# Delayed step/ impulse response

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Daniel Stankowski 2020년 1월 14일
답변: Paul 2020년 9월 20일
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

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### 답변(2개)

Raj 2020년 1월 14일
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.
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Bill Tubbs 2020년 9월 20일
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)

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Paul 2020년 9월 20일
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

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R2019b

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