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Where are magnitude and phase in the output of the FFT?

Posted in: Blog

So you want to know more about your signal.  Putting it through an FFT will tell  you exactly what you want to know.  But the output is not quite what you expected.  What does this strange list of numbers mean? Where have the magnitude and phase graphs gone? They're right here in this list but to properly understand  what the FFT is telling us, we need a little help from my friend  here and right angled triangle.

Are you coming, Pythagoras?  Hi, I’m Mark Newman and this channel is all about  helping you understand the fascinating world   of signals and systems. When I was at university,   everything was about manipulating equations. Lecturers would skip stages in their   derivations and diagrams of what the maths was  actually doing were very few and far between. So in these videos, I prefer to explain things  in a more systematic, visual and intuitive way. You’ve probably heard of my friend  here, the Greek philosopher Pythagoras.   He lived some 2,000 years before Fourier was even  born.

However, ancient though Pythagoras’ ideas   were, his way of looking at right-angled  triangles was very useful to Fourier.  The Fast Fourier Transform  splits your signal into sinusoids   and records the properties of each sinusoid  as a complex number. That’s what this list is. We found out in the last video that the position  of each item in the list tells you its frequency.   But what about the item itself? Each item is a  complex complex number and it is this number that   tells you the amplitudes and phases of the cosine  and sine components making up each sinusoid.

The real part of the complex number  represents the cosine component   and the imaginary part  represents the sine component.  But what have triangles got to do with all this? Think of a triangle rotating within a circle.  As the angle of the triangle increases,   its height increases. Once the angle passes  90°, the height starts to decrease again.   As the angle passes 180°, the height starts  increasing, but in the opposite direction   and once the angle passes 270°, the height starts  to decrease again and everything starts over.

Now if we plot this on a graph with the angle  of the triangle on the x-axis and the height   of the triangle on the y-axis, just look at  which function it describes. It’s a sine wave. Something similar happens with  the width of the triangle too;   only this time, the wave it  describes is a cosine wave. So cosine waves are linked  to the width of the triangle   and sine waves are linked to  the height of the triangle. Now that we know how sines and  cosines are linked to triangles,   let’s randomly choose one of the  items from the FFT output list   to use as an example of how to calculate the  magnitude and phase of the sinusoid it represents. Oh… before we go on, we need to clear up  some terminology. The real and imaginary   parts of this complex number describe the  amplitudes of the cosine and sine waves,   however, we are looking to find the magnitude of  the sinusoid that this complex number represents.

What is the difference between  magnitude and amplitude? Magnitude is a scalar quantity  meaning it has only a numerical value;   it’s always positive. Amplitude on the  other hand can be positive or negative. The magnitude tells us how large the sinusoid is;   whereas amplitude tells us  something about its phase too. Look at these two waves. They’re both sinusoids  and they both have the same magnitudes,   but the amplitude of the first wave is positive  and the amplitude of the second wave is negative.   There is a phase difference  of 180° between the two waves. Let’s make the width of the triangle  equal the real part of the complex number,   representing the amplitude of the cosine wave.  And let’s make the height of the triangle equal   to the imaginary part of the complex number,  representing the amplitude of the sine wave.   Notice the direction of the line. The amplitude  of this particular sine wave is negative. What happens if we add the cosine  wave and the sine wave together? They form a sinusoid with  a new magnitude and phase.   How do we calculate the magnitude  and phase of this new sinusoid? To answer this question, we need to  think of the triangle, not as a triangle,   but as a route in a forest.

These three  trees roughly form a right-angled triangle.   This tree here represents the new sinusoid  and is the tree I’m trying to get to. And this tree here represents my starting point. These two trees form the base of the triangle  representing the amplitude of the cosine wave. And these two trees form  the height of the triangle   representing the amplitude of the sine wave. But how can I get from the starting  point here to my destination over there? Well, there are two possibilities.  I could take the scenic route   and walk first along the cosine  path and then along the sine path Or alternatively, I could take a  shortcut and head straight there.   The angle or heading I set off in,  represents the phase of the sinusoid   and the distance I walked along the hypotenuse  represents the magnitude of the sinusoid.

But how can we calculate the distance  along the hypotenuse of a triangle   if all we know are the lengths  of its base and its height? Pythagoras taught us that if we draw  squares on the three sides of the triangle,   then the area of the square of the hypotenuse is  equal to the sum of the areas of the squares on   the other two sides, and if we know the  area of the triangle on the hypotenuse,   we can work out its length  by taking the square root. So the length of the hypotenuse of this  triangle is the square root of 43.0 squared   plus -28.3 squared which equals 51.5 which  is magnitude of this sinusoid in the list But how do we calculate the phase of the sinusoid?   This is equivalent to the heading  or angle I need to set off in? The angle of a triangle is related to its  width and its height. Increase the height   and the angle increases and vice versa. Increase  the width and the angle decreases and vice versa.   But the relationship between the angle, the  height and the width of a triangle is not linear. The function it describes is an inverse  tangent function.

So to calculate the   angle of the triangle, or the phase of  the sinusoid, we need to divide the height   by the width and take the inverse  tangent function of the result. But you may notice that the inverse tangent  function only covers phases between -90° and +90°.   What happens if the triangle is  in this quadrant of the circle   when both the amplitudes of cosine  and sine components are negative? Or maybe this quadrant when the amplitude  of the cosine component is negative,   but the amplitude of the sine component is  positive? If we want to be able to cover   all quadrants of the circle, we have to  remember the signs of the two amplitudes.   In most modern programming languages  this is taken care of for us by using   the extended inverse tangent function, or as  it is usually referred to: the atan2 function.

Therefore, the phase of the  sinusoid we’ve been looking at   can be calculated by dividing the height  of the triangle, -28.3, by the width, 43.0   and then taking the inverse tangent of  the result giving us an angle of -33°. So if we use Pythagoras and the inverse tangent  function on the real and imaginary parts   of each item in the FFT output list, we can  plot graphs of the magnitudes and the phases   for every frequency in our signal just like  the ones we saw at the beginning of the video. If you would like assistance with any  of the principles I teach in my videos,   or help in applying them to  your specific application,   you can schedule a one-on-one consultation session  with me. See the link at the end of the video. Notice the symmetry in the magnitude graph.   It’s as if someone has put a  mirror at this frequency here.

A similar symmetry exists  in the phase graph as well,   it's just that the phase graph  is flipped as well as reflected. So here's your challenge for this video. Why does this symmetry happen? What does it mean?   Is this something unique to the signal I chose?  or is it due to something the FFT has done? You’ll find the answer in the next video..

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