how to sum unit function and step function

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chae hyeon shin
chae hyeon shin 2020 年 5 月 1 日
回答済み: VIVEK 2022 年 9 月 20 日
I want to calculate the convolution of x(t) and h(t).
Here is my code :
clear;
t = [ -10 : 0.01 : 10 ];
xt = ( t >= -2 ) & ( t <= 2) + (t == 1)
ht = ( t >= -1 ) & ( t <= 2) + 2*(t == 0) + (t == 3)
plot(t,ht);
ylim([-0.5 2]);
yt = conv(xt,ht,'same');
t1 = [-inf, inf];
plot(t1,yt);
I'm not sure that xt and ht are correct.
Moreover, there is an error with last line.
HELP ME
  1 件のコメント
Ameer Hamza
Ameer Hamza 2020 年 5 月 1 日
What does the upward arrow represent?

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回答 (2 件)

Ajay Pattassery
Ajay Pattassery 2020 年 5 月 4 日
I assume you are trying to do the convolution of xt, ht as attached in the image.
t = ( -10 : 0.01 : 10 );
xt = (( t >= -2 ) & ( t <= 2)) + (t == 1);
ht = (( t >= -1 ) & ( t <= 2)) + 2*(t == 0) + (t == 3);
subplot(3,1,1);plot(t,xt);
ylabel('xt');
subplot(3,1,2);plot(t,ht)
ylabel('ht');
yt = .01*conv(xt,ht,'same');
subplot(3,1,3);plot(t,yt);
ylabel('yt');
I have just edited your above code.
If you are doing convolution of continuous signals by approximating as above in MATLAB, you need to multiply the output of conv with dt. In your case .01. What you are basically doing is approximating the continuous signal with boxes of width .01 and doing the discrete convolution. Hence while doing convolution, the integration can be achieved by mulitplying with dt.

VIVEK
VIVEK 2022 年 9 月 20 日
t = ( -10 : 0.01 : 10 );
xt = (( t >= -2 ) & ( t <= 2)) + (t == 1);
ht = (( t >= -1 ) & ( t <= 2)) + 2*(t == 0) + (t == 3);
subplot(3,1,1);plot(t,xt);
ylabel('xt');
subplot(3,1,2);plot(t,ht)
ylabel('ht');
yt = .01*conv(xt,ht,'same');
subplot(3,1,3);plot(t,yt);
ylabel('yt');

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