Suppose if I have a real BPF (band pass iir filter with co-efficients of numerator and denominator), can one get the complex BPF filter coefficients using these ?

Real BPF will be one sided and complex BPF will be two sided.

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    $\begingroup$ i thought it was the other way around. i thought that real BPF is two-sided and complex BPF would be one-sided, $\endgroup$ Feb 15, 2022 at 8:49
  • $\begingroup$ I'd agree with robert, real systems have to be spectrally symmetric, so, the real BPF will be two-sided. Anyway, mano, could you elaborate what specifically you want that conversion to do? Maybe a drawing like "real BPF's amplitude response:" ___/~~~\__.__/~~~\___, "complex BPF's desired amplitude response:": ??? would help! $\endgroup$ Feb 15, 2022 at 10:25
  • $\begingroup$ No I think the poster is right- Real signals can have what we call "One-Sided" spectrums due to the Hermitian Symmetry such that we can represent everything by just showing one side and doubling it. For example phase power spectral densities that are one-sided are $S_\phi(f)$ while two-sided spectrums are $\mathscr{L}_\phi(f)$. And $S_\phi(f) = 2\mathscr{L}_\phi(f)$. For example see: zone.ni.com/reference/en-XX/help/371361R-01/lvanlsconcepts/…. $\endgroup$ Feb 16, 2022 at 0:04
  • $\begingroup$ I am not sure however if the OP wants a complex two-sided spectrum with a different but non-zero result in each side, or a two-sided spectrum where one side (positive or negative) is zero such as the analytic signal representation. $\endgroup$ Feb 16, 2022 at 0:06

1 Answer 1


To convert a real band pass filter to it's closest complex band pass filter equivalent that only passes the positive or negative half of the frequency spectrum, you can remove all poles and zeros in that associated part of the complex plane when the filter consists of complex poles and zeros.

For example consider this simple Butterworth filter given by numerator coefficients:

$$b = [0.07296, 0, -0.072960]$$

and denominator coefficients:

$$a = [1, -1.41421, 0.85408]$$

Such that

$$H(z) = \frac{0.07296z^2- 0.07296}{z^2-1.414z+0.8541}$$

This filter has frequency response as shown in the plot below:

freq response

And the pole zero map for this filter is:

pz map

We see that there are two complex poles defining the center of the bandpass filter, and then two real poles at DC and $f_s/2$. We can eliminate the negative frequency pole, resulting in the following transfer function:

The roots of the denominator are $0.70711 \pm j0.59505$ so the new transfer function is:

$$H_c(z) = \frac{0.07296z^2- 0.07296}{z + 0.70711 - j0.59505}$$

This is an improper transfer function (non-causal) as the numerator is of higher order than the denominator). To make this causal we need to introduce a delay $z^{-1}$ which is a pole at the origin resulting in:

$$H_c(z) = \frac{0.07296z^2- 0.07296}{z^2 + (0.70711 - j0.59505)z}$$

The frequency response of the first form is given below:

complex filter response

With the additional $z^{-1}$ we would get the same magnitude response with an additional linear phase added to the phase response.


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