Different results between Step and Stepinfo

Question

Gerard Carroll 2021-6-3

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回答： Sam Chak 2024-2-22

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Hi,

I'm running R2020b update 5.

This is my code:

s = tf('s');
G3 = (25/5.05)/(s^3+(101/5.05)*s^2+(505.2/5.05)*s+100/5.05);
roots_3 = roots([1 101/5.05 505.2/5.05 100/5.05]);
G3_c = 500.789*((s-roots_3(2))*(s+.0273103))/((s+29.356)*(s+0.01));
G3_final = G3*G3_c;
TS_3 = feedback(G3_final,1);
[y3,t3_2] = step(TS_3);
si = stepinfo(y3,t3_2);
step(TS_3)

This gives me this graph:

However, my stepinfo gives:

        RiseTime: 0.1871
    SettlingTime: 1.0105
     SettlingMin: 0.9213
     SettlingMax: 1.1644
       Overshoot: 18.5698
      Undershoot: 0
            Peak: 1.1644
        PeakTime: 0.4386
        

These are very different. Any help?

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Answer 1

Vidhi Agarwal 2024-2-22

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Hi Gerard,

I understand you encountering discrepancy between the values of percentage overshoot in step-response graph and stepinfo() data.

This is occurring because the response has not fully settled here due to the near cancellation of the slow pole/zero pair at s=-0.027, using the (y,t) data out of STEP to compute the characteristics will be inaccuracies. To fix this, you can either use:

si = stepinfo(sys)

or specify a large enough final time for STEP:

[y,t] = step(TS_3,200);

si = stepinfo(y,t)

I hope this solves your problem.

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Answer 2

Sam Chak 2024-2-22

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Hi @Gerard Carroll

The main reason behind the difference in the step-response characteristics between Approach 1 (from Transfer function) and Approach 2 (from Response data) is that you didn't compare them in a consistent manner. Additionally, it's worth noting that the computation of response characteristics has changed since R2021b.

To ensure accurate comparison, here's the correct procedure:

%% Transfer function

s = tf('s');

G3 = (25/5.05)/(s^3+(101/5.05)*s^2+(505.2/5.05)*s+100/5.05);

roots_3 = roots([1 101/5.05 505.2/5.05 100/5.05]);

G3_c = 500.789*((s-roots_3(2))*(s+.0273103))/((s+29.356)*(s+0.01));

G3_final = G3*G3_c;

TS_3 = feedback(G3_final, 1)

TS_3 = 2479 s^2 + 2.11e04 s + 574.5 ----------------------------------------------------------- s^5 + 49.37 s^4 + 687.7 s^3 + 5443 s^2 + 2.172e04 s + 580.3 Continuous-time transfer function.

%% Steady-state response (need this in Approach #2)

ssr = dcgain(TS_3)

ssr = 0.9900

%% Approach 1: Obtain step-response characteristics from transfer function

si_1 = stepinfo(TS_3);

%% Approach 2: Obtain step-response characteristics from the response data

[y3, t3_2] = step(TS_3);

si_2 = stepinfo(y3, t3_2, ssr);

%% Comparing the characteristics between both approaches

stepInfoTable = struct2table([si_1, si_2]);

stepInfoTable = removevars(stepInfoTable, {...

'SettlingMin', 'SettlingMax', 'Undershoot', 'PeakTime'});

stepInfoTable.Properties.RowNames = {'Approach 1', 'Approach 2'};

stepInfoTable

stepInfoTable = 2×5 table

RiseTime TransientTime SettlingTime Overshoot Peak ________ _____________ ____________ _________ ______ Approach 1 0.18887 1.0563 1.0563 17.614 1.1644 Approach 2 0.18887 1.0563 1.0563 17.614 1.1644

%% Plot results

step(TS_3), hold on

plot(t3_2, y3), grid on