Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to convolve two square functions (literally square pulses). In my opinion following calculation is correct, but my teacher said, it isn't (I did not yet have the time to speak to him again). Is my teacher wrong or am I missing something?

Since I have to express the functions with a dummy variable, flip and shift one I get the following: 1

So far, so good .. When I write down the integral now it looks like $\int_0^t 2*1 dT$ From this I'm implying that the convolution y(t) goes from 0 to 4 while t < 2. Doing this for every section this results in:
2

I can't find an error, I hope someone can tell me whether this is right or wrong.

share|improve this question
    
Could you please add the two functions to your post? Also, if your teacher left any specific comments, maybe you could add those as well? –  chris Jun 3 '12 at 18:07
    
Ok, I could just add 2 hyperlinks in my main post here the functions: 1.) i.imgur.com/jN1tI.png 2.) i.imgur.com/mY8Qf.png My teacher said casually that he thinks the convolution has a maximum of 2, not 4. –  Marco Jun 3 '12 at 18:10
    
Ah, when you said square functions I didn't realize you literally meant a square. Thanks for the clarification. –  chris Jun 3 '12 at 18:13
    
Oh, yes, I realize that this is misleading, i'm editing –  Marco Jun 3 '12 at 18:14
add comment

1 Answer

It looks correct to me.

I am assuming that $x(t) = 2 \cdot 1_{[0,2]}(t)$, $h(t) = 1_{[0,4]}(t)$. Then we have $(x * h)(t) = 2 \int 1_{[0,2]}(\tau) 1_{[0,4]}(t-\tau) \, d \tau = 2 \int 1_{[0,2]}(\tau) 1_{[t-4,t]}(\tau) \, d\tau = 2 \int 1_{[0,2]\cap [t-4,t]}(\tau) d\tau$, from which we get $(x * h)(t) = 2 m([0,2]\cap [t-4,t])$, where $m(\cdot)$ is the length of the interval.

There are five intervals to consider when evaluating the convolution:

$t\in (-\infty,0)$: $(x * h)(t) = 0$.

$t \in [0,2)$: $(x * h)(t) = 2t$.

$t \in [2,4]$: $(x * h)(t) = 4$.

$t \in (4,6]$: $(x * h)(t) = 12-2t$.

$t \in (6, \infty)$: $(x * h)(t) = 0$.

This is the same as your graph.

share|improve this answer
    
Thank you, I will talk to my teacher tomorrow and see what he says. –  Marco Jun 3 '12 at 20:41
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.