# Analytical Solution for the Convolution of Signal with a Box Filter

I have an exercise in which I am trying to filter an input signal $$y(x) = \sin(x)$$. Ideally, I would like to apply a box filter to this signal.

Previously, I successfully convolved the input signal $$y(x)$$ with a decaying response $$h(x) = e^{-x}$$.

I did so by the following the definition of convolution (e.g., integrating $$\int_0^t\sin(x')e^{-(x-x')}\mathrm{d}x'$$ and computing a damped sinusoidal signal.

My box filter is given by $$\frac{1}{\Delta}$$ for$$|x-\xi| \leq \frac{\Delta}{2}$$ and 0 elsewhere, where $$\Delta$$ is the filter width. I understand that a box filter is a local average, and I can implement this numerically, but I do not understand how to analytically integrate this as I did with the damped exponential 'filter'.

I tried to take the Fourier transform of $$y(x)$$ and $$h(x)$$ and multiplying them in Fourier space, but I could not figure out how to do so.

Thanks for any help.

## 1 Answer

DO NOT PASS GO; DO NOT Collect \$200; DO NOT Take Fourier transforms, or worse yet, FFTs

You do the convolution exactly the way you would do any other convolution: start with the basic convolution integral (not what you wrote) and apply the properties of the signals that you are using to come up with an easier calculation. Begin with $$\int_{-\infty}^\infty y(x^\prime)h(x^\prime-x)\mathrm dx^\prime\tag{1}$$ and use the result that $$h(\cdot)$$ is nonzero only when its argument ($$x^\prime-x$$ in this instance) lies in the interval $$\left[-\frac{\Delta}2, +\frac{\Delta}2\right]$$. So, the integrand of the convolution integral is $$0$$ whenever $$x^\prime$$ is such that $$x^\prime-x > \frac{\Delta}2 \implies x^\prime > x + \frac{\Delta}2$$ or that $$x^\prime-x < -\frac{\Delta}2 \implies x^\prime < x - \frac{\Delta}2.$$ This allows us to simplify the convolution integral $$(1)$$ into $$\int_{x - \frac{\Delta}2}^{x + \frac{\Delta}2} y(x^\prime)\frac{1}{\Delta}\mathrm dx^\prime\tag{2}$$ where I have substituted the nonzero value of $$h$$ in the integral to save a step. Can you take it from here?

• Thanks for the help. I think you have the integral bounds wrong. Sep 24, 2019 at 15:08