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having derived the Butterworth Lowpass Time domain response, I am now struggling to find a similar function for a Butterworth Highpass filter. I understand you need to replace s by 1/s. But this leads to a transfer function with as many zeros as there are poles. So you can no longer use the Heaviside 'cover-up' method to decompose the individual exponential functions, to arrive at the time domain response.

Is there perhaps someone who stumbled on the same problem and arrived at a solution ?

The ultimate goal is to arrive at a Butterworth bandpass filter whereby the lowpass and high pass sides can have a different order (slope). So I still need to convolve the LP- with the HP-filter. . .

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$\frac{s^n + b_{n-1}s^{n-1}+\cdots+b_0}{s^n + a_{n-1}s^{n-1}+\cdots+a_0} = 1 + \frac{(b_{n-1}-a_{n-1})s^{n-1}+\cdots+b_0 - a_0}{s^n + a_{n-1}s^{n-1}+\cdots+a_0}$, and the inverse Laplace transform of 1 is $\delta(t)$.

Is that enough to get on with?

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  • $\begingroup$ Thanks for your quick reply. I need to look at this carefully. For your information; for ease of calculation the poles have all been decomposed into (s - ai) factors. So I don't have a slew of ai factors with i = n-1 to 0... As a matter of fact, I arived at the same point, and with a Butterworth HP filter you get all bi = 0. But that leaves you with the daunting task to calculate the individual ai parameters. $\endgroup$
    – BartD
    Feb 24 at 22:09

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