Use the results from question 1, Test your TF functions (1a and 1b are Open Loop

Question

Use the results from question 1, Test your TF functions (1a and 1b are Open Loop TF, 1c is ClosedLoop) using the impulse and step forced responses. (Hint: matlab commands  impulse(Function) andstep(Function) ). Explain your results obtain above in terms of Percent overshoot %S, settling time Ts, steady state SS, Natural Frequency n and Damping Ration . Also explain base on your poles and zeros the behavior of your TF. Is there any difference between OpenLoop (OL) and ClosedLoop (CL)? Explain. What happens to the poles and Zeros of OL and CL TF?

Results:

Question1: To match results with quesiton

Explanation / Answer

>> G=tf([1 1],[1 2 0]);
H=tf([1 3],[1 4]);
sys1=series(G,H)
S1 = stepinfo(sys1,'RiseTimeLimits',[0.05,0.95])
sys2=parallel(G,H)
S2 = stepinfo(sys2,'RiseTimeLimits',[0.05,0.95])
sys3=feedback(G,H)
S3 = stepinfo(sys3,'RiseTimeLimits',[0.05,0.95])

Transfer function:
s^2 + 4 s + 3
-----------------
s^3 + 6 s^2 + 8 s

S1 =

        RiseTime: NaN
    SettlingTime: NaN
     SettlingMin: NaN
     SettlingMax: NaN
       Overshoot: NaN
      Undershoot: NaN
            Peak: Inf
        PeakTime: Inf

Transfer function:
s^3 + 6 s^2 + 11 s + 4
----------------------
s^3 + 6 s^2 + 8 s

S2 =

        RiseTime: NaN
    SettlingTime: NaN
     SettlingMin: NaN
     SettlingMax: NaN
       Overshoot: NaN
      Undershoot: NaN
            Peak: Inf
        PeakTime: Inf

Transfer function:
    s^2 + 5 s + 4
----------------------
s^3 + 7 s^2 + 12 s + 3

S3 =

        RiseTime: 9.1544
    SettlingTime: 12.2882
     SettlingMin: 1.2672
     SettlingMax: 1.3333
       Overshoot: 0
      Undershoot: 0
            Peak: 1.3333
        PeakTime: 33.0216

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