- #1
dfx
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Homework Statement
Right I'm having a lot of problems with convolution in general. I'll give an example of a question that I understand and why I think I understand it... and then one I don't at all.
So:
Consider a system with the impulse response g(t) = 0 for t<0, e^{-5t} for t \geq 0.
Find the output for input f(t) = H(t) (step function).
So y(t) = \int g(t-\tau)f(\tau)d\tau between t and -\infty
= \int e^{-5(t-\tau)}H(\tau)d\tau between t and -\infty
= \int e^{-5(t-\tau)}.1.d\tau between t and 0. This is because the step function takes a value of 1 for t >= 0 hence the limits change to 0 and t?
Now for the same system say you have an input of:
f(t) = (0, t<0) ... (a) ; (v, 0<t<k) ... (b); (0, t>k) ... (c)
To find the output you need to perform 3 integrals: 1 for (a), 1 for (b) and 1 for (c).
According to my course notes these 3 integrals are:
1. part (a):
y(t) = \int g(t-\tau)0d\tau between t and -\infty
This sort of makes sense but why isn't the upper limit 0 as surely (a) is only for (t<0).
2. part (b):
y(t) = \int g(t-\tau)0d\tau between 0 and -\infty
+ \int g(t-\tau)vd\tau between t and 0.
I don't understand this bit. Firstly, why the initial bit between 0 and -infty ... surely it's unnecessary as the amplitude is 0 as in the previous example for the step input we didn't bother with t<0. Secondly for the second bit with amplitude v why on Earth are the limits t and 0. Surely they should be k and 0 as the amplitude is only v between k and 0?
3. part (c):
y(t) = \int g(t-\tau)0d\tau between 0 and -\infty
+ \int g(t-\tau)vd\tau between k and 0.
+ \int g(t-\tau).0.d\tau between t and k.
How are the first 2 lines in this integral even relevant to part (c) which is only for t>k where the amplitude is 0 (ie the way I see it only the third line is relevant).
Sorry for the rather unconventional post but this has been confusing me for a few months and having spent hours on it I just can't seem to understand what on Earth is going on! Any help/feedback much appreciated. Cheers.