Anti-aliasing and practical converters

Here is a plan I might make. I will sample at 8 kHz, so the Nyquist frequency is 4 kHz, and I put a filter in front of the sampler with its corner at 4 kHz. Is a 7 kHz tone safe? Make your guess, then watch the tone’s bar pass the filter and land after sampling.

A corner is not a wall

Sampling and aliasing (10.1) showed that a tone at frequency $f$ above the Nyquist frequency $f_N=f_s/2$ lands at $f_s-f$, and that nothing afterwards can separate it from a real tone there. The sampling theorem (10.2) gave the rule: keep everything above $f_N$ out. This lesson is about the part that does the keeping out. It is a low-pass filter placed in front of the sampler, called an anti-aliasing filter.

The filter in the picture at the top is one RC stage, the circuit from Frequency response and Bode plots (8.4), whose gain at frequency $f$ is

with corner frequency $f_c=4$ kHz. At the corner this gives $1/\sqrt2$, which is $-3.01$ dB, the figure from 8.4.

The filter did take 3 dB off at its corner, but the tone is 3 kHz past the corner and the gain there is still $0.496$. One RC stage falls slowly. A corner marks where the filter has started to act, and it has not stopped anything yet. A cheap sound card built this way records a 7 kHz whistle as a 1 kHz tone.

Where the filter must be strong

The mistake above was looking at $f_N$. The right question is which frequencies fold into the part of the spectrum I want to keep. Suppose I want to keep everything up to $f_\text{pass}$, the top of the passband. A tone at $f$ lands at $f_s-f$, so it lands inside the passband when $f_s-f\le f_\text{pass}$, which is when $f\ge f_s-f_\text{pass}$.

So the filter must be small everywhere from

upward. I call $f_\text{stop}$ the stop edge. It is not $f_N$. Between $f_\text{pass}$ and $f_\text{stop}$ the filter is allowed to fall, and that range of frequencies is the transition band. Everything about practical anti-aliasing is a trade for the width of that band.

How steep must the fall be? One RC stage falls 20 dB for every factor of 10 in frequency above its corner (8.4, section 3). Put $N$ such stages in a row and their slopes add, so the fall is $20N$ dB per decade. The number $N$ is the order of the filter. Starting from a corner at $f_\text{pass}$, the straight-line fall to $f_\text{stop}$ is $20N\log_{10}(f_\text{stop}/f_\text{pass})$ dB. To be $A$ dB down there,

I round $N$ up to a whole number, because stages come whole. Designs in Chapter 20 keep the passband flat, for example the Butterworth filter in Analog prototype filters (20.1), and they need the same order for this spec.

Now keep 0 to 20 kHz, with a 60 dB fall. At the CD rate of 44.1 kHz, the stop edge is $44.1-20=24.1$ kHz. The ratio $24.1/20$ spans only $\log_{10}(1.205)=0.0810$ of a decade, so

and the order is 38.

Before you play the next instrument, guess: how much does the order fall if I double the sample rate? Watch the stop edge move away from 20 kHz, and the order needed fall with it.

The caption at the end says: “At four times, 176.4 kHz: order 4. Sample faster and a cheap analog filter will do; the digital side finishes the job.” When the clip has finished, drag along the frequency axis to try your own rate. At 400 kHz the stop edge is 380.0 kHz and the order is 3.

Here is why. On a log axis, a stop edge at 24.1 kHz is a sliver away from 20 kHz. At 88.2 kHz it is at 68.2 kHz, over half a decade away, and the same 60 dB can be spread over a much longer slope. The orders for the three rates are 38, 6 and 4. Order 38 is a very large analog circuit to build and keep stable. Order 4 is a small one.

The caption said “the digital side finishes the job”. After sampling fast, a digital low-pass filter removes what lies between 20 kHz and the new $f_N$, and then the sample rate is lowered. That is the subject of Downsampling and decimation (22.1). Oversampling and noise shaping (11.3) shows a second gain from sampling fast. This is why audio converters sample at many times 48 kHz inside the chip and only hand you 48 kHz at the end.

A late sample on a steep slope

The filter is half of the story. The other half is the measurement itself. A converter cannot read a voltage that is moving, so a sample-and-hold circuit freezes the input at each tick, and the converter measures the frozen value. The tick has to arrive at the right instant, and a real clock is slightly wrong each time. That timing error is clock jitter, and I write its size as $t_j$.

A sample taken $t_j$ late is wrong by about the slope times $t_j$. A sine $x_c(t)=A\sin(2\pi ft)$ has slope $2\pi fA\cos(2\pi ft)$, which is the derivative rule from Fourier series and LTI systems (7.4), $\tfrac{d}{dt}e^{j\omega t}=j\omega e^{j\omega t}$. The steepest slope is $2\pi fA$. A fast signal is steep, so it suffers most.

Before you play the clip: a 20 kHz sine of size 1 is sampled 1 ns late, at its steepest point. Guess how big the height error is. As the view zooms in, watch the sine turn into a straight line through the sample instant.

The bar is 0.000126 tall, and here is why. The slope is $2\pi\times20\,000=1.2566\times10^{5}$ per second, and a slip of $10^{-9}$ s gives

The SNR limit needs the RMS values from How big is a signal (1.3). A sine of size $A$ has RMS $A/\sqrt2$. Its slope is a cosine of size $2\pi fA$, so the slope’s RMS is $2\pi f$ times the signal’s. A timing error with RMS $t_j$ then gives a height error with RMS $2\pi f\,t_j$ times the signal’s RMS. The signal-to-noise ratio is the signal’s RMS over the error’s:

For 20 kHz and 1 ns this is $-20\log_{10}(1.257\times10^{-4})=78.0$ dB. Adding bits does not move it, because the error comes from the clock and not from the rounding.

Radio receivers sample hundreds of MHz, so they need clocks good to picoseconds. At 100 MHz with $t_j=1$ ps, the limit is only 64.0 dB.

Three ways to convert

An anti-aliasing filter and a good clock serve a converter. The converter itself comes in a few designs, and each one trades something. Here are three.

The sigma-delta row is the oversampling idea of this lesson taken as far as it will go. Oversampling and noise shaping (11.3) explains it.

Worked example

The trap, in numbers. One RC stage with corner 4 kHz has gain $1/\sqrt{1+(f/4)^2}$, with $f$ in kHz. At 5, 6 and 7 kHz it is $0.625$, $0.555$ and $0.496$, which is $-4.09$, $-5.12$ and $-6.09$ dB. At the corner it is $-3.01$ dB.

Orders for 60 dB, $f_\text{pass}=20$ kHz.

The same rule at 48, 96 and 192 kHz gives 21, 6 and 4. The question from Sampling and aliasing (10.1), protecting the band against a tone at 25 kHz, has ratio $25/20$ and gives $60/(20\times0.0969)=30.96$, so 31.

A telephone line. Sample at $f_s=8$ kHz and keep up to 3.4 kHz, so the stop edge is $8-3.4=4.6$ kHz. For 40 dB, $N\ge40/(20\log_{10}(4.6/3.4))=15.23$, so 16.

Jitter.

Where you’ll meet this

Every audio interface has an anti-aliasing filter before its converter, and most of them sample far above 48 kHz inside. Phone lines sample at 8 kHz, so they need the steep filter in the worked example. Software-defined radios and oscilloscopes spend much of their cost on a low-jitter clock. Chapter 22 returns to the digital filter that follows a fast sampler.

Reference card