How to solve a trasfer function absolute value whit the tf function command.

I'm tryng to implement a control system method in matlab but I have a problem in plotting a curve so I don't find a solution to compute the absolute value of a tranfer function declared with the classical command "tf". The function "abs()" is not useful so I ask you which the way to solve the problem. Thank you so much.

[Azzi Abdelmalek]

what does mean: a trasfer function absolute value

[Graz Mand]

An absolute value of a transfer function in the Laplace domain, sorry.

[Azzi Abdelmalek]

Ok, but what does that mean mathematically ?

[Graz Mand]

Not mathematically! I'm tryng to implement a control system method but the curves I want to plot, are in function of the absolute value of other functions.

I want to plot, for example C1=|P1|/(1-|P2|);

the problem is that P1 e P2 are transfer function too! Please help me! It's important for my work..

[Graz Mand]

If I want to plot the Bode diagram of P1 or P2 is not a problem, but that's for C1..

Answer 1

[Azzi Abdelmalek]

Example

N=1;
D=[1 10]
model=tf(N,D)  % your transfer function

Now what do you mean by: An absolute value of a transfer function

[Azzi Abdelmalek]

If you want to plot the abs of your transfer function

p=0:0.1:10;
tf1=1./(p+10)
tf1a=abs(tf1)
plot(p,tf1)

%or for example

N=[1 2]
D=[1 2 3]
p=0:0.1:10
f=abs(polyval(N,p)./polyval(D,p))
plot(p,f)

[Graz Mand]

Thanks Azzi, but I think your way don't consider p like a Laplace domain variable, i.e. p=jw where w is \omega, the frequency..what do you think?

[Azzi Abdelmalek]

I don't know exactly what you want, but in the above case, by replacing p by jw you will get a frequency response. You can do it like this

N=[1 2];
D=[1 2 3];
w=-10:0.01:10;
p=w*j;
f=abs(polyval(N,p)./polyval(D,p));
plot(w,f)

[Graz Mand]

It's the code with your solution: it doesn't work, I don't know the reason the outpout f is a constant.

function [] = plotting1(P1,P2)
[num1,den1]=tfdata(P1,'v');
[num2,den2]=tfdata(P2,'v');
w=0:0.01:100;
p=w*j;
P1=abs(polyval(num1,p)./polyval(den2,p));
P2=abs(polyval(num2,p)./polyval(den2,p));
f=P1/(1-P2)
plot(w,f)

[Graz Mand]

I want to plot the function: C1=|P1|/(1-|P2|);

[Graz Mand]

Anyways with p or w the plot of P1 results the same one..

[Azzi Abdelmalek]

It's not f=P1/(1-P1) but

f=P1./(1-P1)
plot(w,f)

Answer 2

[Negin Abaeian]

Hey Graz.

I can see the post belongs to long ago. But did you ever find the answer to this question? I'm trying to plot the exact same transfer function for an Audio application.