Amplitude modulation and the sampling theorem

Allen DowneyDecember 18, 20156 comments

I am working on the 11th and probably final chapter of Think DSP, which follows material my colleague Siddhartan Govindasamy developed for a class at Olin College.  He introduces amplitude modulation as a clever way to sneak up on the Nyquist–Shannon sampling theorem.

Most of the code for the chapter is done: you can check it out in this IPython notebook.  I haven't written the text yet, but I'll outline it here, and paste in the key figures.

Convolution with impulses

I start with an example that demonstrates convolution with a series of impulses: it adds up shifted, scaled copies of the original wave, as shown here:

You can hear what it sounds like in the IPython notebook.

Amplitude modulation

Then I show an example of amplitude modulation in four steps (from top to bottom):

1) The spectrum of the original wave.

2) After multiplying by a cosine carrier wave at 10 kHz, the spectrum is shifted by plus and minus 10 kHz.

3) After demodulating by multiplying by the same carrier wave, each of the copies splits and shifts again, adding up in the middle.

4) By filtering out the high-frequency peaks at plus and minus 20 kHz, we can recover the original signal.

To understand how that works, we can look at the spectrum of the carrier wave: it has impulses at plus and minus 10 kHz.  When we multiply in the time domain, we are convolving in the frequency domain.  And as we saw in the first example, convolving with impulses makes shifted, scaled copies.  In the first example it makes copies of the wave; in this example it makes copies of the spectrum.


Next we see what happens when we sample a signal.  I start with a recording sampled at 44.1 kHz, keep every 4th sample, and set the other samples to 0.  This simulates the effect of sampling the original signal at about 11 kHz.

Here's what happens in the frequency domain: the top spectrum is the 44 kHz wave; the bottom spectrum is the 11 kHz wave after downsampling:

To understand what happened, we have to realize that sampling is effectively the same as multiplication by a series of impulses.  The following figure shows the process in four steps:

1) The top row shows the spectrum of the 44 kHz wave.

2) The second row shows the spectrum of the impulse train used for sampling.  Downsampling by a factor of 4 corresponds to 4 impulses in the frequency domain, which...

3) ...creates 4 copies of the spectrum in the frequency domain.  The leftmost frequencies wrap around to the right, which is why 4 copies makes 5 peaks.

4) If we apply a low pass filter, we can clobber all but the middle peak.

But the result is not very good because

1) We've lost all of the components of the original wave above 5500 Hz, which is the folding frequency of the downsampled wave.

2) Even the components below 5500 Hz are not right, because they still include contributions from the overlapping copies we just clobbered.

But if we do the same thing again with a bandwidth-limited signal, things are much better:

The original signal contains almost no energy about 5000 Hz, so when we sample it, the copies of the spectrum don't overlap, and when we filter away the copies, we don't lose anything.  The original and the sampled wave are identical (well, almost identical -- the particular example I chose has a tiny bit of energy above 5500 Hz).

To understand how we were able to recover the original signal, even after we threw away 75% of the data, we can look more closely at the filter we used to clobber the unwanted copies of the spectrum.  I used a "brick wall" low-pass filter, which has the shape of a boxcar.  And as I showed in a previous chapter, the Fourier transform of a boxcar function is a sinc.  In this example, the window that corresponds to the filter looks like this:

Multiplying by the boxcar filter in the frequency domain corresponds to convolution with a sinc function in the time domain.  And we can think of convolution as adding up a bunch of shifted, scaled copies.  Here's what that looks like for a short segment of the wave:

The green lines show the shifted, scaled copies of sinc function; the blue line is their sum.  The gray line (which you can only see at the beginning and end of the segment) shows the original signal.In this example, the interpolation doesn't work very well at the beginning and the end because this segment of the wave is not periodic -- there is a discontinuity between the right and left ends which the interpolation is trying to smooth out.

If we zoom in on a shorter segment, we can see more clearly how the interpolation works:

Again, the green lines are the sincs and the blue lines are the total.  The vertical gray lines are the samples.

Each sinc function has a peak at the location of one sample and the value 0 at every other sample location.  That way, when we add them up, the sum passes through each of the samples.  In between the samples, the sincs add up to the values of the original signal.

The sampling theorem

In summary, sampling a wave makes copies of its spectrum.  The distance between the copies is the sampling frequency.  If the wave contains components that exceed half of the sampling frequency, the copied spectrums overlap and distort each other.  And when we filter away the unwanted copies, we lose information.

But if the original wave contains no components that exceed half of the sampling frequency, the copies don't overlap, the filter loses no information, and we can recover the original wave exactly.

And that is the Nyquist-Shannon sampling theorem.  Now you should go watch this amazing video that demonstrates the sampling theorem using an analog-to-digital-to-analog converted and two oscilloscopes.

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Comment by Rick LyonsDecember 24, 2015
Ah, OK. Thanks. It seems to me that what you've done is the same as sampling an analog signal at 11.25 kHz and then inserted three zero-valued samples between each of those original samples. I believe that process is typically called "upsampling by four" because the result is a sequence whose sample rate is four times 11.25 kHz = 44.1 kHz. And that upsampling operation results in three (typically unwanted) spectral images in your final signal's spectrum, as you have shown. I also touched on this topic in my blog at: http://www.dsprelated.com/showarticle/761.php
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Comment by AllenDowneyDecember 24, 2015
That's excellent. Thanks!
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Comment by Rick LyonsDecember 19, 2015
Hi. I'm trying to understand something. In the 1st paragraph of your 'Sampling' section you wrote:

"I start with a recording sampled at 44.1 kHz, keep every 4th sample, and set the
other samples to 0."

I'm trying to understand what it means to "keep every 4th sample, and set the other samples to 0." If you sampled an analog signal at 44.1 kHz then the time period between your discrete samples is 1/44,100 seconds. What is the time period between your discrete samples after you "keep every 4th sample, and set the other samples to 0"? Thanks.
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Comment by AllenDowneyDecember 21, 2015
Thanks for the question; I can see why this is not clear. The sampling rate is the same after I downsample, so "downsample" is not really the right name for this operation. But I don't know if there is a right name.

What I am trying to do is simulate the effect of sampling the original signal at a lower sampling rate. By keeping every 4th sample (without lowering the sampling rate) I make the copies/images of the spectrum visible.

I might consider a better way to explain this.
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Comment by Rick LyonsDecember 23, 2015
Hi Allen.
Let's say your original Fs = 44.1 kHz sample rate data sample values are [1,2,3,4,5,6,7,8,9,10,11,12,13]. My question is, after your "keep every 4th sample, and set the other samples to 0" operation do you produce the samples:




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Comment by AllenDowneyDecember 23, 2015
The second one. The first would not show the copies in the frequency domain.

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