Linear-phase systems

One pulse through two systems with the same gain and the same average delay. Watch which output is still the pulse.

Straight phase keeps the shape

In Frequency response of discrete-time systems (12.4) you found that a straight phase line, $\angle H=-\Omega n_0$, delays every frequency by the same $n_0$ samples. This page shows what that buys you, and how to build it. The answer is a pulse that arrives late but unchanged.

I will use one pulse and two systems. The pulse has the five samples $0.25$, $0.75$, $1$, $0.75$, $0.25$ at $n=4$ to $8$. By centre I mean its balance point, $\sum_n n\,y[n]\big/\sum_n y[n]$. For this pulse the sum of the samples is 3, the sum of $n\,y[n]$ is 18, and the centre is 6.

The first system is a delay of three samples, $z^{-3}$. Its response is $e^{-j3\Omega}$, so its gain is 1 at every frequency and its phase is the straight line $-3\Omega$. The second is the first-order all-pass of All-pass systems (17.1), with $a=0.5$:

It also has gain 1 at every frequency. Its group delay is $\tau_g=(1-a^2)\big/(1-2a\cos\Omega+a^2)$, which is 3 samples at $\Omega=0$ and falls to $0.33$ at $\Omega=\pi$.

So the two systems agree on two things: they do not change any size, and they delay a slow signal by 3 samples. Only their phase differs, one straight and one bent. That makes the comparison fair, because only the phase can explain what you see in the picture at the top of this page.

Both outputs are centred at $n=9$: on average both delay by 3 samples. Moved 3 samples, the input fits $z^{-3}$‘s output exactly, and only that output is the pulse itself. The all-pass output dips to $-0.19$ first and trails a tail: its phase delays slow parts 3 samples and fast ones 0.33, so the pulse comes apart.

Notice that the all-pass output goes below zero at $n=4$ and $n=5$, before the delayed pulse could have arrived. The delay’s output only starts at $n=7$. Afterwards the all-pass output does not stop: from $n=12$ it halves at every sample, $0.092$, $0.046$, $0.023$, and so on. Both outputs have centre 9.00, so the mean lateness is the same. The shape is not.

When the clip has finished, drag across the input to change the pulse’s width, from 3 to 11 samples. At width 9 both outputs still centre at 9.00, and the all-pass output dips to $-0.10$. The dip is $-0.25$ at width 3, $-0.19$ at 5, $-0.14$ at 7, $-0.10$ at 9 and $-0.07$ at 11. A wide pulse has most of its energy at low frequencies, where the all-pass delay is nearly constant at 3 samples. A narrow pulse reaches higher frequencies, where the delay is shorter, and it is torn apart more.

Think of a marching band crossing a bridge in step. It arrives in formation, a little later. A crowd walking at different paces arrives strung out, even if the average arrival time is the same. A straight phase is the band, and a bent phase is the crowd.

Why symmetric taps give a straight phase

A finite impulse response with $N_h$ taps has $H(e^{j\Omega})=\sum_{k=0}^{N_h-1}h[k]e^{-j\Omega k}$. Suppose the taps are symmetric, $h[n]=h[N_h-1-n]$. This is the reversal $x[-n]$ of Shifting, reversing and scaling time (2.1), moved so that it fits the taps: reversing them leaves them unchanged. Their centre is at $\alpha=(N_h-1)/2$.

Write each exponent as $k=\alpha+m$, where $m$ runs from $-\alpha$ to $\alpha$. Then $h[\alpha+m]=h[\alpha-m]$, and each tap and its mirror add up to

This is Euler’s pair from Complex exponentials & phasors (3.4). After factoring $e^{-j\alpha\Omega}$ out of the whole sum, what remains is a sum of cosines with real weights, so it is real:

I call the real factor $H_\text{zp}(\Omega)$ the zero-phase response. The prefix $e^{-j\alpha\Omega}$ is a straight-phase delay of $\alpha$ samples, so $\tau_g=\alpha$ at every $\Omega$. The real factor can change sign. Where it does, the phase jumps by $180°$, which is a real jump and not an unwrapping artefact (The DTFT, 12.2).

The triangle $(1,2,3,2,1)/9$ is the example from Properties of the DTFT (12.3). It has five taps, so $\alpha=2$, and $H_\text{zp}=(3+4\cos\Omega+2\cos2\Omega)/9$. At $\Omega=0$ this is 1, and at $\Omega=\pi$ it is $0.111$. It never changes sign, so the phase is exactly $-2\Omega$.

A symmetric pulse therefore comes out symmetric. The triangle moves your pulse from centre 6 to centre 8, which is the pulse’s centre plus $\alpha=2$. Its output rises from $0.028$ at $n=4$ to $0.722$ at $n=8$ and falls through the same values on the way back to $0.028$ at $n=12$. It reads the same in both directions.

Now suppose the taps are antisymmetric, $h[n]=-h[N_h-1-n]$. The same pairing gives a difference instead of a sum, $e^{-j\Omega m}-e^{j\Omega m}=-2j\sin(m\Omega)$, the partner of Euler’s pair. So the response is $e^{-j\alpha\Omega}\,j\,H_\text{zp}(\Omega)$, with $H_\text{zp}$ real again. Where $H_\text{zp}$ is positive, its angle is

This is generalised linear phase: a straight line that no longer passes through zero. Its slope is still $-\alpha$, so the group delay is still $\alpha$ samples at every frequency. The constant $\beta$ is a quarter turn that adds to every frequency alike.

Four kinds of symmetry, four sets of forced zeros

Symmetric or antisymmetric, odd length or even length: that gives four types. I number them I to IV. Each type, by its symmetry alone, forces zeros of $H$ at $z=1$ or $z=-1$, and a filter with a zero there has gain 0 at $\Omega=0$ or $\Omega=\pi$, as long as the taps keep their symmetry.

Watch the forced zeros, the ones in a second ring at $z=\pm1$, and the gain at the two ends of the band.

Notice that types III and IV give 0 at $\Omega=0$, and types II and III give 0 at $\Omega=\pi$. Type IV is the first difference from Oversampling and noise shaping (11.3), halved: it has gain 0 for a constant, and no filter of this type can do otherwise.

Here is why the zeros are forced. Take a symmetric filter of even length at $z=-1$. The tap $h[k]$ contributes $h[k](-1)^k$ and its mirror contributes $h[k](-1)^{N_h-1-k}$. Because $N_h-1$ is odd, the two powers of $-1$ have opposite signs, so every pair cancels and $H(-1)=0$. The other cases work alike. At $z=1$ an antisymmetric filter cancels pair by pair. An antisymmetric filter of odd length also cancels at $z=-1$, because its centre tap is 0 and its other pairs cancel.

The group delays follow from $\alpha=(N_h-1)/2$. They are 2, 1.5, 1 and 0.5 samples for the four example filters. An even length delays by a whole number plus a half, which is the price of having no centre tap.

Where do the other zeros go? Every zero of a linear-phase FIR is either on the unit circle or paired with its mirror point $1/z_k^*$. This is the flip of Minimum phase (17.2): flipping one zero keeps the gain, but it keeps a symmetric $h$ symmetric only if the mirror zero is there too. The antisymmetric filter $(1,2,-2,-1)$ shows it. Its zeros are $1$, $-0.382$ and $-2.618$, and $0.382=1/2.618$.

The SciPy function firwin refuses an even length together with a high-pass, for exactly this reason. A type II filter has a zero at $\pi$, so it cannot pass the top of the band. The design pages come in Window-method FIR design (19.1) and Optimal FIR design (19.3).

Worked example

Check the generalised phase of type III, $(1,0,-1)/2$, which has $\alpha=1$ and $H_\text{zp}=\sin\Omega$. At $\Omega=0.25\pi$ the angle should be $90°-45°=45°$, and at $0.5\pi$ it should be $90°-90°=0°$. Evaluating the response gives $45°$ and $0°$. Type IV, $(1,-1)/2$, has $\alpha=0.5$ and $H_\text{zp}=\sin(\Omega/2)$. Its angles are $90°-22.5°=67.5°$ at $0.25\pi$ and $90°-45°=45°$ at $0.5\pi$, which the response confirms.

Now go back to the all-pass of the first section. Its group delay is 3 samples at $\Omega=0$, $2.509$ at $0.1\pi$, $1.381$ at $0.25\pi$ and $0.600$ at $0.5\pi$. The delay $z^{-3}$ has 3 at all of them. The difference between those two lists is the whole difference between the two outputs, because their gains are both 1.

The pulse widths 3, 5, 7, 9 and 11 give the lowest all-pass samples $-0.250$, $-0.188$, $-0.140$, $-0.101$ and $-0.075$. The pulse through the triangle keeps its shape, with centre 8.

Where you will meet this

Whenever the shape of a waveform matters, a linear-phase FIR is a common choice, for example for pulses in communications and for edges in images. The price is delay, $\alpha$ samples, and a long filter has a long delay. Where the shape does not matter, a short IIR filter with a bent phase is cheaper, and Choosing FIR or IIR (20.6) weighs the two. Type III and IV filters give differentiators and Hilbert filters, in Special FIR filters (19.4) and The Hilbert transform and the analytic signal (27.2).

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