# Continuous convolution between two functions

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.

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?

• It would became: $\int_{-\infty }^{+\infty} (δ(y-x_0)+δ(y-2x_0)) Λ(x-y) dy$ Aug 29 '19 at 18:33
• Perfect! Now what does the sifting property of the delta say that integral should be? Aug 29 '19 at 18:34
• $Λ(x- x_0)+Λ(x-2x_0)$ ? Aug 29 '19 at 18:41
• 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? Aug 29 '19 at 18:45
• 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! Aug 29 '19 at 18:51

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:

$$Λ(x_0)+Λ(2x_0)$$