I have a basic question.

The "well known" lowpass to bandpass transformation is $$ s \longmapsto \frac{\bar{s}^2 + \omega_1\omega_2}{\bar{s}(\omega_1 - \omega_2)}, $$ which gives a bandpass transfer function of $$ \frac{1}{s + 1} \longmapsto \frac{\bar{s}(\omega_1 - \omega_2)}{\bar{s}^2 + \bar{s}(\omega_1 - \omega_2) + \omega_1 \omega_2}. $$

My intuition is that a bandpass should be the product of a lowpass and a highpass. However, this product gives a different transfer function: $$ \frac{\omega_1}{s + \omega_1} \frac{s}{s + \omega_2} = \frac{\omega_1 s}{s^2 + (\omega_1 + \omega_2) s + \omega_1 \omega_2}, $$ which indicates that the bandpass transformation does not give this cascade of lowpass and highpass.

  • My question is, how is the bandpass transformation designed, in terms of either combining lowpass filters or by pole placement?

  • Related question, but using a different derivation technique, and reference is made to the lowpass/highpass derivation, but it is not shown: How is the lowpass to bandpass transformation derived?

  • $\begingroup$ You may want to reference the source for the transformation you're using. There is probably more than one, and the transformation may not be unique. I'd suspect it is done by stretching the low pass filter to get the correct bandwidth (frequency dilation) and then frequency shifting it to the correct center frequency. I think this used to be in the old version of Oppenheim and Schafer but it is omitted in the new "Discrete-Time Signal Processing" $\endgroup$
    – David
    Commented Dec 4, 2020 at 15:09
  • $\begingroup$ Yes, good point David. It is the transformation given on Wikipedia, and the reference they cite. And thanks for the reference - yes I couldn't see it in the new edition. $\endgroup$
    – chaffdog
    Commented Dec 5, 2020 at 19:58
  • $\begingroup$ @chaffdog please correct the last equasion $\endgroup$ Commented Nov 10, 2021 at 14:49

1 Answer 1


You're right that the multiplication of a low pass and a high pass filter results in a band pass filter, as long as the cut-off frequency of the low pass is higher than the cut-off frequency of the high pass. The problem with that approach is that low pass and high pass filters with magnitude responses that are optimal according to some chosen criterion (Butterworth, Chebyshev, Cauer) will not result in an optimum band pass filter.

On the other hand, mapping a single optimal filter will result in another optimal filter. Using $\omega_l\omega_u=\omega_0^2$, where $\omega_l$ and $\omega_u$ are lower and upper band edges, respectively, and $\omega_0$ is the center frequency of the band pass filter, and leaving out constants for the sake of simplicity, the transformation can be written as

$$s\longmapsto \frac{s^2+\omega_0^2}{s}\tag{1}$$

[Note that $\omega_l$ and $\omega_u$ are denoted as $\omega_1$ and $\omega_2$ in the OP, but they're used in a different way in the figure below.]

The mapping $(1)$ maps DC ($\omega=0$) to the desired center frequency $\omega_0$. Furthermore, $s=\pm\infty$ is mapped to $s=0$ and $s=\infty$. So the whole frequency axis of the low pass filter is mapped to the positive frequency axis of the band pass filter. (The same is true for the negative half axis of the band pass filter):

enter image description here

(from: Digital Filter Design by Parks and Burrus)

  • $\begingroup$ I don't see how you get $\omega_1\omega_2=\omega_0^2$ since $\omega_1<\omega_0$ and $\omega_2<\omega_0$. Is this an approximation? $\endgroup$
    – David
    Commented Dec 4, 2020 at 19:29
  • $\begingroup$ @David: I used $\omega_1$ and $\omega_2$ as used in the OP; they're different from the ones in the figure. I'll clarify that ... $\endgroup$
    – Matt L.
    Commented Dec 5, 2020 at 9:58
  • $\begingroup$ Thanks for the wonderfully clear explanation! $\endgroup$
    – chaffdog
    Commented Dec 5, 2020 at 19:59

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