# Thread: Convolution integral - how to take into account the t and tau

1. ## Convolution integral - how to take into account the t and tau

Hello,

how to I integrate something like this?: (I do understand how to integrate something just of $t$, $x$ or some other variable)

$\int_{-\infty}^{\infty}3cos(\omega\tau)1(\tau)1(t-\tau)d\tau$

Can someone give some guidlines? Maybe I will post more questions later, if I will not completly understand.
Thank you.

2. ## Re: Convolution integral - how to take into account the t and tau

Hey Nforce.

You treat the non-integrand variable [i.e. the t variable] as a constant and integrate with that if it is independent [which in this case it is].

3. ## Re: Convolution integral - how to take into account the t and tau

Thank you. So I need to solve this:

$3\int_{-\infty}^{\infty}cos(\omega\tau)1(\tau)1(t-\tau)d\tau$

But what do I do with unit step function of tau, and the same thing with a delay of t?

4. ## Re: Convolution integral - how to take into account the t and tau

so that's what that is 'the unit step function'

so just integrate

$\int _0^13\text{cos}(\omega \tau )d\tau$

5. ## Re: Convolution integral - how to take into account the t and tau

I replaced $\omega$ with $a$ and $\tau$ with $x$:
Wolfram|Alpha: Computational Knowledge Engine

6. ## Re: Convolution integral - how to take into account the t and tau

Yes, i know what is the answer, I know how to use wolfram alpha.
I just want to learn this, to do with my hands.

Let's go back to the point:

If I integrate just the cosine function (without the unit step), the integral have to be larger then the integral of cosine function multiplied with unit step or delayed unit step. Why do I think that? Because we cut the left part, the x < 0 is zero. So the integral on the left part zero. Unless we consider the unit step as just one. So:

$1(t) = 1(t-\tau) = 1$

7. ## Re: Convolution integral - how to take into account the t and tau

$\displaystyle 3\int_{-\infty}^\infty \cos (\omega \tau ) 1(\tau ) 1 (t-\tau ) d\tau = \begin{cases}\displaystyle 3\int_0^t \cos (\omega \tau ) 1(\tau ) 1(t-\tau ) d\tau & t\ge 0 \\ 0 & t < 0\end{cases}$

For $t\ge 0$, this is:

$\dfrac{3}{\omega} \left. \sin ( \omega \tau ) \right]_0^t = \dfrac{3}{\omega} \sin ( \omega t)$

But, the condition on $t$ is given by multiplying by $1(t)$.

8. ## Re: Convolution integral - how to take into account the t and tau

Why do we integrate then between 0 and t?
Yes because of the condition.
On the start we have minus infinity to infinity.

How do we know that? How do we know that when we multiply with unit step function we must integrate between 0 and t?

9. ## Re: Convolution integral - how to take into account the t and tau

If $t < 0$ then for all $\tau < 0$, we have $1(\tau ) = 0$ and for all $\tau \ge 0$ we have $1(t-\tau ) = 0$. For $t \ge 0$ we have:

For $\tau \in (-\infty, 0]$, $1(\tau ) = 0$. On $\tau \in (t,\infty)$, $1(t-\tau ) = 0$. So, $1(\tau )1(t-\tau ) = 1$ only on the interval $0 \le \tau \le t$. This gives us our bounds of integration.

For multiplying by the unit step function, we found that for $t < 0$, the integral is zero. For $t \ge 0$, the integral is times 1. That gives us the multiplication by $1(t)$.

10. ## Re: Convolution integral - how to take into account the t and tau

But how? Let's say $\tau = 5$.
The graph is: https://www.wolframalpha.com/input/?i=H(5)*H(t-5).

So it's one from 5 to $\infty$. So here we must integrate between 5 to $\infty$.

Maybe I am missing something.

11. ## Re: Convolution integral - how to take into account the t and tau

Originally Posted by Nforce
But how? Let's say $\tau = 5$.
The graph is: https://www.wolframalpha.com/input/?i=H(5)*H(t-5).

So it's one from 5 to $\infty$. So here we must integrate between 5 to $\infty$.

Maybe I am missing something.
$t$ is a constant. $\tau$ is a variable. So, at $t=5$, it is 1 on $\tau \in (0,5)$.

https://www.wolframalpha.com/input/?i=H(x)*H(5-x)

12. ## Re: Convolution integral - how to take into account the t and tau

Thanks Slipeternal, now I fully understand the unit step part.

But I have another question: If we had some integral like this:

$\int_{0}^{t}3cos(t-\tau)d\tau$

How can I here expose the $t$ as a constant? It's inside the cosine function.

13. ## Re: Convolution integral - how to take into account the t and tau

$u = t-\tau$
$du = -d\tau$

$\displaystyle \int_0^t 3\cos (t-\tau) d\tau = -3\int_t^0 \cos u \,du = 3\int_0^t \cos u \, du = 3\sin t$

14. ## Re: Convolution integral - how to take into account the t and tau

I think you forgot the minus sign. But ok. Now I know.

Thank you.

15. ## Re: Convolution integral - how to take into account the t and tau

Originally Posted by Nforce
I think you forgot the minus sign. But ok. Now I know.

Thank you.
$u(\tau) = t-\tau$
$u(0) = t-0 = t$
$u(t) = t-t = 0$
$du = -d\tau$

$\displaystyle \int_0^t 3\cos(t-\tau) d\tau = \int_{u(0)}^{u(t)} 3\cos u (-du) = -3 \int_t^0 \cos u \, du = 3\int_0^t \cos u \, du = 3\sin t$

No, I did not miss a minus sign. When you swap the bounds of integration, it gives an extra negative sign.