The Fourier Transform Part V – Convolution

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Convolution

Last time, we developed a notation, a language through which we could express Fourier’s theory of modelling signals like sound signals. That language was called “Complex Numbers”. Now we’re going to use that language to begin our journey into exactly how the Fourier Transform does what it does.

Lets first remind ourselves what the Fourier transform actually does:

“The Fourier Transform decomposes a function of time into the frequencies that make it up” -Wikipedia

As we saw said in Part II of this blog, Fourier said that any signal is simply a collection of sine waves of different frequencies, amplitudes and phases added together. He needed some way to seperate out all those different sine waves from his signal. The method by which he managed to do this is called “Convolution”.

Convolution is where we take a wave with a known freqency and amplitude (I’m going to call it our convolving wave) and slide it over our signal. You can see this happening in the animation below. Our signal is represented by the blue line and our convolving wave is represented by the red line.

You can see how the convolving wave slides over our signal. I’ve purposefully chosen a very simple signal to begin with. In fact, the signal is a sine wave with frequency and amplitude identical to our convolving wave. In order to make the convolving wave slide along the x-axis, we change its phase. As our convolving wave slides, it sometims manages to completely cover up our signal. This happens when both waves are “in phase” with each other meaning that both waves have the same phase value. We know the phase of our convolving wave, as we are controlling its position by changing its phase. If we could somehow identify mathematically when our convolving wave completely covers our signal, we would be able to find out the phase of our signal.

A good way to start is to see what would happen if we were to multiply the two waves together as our convolving wave slides over our signal. The animation below shows just that.

In the lower graph we have our two waves from before. The upper graph shows what happens if we multiply each point on these two waves together. You can see the multiplied waveform moving up and down. It reaches its highest point when our signal and convolving wave are both in phase with each other and its lowest point when our signal and convolving wave are completely out of phase with each other. So we’re well on our way to a mathematical solution to our problem. However, rather than have to look at a whole wave, it would be nice if we could whittle this all down to one number which would tell us when the two waves are in phase. A good way to do this would be to add all the points on the upper graph together.

This is exactly what I have done in the top graph in the animation below.

When the two waves on the bottom graph are in phase, all the multiplied points in the middle graph have values that are greater than 0, i.e. they are all in the upper half of the graph. This means that if we add them together, the result they will give will be the highest it can be. You can see this on the top graph. The height of the line as it moves along the x-axis is at its greatest when both waves on the bottom graph are in phase. It is at its lowest when both waves are completely out of phase.

This is all very well for the simple signal  such as a single sine wave. What about something a little more complicated. Well, lets try a new signal. One that is made up of a number of sine waves added together. I’m not going to show you the formula for this signal just yet. We’re going to work it out using what we have learned about the convolution up till now. So here is my new signal:

What frequencies are present in our signal, how much do they contribute to our signal (relative amplitude) and what are their phases?

Lets try our first convolving wave with a frequency of 1Hz.

Something strange has happened. When we multiply our signal and convolving wave together, we get a wave (shown in the middle graph) that, no matter what the phase of our convolving wave, always has the same number of values below zero as above zero. If we add together all the values in the middle graph, the result is going to be zero. You can see this in the top graph. No matter what the phase, the point is always at zero. This means that the frequency of 1Hz does not exist in our signal at all.

I’m going to save a little on internet bandwidth and tell you that not only does a frequency of 1Hz not exist in our signal, neither does 2, 3 or 4Hz. The first frequency that actually exists in our wave is 5Hz as you can see in the animation below.

You can see on the bottom graph how the distance between the high peaks of our signal is covered quite nicely when the phase of our convolving wave is a certain value. This is a good indication that the frequency we are trying is present in the signal. This is confirmed by the top graph as we can see the point moving up and down indicating some positive result from adding all the points in the middle graph together. The point in the top graph is at its peak when at a phase of 0 and 180 degrees and so on. This would seem to indicate that the phase of this first frequency component is zero. The greatest amplitude our point reaches is 72 so we now know all the details of our first sine wave:

Frequency = 5Hz

Relative Amplitude = 72

Phase = 0 degrees

Again, I’ll save on your internet bandwidth and tell you that the frequencies of 6, 7, 8 and 9Hz are also not present in our signal and colvolving waves at those frequencies would only achieve the same result as when we convolved 1Hz with our signal. The next frequency present is 10Hz.

The point in the top graph is at its peak when at a phase of 30 and 210 degrees and the greatest amplitude it reaches is 54.  Therefore the important things we need to know about our second sine wave are:

Frequency = 10Hz

Relative Amplitude = 54

Phase = 30 degrees

 If we continue increasing the frequency of our convolving wave and try out all the other frequencies using the same method as above, we will find that there is only one more frequency in our signal: 15Hz. Here it is:

Now we see that the point on the top graph is at its greatest when the phase is 60 and 240 degrees meaning that our signal’s third component frequency has a phase of 60 degrees. The point reaches a peak of 72 which gives us our amplitude. So, for the third and final frequency in our wave, we can say the following:

Frequency = 15Hz

Relative Amplitude = 72

Phase = 60 degrees

So our signal is made up of 3 seperate waves. The details of the three waves are as follows:

	Frequency (Hz)	Relative Amplitude	Phase (degrees)
First wave	5	72	0
Second wave	10	54	30
Third wave	15	72	60

So our signal can be described by the formula:

So we’ve managed to reverse engineer the equation for a signal by convolving lots of different sine waves with known frequencies, phases and amplitudes. However there is something you may notice about our reverse engineered signal if you were to plot it on a graph. The signal we end up with, although exactly the same shape as our original signal, is much larger in amplitude. In audio terms, if this was a piece of music we were listening to, the music from the recovered signal would be much louder than our original signal.

This is why, when I asked: “How much does each wave contribute to the overall signal?” I called the result the wave’s relative amplitude. In audio terms, so long as the relative amplitude of each frequency component is known, then that is enough as it doens’t matter to us whether the signal is too loud (we can always turn it down if needed), what is important is that the relative contribution of each frequency to the overall signal remains the same, which indeed it does.

How Fourier made use of convolution

Fourier was a wily old mathematician. Although he knew he could get his answer using convolution, he noticed that there was a slightly quicker way to do it.

What I am about to show you is not the “Fast Fourier Transform”, we’ll take a look at that in a later blog post. What I want to demostrate here is how we don’t actually need to do the whole convolution operation, i.e. slide our convolving wave all the way over our signal, to get our result. All we need to do is look at the result when our convolving wave is at 0 degrees phase and -90 degrees phase, or to put it another way, to multiply our signal by one known sine wave and one known cosine wave.

We know from before that there are only 3 frequencies in our wave, so, to save time, I’m not going to try any others.  What I’m going to do is multiply our signal by 6 waves. 3 sine waves and 3 cosine waves at 3 frequencies each, 5Hz, 10Hz, and 15Hz. The left hand graph shows what happens when I multiply my signal by a cosine wave, and the right hand column shows what happens when I multiply my signal by a sine wave.

5 Hz Component

Starting with the 5Hz component, looking at the left hand graph, we see that the contribution of the cosine wave is zero. This is logical as we know from before that the 5Hz frequency is a pure sine wave. We can see this in the right hand graph as the amplitude of the sine wave is equal to the total amplitude of this frequency’s contribution to the overall wave. So the fomula for our 5Hz component is the same as we found before:

10Hz Component

Now looking at the 10Hz component. We can see that there is a non-zero result from multiplying both a sine wave and a cosine wave with the signal. This means that the 10Hz component also has a shift in phase. Remember from part III of this blog that we can represent a phase shifted wave by adding together a sine and cosine wave of differing amplitudes but with the same frequency. Multiplying our signal with a cosine wave gives us an overall amplitude of -27. Multiplying our signal with a sine wave gives us an overall amplitude of 47. This means that the formula for our 10Hz component is:

which, if you work it out for all values of θ, is exactly the same as the equation we had before for the 10Hz component…

15Hz Component

Now looking at the 15Hz component. We can see that there is a non-zero result from multiplying both a sine wave and a cosine wave with the signal which again means that the 15Hz component also has a shift in phase. Multuplying our signal with a cosine wave gives us an overall amplitude of -63. Multiplying our signal with a sine wave gives us an overall amplitude of 36. This means that the formula for our 15Hz component is:

which, if you work it out for all values of θ, is exactly the same as the equation we had before for the 15Hz component…

So to summarize:

The Fourier Transform identifies the component frequencies of a signal via convolution using the following series of operations:

1. Multiply every point of our signal by every point of a cosine wave of known frequency and amplitude.
2. Add together all of the results from our multiplication operation. This gives us the cosine contribution for this frequency.
3. Multiply every point of our signal by every point of a sine wave of known frequency and amplitude.
4. Add together all of the results from our multiplication operation. This gives us the sine contribution for this frequency.
5. Add the Sine and Cosine components together.
6. Go back to (1) only now using a slightly higher frequency.
7. Repeat steps 1-6 until we have covered every possible frequency in our signal.

This is how the Fourier transfrom works. By multiplying our signal by lots of cosine and sine waves at known amplitudes and frequencies and adding the results of the multiplication together. This way we can find which frequencies are present in our signal and how much they contribute to the overall signal.

In part II, we learned about how sound is made up of lots of sine waves.

In part III, we learned about how a cosine wave is simply a phase shifted sine wave and how by adding them together, we can represent any phase shifted wave.

In part IV, we learned about complex numbers, a notation to help us represent sine and cosine waves in an ink saving way.

In part V, we learned about convolution and how it helps us identify the different frequency components and what their contributions to our signal are.

Now we have all the tools necessary to try and understand the formula we saw at the beginning of part II – the Fourier transform formula itself! This is what we will be doing in the next post. I will also be admitting to a little cheat that I have used to obtain the above results. Sorry about that! All will become clear next time.

I would like to thank Professor Robert Cottis, Emeritus Professor in Corrosion Science and Engineering at Manchester University who read this post and sent me some very useful suggestions on how I might improve it. I have tried incorporate those suggestions in this updated post.

 

Next Time: The Fourier Transform Formula >>

Tags: Complex NumbersConvolutionCosine WavesFourierFourier SeriesFourier TransformFrench mathematicianJoseph FourierPhaseSine Waves