March 23rd, 2017, 12:21 PM  #1 
Senior Member Joined: Oct 2014 From: I don't know... Posts: 220 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: CA Posts: 1,264 Thanks: 650 
convolving a 1D and a 2D function together is going to results in a 2D function. also $x(t) * \delta(tt_0) = x(t_0)$ not $x(tt_0)$ 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  
Senior Member Joined: Oct 2014 From: I don't know... Posts: 220 Thanks: 26 Math Focus: Calculus 
Hello romsek, Thank you for being interested in helping me. Quote:
An example here (Unit Impulse Function) but also on Oppenheim's book Signal and Systems example 2.11 clarifies this concept. Quote:
$\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: CA Posts: 1,264 Thanks: 650 
you're right. Let me take another look. 
March 23rd, 2017, 02:52 PM  #5 
Senior Member Joined: Sep 2015 From: CA Posts: 1,264 Thanks: 650 
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, tu)~du = \\ &\int_{\infty}^\infty R(u) \phi(tu)\delta(\tau\rho(tu))~du =\\ &R(t\rho^{1}(\tau))\phi(\rho^{1}(\tau)) \end{align*} $ The final line because $\delta(\tau  \rho(t  u))=1$ only for $\tau = \rho(tu) \Rightarrow u = t\rho^{1}(\tau)$ 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 $R(t) * h(\tau,t) = R(t) + R(t\rho^{1}(\tau))\phi(\rho^{1}(\tau))$ 

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continuous, convolution, time 
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