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One of the main disadvantages of realizing digital filters using impulse invariance is aliasing. According to the Nyquist sampling criterion, in order for the frequency response of the digital filter to match that of the corresponding analog filter with impulse response $h(t)$, $h(t)$ has to be band-limited. That is, the analog filter has to be low-pass. Does this mean we cannot map high-pass analog filters to high-pass digital filters using impulse invariance? Or is it just a matter of mapping a low-pass analog filter to a low-pass digital filter with frequency response $H(e^{j\Omega})$, and then finding the frequency response of the corresponding high-pass filter simply by substracting it from 1?

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  • $\begingroup$ that's a good question that i had never wondered about. i think, you're right. if you sample the continuous-time impulse response of a theoretical HPF, i think there is trouble. $\endgroup$ – robert bristow-johnson Apr 7 '18 at 1:01
  • $\begingroup$ I agree with Robert, however, the devil in me tells me to just do it (not without thinking, of course). After all, even lowpass filters can't have a clear bandwidth, which gives the risk of aliasing, so why not just ride the shotgun and see what you get. Of course, you'll have to be reeeally careful with input signal, they should not have anything going above Nyquist, or the least possible. I see this as type II highpass FIR: nobody says you can't do it, but if you do, prepare for an inherent zero that will render your filter a bandpass. That is, it's not forbidden, but there are consequences. $\endgroup$ – a concerned citizen Apr 7 '18 at 7:05
  • $\begingroup$ @robertbristow-johnson: The problem is you can't even sample the impulse response because it has a Dirac delta, and nobody has yet figured out how to sample a Dirac delta ... $\endgroup$ – Matt L. Apr 7 '18 at 12:33
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You're right, you can't use the impulse invariance method for transforming continuous-time filters with a transfer function that is not strictly proper, i.e., for which the degree of the numerator is not strictly smaller than the degree of the denominator. High-pass and band-stop filters have transfer functions with numerator and denominator polynomials of the same degree, which means that the corresponding partial fraction expansion has a constant term. The latter corresponds to a (scaled) Dirac delta impulse in the impulse response, which obviously cannot be sampled.

Take a first-order high-pass filter as an example:

$$H(s)=\frac{s}{s+a}=1-\frac{a}{s+a}\Longleftrightarrow h(t)=\delta(t)-ae^{-at}u(t)\tag{1}$$

Now you can't determine $h(nT)$, $n\in\mathbb{Z}$, at $n=0$.

In sum, for the impulse invariance method to be applicable we require

$$\lim_{s\to\infty}H(s)=0\tag{2}$$

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  • $\begingroup$ What about designing a digital lowpass filter based on an analog prototype through impulse invariance, and then somehow implementing a filter with a transfer function $1 - H_{LP} (z)$? Would that work? $\endgroup$ – 0MW Apr 8 '18 at 1:11

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