The moving of the poles and zeroes during the low-pass tranform

Question

Usually, we start the design of a filter using a low-pass prototype. Then we transform it to the required high-pass, band-pass or band-stop form as required. I'm looking for a good explanation on how the poles and zeroes of the low-pass prototype are transformed to the poles and zeroes of the final filter.

Answer 1

Robert's answer shows you the transformations for highpass, bandpass, and bandstop filters, but in the case of bandpass and bandstop it's not immediately obvious how the poles and zeros move. This is shown below.

Let $p$ denote the complex frequency variable for the lowpass prototype filter, and let $s$ be the complex frequency variable for the transformed filter.

For the highpass filter, we have

$$p=\frac{1}{s}\tag{1}$$

Consequently, any complex point $p$, specifically a pole or zero of the lowpass prototype, is transformed to $s=1/p$. I.e., if $p=re^{j\theta}$ is a pole (or zero) of the prototype filter, it is transformed to a pole (or zero) at $1/p=e^{-j\theta}/r$. Clearly, the left half-plane is mapped to the left half-plane, and the imaginary axis is mapped to the imaginary axis. Hence, if the prototype is stable, the transformed filter is stable too, and any zeros on the imaginary axis will remain on the imaginary axis.

The transformation for a bandpass filter with center frequency $\omega_0$ is

$$p=\frac{s^2+\omega_0^2}{s}\tag{2}$$

which maps the origin (DC) to $s=\pm j\omega_0$. Note that this transformation doubles the number of poles and zeros.

Solving $(2)$ for $s$ gives

$$s=\frac{p}{2}\pm\sqrt{\frac{p^2}{4}-\omega_0^2}\tag{3}$$

I.e., given a pole (or zero) location $p$, the resulting two poles (or zeros) are given by $(3)$.

Finally, for a bandstop filter, we have the following transformation:

$$p=\frac{s}{s^2+\omega_0^2}\tag{4}$$

which maps the origin (DC) to the origin and to infinity, and infinity is mapped to the center frequency $s=\pm j\omega_0$. Solving $(4)$ for $s$ gives

$$s=\frac{1}{2p}\pm\sqrt{\frac{1}{4p^2}-\omega_0^2}\tag{5}$$

Each pole (or zero) $p$ is transformed to two poles (or zeros) according to $(5)$.

Note that also the transformations $(2)$ and $(4)$ map the left half-plane to the left half-plane, and the imaginary axis to the imaginary axis. Consequently, a stable lowpass prototype with zeros on the imaginary axis will be transformed to a stable bandpass or bandstop filter with zeros on the imaginary axis.

Answer 2

This is in the Laplace $s$-domain. To do this in the discrete-time $z$-domain, we have to map $s$ to $z$ and the most common mapping used is the Bilinear Transform (a.k.a. Tustin's Method).

Let $H_\mathrm{LP}(s)$ be the low-pass prototype. This is what we get when we normalize the low-pass cutoff frequency $\Omega_0$ to 1.

Remember that when we compute the frequency response that $s \leftarrow j \Omega$. (I am saving small-case $\omega$ for the discrete-time case as do Oppenheim and Schafer.)

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The lowpass-to-highpass transformation is:

$$ H_\mathrm{HP}(s) = H_\mathrm{LP}\left(\frac{1}{s}\right) $$

This maps the frequency at $\Omega=\infty$ in the highpass to $\Omega=0$ (or DC) in the lowpass filter.

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The lowpass-to-bandpass transformation is:

$$ H_\mathrm{BP}(s) = H_\mathrm{LP}\left(s+\frac{1}{s}\right) $$

This maps the frequency at $\Omega=1$ in the bandpass to $\Omega=0$ (or DC) in the lowpass filter.

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The lowpass-to-bandreject transformation is:

$$ H_\mathrm{BR}(s) = H_\mathrm{LP}\left(\frac{1}{s+\frac{1}{s}}\right) $$

This maps the frequency at $\Omega=1$ in the bandpass to $\Omega=\infty$ in the lowpass filter.

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Note that for the bandpass and bandreject transformation, we end up doubling the order of the filter. And we need to introduce a bandwidth parameter (which can also translate to a $Q$ parameter). When I return to this, I'll try to do that in the most concise manner.