The QMF symmetry of a two-channel filter bank is given by

$H_1(z) = H_0(-z)$

I don't understand why such a constraint is quadrature and mirror symmetry. Are $H_1$ and $H_0$ quadrature and mirror symmetry? How to interpret it visually or intuitively?


2 Answers 2


The complementary frequency response is easy to understand since the frequency response is given as $H(e^{j\omega})$ so $H(-z)=H(e^{-j\omega})$ is simply replacing $\omega$ with $-\omega$ which would transform a low-pass into a complementary high-pass (and swap any spectrum even if not low-pass / high-pass).

To understand the quadrature relationship, Julius Smith shows at this link that for the case of causal QMF FIR filters, they can only be done with two taps of the general form:

$$H_0(z) = c_0z^{-2n_0} + c_1z^{-2n_1+1}$$ $$H_1(z) = c_0z^{-2n_0} - c_1z^{-2n_1+1}$$

Where $c_0$ and $c_1$ can be any complex number and $n_0$ and $n_1$ are non-negative integers.

This is derived from the combined property for QMF filters that $H_1(z) = H_0(-z)$ (symmetry constraint) AND importantly that the sum of the two filters provide perfect reconstruction:$|H_1(z)+H_0(z)| = K$ (where K is a positive real constant).

When $c_0=c_1=1$ and $n_0=n_1=0$ we get the simple case of a 2 point DFT which is a low-pass high-pass pair:

$$H_0(z)=1+z^{-1}$$ $$H_0(z)=1-z^{-1}$$

He doesn’t mention this, but I would contend that $c_0$ must also be equal to $c_1$ as this is the only condition in which the two frequency responses will be in quadrature across the entire frequency range. I will show how for the cases when $c_0=c_1$, the angle between $H_0(z)$ and $H_1(z)$ for all $z$ on the unit circle (the frequency response, as given by $z = e^{j\omega}$) will always be either $\pm \pi/2$. This is because the sum and difference of any two complex phasors with the same magnitude and different angles will always be in quadrature. (above $c_0z^{-2n_0}$ and $c_1^{-2n_1+1}$ each can be represented by complex phasors on the z-plane at any given value for complex $z$). We can see how this can result graphically:

quadrature angles

This property can be used in the creation of analog quadrature sources to remove residual quadrature error by first hard limiting to ensure equal amplitude, then adding and subtracting the two signals, and finally hard limiting again to remove the introduced amplitude error. I also used this in an algorithm to efficiently compare the magnitude of complex numbers using a "Sigma Delta Argument Test", with the following mathematical derivation for all cases:

Mathematical Derivation

Consider two complex phasors:

$$z_1 = A_1e^{j\phi_1}$$ $$z_2 = A_2e^{j\phi_2}$$

Where $A_1$ and $A_2$ are positive real quantities representing the magnitude of $z_1$ and $z_2$ and $\phi_1$ and $\phi_2$ are the phase in radians.

Divide both by $z_1$ to have expression for $z_2$ relative to $z_1$

$$z_1' = \frac{z_1}{z_1} = 1$$ $$z_2' = \frac{z_2}{z_1} = \frac{A_2}{A_1}e^{j(\phi_2-\phi_1)} = Ke^{j\phi}$$

Such that if $K>1$ then $z_2>z_1$

The sum and the difference of the $z_1'$ and $z_2'$ would be:

$$\Sigma = z_1' + z_2' = 1 + Ke^{j\phi}$$

$$\Delta = z_1' - z_2' = 1 - Ke^{j\phi}$$

The complex conjugate multiplication of two vectors provides for the angle difference between the two; for example:

Given $$v_1= V_1e^{j\theta_1}$$ $$v_2= V_2e^{j\theta_2}$$ The complex conjugate product is: $$v_1v_2^*= V_1e^{j\theta_1}V_2e^{-j\theta_2}= V_1V_2e^{j(\theta_1-\theta_2)}$$

So the complex conjugate product of $\Sigma$ and $\Delta$ with a result $Ae^{j\theta}$ is:

$$ \begin{aligned} Ae^{j\theta} &= \Sigma \cdot \Delta^* \\ &= (1+Ke^{j\phi})(1-Ke^{j\phi})^* \\ &= (1+Ke^{j\phi})(1-Ke^{-j\phi)}) \\ &= 1 +K(2jsin(\phi))-K^2 \\ &= (1 - K^2) +j2Ksin(\phi) \\ \end{aligned} $$

Noting that the above reduces to $2jsin(\phi)$ when K= 1, and when K < 1 the real component is always positive and when K > 1 the real component is always negative such that:

for $K < 1, |\theta| < \pi/2$

for $K = 1, |\theta| = \pi/2$

for $K > 1, |\theta| > \pi/2$

Thus we also see from this that for all other cases when $c_0 \neq c_1$ that the two frequency responses cannot be in quadrature.


You can substitute $z=e^{\mathrm j\omega}$ to get a frequency response of of your filter. Notice that $-z=e^{\mathrm j(\omega+\pi)}$ and that a purely real filter has $H(z) = H(z^*)^*$.

From this, the constraint ensures that $H_0$ and $H_1$ have a frequency response that has even symmetry around $\frac\pi2$.

  • $\begingroup$ How about the quadrature property? Does it mean $H_0$ and $H_1$ occupy different portions of the frequency axis? $\endgroup$
    – ecook
    Feb 28, 2022 at 8:30
  • $\begingroup$ One's a low pass, one's a high pass. Both with 3 dB cut-off of pi/2. $\endgroup$
    – Uroc327
    Feb 28, 2022 at 8:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.