System impulse response and Convolution by matlab
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Hi everyone, i am very new to matlab, and would like to konw how to obtain y(t) from below
Generate a unit step function as the input function, x(t), and an exponentially decay function as the impulse response function, h(t), such as h(t)=exp(-t/2) (note: 2 is the time constant of the system dynamic response). Using MATLAB to calculate the output of the system, y(t).
Thank you so much in advance
2 Kommentare
Seikh Rana
am 22 Apr. 2018
Convolution (random position zero) in matlab plz help me
Seikh Rana
am 22 Apr. 2018
</matlabcentral/answers/uploaded_files/114156/convooooo.JPG> this picture problem solved in matlab plz help me
Antworten (3)
psyx21
am 29 Apr. 2011
1 Stimme
hey friends what will be the peak acceleration response for 100g 6ms half sine pulse with zeta=0. I need a matlab script with natural frequency on x axis..thanks
Paulo Silva
am 20 Feb. 2011
t=0:0.1:10;
u=0*t;
u(t>=0)=1;
h=exp(-t/2);
y=u.*h;
plot(t,y)
8 Kommentare
gedaa
am 21 Feb. 2011
Paulo Silva
am 21 Feb. 2011
it's the same as
u=zeros(size(t))
It creates a variable with the same size as t but full of zeros. I learned that trick here at matlabcentral :)
David Young
am 21 Feb. 2011
Paulo - I just wonder if this is right. In the time domain, don't you need to convolve the input and the impulse response rather than multiplying them?
Paulo Silva
am 21 Feb. 2011
Yes you are correct, try
t=0:0.1:10;
u=0*t;
u(t>=0)=1;
h=exp(-t/2);
y=conv(u,h);
plot(y)
Paulo Silva
am 21 Feb. 2011
Correction for the amplitude and time scale
T=0.1;
t=0:T:10;
u=0*t;
u(t>=0)=1;
h=exp(-t/2);
y=conv(u,h);
plot(t,T*y(1:numel(t)))
gedaa
am 21 Feb. 2011
Paulo Silva
am 21 Feb. 2011
I'm not sure about the amplitude at
plot(t,T*y(1:numel(t)))
with the step function is good compared to
step(tf([1],[1 1/2])) but with the impulse it's plot(t,y(1:numel(t))) without the T, I can't figure out why that happens.
gedaa
am 23 Feb. 2011
Arpan Patel
am 25 Feb. 2021
0 Stimmen
t=0:0.1:10;
u=0*t;
u(t>=0)=1;
h=exp(-t/2);
y=u.*h;
plot(t,y)
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