# Finding the convolution analytically (or graphically)

• Apr 18th 2013, 05:56 PM
iqbalmmz
Finding the convolution analytically (or graphically)
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.

Quote:

What is the convolution of the follwoing signals?
x(t) = (exp(-t) + exp(-3t))u(t)
h(t) = (u(t) - u(t-1))

I've attached a screenshot picture of these signals in case what I wrote is unclear:
Attachment 28013

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

Attachment 28015
• Apr 19th 2013, 05:45 AM
chiro
Re: Finding the convolution analytically (or graphically)
Hey iqbalmmz.

What is the u(t) function?
• Apr 19th 2013, 02:25 PM
iqbalmmz
Re: Finding the convolution analytically (or graphically)
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(t-tau) 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
• Apr 19th 2013, 07:23 PM
chiro
Re: Finding the convolution analytically (or graphically)
Are you looking at the convolution with respect to the frequency space or respect to the time space?
• Apr 19th 2013, 07:38 PM
iqbalmmz
Re: Finding the convolution analytically (or graphically)
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...
• Apr 19th 2013, 07:44 PM
chiro
Re: Finding the convolution analytically (or graphically)
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 non-zero 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).
• Apr 19th 2013, 09:21 PM
iqbalmmz
Re: Finding the convolution analytically (or graphically)
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.
• Apr 19th 2013, 09:32 PM
chiro
Re: Finding the convolution analytically (or graphically)
Hint: When does h(t) = 0 and when does it equal 1?
• Apr 19th 2013, 09:48 PM
iqbalmmz
Re: Finding the convolution analytically (or graphically)
h(t)=0 when t is 0 and h(t)=1 when t is 1? i don't get what your saying, sorry
• Apr 19th 2013, 10:00 PM
chiro
Re: Finding the convolution analytically (or graphically)
No: h(t) should be 1 in between 0 and 1 and 0 outside of this.

Try graphing h(t) in terms of u(t).
• Apr 19th 2013, 10:19 PM
iqbalmmz
Re: Finding the convolution analytically (or graphically)
okay I attached a picture of the graph you suggested I attempt, so what should I do now?

Attachment 28039
• Apr 19th 2013, 10:25 PM
chiro
Re: Finding the convolution analytically (or graphically)
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.
• Apr 19th 2013, 10:52 PM
iqbalmmz
Re: Finding the convolution analytically (or graphically)
okay attached is the final integral i came up with, honestly I'm still fuzzy about this. Is what I did correct or almost there? Please let me know

Attachment 28042
• Apr 19th 2013, 11:26 PM
chiro
Re: Finding the convolution analytically (or graphically)
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 [tau-1,tau], which means this becomes your limits for the integral at stage 1.

So far this means that convolution = Integral [tau-1,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 [tau-1,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.