Results 1 to 15 of 15
Like Tree8Thanks
  • 1 Post By chiro
  • 1 Post By Idea
  • 1 Post By SlipEternal
  • 1 Post By SlipEternal
  • 1 Post By SlipEternal
  • 1 Post By SlipEternal
  • 1 Post By SlipEternal
  • 1 Post By SlipEternal

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

  1. #1
    Member
    Joined
    Mar 2010
    Posts
    136

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    6,577
    Thanks
    1711

    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].
    Thanks from Nforce
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2010
    Posts
    136

    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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    718
    Thanks
    329

    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
    Thanks from Nforce
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2010
    Posts
    2,837
    Thanks
    1089

    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
    Thanks from Nforce
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Mar 2010
    Posts
    136

    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2010
    Posts
    2,837
    Thanks
    1089

    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 $1(t)$.
    Thanks from Nforce
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Mar 2010
    Posts
    136

    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?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Nov 2010
    Posts
    2,837
    Thanks
    1089

    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 $1(t)$.
    Thanks from Nforce
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Mar 2010
    Posts
    136

    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.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Nov 2010
    Posts
    2,837
    Thanks
    1089

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

    Quote Originally Posted by Nforce View Post
    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)
    Thanks from Nforce
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Mar 2010
    Posts
    136

    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.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Nov 2010
    Posts
    2,837
    Thanks
    1089

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

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

    $\displaystyle \int_0^t 3\cos (t-\tau) d\tau = -3\int_t^0 \cos u \,du = 3\int_0^t \cos u \, du = 3\sin t$
    Thanks from Nforce
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Mar 2010
    Posts
    136

    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.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor
    Joined
    Nov 2010
    Posts
    2,837
    Thanks
    1089

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

    Quote Originally Posted by Nforce View Post
    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.
    Thanks from Nforce
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Convolution Integral ...and not
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Dec 16th 2016, 01:02 AM
  2. Convolution Integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jun 23rd 2014, 07:11 AM
  3. Convolution integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Mar 27th 2011, 11:12 AM
  4. Convolution integral
    Posted in the Advanced Math Topics Forum
    Replies: 3
    Last Post: Dec 4th 2009, 02:56 AM
  5. convolution integral
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Nov 18th 2009, 02:56 AM

/mathhelpforum @mathhelpforum