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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. $\endgroup$ Apr 7, 2018 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. $\endgroup$ Apr 7, 2018 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, 2018 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? $\endgroup$
    – user33568
    Apr 8, 2018 at 1:11
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Does this mean we cannot map high-pass analog filters to high-pass digital filters using impulse invariance?

In my opinion this is not correct albeit it is stated everywhere regarding this topic. The reason this is not correct is due to the lemma of riemann-lebegue which states that any fourier transform converges to zero towards infinity (for L1 functions that is). Thus there exist no high pass filters in the sense that the fourier transform does not converge towards zero at some point. Thus, if the fourier transform converges towards zero faster than 1/Omega (at some point), for such an L1 impulse response you should be able to find a sampling period (which can be very small but greater than zero) that gives you arbitarily exact (aliasing can be made arbitarily small) the corresponding discrete time filter.

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    $\begingroup$ A continuous-time highpass filter has a Dirac delta impulse in its impulse response, which is not $L_1$ integrable, so your argument doesn't work in that case. Also take a look at this. $\endgroup$
    – Matt L.
    Jan 13, 2022 at 21:09

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