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 Commenti
Seikh Rana
il 22 Apr 2018
Convolution (random position zero) in matlab plz help me
Seikh Rana
il 22 Apr 2018
</matlabcentral/answers/uploaded_files/114156/convooooo.JPG> this picture problem solved in matlab plz help me
Risposte (3)
psyx21
il 29 Apr 2011
1 voto
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
il 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 Commenti
gedaa
il 21 Feb 2011
Paulo Silva
il 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
il 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
il 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
il 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
il 21 Feb 2011
Paulo Silva
il 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
il 23 Feb 2011
Arpan Patel
il 25 Feb 2021
0 voti
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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