All-pass systems

A pole sits at 0.5 and a zero at 2, on either side of the unit circle, and a test point walks round the circle. Watch the arrow from each of them to the test point, and the gain the two arrows give.

Twice as far, all the way round

On the page Transfer functions, poles and zeros (16.3) you read a gain from arrows: one from each pole and each zero to the test point $e^{j\Omega}$ on the unit circle. A pole’s arrow divides the gain and a zero’s arrow multiplies it. A pole near the circle makes a peak, and a zero near it makes a dip. Is there a place for a zero where it cancels the pole’s effect on the gain, for every $\Omega$ at once?

There is, and it is the pole’s mirror image across the circle. For a pole $p=re^{j\theta}$, the mirror point is $1/p^*$. It has the same angle $\theta$ and the reciprocal radius $1/r$, as the mirror image $z^*$ of Complex numbers for signals (3.3) keeps the radius and flips the angle. For $p=0.5$ the mirror point is 2.

Take the pole at 0.5 and the zero at 2. Multiply the top and the bottom by $z$ and factor:

On the unit circle the gain is therefore $0.5$ times the length of the zero’s arrow divided by the length of the pole’s arrow. The claim to check is that, from every point of the circle, the zero at 2 is exactly twice as far away as the pole at 0.5. Then the gain is $0.5\times2=1$ at every frequency. Imagine two lamps placed so that wherever you stand on a round path, one lamp is twice as far away as the other.

The picture at the top of the page is this pair of lamps. Notice that both arrows change and their ratio does not. When the clip has finished, drag the test point round the circle yourself, or use the arrow keys.

Now the reason. For a real pole $a$, the factor $e^{-j\Omega}-a$ is the zero’s arrow turned and scaled, and for a complex pole the same steps work with $a^*$. Pull out $e^{-j\Omega}$, which has size 1:

The last step uses $\lvert w^*\rvert=\lvert w\rvert$ from 3.3. So the top and the bottom of $H_\text{ap}(z)=(-a^*+z^{-1})/(1-az^{-1})$ have the same size on the circle, and $\lvert H_\text{ap}\rvert=1$. A system with this property is called all-pass: it changes no size, only phase. The pole $a$ must lie inside the circle for the system to be causal and stable, as in Stability and causality (16.4), so its mirror zero lies outside.

The same steps work for a complex pole $p$, paired with the zero $1/p^*$. A real system needs the conjugate pair as well, and then the numerator holds the denominator’s coefficients in reverse order. For $p=0.5e^{j\pi/3}$ the denominator is $1-0.5z^{-1}+0.25z^{-2}$ and the numerator is $0.25-0.5z^{-1}+z^{-2}$. Because gains multiply, a cascade of all-pass systems is all-pass too.

The points that are twice as far from 2 as from 0.5 form a circle, known as a circle of Apollonius. Here that circle is the unit circle: $\lvert z-2\rvert=2\lvert z-0.5\rvert$ reduces to $\lvert z\rvert^2=1$.

Flat gain, shaped delay

The gain cannot move, but the phase can. Write the real first-order all-pass as $H_\text{ap}=e^{-j\Omega}D^*/D$, with $D=1-ae^{-j\Omega}$. This is the same expression as before, since $e^{-j\Omega}D^*=e^{-j\Omega}-a$. The ratio $D^*/D$ has size 1 and angle $-2\angle D$, so the phase is

The group delay is minus the slope of this unwrapped phase (12.4, Frequency response of discrete-time systems). The slope of the arctangent is $(a\cos\Omega-a^2)/(1-2a\cos\Omega+a^2)$. Adding it up gives

At $\Omega=0$ this is $(1+a)/(1-a)$, and at $\Omega=\pi$ it is $(1-a)/(1+a)$. With $a=0$ the pole is at 0, the zero is infinitely far away, and $H=z^{-1}$: a plain delay of one sample at every frequency. With $a$ near 1 the low frequencies wait much longer than one sample, and the high ones wait a fraction of one. A negative $a$ does the reverse. I think of a prism: every colour gets through, and each is delayed by its own amount.

Below, the pole moves along the real axis. Notice that the gain line never moves while the delay curve does.

When the clip has finished, drag the pole along the real axis yourself, or use the arrow keys.

One fact holds for every $a$. At $\Omega=0$ the system gives $H=(1-a)/(1-a)=1$, and at $\Omega=\pi$ it gives $H=(-a-1)/(1+a)=-1$. The phase therefore falls from $0$ to $-180°$, and it falls all the way, because $\tau_g$ is positive. The area under the delay curve between $0$ and $\pi$ is $\pi$ radians’ worth of phase, so the average delay is one sample for every $a$. Moving the pole only moves where the delay sits.

Fractional delay

At low frequencies $\tau_g(0)=(1+a)/(1-a)$. A delay of $d$ samples at low frequencies needs $d(1-a)=1+a$, so

For half a sample, $d=0.5$ gives $a=-1/3$: the pole at $-0.333$ and the zero at $-3$. The group delay is 0.500, 0.562, 0.800, 1.390 and 2.000 samples at $\Omega=0$, $0.25\pi$, $0.5\pi$, $0.75\pi$ and $\pi$. So the half-sample delay holds only at low frequencies, and it is already 0.8 at $0.5\pi$. A sampled signal has no value between its samples, so this is how a fractional delay is built: not as a shifted copy, but as a system whose delay is close to $d$ over the band you care about.

Equalising a phase

A phase equaliser is an all-pass placed after a filter. The filter’s group delay is rarely flat. The all-pass adds delay where the filter has little, so the total delay is flatter, and the gain is untouched. Designing one is work for the IIR design pages (20.x).

Echoes with a flat gain

Replace $z^{-1}$ by $z^{-D}$, a delay of $D$ samples, and the same argument gives the Schroeder all-pass of 1962:

On the unit circle $e^{-jD\Omega}$ plays the part of $e^{-j\Omega}$, so the gain is still 1 at every $\Omega$. The difference equation is $y[n]=ay[n-D]-ax[n]+x[n-D]$. Feed it a single impulse. The first output is $y[0]=-a$, then $y[D]=a\cdot(-a)+1=1-a^2$, and every output after that is $a$ times the one $D$ samples before.

Take $D=5$ and $a=0.7$. The first stem is $-0.7$, then come $0.51$, $0.357$, $0.250$, $0.175$, $0.122$ and $0.086$, every 5 samples.

The total energy of the stems is $a^2+(1-a^2)^2(1+a^2+a^4+\cdots)=a^2+(1-a^2)=1$, the same as the energy of the impulse that went in. Every echo is smaller than the one before, but the echoes together carry all of the impulse’s energy.

Worked example

Check the gain of $H_\text{ap}$ with $a=0.5$ by hand at $\Omega=0.25\pi$, where $\cos\Omega=0.7071$. The pole’s arrow has length $\sqrt{1-2(0.5)(0.7071)+0.25}=\sqrt{0.5429}=0.737$. The zero’s arrow has length $\sqrt{1-2(2)(0.7071)+4}=\sqrt{2.1716}=1.474$. Their ratio is 2, so the gain is $0.5\times2=1$.

For the delay, $a=0.5$ at $\Omega=0.5\pi$ has $\cos\Omega=0$, so $\tau_g=(1-0.25)/(1+0.25)=0.6$ samples. For $a=0.8$ at $\Omega=0$, $\tau_g=0.36/0.04=9$ samples.

Now design for a quarter of a sample, $d=0.25$. Then $a=(0.25-1)/(0.25+1)=-0.6$, with the pole at $-0.6$ and the zero at $-1/0.6=-1.667$. Check the ends: $\tau_g(0)=(1-0.6)/(1+0.6)=0.25$ and $\tau_g(\pi)=(1+0.6)/(1-0.6)=4$. In between, $\tau_g$ is 0.290, 0.471 and 1.251 samples at $0.25\pi$, $0.5\pi$ and $0.75\pi$. So the delay is a quarter of a sample only near $\Omega=0$, and it has nearly doubled by $0.5\pi$.

Where you will meet this

A phase equaliser corrects the phase of a loudspeaker crossover or an anti-aliasing filter. A fractional delay lines up two sampled signals that arrived a fraction of a sample apart, as described in Laakso et al., “Splitting the unit delay”, IEEE Signal Processing Magazine, 1996, and it is the building block of Resampling by any factor (22.3). The Schroeder all-pass is a part of artificial reverberation, which returns in Audio effects (29.2).

Next, Minimum phase (17.2) splits any stable system into a part that has the smallest delay and an all-pass. Linear-phase systems (17.3) take the opposite goal: a delay that is the same at every frequency.

Reference card