Hey iqbalmmz.
What is the u(t) function?
Hi everyone
This question is from my Signals and Systems class, its an EE course so I'm not sure if this question is appropriate for this "Applied Math" section but maybe someone here can answer me with a concise answer.
Now I've done a bunch of convolution problems before but I always seem to have trouble with the u(t) term, I understand its related to the limits of integration but I don't understand how to get rid of it or multiply it out with other u(t) or u(t tau) terms. There's so much I'd like to say here but I know the human attention span is limited so if someone could try to explain this to me, preferably solve it out in detail (please don't skip any steps). I've also attached my work (#58)to this post which has the equation and a worked out similar problem (#48), this similar problem deals with cos(x) and just 1 u(t) term so I solved it relatively quickly.
Any help or assistance will be much appreciated.

iqbalmmz
the u(t) function is the unit step function used to manipulate limits of integration.
for convolution the u(t) and the u(t tau) change the limits of integration,
if you look at number 48 in the attachement I posted, it shows how:
u(t) is 1 when t is 0 or greater, so the lower limit changes from infinity to zero and
u(ttau) is 1 when t is greater than tau and zero when t is less than tau so the upper limit changes from +infinity to t
I hope I'm making sense perhaps I should repost this question in a different section but I'm not sure where
Well it's in respect to tau, tau is just the displacement between the 2 signals when convoluting them. So it's all in respect to the time domain. By your question I can tell you are familiar with this, please help...
If its in the time domain I would suggest using h(tau  t) = (u(tau  t0  u(tau  t + 1)) and finding integral (0,t) h(tau  u)x(u)du.
Hint: Find the part of h(t) when it is zero and nonzero and then use this to get the integral in terms of Integral x(u)du over the right region. (Draw a diagram if you have to).
thanks chiro for your suggestion but i must admit, i don't understand what you're saying. Like how does t0 come into play (is t0 just the initial limit)? Why are you displacing h(t) as your convoluting signal as opposed to x(t)?
my initial question was that I don't know to deal with the u(t) term so drawing a diagram would not be possible, because I don't know what I am drawing.
Now use this information to find the convolution integral. If you have h(tau  u)*x(u)du with limits from 0 to tau, try and find where h(tau  u) = 1 and replace the h(tau  u)*x(u) with x(u) for that interval and integrate that function.
Note that x(t) involves h(t) so do the same thing again (find when h(t) = 1 for this interval and remove the h(t) term once you do this).
The final integral will involve only exponential functions which are easy to integrate.
Lets look at this in detail.
h(tau  t) = u(tau  t)  u(tau  t  1) which will be 0 when tau  t < 0, 1 when tau > t and tau  t  1 < 0 and 0 when tau  t  1 >= 0. So t > tau  1 < and t < tau has a common intersection of [tau1,tau], which means this becomes your limits for the integral at stage 1.
So far this means that convolution = Integral [tau1,tau] (exp(v) + exp(3v))*u(v)dv. Since u(v) = 1 for v > 0, if tau >= 1 then the integral is just (exp(v) + exp(3v))dv but if tau < 1 then you have to calculate an extra term. Lets just for now, assume tau > 1.
This means our final integral will be Convolution_Function(tau) = Integral [tau1,tau] [exp(v) + exp(3v)]dv which is easy to calculate.
If tau < 1, then you have to find when u(v) = 0 and when u(v) = 1.