Bandpass sampling

Take a signal that lives only between 17.5 MHz and 22.5 MHz, such as one radio channel. Watch the copies of its spectrum as the sample rate slides down from 50 MHz to 10 MHz, and note each rate where they stop overlapping.

Twice the top edge is not the rule

The sampling theorem (10.2) asked for $f_s>2f_\text{max}$. The channel’s top edge is 22.5 MHz, so that rule asks for more than 45 MHz. But the rule was a shortcut for something else: the copies of the spectrum must not overlap. This signal has nothing from 0 to 17.5 MHz, so copies can sit in that gap.

I call the band’s lower and upper edges $f_L$ and $f_H$, and its bandwidth $\mathrm{BW}=f_H-f_L$. Here that is $\mathrm{BW}=5$ MHz, and $f_H/\mathrm{BW}=4.5$.

A real signal has a second half. Its spectrum is mirrored at negative frequency, from $-f_H$ to $-f_L$ (a cosine is a pair of arrows, one at $+f$ and one at $-f$, as in Complex exponentials (3.4), and From series to transform (8.1) says the same of every real signal). Sampling copies both halves to every multiple of $f_s$. The mirror’s copy of a frequency $f$ lands at $kf_s-f$, which is the fold from Sampling and aliasing (10.1). So there are two sets of copies, and a rate is allowed when none of them overlap.

I made the band lopsided: its spectrum rises from $f_L$ to $f_H$, so the mirror falls, and you can tell which way round a copy is.

Allowed rates come in islands

Go back to the picture at the top of the page and drag the sample rate yourself, between 10 and 50 MHz. The solid copies are the band and the hatched ones its mirror, and the chart under them keeps a mark for each rate you visit. Notice that 30 MHz and 16 MHz work, while 40 MHz and 20 MHz do not.

Below 45 MHz the copies collide, but at 35 MHz they slip apart again. Allowed rates come in islands: 45 MHz and up, 22.5 to 35, 15 to 17.5, and 11.25 to 11.67 MHz. Between them the copies overlap.

Where the islands come from

Mark the multiples of the Nyquist frequency $f_N=f_s/2$: $0,\,f_N,\,2f_N,\,\dots$. Between two neighbours lies a stretch, and I number the stretches $m=1,2,3,\dots$, so stretch $m$ runs from $(m-1)f_N$ to $mf_N$. The rule from 10.1 moves a whole stretch into 0 to $f_N$ as one piece, either shifted or turned over. A band that fits inside one stretch is moved as one piece, so its copy and its mirror’s copy do not touch. A band that straddles a multiple of $f_N$ is folded onto itself.

So the condition is $(m-1)f_N\le f_L$ and $f_H\le mf_N$. With $f_N=f_s/2$ it becomes the rule for an allowed zone:

For $m=1$ there is no upper limit, and the rule is 10.2’s $f_s\ge2f_H$. An island exists only if $2f_H/m\le2f_L/(m-1)$. That simplifies to $m\,\mathrm{BW}\le f_H$, so $m$ runs up to the largest whole number not above $f_H/\mathrm{BW}$. Here that is $m=1$ to $4$.

Check it against the clip. 30 MHz is in zone 2 and 16 MHz is in zone 3, so both work. 40 MHz lies between 35 and 45, and 20 MHz between 17.5 and 22.5, so neither is in any zone.

The band can arrive flipped

Inside stretch $m$, 10.1’s rule gives a simple landing place. If $m$ is odd, a frequency $f$ lands at $f-(m-1)f_N$, shifted down and still in order. If $m$ is even, it lands at $mf_N-f$, turned over, so the order is reversed. That is why the table alternates.

Take $f_s=16$ MHz, so $f_N=8$ MHz and the band is in zone 3. It lands at $17.5-16=1.5$ MHz up to $22.5-16=6.5$ MHz, with the same rise. Take $f_s=30$ MHz, so $f_N=15$ MHz and the band is in zone 2. Now 22.5 MHz lands at $30-22.5=7.5$ MHz and 17.5 MHz lands at $30-17.5=12.5$ MHz.

If the band were symmetric, a flip would not show. A lopsided spectrum is different: you must know whether the zone you chose is odd or even before you read the spectrum back.

Stay away from the edges

An island’s edge is allowed, but only just. Take $f_s=15$ MHz, the lower edge of zone 3. The mirror’s copy, from $3\times15-22.5=22.5$ MHz to $3\times15-17.5=27.5$ MHz, touches the band at 22.5 MHz. Now let the clock run 1 % slow, to 14.85 MHz. The mirror’s copy sits at $3\times14.85-22.5=22.05$ MHz to $27.05$ MHz, which is 0.45 MHz inside the band.

The copy moves by three times the clock’s error because it is the third copy. A copy of higher number moves more, so a high zone needs a better clock.

So pick the middle of an island. Zone 3 runs from 15 to 17.5 MHz, whose middle is 16.25 MHz, and from there both edges are 1.25 MHz away. The band’s own edges also need care. The gap from 0 to 17.5 MHz must be truly empty, so a filter has to remove everything outside the band before the sampler, as in Anti-aliasing and practical converters (10.4).

How low can the rate go? Never below $2\,\mathrm{BW}$, which is 10 MHz here: the lowest edge of any zone is $2f_H/m$ with $m\le f_H/\mathrm{BW}$, and that is at least $2\,\mathrm{BW}$. The rate $2\,\mathrm{BW}$ itself is reached only when $f_H/\mathrm{BW}$ is a whole number. Then the island has no width at all, so there is no margin. Our band has $f_H/\mathrm{BW}=4.5$, so its lowest rate is 11.25 MHz, not 10 MHz.

Worked example

1. Zones for 17.5 to 22.5 MHz, from $2f_H/m\le f_s\le2f_L/(m-1)$. $m=1$: $f_s\ge45$. $m=2$: 22.5 to 35 (flipped). $m=3$: 15 to 17.5 (upright). $m=4$: 11.25 to 11.667, only 0.417 MHz wide (flipped).
2. Where the band appears. At $f_s=16$ MHz: 1.5 to 6.5 MHz, upright. At $f_s=30$ MHz: 7.5 to 12.5 MHz, flipped, with $17.5\to12.5$ and $22.5\to7.5$.
3. Guard. At 14.85 MHz the zone-3 mirror copy spans 22.05 to 27.05 MHz, overlapping the band by 0.45 MHz. At the island’s middle, 16.25 MHz, the margins are 1.25 MHz.
4. Whole-number ratio. Take a band from 20 to 25 MHz, so $f_H/\mathrm{BW}=5$. Zone $m=5$ gives exactly $f_s=10$ MHz $=2\,\mathrm{BW}$, from $2\times25/5=10$ to $2\times20/4=10$, an island of zero width.
5. A slow replica. Sample 60 Hz mains at 19.9 Hz. Three times the sample rate is $3\times19.9=59.7$ Hz, so the tone lands at $60-59.7=0.3$ Hz: the same waveform, played $60/0.3=200$ times slower.

Where you’ll meet this

A software radio can sample a high band directly, at a rate set by the bandwidth instead of by the top edge, and then do the rest in numbers. The islands tell the designer which rates are legal, and the margin rule tells which one to pick. The same idea runs through Amplitude modulation (27.1), where a signal rides on a carrier, and through The Hilbert transform and the analytic signal (27.2), which handles the band without its mirror half.

Reference card