フィルターのクリア

using hold on function in step response with subplot

13 ビュー (過去 30 日間)
BALA GUGA GOPAL S
BALA GUGA GOPAL S 2019 年 2 月 24 日
コメント済み: BALA GUGA GOPAL S 2019 年 2 月 26 日
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')

採用された回答

Walter Roberson
Walter Roberson 2019 年 2 月 25 日
Code attached. Mostly it was moving the hold()
  1 件のコメント
BALA GUGA GOPAL S
BALA GUGA GOPAL S 2019 年 2 月 25 日
編集済み: BALA GUGA GOPAL S 2019 年 2 月 25 日
Thank you for your help, It got worked with your method.

サインインしてコメントする。

その他の回答 (1 件)

GMD Baloch
GMD Baloch 2019 年 2 月 24 日
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
  6 件のコメント
GMD Baloch
GMD Baloch 2019 年 2 月 26 日
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
BALA GUGA GOPAL S
BALA GUGA GOPAL S 2019 年 2 月 26 日
thank you this too got worked out.Thank you for your help.

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeLine Plots についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by