Frequency conversion in IIR lowpass-to-lowpass transformation

Question

We use
$ z^{-1} xrightarrow{} frac{z^{-1}-alpha}{1-{alpha}z^{-1}}$
and
$ alpha = frac{sin(omega_c^{'}-omega_c)/2}{sin(omega_c^{'}+omega_c)/2}$
for a lowpass-to-lowpass frequency transformation, where $omega'_c$ is the cut-off frequency of the prototype lowpass filter, and $omega_c$ is the cut-off frequency of the transformed lowpass filter.
In practice we know the desired edge frequencies $ omega_s$ and $ omega_p$ and we are required to find the prototype lowpass cutoff frequencies $ omega_s^{'}$ and $ omega_p^{'}$.
For calculating $ alpha$ we have to know $ omega_c^{'}$ . If we calculate $ alpha$ we can have $ omega_s^{'}$ by $ omega_s^{'} = angle{(frac{e^{-jomega_s}+ alpha}{1-alpha e^{-jomega_s}})}$
But I don't know how select a suitable $ omega_p^{'}$.

Matt L. · Answer

The prototype filter is not uniquely defined by the two band edges of the filter to be designed. So you have to choose one of the two edge frequencies of the prototype filter. E.g., if you choose $omega'_p$, you can compute the value of $alpha$ according to
$$alpha=frac{sin[(omega'_p-omega_p)/2]}{sin[(omega'_p+omega_p)/2]}tag{1}$$
where $omega_p$ is the desired pass band edge. Now that you have $alpha$, the stop band edge of the prototype filter can be computed from the desired stop band edge $omega_s$:
$$omega'_s=-angleleft(frac{e^{-jomega_s}-alpha}{1-alpha e^{-jomega_s}}right)tag{2}$$

EDIT: the lowpass-to-lowpass transformation
$$z^{-1}longrightarrowfrac{z^{-1}-alpha}{1-alpha z^{-1}}tag{3}$$
transforms a prototype edge frequency $omega'_s$ in the following way:
$$e^{-jomega'_s}=frac{e^{-jomega_s}-alpha}{1-alpha e^{-jomega_s}}tag{4}$$
From $(4)$ it follows that
$$-omega'_s=angleleft(frac{e^{-jomega_s}-alpha}{1-alpha e^{-jomega_s}}right)tag{5}$$
which is equivalent to $(2)$.
Note that Eq. $(5)$ is different from Eq. $(8.71)$ in the book because $(8.71)$ is derived from a lowpass-to-highpass transformation.

Frequency conversion in IIR lowpass-to-lowpass transformation

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