Design a Compensator with and without system bandwidth constraints GH ( z ) is a

Question

Design a Compensator with and without system bandwidth constraints GH ( z ) 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 node plot for the resulting closed loop system. Confirm that the specification are

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)=kc((1+Ts)/(1+Ts))

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 1 =150

step 3:

phase margin required s=450  

m=s-1+

(where is the margin of safety as cross over frequency may shift due to compensation,its value is =50 to150)

m=s-1+

   =450-150+50

   =350

Step 4:

  sin m =(1-)/(1+)

sin 350 =(1-)/(1+)

0.5735= (1-)/(1+)

1-=0.5735(1+)  

=0.271  

step 5:

-10log(1/)=-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/(T)

1/T=4.1646

T=0.2401

step 6:

The two corner frequencies are c1=1/T and c2=1/T

step 7:

assume k=10(as the picture is not clear)

kc=k/=10/.271=36.9

step 8:

Gc(s)=kc((1+Ts)/(1+Ts))

   =36.9x.271x(1+.2401s)/(1+0.065s)

step 8:

By taking the exact values draw the bode plot

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---