Convolution integrals

[dfx]

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.

 

[dfx]

I think I understand now. For future reference, treat each each bit individually and remember a system in time remembers the past but doesn't know the future! Each separate bit is the sum of the previous responses, except the system knows what's happened before... if that makes sense.