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 March 23rd, 2017, 01:21 PM #1 Senior Member   Joined: Oct 2014 From: EU Posts: 224 Thanks: 26 Math Focus: Calculus 2D continuous time convolution Hello, I am solving an exercise which involves a convolution between a function of one variable and a function of two variables. The convolution is: \displaystyle \begin{aligned} R(t) *h (t, \tau) = R(t) * (\delta(\tau) + \phi(t)\delta(\tau - \varphi(t)) \end{aligned} where $\displaystyle \phi(t)$ & $\displaystyle \varphi(t)$ are (as the notation suggests) function of $\displaystyle t$. My reasoning is the following: since $\displaystyle \delta(t,\tau) = \begin{cases} 1 & t = \tau = 0\\ 0 & \text{otherwise} \end{cases}$ and $\displaystyle x(t) * \delta(t - t_0) = x(t - t_0) * \delta(t) = x(t - t_0)$ by inspection I end up with: \displaystyle \begin{aligned} R(t) *h (t, \tau) & = R(t) * (\delta(\tau) + \phi(t)\delta(\tau - \varphi(t)) \\ & = R(t) + \phi(t)R(t - \varphi(t)) \end{aligned} At first glance I see my result as correct, but perhaps it isn't. So is my reasoning correct ? Thank you in advance, szz March 23rd, 2017, 01:37 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,648 Thanks: 1476 convolving a 1D and a 2D function together is going to results in a 2D function. also $x(t) * \delta(t-t_0) = x(t_0)$ not $x(t-t_0)$ as you've written. I think the answer is $R(t)*(\delta(\tau) + \phi(t)\delta(\tau - \rho(t))) = R(t) + \phi(\rho^{-1}(\tau))R(\rho^{-1}(\tau))$ March 23rd, 2017, 02:50 PM   #3
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Hello romsek,

Thank you for being interested in helping me.

Quote:
 Originally Posted by romsek $\displaystyle x(t)∗\delta(t−t0)=x(t0)$ not $\displaystyle x(t−t_0)$ as you've written.
Why ? This is not what I have been taugh about the sifting property of convolution with a unit-impulse response.
An example here (Unit Impulse Function) but also on Oppenheim's book Signal and Systems example 2.11 clarifies this concept.

Quote:
 Originally Posted by romsek I think the answer is $\displaystyle R(t)∗(\delta (\tau)+\phi(t) \delta (\tau −\rho(t))) = R(t)+\phi(\rho^{−1}(\tau))R(\rho^{−1}(\tau))$
This is your resolution broken up in three steps:

\displaystyle \begin{aligned} R(t) * (\delta(\tau) + \phi(t)\delta(\tau - \rho(t)) & = R(t) * \delta(\tau) + R(t) * \phi(t)\delta(\tau - \rho(t))\\ & = \underbrace{R(t)}_{\text{1st therm}} + \underbrace{\phi(\rho^{-1}(\tau))R(\rho^{-1}(\tau))}_{\text{2nd therm}} \end{aligned} \qquad (1)

Here you have applied your result I quoted (and which I repeat here):

$\displaystyle x(t)∗\delta(t−t0)=x(t0)$

also, for the 1st therm you obtain $\displaystyle R(t)$ from a convolution of two different domain functions. OK.
But for the 2nd therm you invert the function $\displaystyle \rho(t)$ to $\displaystyle \rho^{-1}(\tau)$

Could you clarify this ?
Thank you,
szz

Last edited by szz; March 23rd, 2017 at 02:55 PM. March 23rd, 2017, 03:05 PM #4 Senior Member   Joined: Sep 2015 From: USA Posts: 2,648 Thanks: 1476 you're right. Let me take another look. Thanks from szz March 23rd, 2017, 03:52 PM #5 Senior Member   Joined: Sep 2015 From: USA Posts: 2,648 Thanks: 1476 as $R(t)$ is only a function of $t$ we are basically convolving $R(t)$ with a $\tau$ slice of $h(\tau,t)$ for every $\tau$ So this is more an infinite number of 1D convolutions than a 2D convolution. We're not convolving in the $\tau$ direction. Let's just look at the second piece of $h(\tau,t)$ $h(\tau,t) = \phi(t) \delta(\tau - \rho(t))$ \begin{align*}&R(t) * h(\tau, t) =\\ &\displaystyle \int_{-\infty}^\infty R(u) h(\tau, t-u)~du = \\ &\int_{-\infty}^\infty R(u) \phi(t-u)\delta(\tau-\rho(t-u))~du =\\ &R(t-\rho^{-1}(\tau))\phi(\rho^{-1}(\tau)) \end{align*} The final line because $\delta(\tau - \rho(t - u))=1$ only for $\tau = \rho(t-u) \Rightarrow u = t-\rho^{-1}(\tau)$ and so you're "2D" convolution is just $R(t) * h(\tau,t) = R(t-\rho^{-1}(\tau))\phi(\rho^{-1}(\tau))$ and of course $R(t)$ is added back to this when we include the $\delta(\tau)$ back into $h(\tau, t)$ for a final answer of $R(t) * h(\tau,t) = R(t) + R(t-\rho^{-1}(\tau))\phi(\rho^{-1}(\tau))$ Thanks from szz Tags continuous, convolution, time Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post szz Applied Math 3 November 3rd, 2015 02:05 PM Rina Economics 0 December 20th, 2012 06:00 AM trojsi Real Analysis 1 November 12th, 2012 08:43 PM tottijohn Advanced Statistics 0 September 6th, 2011 07:55 AM thedoctor818 Real Analysis 17 November 9th, 2010 09:19 AM

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