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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.

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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?

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  • $\begingroup$ Thanks for the help. I think you have the integral bounds wrong. $\endgroup$ – coffeecake Sep 24 at 15:08

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