# Convolution Dirac impulse and periodic signal

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1. Nov 2, 2014

### ellosma

Hi ☺️ i have to do a convolution with a periodic signal and a dirac impulse:
x(t)=sen(πt)(u(t)−u(t−2))
h(t)=u(t−1)−u(t−3)
The first is a periodic graph that intersect axis x in points 0 , 1 and 2 (ecc)
The se ing is a rectangle ( Dirac impulse ) that intersect AxiS x in points 1 and 3.
For t−1<0 and t−3>0 convolution doesn't exist
For me is ∫to 0 from 1 ((senπ(t−τ)dτ) − ∫to 1 from t−1 (senπ(t−τ)dτ))) but my book write ∫to t-1 from 0 (senπtdτ)
Could someone tell me why he cancel -τ and how to do this ex without Fourier ? Thank you very much and Sorry for my terrible english but i'm italian

2. Nov 3, 2014

### vela

Staff Emeritus
I wouldn't say the convolution doesn't exist. I think you mean x*h(t) = 0. If so, it's not true that it's equal to 0 for t-3>0.

I suspect the book simply interchanged the roles of x and h compared to what you did. The convolution is given by
$$(x*h)(t) = \int_{-\infty}^\infty x(\tau)h(t-\tau)\,d\tau,$$ and since x(t)=0 for t<0 and t>2, you can write
$$(x*h)(t) = \int_{0}^2 (\sin\pi\tau)h(t-\tau)\,d\tau.$$ The variable that's in the argument of sine should be $\tau$, not $t$ as you said the book has.

3. Nov 3, 2014

### rude man

What is "sen(πt)" ? Is "sen" Italian for "sin"? I will assume so.
So we have
f(t) = sin(πt){u(t) - u(t-2)}
g(t) = u(t-1) -u(t-3)
You are probably best off with
f*g = integral from T = -∞ to +∞ of f(t-T)g(T)dT.

The trick here is to determine how the various u(t) combine. You will need the fact that u(t-T) = u{-(T-t)}.
If you graph the two u terms (one in f, the other in g) then you will see that the limits of integration change from (-∞, +∞) to (0, t-1).

The final answer should include a coefficient {u(t-1) - u(t-3)} which has the effect of making the convolution = 0 for t < 1 and > 3.

4. Nov 3, 2014

### rude man

???
The convolution g*t is a function of t. After you've performed the integration you need a "t" in the sine function I think.
You also used 2 instead of 3 for the upper limit on t.

5. Nov 3, 2014

### vela

Staff Emeritus
That's wrong.

The sine I was referring to is inside the integral, so no, it's not supposed to be a function of t. If it were, it would be a constant and you could pull it out of the integral.

Because x(t) is non-zero only between 0 and 2. And it's the upper limit on $\tau$, not t.

6. Nov 3, 2014

### rude man

Oh? How?

After integrating to constant limits it is a constant except for modulation by u(t) functions. I never suggested it could be "pulled out of the integral".

The upper limit on T is t-1.

The OP will have to decide I guess which of us is right. For the record, my answer agrees with what is in his/her textbook.

Last edited: Nov 3, 2014
7. Nov 3, 2014

### vela

Staff Emeritus
Here's a plot of $x(\tau)$ and $h(t-\tau)$ for $t=4$. Still think the convolution is 0 when t>3?

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8. Nov 3, 2014

### rude man

It is if the problem states it is, which it does: "For t−1<0 and t−3>0 convolution doesn't exist."

9. Nov 3, 2014

### vela

Staff Emeritus
I don't know if you're trying to be deliberately obtuse. The OP wrote that the book said the integral was
$$\int_0^{t-1} \sin \pi t\,d\tau.$$ $\sin\pi t$ is independent of the variable of integration $\tau$, so that integral is equal to
$$\int_0^{t-1} \sin \pi t\,d\tau = \sin\pi t \int_0^{t-1}d\tau = (t-1)\sin\pi t.$$ Are you claiming that's correct?

10. Nov 3, 2014

### rude man

Of course not.
I think this discussion has borne all the fruit it can. Thanks for your patience.
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