Step Response of a transfer Function with a Multiple Term Numerator.

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Salik Mallick
Salik Mallick 2020 年 3 月 13 日
回答済み: Mahesh Taparia 2020 年 3 月 16 日
Obtain the step response of the transfer function given below:
s s + 3 s + 2
T(S) = ------------------------------------
s + 1 s + 11 s + 2 s + 7
num = input('enter the numerator of the transfer function') %% -- Obtain the step response of the transfer function given below: --
den = input('enter the denominator of the transfer function')
step(num,den)
Is there a function that converts the numerator into a single term? The question says to Obtain the step response of the transfer function given.
  2 件のコメント
Star Strider
Star Strider 2020 年 3 月 13 日
Please supply all the missing multiplication operators, and any necessary parentheses.
Salik Mallick
Salik Mallick 2020 年 3 月 13 日
(s) (s + 3) (s + 2)
T(S) = ------------------------------------
(s + 1) (s + 11) (s + 2) (s + 7)

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回答 (1 件)

Mahesh Taparia
Mahesh Taparia 2020 年 3 月 16 日
Hi
You can use 'conv' function to find the terms in polynomial form (or in a single term). For your case, you can follow the following line of code
a=[1 0];
b=[1 3];
c=[1 2];
Num1=conv(a,b);
Num=conv(Num1,c)
p=[1 1];
q=[1 11];
r=[1 2];
s=[1 7];
Den1=conv(p,q);
Den2=conv(r,s);
Den=conv(Den1,Den2);
TF=tf(Num,Den);
step(TF) %%%%% It will give step response
Hope it will help.

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