1
$\begingroup$

I'm studing for an exam and I'm stuck on a simple exercise about convolution between two functions. It says: A system has a triangular impulse response (LSF) centered at the origin of the plane $h(x)=Λ(x)$. We input an image with two impulses, defined as $f(x)=δ(x-x_0)+δ(x-2x_0)$. Get the output of the system.

I was thinking to use this formula: $\int_{-\infty }^{+\infty} f(x-x_0) h(x_0) dx_0$

So it would become: $\int_{-\infty }^{+\infty} (δ(x-2x_0)+δ(x-3x_0)) Λ(x_0) dx_0$

I don't know if it makes sense, so i'll appreciate any help.

Thank you.

$\endgroup$
0
$\begingroup$

I think you've misunderstood what $x_0$ is. It is a constant, not a variable, so you should not be integrating over it. The variable you integrate over in the convolution is a dummy variable that won't appear in the final answer, so it give a name that doesn't match any names you already have for variables. In this case, the convolution formula would written as $$(f\star g)(x) = \int_{-\infty}^\infty f(x-y)g(y)dy = \int_{-\infty}^\infty f(y)g(x-y)dy$$

Can you try writing the convolution again?

$\endgroup$
  • $\begingroup$ It would became: $\int_{-\infty }^{+\infty} (δ(y-x_0)+δ(y-2x_0)) Λ(x-y) dy$ $\endgroup$ – Rita Albertini Aug 29 '19 at 18:33
  • $\begingroup$ Perfect! Now what does the sifting property of the delta say that integral should be? $\endgroup$ – Ninad Munshi Aug 29 '19 at 18:34
  • 1
    $\begingroup$ $Λ(x- x_0)+Λ(x-2x_0)$ ? $\endgroup$ – Rita Albertini Aug 29 '19 at 18:41
  • $\begingroup$ That's exactly right. And in fact it shouldn't be unexpected at all because our system $H$ was LTI. So if the impulse response $H(\delta(x)) = \Lambda(x)$, then what is the output for a linear combination of shifted deltas? $\endgroup$ – Ninad Munshi Aug 29 '19 at 18:45
  • $\begingroup$ It should be a linear combination of the impulse response, so a combination of the triangular function shifted like the deltas, right? I think I got it, thank you so much for your help! $\endgroup$ – Rita Albertini Aug 29 '19 at 18:51
0
$\begingroup$

Since response of system to $input=\delta(x)$ is impulse response and equal to $$h(x)=Λ(x)$$

Input to the system is a summation of two shifted impulses $$f(x)=\delta(x-x_0)+\delta(x-2x_0)$$

What's the Output of system ?

$$Output=f(x)\star h(x)=\int_{-\infty }^{+\infty} \Big(\delta(x-x_0)+\delta(x-2x_0)\Big) Λ(x)dx$$ By definition and properties of $\delta$ function we have:

$$\int f(x) \delta(x-x_0) dx=f(x_0)$$

So the final result will be:

$$Λ(x_0)+Λ(2x_0)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.