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March 23rd, 2017, 01:21 PM   #1
szz
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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
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March 23rd, 2017, 01:37 PM   #2
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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

$R(t)*(\delta(\tau) + \phi(t)\delta(\tau - \rho(t))) = R(t) + \phi(\rho^{-1}(\tau))R(\rho^{-1}(\tau))$
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March 23rd, 2017, 02:50 PM   #3
szz
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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 02:55 PM.
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March 23rd, 2017, 03:05 PM   #4
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you're right.

Let me take another look.
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March 23rd, 2017, 03:52 PM   #5
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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

$R(t) * h(\tau,t) = R(t) + R(t-\rho^{-1}(\tau))\phi(\rho^{-1}(\tau))$
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