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using hold on function in step response with subplot

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BALA GUGA GOPAL S
BALA GUGA GOPAL S 2019-2-24
评论： BALA GUGA GOPAL S 2019-2-26
I am trying to plot step reponse and impulse reponse of a sys in a single plot,using suplot and lengend, but i too have different values of z to be ploted (means i am using hold on fun). when i try to use subplot with the created code it's plotting for only one value of z, but i want such that all the three values of z has to plotted for both the step and impulse function in subplot, I apperciate your help.
s=tf('s');
z=[3 6 12];
hold on
for i=1:3
zc=z(i);
g=(15/zc)*((s+zc)/(s^2+3*s+15));
subplot(2,1,1)
step(g)
subplot(2,1,2)
impulse(g)
end
legend('z = 3','z = 6','z = 12')

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采纳的回答

Walter Roberson
Walter Roberson 2019-2-25
Code attached. Mostly it was moving the hold()

更多回答（1 个）

GMD Baloch
GMD Baloch 2019-2-24
You just have to a little bit of modification in your code. I have done it and it is producing the following result
untitled.png
I hope that you wanted the above output. The modified code is given below
s=tf('s');
z=[3 6 12];
for i=1:3
zc=z(i);
g=(15/zc)*((s+zc)/(s^2+3*s+15));
subplot(2,2,i)
step(g)
hold on
impulse(g)
legend('z = 3','z = 6')
end
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GMD Baloch
GMD Baloch 2019-2-26
You can try the following code
s=tf('s');
z=[3 6 12];
subplot(2,1,1)
hold on
for i=1:3
zc=z(i);
g=(15/zc)*((s+zc)/(s^2+3*s+15));
step(g)
end
legend('z = 3','z = 6','z = 12')
subplot(2,1,2)
hold on
for i=1:3
zc=z(i);
g=(15/zc)*((s+zc)/(s^2+3*s+15));
impulse(g)
end
legend('z = 3','z = 6','z = 12')
The output for this code is shown below
untitled.png

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