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Plotting responses other than step responses to transfer function

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Aleem Andrew
Aleem Andrew on 10 Apr 2021
Edited: Paul on 10 Apr 2021
I am trying to plot the response of a transfer function to an arbitrary input. The step response can be plotted as shown below.
T = tf(164.6,[1 13 32 184.6])
The step response plot is a plot of the inverse Laplace of 1/s * tf(164.6,[1 13 32 184.6]) , but trying to plot this directly without the step function using the command
syms s
T = tf(164.6,[1 13 32 184.6])
generates a blank plot. Is there a way
to plot the response directly without using the step function that would allow you to plot the response to any input?

Answers (1)

Paul on 10 Apr 2021
Edited: Paul on 10 Apr 2021
Looks like an fplot() confusion issue. This code results in the expected plot:
>> syms t; y(t) = ilaplace(1/s*poly2sym(T.Numerator{:},s)/poly2sym(T.Denominator{:},s));
>> plot(0:.01:9,double(y(0:.01:9)))
However, the result of that ilaplace() operation results in fairly complicated expression in terms of roots(). So this approach might not be practical for even moderately complicated systems.
A partial fraction approach may be feasible to get a cleaner representation of y(t):
syms s t
T(s) = 164.6/(s^3 + 13*s^2 + 32*s + 184);
U(s) = 1/s;
Y(s) = U(s)*T(s);
Y(s) = partfrac(Y(s),'FactorMode','real');
y(t) = sum(real(ilaplace(children(Y(s)))));
For the response of an LTI system to an arbitrary input specificed in the time domain, check out:
doc lsim
If the input is speicified in terms of its Laplace transform, then the output can be plotted with
where U is the Laplace transform of u(t) (expessed in tf or zpk form, most typically).
Sometimes it's best to take advantage of the linearity of LTI systems: break the input up into a sum of simpler inputs, get the output to each simple input, and then sum those ouputs to get the system output to the original input.

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