Generating Smooth Curves Using Simple Digital Filters

Suppose we have a function of a variable tt, x(t)x(t), which consists of multiple connected line segments.

What if we wanted to make a version of this curve with the jagged edges smoothed out? This is fairly easy to do using a digital filter. Define the smoothed function, yy, in the following way.

yk=αyk1+(1α)xk\large{y_k = \alpha y_{k-1} + (1-\alpha) x_k}

If we discretize tt, the kk subscript denotes the present iteration, and the k1k-1 subscript denotes the previous iteration. The equation above is known as an "infinite impulse response" (IIR) filter, because the present filter output is a function of the previous output, and so on, and so on.....The filter will take infinitely many iterations to fully catch up to any step change in the input.

The present value of yy ( the smoothed version of xx), is equal to a weighted sum of the previous value of yy and the present value of xx. Thus, by tweaking the value of the "inertia constant" α\alpha, we can dictate how fast the value of yy can change, and thus, how much smoothing we get.

As an extreme case, suppose α=1\alpha = 1. This would mean that yy would be frozen at a static value forever (infinite inertia). For α=0\alpha = 0, the smoothed and un-smoothed curves are exactly equal, corresponding to zero inertia.

Some graphs for intermediate values of α\alpha are shown below. The third one is particularly reminiscent of the Fourier series. We could create a saw-tooth wave by summing up sine waves with various frequencies (from low to high). By increasing the inertia, we essentially implement a low-pass filter which removes the higher harmonics from the saw-tooth, revealing lower-frequency sinusoids.

α=0.97\alpha = 0.97

α=0.99\alpha = 0.99

α=0.998\alpha = 0.998

#Calculus

Note by Steven Chase
4 years, 1 month ago

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