I'm using a lowpass to bandpass filter transformation:

$$ s \leftarrow Q \left( \frac{s}{\omega_0} + \frac{\omega_0}{s} \right) $$

Based on the wiki article: http://en.wikipedia.org/wiki/Prototype_filter which basically is a change of variable. But I was wondering how this transformation is derived?

The reason being I need to create my own transformation so I can combine a low pass with a shelf using a single transformation from a low pass (to use in the Nth order butterworth equation). Knowing how the lowpass to bandpass transformation is derived would take me a step further to experimenting combining other filters.


1 Answer 1


In the $s$-domain, the LPF-to-BPF transformation doubles the order of the filter. that is because the LPF has one transition from passband to stopband, but the BPF has two such transitions.

Remember that when we do frequency response, we substitute $s=j\omega$. That is, we evaluate the $s$-plane transfer function $H(s)$ as $H(j\omega)$. The magnitude of $H(j\omega)$ or $|H(j\omega)|$ is the gain of the filter when the input frequencey is $\omega$. Also keep in mind that the reciprocal of $j$ is its negative: $ \frac{1}{j} = - j $.

In the LPF, the middle of the passband is at DC, $\omega=0$. The transformation to BPF is such that maps, using a nice continuous function, the middle of the BPF passband at $\omega=\omega_0$ to $\omega=0$. A simple, mathematically continuous mapping as such is:

$$ \omega \leftarrow A ( \omega^2 - \omega_0^2 ) \quad A \ne 0 $$

So whenever the $\omega$ on the right (which is the BPF $\omega$) is equal to either $+\omega_0$ or $-\omega_0$, the $\omega$ on the left (which is the LPF $\omega$) is zero.

now suppose we set $A \triangleq \frac{Q}{\omega_0 \omega} $ which is still non-zero (and non-infinite for $\omega \approx \omega_0$). now you have

$$ \omega \leftarrow Q \left( \frac{\omega}{\omega_0} - \frac{\omega_0}{\omega} \right) $$

$$ \begin{align} j\omega & \leftarrow Q \left( j\frac{\omega}{\omega_0} - j\frac{\omega_0}{\omega} \right) \\ & = Q \left( \frac{j\omega}{\omega_0} + \frac{\omega_0}{j\omega} \right) \\ \end{align} $$


$$ s \leftarrow Q \left( \frac{s}{\omega_0} + \frac{\omega_0}{s} \right) $$

so this means, regarding frequency response:

$$ \begin{align} H_\text{BPF}(j\omega) & = H_\text{LPF}\left(j Q \left( \frac{\omega}{\omega_0} - \frac{\omega_0}{\omega} \right) \right) \\ & = H_\text{LPF}\left(Q \left( \frac{j\omega}{\omega_0} + \frac{\omega_0}{j\omega} \right) \right) \\ \end{align} $$

or, regarding transfer function:

$$ H_\text{BPF}(s) = H_\text{LPF}\left(Q \left( \frac{s}{\omega_0} + \frac{\omega_0}{s} \right) \right) $$

  • $\begingroup$ Thanks Robert, out of interest is there another derivation that takes as the starting assumption that the transfer function of a bandpass is the product of a low pass and high pass transfer function and works backwards to algebraically deduce the mapping? $\endgroup$
    – keith
    Commented Mar 13, 2015 at 7:59
  • $\begingroup$ well, first you have to start with the knowledge that the HPF is a LPF with another mapping: $$ H_\text{HPF}(s) = H_\text{LPF}\left(\frac{1}{s}\right) $$ then multiply these together and you will see your order double. and when you see that, you might come upon the $\omega^2-\omega_0^2$ expression above. $\endgroup$ Commented Mar 13, 2015 at 18:07

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