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 March 23rd, 2017, 12:21 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, 12:37 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372 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, 01:50 PM   #3
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Joined: Oct 2014
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Math Focus: Calculus
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 01:55 PM.

 March 23rd, 2017, 02:05 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372 you're right. Let me take another look. Thanks from szz
 March 23rd, 2017, 02:52 PM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372 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

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