Why is the convolution so different from the analytical answer?

5 次查看（过去 30 天）
Bradlee Harrison
Bradlee Harrison 2021-4-10
评论： Bradlee Harrison 2021-4-10
I am trying to graph the convolution of two functions.
t = linspace(0,5,1001);
xt = (exp(-t) + exp(-3*t)).*heaviside(t);
ht = (1.5.*exp(-2.*t) + 1.5.*exp(-4.*t)).*heaviside(t);
yt_n = conv(xt,ht);
yt_a = (2.*exp(-1.*t)-2.*exp(-4.*t)).*heaviside(t)
When comparing yt_n and yt_a, the results are very different. Though, they should be the same since yt_a is just the analytically derived output response for x(t) and y(t), Why is this?

请先登录，再回答此问题。

采纳的回答

Paul
Paul 2021-4-10
编辑：Paul 2021-4-10
When approximating the continuous convolution with a discrete convoluton, the discrete convolution needs to be multilplied by dt, which in this case is 0.005
plot(t,yt_a,t,yt_n(1:numel(t))*.005,'o'),grid

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Logical 的更多信息

标签

产品


版本

R2019b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by