Matlab PI control first order system

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MikeSv 2017-11-13

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评论： Robert U 2019-8-14

在 MATLAB Online 中打开

Hi everyone. Iam quite new to Control theory and I have a question regarding PI Control using Matlab.

I have a first order transfer function :

G(s) = 5/(s+1)

and I try to use Matlab to form a feedback loop using a PI controller with the following code:

 s = tf('s');
Kp = 1;
Ki = 1;
C = Kp + Ki/s
G = tf([5],[1 1])
figure(1)
step(G)
H = feedback(C*G,1)
figure(2)
step(H)

When I plot the step response of G, it goes from zero to five like expected, but when I plot the feedback response it only goes from zero to 1. Changing Kp and Ki does not have any effect.

Any help would be great!

Thanks in advance,

/M

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Divyani Rathod 2019-2-1

How can i calculate Kp ki value for Peak overshoot <=11%, settling time<=5s. Transfer function is 3.24/15.24s+1 (My calculated value is, kp = 7.2152 and ki = 9.1061, zeta = 0.5749, w_n = 1.3916); (using state space and without state space) and also How can i calculate Fractional kp and ki.

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Answer 1

Robert U 2017-11-13

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编辑：Robert U 2017-11-13

Hello MikeSv:

The system response is correct from my point of view. The desired response value of step-function is "1". The uncontrolled system G is responding with a gain of "5". The controlled system instead is responding with "1" as desired.

Edit:

The change of Kp and Ki have - of course - an effect. Due to automatic scaling it might be you did not mention.

Kind regards,

Robert

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bryce howard 2019-7-24

Hey Robert, how did you get Ki and Kp to display like that?

Robert U 2019-8-14

在 MATLAB Online 中打开

% define transfer functions
s = tf('s');
C =@(Kp,Ki) Kp + Ki/s;
G = tf([5],[1 1]);
% define parameter combinations
Kp = [1,2,4,1,1];
Ki = [1,1,1,2,4];
% create figure
figure
ah = axes;
hold(ah,'on');
% loop through given combinations of Kp and Ki
for ik = 1:length(Kp)
    H = feedback(C(Kp(ik),Ki(ik))*G,1);
    step(H)
end
% create cell array of legend entries
cLegend = arrayfun(@(Kp,Ki) sprintf('Kp = %i, Ki = %i',Kp,Ki),Kp,Ki,'UniformOutput',false);
% add legend to figure
legend(ah,cLegend)

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Answer 2

MikeSv 2017-11-13

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Hi and thank you so much for the replies!

What I don't understand is, when I have my open loop system, and I want the output to be 5 when the input is 1 (a unit step). What happens to my DC Gain in the closed loop? I expected my output to be 5 as well in the closed loop, just with in example a faster rise time when changing the value of the P controller.

Thanks again,

M

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Answer 3

partha das 2019-2-5

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编辑：partha das 2019-2-5

OLTF is 5/(s+1). In this case DC gain is 5 (obtained by setting s=0 in the OLTF).

Your CLTF is 5/(s+1) when you use Kp=1 and Ki=1. So in this case DC gain is 1. As a result when you apply an unit step input to this CLTF, your steady state output will be 1 i.e. the steady state error (reference step input of magnitude 1 - steady output of 1) is 0. In fact for any values of Kp and Ki, you will get a steady state error of 0 because the PI controller always tracks the reference input perfectly.