Another common approach for filter design is to start with a prototype filter, often a low-pass filter due to their analytical simplicity. We can then use Z-transform/Fourier properties to manipulate the into another filter that we desire. This section discusses these transformation properties and how they are used.
Assume we have a low pass filter represented by $H_{\textrm{low}}(z)$. We can express a high-pass filter $H_{\textrm{high}}(\omega)$ by shifting the low pass filter, resulting in $$H_{\textrm{high}}(\omega) = H_{\textrm{low}}(\omega - \pi) \;. $$
In the time domain, this is equivalent to $$h_{\textrm{high}}[n] = h_{\textrm{low}}[n] \cos(\pi n) \;. $$ Hence, by multiplying by a cosine, we increase the frequency of our impulse response.
In the Z-domain, this is equivalent to $$H_{\textrm{high}}(z) = H_{\textrm{low}}(-z) \;. $$ Hence, a pole placed at $z=1$ would now be placed at $z=-1$.
Assume we have a low pass filter represented by $H_{\textrm{low}}(z)$. We can express a band-pass filter $H_{\textrm{high}}(\omega)$ by shifting the low pass filter, resulting in $$H_{\textrm{pass}}(\omega) = H_{\textrm{low}}(\omega - \omega_0) + H_{\textrm{low}}(\omega + \omega_0) \;, $$ where $\omega_0$ is the center frequency of the band-pass filter.
In the time domain, this is equivalent to $$ \begin{align*} h_{\textrm{pass}}[n] &= h_{\textrm{low}}[n] \left[ e^{- j \omega_0 n } + e^{+ j \omega_0 n } \right] \\ &= 2 h_{\textrm{low}}[n] \cos(\omega_0 n) \;. \end{align*} $$ Again, by multiplying by a cosine, we increase the frequency of our impulse response.
In the Z-domain, this is equivalent to $$H_{\textrm{pass}}(z) = H_{\textrm{low}}(z e^{j \omega_0}) + H_{\textrm{low}}(z e^{-j \omega_0}) \;. $$ Hence, a pole placed at $z=1$ would now be placed at $z=e^{\pm j \omega_0}$.
Assume we have a low pass filter represented by $H_{\textrm{low}}(z)$. We can express another low-pass filter $H_{\textrm{low}}'(z)$ through the transformation $$H_{\textrm{low}}'(z) = H_{\textrm{low}}\left( \frac{z-a}{1-a z} \right) \;. $$
This transformation will work for any $a$, but if you know the cut-off frequency of the current filter $w_c$ and the cut-off frequency of the new filter $w_c'$, we can set $$ \begin{align} a = \frac{\sin[(\omega_c - \omega_c')/2]}{\sin[(\omega_c + \omega_c')/2]} \end{align} $$ Note that when $\omega_c = \omega_c'$, the filter is unchanged.
Assume we have a low pass filter represented by $H_{\textrm{low}}(z)$. We can express a high-pass filter $H_{\textrm{high}}(z)$ through the transformation $$H_{\textrm{high}}(z) = H_{\textrm{low}}\left( - \, \frac{z-a}{1-a z} \right) \;. $$
This transformation will work for any $a$, but if you know the cut-off frequency of the current filter $w_c$ and the cut-off frequency of the new filter $w_c'$, we can set $$ \begin{align} a = \frac{\cos[(\omega_c + \omega_c')/2]}{\cos[(\omega_c - \omega_c')/2]} \end{align} $$ Note that when $\omega_c = \omega_c'$, we get $H_{\textrm{high}}(z) = H_{\textrm{low}}\left( - z \right)$.
Assume we have a low pass filter represented by $H_{\textrm{low}}(z)$. We can express a band-pass filter $H_{\textrm{pass}}(z)$ through the transformation $$H_{\textrm{pass}}(z) = H_{\textrm{low}}\left( - \, \frac{z^2 - a_1 z + a_2}{a_2 z^2 - a_1 z + 1} \right) \;. $$
If you know the cut-off frequency of the current filter $w_c$, the upper frequency edge of the band-pass filter $\omega_u$ and the lower frequency edge of the band-pass filter $\omega_\ell$, we can set $$ \begin{align} a_1 &= \frac{-2 \alpha K}{K +1} \\ a_2 &= \frac{K-1}{K+1} \\ \alpha &= \frac{\cos[(\omega_u + \omega_\ell)/2]}{\cos[(\omega_u - \omega_\ell)/2]} \\ K &= \cot[(\omega_u - \omega_\ell)/2] \tan[\omega_c/2] \end{align} $$ Note that when $\omega_u - \omega_\ell = \omega_c$, then $K=1$ and the transformation becomes a low-pass to high-pass conversion (with an additional shift).
Assume we have a low pass filter represented by $H_{\textrm{low}}(z)$. We can express a band-stop filter $H_{\textrm{stop}}(z)$ through the transformation $$H_{\textrm{stop}}(z) = H_{\textrm{low}}\left( - \, \frac{z^2 - a_1 z + a_2}{a_2 z^2 - a_1 z + 1} \right) \;. $$
If you know the cut-off frequency of the current filter $w_c$, the upper frequency edge of the band-pass filter $\omega_u$ and the lower frequency edge of the band-pass filter $\omega_\ell$, we can set $$ \begin{align} a_1 &= \frac{-2 \alpha K}{K +1} \\ a_2 &= \frac{1-K}{1+K} \\ \alpha &= \frac{\cos[(\omega_u + \omega_\ell)/2]}{\cos[(\omega_u - \omega_\ell)/2]} \\ K &= \tan[(\omega_u - \omega_\ell)/2] \tan[\omega_c/2] \end{align} $$