# Delayed step/ impulse response

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Daniel Stankowski on 14 Jan 2020
Answered: Paul on 20 Sep 2020
Hi everyone,
I would like to plot an impulse and step response of some arbitrary system sys1.
Normally I would use just a command impulse(sys1,t) or step(sys1,t) however the signals in my case are delayed.
1) The impulse is represented as: diract(t-5)
2) step is represeneted as: 1(t)-2*1(t)(t-tsw)
I know some people use an exponential function for it, but I am unsure why and how,
I would greatly appreciate any help

Raj on 14 Jan 2020
You have your system model and input signals equation. You can just generate your input signals and get the system response for those signals using lsim command.
##### 2 CommentsShowHide 1 older comment
Bill Tubbs on 20 Sep 2020
It seems a shame that this wouldn't be possible as an option in stepDataOptions, similar to specifying the step amplitude. Something like this perhaps:
opt = stepDataOptions('StepTime',10,'StepAmplitude',2);
step(sys,opt)

Paul on 20 Sep 2020
Use the time invariance and lineraity properties of an LTI system. One approach is to generate the the nominal impulse or step response and then use apppropriate time shifing and superposition of the nominal outputs.
sys1 = tf(1,[1 1]);
tsw = 4.3;
t = 0:.01:10; % Important to define tnom such that tsw is one element of t. verify this
sum(t == tsw);
% nominal impulse response
ynom = impulse(sys1,t);
% delayed impulse response
yimp = 0*ynom;
yimp(t >=tsw) = ynom(1:sum(t >= tsw));
% nominal step response
ynom = step(sys1,t);
% first term
y1 = ynom;
% delayed and scaled second term
y2 = 0*ynom;
y2(t >=tsw) = 2*ynom(1:sum(t >= tsw));
% total output
ystep = y1 + y2;
plot(t,[yimp ystep]),grid

R2019b

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