Choosing FIR or IIR

One spec can be met by an FIR filter or by an IIR filter. Below, both are drawn against the same spec, and then their delays. Watch the delay panel: one line stays flat, and the other climbs at the band edge.

One spec, two filters

You now know two ways to build a low-pass filter. Chapter 19 built FIR filters, with a finite list of taps. Chapter 20 built IIR filters, which feed their own outputs back. Both can meet the same spec, so which one should you use? This page puts the two side by side, on the spec you know best.

That is the running spec of Filter specifications (18.1). The sample rate is 8 kHz. From 0 to 1 kHz the gain must stay within $1\pm0.05$, and from 1.5 kHz to 4 kHz it must stay at or below 0.01, which is 40 dB down.

The FIR candidate is the shortest design of Optimal FIR design (19.3): the equiripple filter with 26 taps and stop-band weight 5. Its pass band strays from 0.9565 to 1.0436, and its stop band reaches −41.08 dB.

The IIR candidate is the elliptic filter of Analog prototype filters (20.1), turned digital by IIR design by the bilinear transform (20.2). Analog frequency is ω (rad/s), digital frequency is Ω (rad/sample); Oppenheim and Schafer use the opposite letters. An IIR filter’s gain peaks at 1, so its pass band is written as a loss: at most $-20\log_{10}0.95=0.4455$ dB. SciPy’s ellipord asks for order 5, and ellip(5, 0.4455, 40, 1000, fs=8000) designs it.

How do we compare their cost? Each output sample takes a fixed number of multiplications, so I count multiplies per sample. The FIR filter computes $y[n]=\sum_{k=0}^{25}h[k]\,x[n-k]$, so it needs 26.

Its taps are symmetric, $h[k]=h[25-k]$ (17.3). So you can add each pair of inputs first and multiply once:

That is 13 multiplies.

The IIR filter is a difference equation, as in Difference equations (6.1). With $b_k$ the numbers that multiply inputs and $a_k$ the numbers that multiply past outputs,

There are six $b_k$ and five $a_k$ ($a_0=1$ needs no multiply), so 11 multiplies. Computed this way, straight from the difference equation, the filter is in what Filter structures (21.1) calls direct form.

So the IIR filter costs 11 multiplies, against 26 for the FIR, or 13 if you use the symmetry. That is less than half. What does it cost in other ways? The picture at the top of the page draws both filters against the running spec, and then their delays.

After the clip, a dotted cursor on the delay panel reads both delays; drag it, or use the arrow keys. Move it from 0 Hz to the band edge. The dashed FIR line never moves from 12.5. The solid IIR curve starts at 3.7 samples, dips a little, passes 4.2 at 500 Hz and 9.4 at 900 Hz, then climbs fast: 14.0 at 950 Hz and 23.1 at 1000 Hz.

Notice the band edge. There the IIR filter is the slower of the two, by more than 10 samples. Below about 940 Hz the solid curve is the lower one, and above it the dashed line is.

Think of two ways to drive across a city. The short cut through town is faster on average, but its time depends on the traffic. The motorway always takes the same time. The FIR filter is the motorway: every frequency arrives 12.5 samples late. The IIR filter is the short cut.

Why does the delay climb at the edge? In 20.1 you saw that a steep edge bends the delay near the cutoff. The elliptic filter has the steepest edge for its order, and it buys it with a pole pair at radius 0.9527, close to the unit circle. Their angle, ±45.64°, is the frequency 1014 Hz, just past the pass-band edge.

What does a changing delay do to a signal? Linear-phase systems (17.3) showed it: a bent phase delays the slow parts of a pulse by one amount and the fast parts by another, so the pulse comes apart. A straight phase only moves it.

So the IIR filter is cheaper and, at low frequencies, quicker, but it changes the shape of a waveform. The FIR filter keeps the shape and is late by the same amount everywhere.

Run it forwards, then backwards

Is there a way to have the IIR filter’s low cost and still keep the shape? If the whole signal is already stored, yes. Run the filter over the signal, then run it again over the result, from the last sample back to the first.

The backwards run is the reversal of Shifting, reversing and scaling time (2.1). I reverse the signal, filter it as usual, and reverse the output.

Call the first output $v[n]$, so $V(e^{j\Omega})=H(e^{j\Omega})\,X(e^{j\Omega})$. Reversing $v$ gives $V(e^{-j\Omega})$, as in Properties of the DTFT (12.3). Filtering multiplies by $H(e^{j\Omega})$, and the last reversal replaces $\Omega$ by $-\Omega$ again:

The impulse response $h[n]$ is real, so by 12.3, $H(e^{-j\Omega})=H^*(e^{j\Omega})$. A number times its conjugate is its size squared:

The gain $\lvert H\rvert^2$ is real and never negative, so it adds no phase at all. I call this forward–backward filtering, or zero-phase filtering. Every frequency is delayed by zero samples, and nothing changes shape through the phase. The price is that the gain is squared: the pass band now lies between $0.95^2=0.9025$ and 1, and the stop band at $0.01^2$, 80 dB down.

In SciPy this is scipy.signal.filtfilt(b, a, x). The instrument below does it by hand on a pulse, with the elliptic filter of the first instrument. It marks each output’s centre with a dot; by centre I mean 17.3’s balance point, $\sum_n n\,y[n]\big/\sum_n y[n]$. Watch where the dot sits after each pass.

Look at the first panel first. The forward output is the pulse with its corners rounded and a ripple after each edge. Before $n=188$ it is exactly zero: a filter that runs in real time cannot react to an edge before the edge arrives.

The centre moved by 3.72 samples, and that is no accident. The centre of a filter’s output moves by the centre of its impulse response, $\sum_n n\,h[n]\big/\sum_n h[n]$, and for a real $h$ that equals the group delay at 0 Hz, 3.717 samples.

Now the second panel. The output is symmetric about $n=199.5$, the centre of the input. Its ripples dip to −0.047 before the pulse and again after it, and its peak, 1.065, is lower than the forward peak.

That ripple before the pulse is the thing to notice. No real-time filter can produce it, because it is a reaction to an edge that has not arrived yet. The backward pass saw the future, because the whole signal was stored.

Think of combing tangled hair. Comb only from the left, and the hair ends up pushed to the right. Comb from the left and then from the right, and no side is favoured.

So forward–backward filtering suits recordings: a stored ECG, offline audio, a column of data in a data-science notebook. It never suits a live stream, because the backward pass needs samples that have not arrived yet.

FIR or IIR: a checklist

Here is how I decide. Each row names a need, the filter it points to, and the reason.

For the running spec, in one line each: the FIR filter costs 26 multiplies and delays every frequency by 12.5 samples. The IIR filter, order 5, costs 11 multiplies and delays from 3.5 to 23.1 samples across the pass band.

A word on stability. An FIR filter has no feedback, so it cannot become unstable. The IIR filter is stable because all its poles are inside the unit circle, as Stability and causality (16.4) showed. The largest radius is 0.9527. Rounding the coefficients in fixed point moves the poles, and a pole that close to the circle has little room. 21.2 and 21.3 show how to keep it inside.

One more choice sits between the two columns. A minimum-phase FIR filter, as in Minimum phase (17.2), has a shorter delay than a linear-phase one, but it gives up the constant delay.

Worked example

Let’s write out both designs for the running spec and check the numbers of this page.

1. The elliptic design. ellip(5, 0.4455, 40, 1000, fs=8000) gives these coefficients, to six decimals:

• $b_k$: 0.022446, −0.012274, 0.021538, 0.021538, −0.012274, 0.022446
• $a_k$: 1, −3.317488, 5.080257, −4.269604, 1.962663, −0.392408

Its five poles are 0.6659, $0.8058\angle\pm35.05°$ and $0.9527\angle\pm45.64°$. All five lie inside the unit circle, so the filter is stable. Its five zeros lie on the circle, at $\pm57.44°$, $\pm76.40°$ and at $-1$. As frequencies, those zeros are 1276 Hz, 1698 Hz and 4000 Hz, where the gain is exactly zero.

2. Other IIR families on the running spec. SciPy’s order functions give Butterworth 12, Chebyshev I 7 and Chebyshev II 7. Their delays in the pass band differ a lot:

The Chebyshev II filter is interesting: its delay never reaches 12.5 samples in the pass band, so it is quicker than the FIR filter at every frequency there. It costs 15 multiplies in direct form, against the elliptic filter’s 11.

3. Cost at 8 kHz. The FIR filter needs $26\times8000=208\,000$ multiplies a second, or 104 000 with the symmetry. The IIR filter needs $11\times8000=88\,000$.

4. Delay in time. At 8 kHz one sample is 0.125 ms. The FIR filter’s 12.5 samples are 1.5625 ms. The IIR filter’s 3.717 samples at 0 Hz are 0.46 ms, and its 23.14 samples at 1000 Hz are 2.89 ms.

5. Forwards and backwards. The gain is squared. The pass band then lies between 0.9025 and 1, which is −0.89 dB at worst, and the stop band is at −80.00 dB. The worst pass-band loss doubles, from 0.4455 dB to 0.8911 dB, because squaring a gain doubles its dB value.

Where you’ll meet this

Audio plug-ins often offer both kinds. An equaliser may have a “linear phase” mode, which keeps the shape of drum hits at the cost of delay, and a “minimum phase” mode with almost no delay, for live use.

Biomedical toolkits filter stored recordings forwards and backwards. An ECG’s waves keep their timing and shape, so a doctor can still measure them. Real-time control loops, such as a motor controller, use IIR filters for their short delay, because every sample of delay in a feedback loop makes it harder to keep stable.

How do you actually run an IIR filter, and how do you keep it stable with rounded numbers? That is chapter 21: Filter structures (21.1), Second-order sections (21.2) and Finite word-length effects (21.3).

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