the lead compensator satisfies the above given specifications and now we will design a lead compensator for the above given question step 1: Assume that lead compensator as,Gc(s)=kca((1+Ts)/(1+aTs)) step 2:(as the picture is not clear we assume the nearest values and by substituting the exact values we get exact answer) from the bode plot itself we can notice that ?gc=4 rad/sec, Phase margin P.M f1 =150 step 3: phase margin required fs=450 fm=fs-f1+e (where e is the margin of safety as cross over frequency may shift due to compensation,its value is e=50 to150) fm=fs-f1+e =450-150+50 =350 Step 4: sin fm =(1-a)/(1+a) sin 350 =(1-a)/(1+a) 0.5735= (1-a)/(1+a) 1-a=0.5735(1+a) a=0.271 step 5: -10log(1/a)=-10log(1/0.271)=-5.67db find the frequency at which the gain of the uncompensated system is -5.67db this is ?m=8rad/sec ?m=1/(Tva) 1/T=4.1646 T=0.2401 step 6: The two corner frequencies are ?c1=1/T and ?c2=1/aT step 7: assume k=10(as the picture is not clear) kc=k/a=10/.271=36.9 step 8: Gc(s)=kca((1+Ts)/(1+aTs)) =36.9x.271x(1+.2401s)/(1+0.065s) step 8: By taking the exact values draw the bode plot