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Plotting the convolution of two signals

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Dennis
Dennis 2014 年 10 月 23 日
コメント済み: Celeste MacNeil 2018 年 10 月 14 日
I am given two functions x(t) =5[u(t+1)-u(t-1)] and h(t)=u(t-1)-u(t-7). I am asked to convolve these two signals and plot the result in the range -3 to 10. Here is the code that I wrote:
t=-3:0.1:10; t_c=-3:0.05:10; h_t=heaviside(t-1)-heaviside(t-7); x_t1=5.*(heaviside(t+1)-heaviside(t-1)); c_x_h=conv(x_t1,h_t); figure(1) plot(t_c,c_x_h)
However, since this is a simple convolution I verified it by hand and it does not look like the plot i get in MATLAB. The max value should be 10 but in MATLAB i get 100. Also, the duration of the function should be 8 but i get 4 with MATLAB. Please help.

採用された回答

Rick Rosson
Rick Rosson 2014 年 10 月 23 日
Please try:
t_c = -6:0.1:20;
and:
dt = t(2) - t(1);
c_x_h = dt*conv(x_t1,h_t);
  4 件のコメント
Celeste MacNeil
Celeste MacNeil 2018 年 10 月 14 日
What do you mean dt = t(2)-t(1)?
Celeste MacNeil
Celeste MacNeil 2018 年 10 月 14 日
I assume you just mean the step?

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その他の回答 (1 件)

IKRAM jebali
IKRAM jebali 2017 年 3 月 31 日
Hello! Thank you. I'had the same problem. Me too amnclear as to why the dt is needed though and why t_c needed to be extended to -6 to 20. I use it but i don't know why? There is any answer please. Thank you.
  1 件のコメント
Aswin Farzana Mohamed Ansar
Aswin Farzana Mohamed Ansar 2017 年 10 月 20 日
Because the length of the convolution of two matrices of ,length m and n will be m+n-1. inorder to get the same vector size, you need to do this.

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