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Discrete Inverse Laplace Transform      Home

This page shows an easy way to get a numerical solution to the Inverse Laplace Transform of H(s). This method is useful for obtaining the time domain response of an analog filter, phase lock loop, power control circuit, and other types of control loops.

It shows delay times and settling times and allows you to see the overshoot and ringing caused by a pulse input.

A Low Pass Filter Example
In general, low pass responses are the most important to us. In this example we use a low pass filter with a corner frequency (Fc) of 4 kHz.

The following is a screen shot from a MathCAD file which shows our definition of H(s) and its frequency response. Next we show the how we obtain the pulse response for H(s). Note that the time span covered is defined in terms if Fc. We do this because the time span of interest is usually directly related to Fc. For example, if H(s) has a 1 kHz  bandwidth, there isn't anything interesting happening beyond 10 / Fc = 10 ms. It is important to note that while this method works fine for a pulse input, but it does not work for an infinitely long step response. The width of the pulse can be as wide as N T seconds, but beyond that this method fails. We show a plot of an infinitely long step response below.

This method also works well for an impulse response. First set g = H(k) (i.e. stop multiplying H(k) by the Laplacian Pulse). Then set MaxFreq to 100 Fc and increase N. Then adjust the vertical scale on the time domain plot.

Although we used a low pass H(s) in this example, this method works well for high pass, band pass, and notch responses as well.

We set MaxFreq to 10 Fc, but a wide range of values works fine. MaxFreq needs to be greater than about 5 Fc and less than N / PW. The size of the FFT can be almost any value, so long as the MaxFreq < N / PW constraint is met.

A way to validate this method is to observe that we can set the width of the pulse and see it drop at the correct time. Another way is to set H(s) to 1 and see the resulting square pulse output as shown here. As another validation of this method, we compare the results from this method to that obtained for an IIR filter. We use an IIR filter's response because obtaining its pulse response is trivial. The filter is a 4 pole Butterworth high pass.

We offset the analog response on the x axis a bit for a better comparison. The Laplace pulse rises on the first grid line, and falls on the 5th grid line. This shows that the method described here can give a good estimate of waveforms that are rather complicated.  