Frequency transformations

One low-pass design can be turned into a high-pass, a band-pass or a band-stop. Below, an order-3 Butterworth takes the four shapes in turn. Watch the zeros, the open circles, more than the poles.

One prototype, four shapes

In Filter specifications (18.1) a spec came in four shapes: low-pass, high-pass, band-pass and band-stop. Analog prototype filters (20.1) gave us only low-passes. IIR design by the bilinear transform (20.2) turned them into digital filters. So where do the other three shapes come from?

I don’t design them from scratch. I take the low-pass and replace its variable by something else, and that move is called a frequency transformation. Think of one stencil: flip it, or repeat it, and the same cut-out makes new patterns.

Analog and digital frequencies meet again on this page, so here are the letters once more. Analog frequency is $\omega$ (rad/s), digital frequency is $\Omega$ (rad/sample); Oppenheim and Schafer use the opposite letters.

Start from 20.1’s analog Butterworth prototype, the low-pass with its cutoff at 1 rad/s. Its squared gain is $\lvert H(j\omega)\rvert^2=1/(1+\omega^{2N})$, and its $N$ poles sit evenly on the left half of the unit circle. For $N=3$ they are $-0.5\pm j0.866$ and $-1$.

Four substitutions

The low-pass is the plainest case. Replace every $s$ by $s/\omega_c$. The new filter does at $\omega$ what the prototype did at $\omega/\omega_c$, so the cutoff moves from 1 rad/s to $\omega_c$.

For a high-pass, replace $s$ by $\omega_c/s$. At low frequencies $\omega_c/(j\omega)$ is large, so the filter acts like the prototype’s stop band. At high frequencies it is small, like the prototype’s pass band. At $\omega=\omega_c$ it equals $-j$, of size 1, so that is the cutoff, at −3 dB.

A band-pass from $\omega_1$ to $\omega_2$ needs a substitution with its own centre, $\omega_0$, and width, $\mathrm{BW}$:

On the axis, $s=j\omega$ becomes $-j(\omega_0^2-\omega^2)/(\omega\,\mathrm{BW})$. At $\omega=\omega_0$ that is 0, the prototype’s zero frequency, so the band’s centre passes. Near 0 and far above the band it is large, which is the prototype’s stop band. At $\omega_1$ and at $\omega_2$ its size is 1, so both edges sit at −3 dB.

A band-stop uses the reciprocal, $s\to s\,\mathrm{BW}/(s^2+\omega_0^2)$. Now the band’s centre goes where the prototype’s frequency is infinite, deep in its stop band, and both far ends pass.

Each band substitution is quadratic in $s$. A prototype pole $p$ becomes the two roots of $s^2-p\,\mathrm{BW}\,s+\omega_0^2=0$, so each pole becomes two. An order-3 prototype gives an order-6 band filter.

The digital design adds 20.2’s two steps. Pre-warp the edges with $\omega_\text{pw}=\frac2{T_s}\tan\frac\Omega2$, substitute, and map with the bilinear transform. SciPy does all of it in one call, such as butter(3, 1000, btype='high', fs=8000).

The picture at the top of the page shows the four results in the z-plane, where the filter is used. Watch where the poles, drawn as crosses, and the zeros, drawn as open circles, go.

After the clip, choose a shape with its button to play the move to it again. Notice that both band filters have six poles, twice the prototype’s three.

Why the poles stay where they are

The high-pass surprised me the first time I drew it: its poles are the low-pass’s poles. The prototype’s poles lie on the unit circle, where $1/p=p^*$. So the high-pass substitution sends a pole $p$ to $\omega_c/p=\omega_c\,p^*$. That is the low-pass pole $\omega_c\,p$ reflected across the real axis, and poles come in such pairs, so the set does not change.

The zeros are what moves. The prototype has no zeros at finite $s$, but its gain falls like $1/\omega^3$ as $\omega$ grows: three zeros “at infinity”. By 20.2 the bilinear map lands those at $z=-1$, which is why the low-pass’s gain is 0 at 4 kHz. The high-pass substitution swaps 0 and infinity, so its three zeros sit at $s=0$, which the map sends to $z=1$.

The band-pass substitution is infinite both at $s=0$ and at infinity. So its six zeros split three and three: three at $z=1$ and three at $z=-1$, blocking both ends. The band-stop substitution is infinite at $s=\pm j\omega_0$, so its zeros sit there, three on each, and the map puts them on the unit circle at the band’s centre. By the same $1/p=p^*$ argument, its poles are the band-pass’s poles.

On the circle, the band’s centre is not 1500 Hz, the middle of 1000 and 2000 Hz. The pre-warp bends the axis, and the centre lands at 1456 Hz. The worked example below does the arithmetic.

One number retunes the filter

There is a second route. Instead of substituting for $s$ before the map, substitute for $z^{-1}$ after it. Use the first-order all-pass of All-pass systems (17.1), with a real number $a$ between −1 and 1:

Why an all-pass? By 17.1 it has gain 1 at every frequency. So when $z^{-1}=e^{-j\Omega}$, it gives back another point on the circle, which I’ll write $e^{-j\theta}$. The new filter at $\Omega$ then gives what the old one gave at $\theta$.

The gain keeps its values; they only move along the frequency axis. As $\Omega$ runs from 0 to $\pi$, $\theta$ also runs from 0 to $\pi$, always climbing, so nothing is folded or repeated.

Constantinides worked out an all-pass for each shape, and Oppenheim and Schafer print his table in their chapter 7. For the band shapes it is a second-order all-pass, so the order doubles there too. To move a low-pass’s cutoff, the first-order one above is enough, and it needs one number, $a$.

The tuning constant

Let’s find $a$. I want the new cutoff $\Omega_\text{new}$ to land on the old cutoff $\Omega_c$:

Multiply out and collect the terms with $a$. Then take half of each angle out of the top and the bottom, and use $e^{jx}-e^{-jx}=2j\sin x$ from Complex exponentials & phasors (3.4). That gives the tuning constant:

A lower cutoff gives a positive $a$ and a higher one a negative $a$. With no change, $a=0$ and the all-pass is $z^{-1}$ itself.

Each pole and zero moves too. Put the all-pass into a factor $1-rz^{-1}$ of 16.3’s factored form, and the root $r$ goes to $(r+a)/(1+ar)$. A zero at $-1$ goes to $(-1+a)/(1-a)=-1$, so it stays, and the gain at 4 kHz stays 0.

The instrument below applies this to the 1 kHz low-pass of the first instrument. Think of a synthesiser’s cutoff knob: you turn one thing, and the whole filter follows. Watch the poles slide together while the curve keeps its shape.

Notice where the real pole goes at 2 kHz. There $a=1-\sqrt2=-0.4142$ and the old real pole is $\sqrt2-1=0.4142$, so $r+a=0$: the pole sits at the centre of the circle.

After the clip, drag the ring at the −3 dB point along the curve, or use the arrow keys, to retune the cutoff yourself. Try 100 Hz: $a$ climbs to 0.8267 and all three poles crowd near $z=1$. Then try 3900 Hz, where $a=-0.9680$ and the poles crowd near $z=-1$.

The same Butterworth either way

Is the retuned filter a real Butterworth, or only an imitation? Here it is the real thing. The all-pass multiplies every pre-warped frequency $\tan(\Omega/2)$ by the same number, $\tan(\Omega_c/2)/\tan(\Omega_\text{new}/2)$, which is the analog step $s\to s/\omega_c$ in disguise.

A Butterworth’s gain depends only on $\omega/\omega_c$. So the analog route and the digital route give the same filter: SciPy’s butter(3, f, fs=8000) at the new cutoff. This holds for every low-pass made by the bilinear map, which is why the reference card says “exact for bilinear designs”.

Worked example

Let’s redo the page’s numbers. All three designs are order-3 Butterworths at $f_s=8$ kHz, so $\omega_\text{pw}=2f_s\tan(\Omega/2)=16\,000\tan(\Omega/2)$ rad/s.

1. High-pass from low-pass, 1 kHz. butter(3, 1000, fs=8000) has poles $0.6911\angle{\pm40.89°}$, that is $\pm0.7137$ rad, and $0.4142$. Its three zeros sit at $-1$. The high-pass has the same poles, and its zeros move to $1$.

What changes besides the zeros is the gain constant $K$ of 16.3’s factored form. It is set so that the pass band has gain 1. For the low-pass, at $z=1$, $K\cdot2^3/\prod(1-p_k)=1$ gives $K=0.2535/8=0.031689$. For the high-pass, at $z=-1$, $K=\prod(1+p_k)/8=3.5672/8=0.445903$.

2. Band-pass, 1 to 2 kHz. The edges are $\Omega_1=2\pi\cdot1000/8000=\pi/4$ and $\Omega_2=\pi/2$. Pre-warped, they become $\omega_1=16\,000\tan(\pi/8)=6627.4$ rad/s and $\omega_2=16\,000\tan(\pi/4)=16\,000.0$ rad/s.

So the centre is $\omega_0=\sqrt{6627.4\cdot16\,000}=10\,297.5$ rad/s, and the width is $\mathrm{BW}=9372.6$ rad/s. The map sends the centre to $\Omega_0=2\arctan(10\,297.5/16\,000)=1.1437$ rad, which is $65.53°$ or 1456.2 Hz.

The six poles are $0.8560\angle{\pm46.31°}$, $0.8073\angle{\pm87.21°}$ and $0.6436\angle{\pm62.93°}$. The prototype’s real pole $-1$ gives the last pair. The band-stop has the same poles, and its zeros are $e^{\pm j65.53°}$, three each. Its gain is −65.7 dB at $\sqrt{1000\cdot2000}=1414.2$ Hz, and the gain itself is zero at 1456.2 Hz.

3. The tuning table. For 500 Hz, $\Omega_c=\pi/4$ and $\Omega_\text{new}=\pi/8$, so $a=\sin(\pi/16)/\sin(3\pi/16)=0.19509/0.55557=0.3512$. The table gives $a$ for eight cutoffs.

At 3000 Hz the poles are $0.6911\angle{\pm139.11°}$ and $-0.4142$. These are the 1 kHz poles reflected across the imaginary axis: $180°-40.89°=139.11°$, and $0.4142$ becomes $-0.4142$.

Where you’ll meet this

A tunable filter has to follow a knob quickly: the cutoff of a synthesiser’s filter, or a band of a tunable equaliser. With the all-pass substitution the knob sets one number, $a$, and the new coefficients follow from it without a new design. Audio equalisers and biquads (20.5) builds the equaliser shapes from second-order filters.

In SciPy the analog route sits behind the btype argument. butter(3, [1000, 2000], btype='bandpass', fs=8000) pre-warps the edges, substitutes into the prototype and applies the bilinear map, as above. Band-pass filters of this kind split a signal into bands in Filter banks (23.1).

Reference card