When is aliasing a good thing?

Question

In Hamming's book, The Art of Doing Science and Engineering, he relates the following story:

A group at Naval Postgraduate School was modulating a very high frequency signal down to where they could afford to sample, according to the sampling theorem as they understood it. But I realized if they cleverly sampled the high frequency then the sampling act itself would modulate (alias) it down. After some days of argument, they removed the rack of frequency lowering equipment, and the rest of the equipment ran better!

Are there any other ways to use aliasing as a primary technique for processing a signal, as opposed to a side-effect to be avoided?

Answer 1

The quoted text in the question is a case of using bandpass sampling or undersampling.

Here, to avoid aliasing distortion, the signal of interest must be bandpass. That means that the signal's power spectrum is only non-zero between $f_L < |f| < f_H$.

If we sample the signal at a rate $f_s$, then the condition that the subsequent repeated spectra do not overlap means we can avoid aliasing. The repeated spectra happen at every integer multiple of $f_s$.

Mathematically, we can write this condition for avoiding aliasing distortion as

$$\frac{2 f_H}{n} \le f_s \le \frac{2 f_L}{n - 1}$$

where $n$ is an integer that satisfies

$$1 \le n \le \frac{f_H}{f_H - f_L}$$

There are a number of valid frequency ranges you can do this with, as illustrated by the diagram below (taken from the wikipedia link above).

In the above diagram, if the problem lies in the grey areas, then we can avoid aliasing distortion with bandpass sampling --- even though the sampled signal is aliased, we have not distorted the shape of the signal's spectrum.

Answer 2

One instance that jumps to mind is digital demodulation. The optimal detector for a linear modulation scheme is matched filtering and decimating to the middle sample of each symbol.

The matched filtering may not do a very good job of reducing the bandwidth, but we still want to make decisions at the symbol rate.

The aliasing of energy in this case is part of reconstructing the modulated symbols.

The key point is that the energy must alias coherently in the correct phase i.e. timing is crucial.

Answer 3

Super-resolution is another area where aliasing is necessary, or to put it better, the optical system should not be the weakest link in the chain (and optical components that effectively anti-alias such as anti-moire optical filters should not be part of the chain)

Answer 4

One other time when aliasing isn't a problem is when designing lowpass filters used for decimation. You can allow some amount of aliasing after the decimation operation to relax the constraints on the filter's performance, resulting in a lower-order design. Instead of placing the stopband edge at the post-decimation Nyquist frequency, you can slide it out just far enough that it doesn't alias back into the filter's passband (and therefore corrupt your signal of interest).

Stated more mathematically, assume that your original signal is sampled at rate $f_s$ and you are decimating by a factor $D$. The filter's passband edge $f_p$ is defined by the bandwidth of the signal of interest. The Nyquist frequency after decimation will be $\frac{f_s}{2 D}$, so obviously the passband edge frequency must be less than this.

Since I've asserted that you can allow the stopband to bleed past the decimated Nyquist frequency, recall how aliasing in the second Nyquist zone works: any content at frequency $\frac{f_s}{2 D} + \Delta f$ before the decimation operation will appear to be located at $\frac{f_s}{2 D} - \Delta f$ afterward. Thus, we can place the stopband edge at some frequency $f_{stop} = \frac{f_s}{2 D} + \Delta f$ and select $\Delta f$ in such a way that the filter's transition band doesn't overlap with the passband after we decimate. In order for this to be true:

$$ f_{stop_{aliased}} = \frac{f_s}{2 D} - \Delta f \ge f_p $$

$$ \Delta f \le \frac{f_s}{2 D} - f_p $$

The takeaway from this is if there is still a decent amount of oversampling present in the post-decimated signal (there are some reasons why you would do this), then you can push the stopband out by a nontrivial amount. As a quantitative measure, you can look at the transition ratios of the "naive" and "relaxed" filter specifications:

$$ T_{naive} = \frac{f_p}{f_{stop_{naive}}} = \frac{f_p}{\frac{f_s}{2 D}} = \frac{2 D f_p}{f_s} $$

$$ T_{relaxed} = \frac{f_p}{f_{stop_{relaxed}}} = \frac{f_p}{\frac{f_s}{2 D} + (\frac{f_s}{2 D} - f_p)} = \frac{f_p}{\frac{f_s}{D} - f_p} $$

$$ \frac{T_{relaxed}}{T_{naive}} = \frac{\frac{f_p}{\frac{f_s}{D} - f_p}}{\frac{2 D f_p}{f_s}} $$

$$ \frac{T_{relaxed}}{T_{naive}} = \frac{1}{2 - \frac{f_p}{\frac{f_s}{2 D}}} $$

This last expression gives you a compact representation of the improvement in transition ratio that can be obtained by relaxing the filter specification in this way, parameterized by the ratio of the filter's passband (i.e. the signal of interest's bandwidth) to the post-decimation Nyquist frequency. Plotting this ratio as a function of the passband frequency (normalized by the post-decimation sample rate), you get:

So in summary, if your signal is still decently oversampled after the decimation operation, then you can reduce the filter's transition ratio by a factor of up to 2 by relaxing its specification in this way. As a rule of thumb, the number of required taps for an FIR filter is roughly proportional to the transition ratio. It does allow some aliasing when performing the decimation, but the specifications are designed such that the aliasing does not overlap with the desired signal. This allows it to be removed later if needed, by a filter that operates at the decimated sample rate $\frac{f_s}{D}$.

Answer 5

Aliasing can indeed be a good thing under certain conditions.

Look at it this way: lets say your sampling rate is 100 Hz. Lets also say you have a signal somewhere out there, that is sitting from, say, 990 to 1010 Hz. (So its total bandwidth is 20 Hz, and it is centered at 1000 Hz).

Ok great, now what?

Lets suppose you sampled this signal at your 100 Hz rate. All that happens, is that your signal (sitting from 990-1010, centered at 1000Hz) is copied and shifted at integer multiples of 100 right?

So now all the sudden you have a copy of your original 990-1010 signal, except now you have one centered at 900, 800, 700, 600, etc etc, and also 1100, 1200, 1300, etc etc. The BW is the same of course. So your copy of your signal centered at 900 occupies 890-910 Hz. The copy sitting at 800 Hz occupies 790-810 Hz, and so on and so forth. You will also have a copy at 'baseband', (meaning it is centered at 0Hz, and so occupues -10 to 10Hz).

So when is this useful? Well, look at what you just did - you just managed to take your signal sitting at 1000Hz, put it down at baseband, and all this with a sampler running at just 100Hz! And guess what! You did this all legally according to Nyquist!

This is because Nyquist doesn't say you have to sample at at least twice of the maximum freuqnency - wrong wrong wrong wrong wrong! (But very common misconception.) He says you have to sample at least twice the maximum bandwidth of your signal, which in this case, is 20Hz.

Applications? Well, a lot of basestations for cell phones actually use this 'undersampling' technique. So your cell phone signal is sitting at some high up Ghz range, and the basestation is sampling in the hundreds of Mhz range.

And by the way, seeing as how Nyquist actually works, I dont like the term 'undersampling' - because that implies that we are, well, under-sampling. But we arent! We are completely following Nyquist, and always sampling at least twice the maximum bandwidth of the signal in question.