The intuition behind Fourier and Laplace transforms I was never taught in school

this video is sponsored by brilliant here we have a rectangular function and here's its associate Fourier transform just a sinc function which is found through this equation here what makes this integral kind of scary-looking especially the people first learning this is of course the exponential term since it's got I in it but you know what let's just get rid of that completely so we can see the geometric intuition behind this transform first we can expand e to the minus I Omega T by using Euler's formula by basic trig rules the negative in the cosine can just be removed and the negative and the sine function can be moved to the front as shown just a little simplification there then I'm going to replace the ex menschell in the integral with that and lastly we'll distribute the f of t inside so we can split this into two integrals one with the cosine and won't the sine where the eye has been pulled out of the right one so how can we visually understand this equation just take your function and multiply it by a cosine and sine curve both with the same arbitrary angular frequency then find the area under both those curves which is what the integral tells us those areas represent the real and imaginary component respectively of some complex number since I is multiplied by one of those terms the magnitude of that number is the magnitude of the Fourier transform at that specific Omega and the angle is the face as we sweep the angular frequency and keep track of that magnitude and phase however they may change you get the entire Fourier transform that's pretty much it we just find some areas put those on a right triangle and the magnitude and phase come from that and by the way the areas can be negative senses and integrals so we need to think of this more like the unit circle where the phase can be positive or negative or greater than 90 degrees and so on this will tell us if any of those areas are in fact negative which we'll see later but here let's just go to an example I'll put the rectangular function back on top here but what we're going to do is multiply that by cosine of Omega T and also sine of Omega T so we can see why this is the magnitude of the Fourier transform which is a function of Omega by the way making sure there's room on the screen the tougher part for this video but I'll keep the rectangular function plotted on both the left and right but use dots instead while the trig functions will be solid I'll start with Omega equals 3 cosine and sine of 3t look like this but since our rectangular function is 0 everywhere besides these regions then multiplying the equations together leaves us with the same things just chopped off on the sides as shown now the Fourier transform says find the area under these curves and just get the magnitude from there since areas beneath the x-axis are negative then the plot on the right has an area of 0 the region on the left has an area of about point 6 6 and the magnitude of these is of course 0.66 no Pythagorean theorem needed when one side is 0 which means that Omega equals 3 the magnitude of the Fourier transform is 0.66 so all we need to do from here is sweep Omega and keep track of those associated areas and this will get us the entire Fourier transform magnitude by the way notice as I rewind a bit that the graph on the right always has an area of 0 that'll be the case for even function symmetric about the y-axis because the positive and negative areas will always cancel when x sine of anything T so really we are plotting this area as a function of Omega that entire time in this case that's really all the Fourier transform is and let me also show you two points of interest one is here at Omega equals 0 because this y-coordinate actually tells us the area under the curve reason for this is when Omega is 0 this here just becomes 1 so we have the graph of f of T and the function on the right becomes zero regardless of what your f of T is so we ignore that this means we're just playing the area of f of T the original curve you can see that here the original rectangular function and the area beneath it then let's move to Omega equals 2 pi because that has a value of 0 on the magnitude plot if you ever see a magnitude of 0 what that means is that if you multiply your original function by cosine or sine with that angular frequency 2 pi in this case the area under both those curves will zero this is where the left plot finally got enough negative area to cancel out the positive area then as we keep increasing Omega the area on the Left oscillates from net negative to net positive while converging to zero so we've constructed the magnitude plot by just using real areas of real functions didn't have to really think about imaginary numbers to see the intuition here and just to summarize all that with a slightly more complicated equation let's say this is our f of T in order to find the Fourier transform multiply that by cosine and sine of Omega T starting at Omega equals zero which leaves us with just the function itself and y equals zero then find the area under both those curves and use the Pythagorean theorem to get the magnitude and plot that point at Omega equals zero then just sweep Omega and keep track of that magnitude to create the Fourier transform magnitude plot for any Omega like 10 in this case you'll see that the magnitude of those areas is equal to the y-coordinate on the bottom graph at that specific Omega and we see the same thing for negative values of Omega meaning this is the final Fourier transform magnitude and real quick in regards to phase remember that was just the angle in the triangle with side lengths equal to those areas that we see here if that phase is close to 90 like it is right now then the times cosine Omega T plot on the left has much less area than the sine Omega T 1 on the right visually just means that the bottom length is really small compared to the height I'll show the actual phase plot now which has a value of roughly negative 80 degrees at Omega equals 10 why is it negative low well because that imaginary term from before aka the height of the triangle had a negative sign in front of it essentially we're in the fourth quadrant of the unit circle where negative 80 degrees is because of the negative Y value then as I change Omega we can see some of the intuition behind the face like when it's roughly 45 or negative 45 degrees for example then the absolute value of the two areas is roughly the same as this corresponds to a 45-45-90 triangle with two equal side lengths and if the angle goes to zero or close to it then the sine omega-t plot on the right will have zero or nearly zero area essentially our triangle no longer has any height then when we have positive phase values at least between 0 and 180 this corresponds to a negative area for the sine omega-t plot so in the end the magnitude plot tells you how large the areas are when sort of combined but really when just put through the Pythagorean theorem the phase plot on the other hand tells you relatively how big one area is compared to the other as well as if either of them are negative put these together and you get the full picture of the Fourier transform now let me ask this what is the area under the entire curve of just cosine X from negative infinity to infinity well I guess we can say it doesn't exist because the area oscillates as the area moves away from the origin it goes from 0 to 2 to negative 2 and does this forever it also H around 0 though so it might not be right but we're just going to call that infinite area 0 but now what about two cosine functions multiplied together I'm using two curves with very different periods making for a strange-looking plot but even here the area is going to oscillate between two finite values in fact I'll move to the right and we'll see there are some regions with more green area and others with more blue but just keeps going back and forth so again I'm going to call this area 0 as I change the period of one of those cosine curves that area remains 0 in that it oscillates and does not diverge even right now where it looks like the area of may diverge as we move over we still see that periodicity however once the two equations have the same period and only at this time does the graph snap up above the x axis leaving us with an infinite area this is just cosine of PI X all squared which is why we only get positive values now if you were asked to find the Fourier transform of let's say cosine of PI X from before you know you're going to multiply that by cosine of Omega X and sine of Omega X and track the areas of each of those the second one doe will always have zero area as I increase Omega you'll see that the green and blue areas are always equal and thus cancel which happens because the curve is always symmetric about the origin even when the coefficients are the same we still see that symmetry and zero area so I'm going to completely ignore this plot for this other one when Omega equals zero The Associated graph has zero net area which a plot to begin our Fourier transform magnitude but even when we change Omega we'll see the same thing as before a total area that just oscillates which we say is zero but at Omega equals PI the area all the sudden jumps to infinity which I'll represent on the magnitude plot with an arrow and this only happens at Omega equals PI the angular frequency of our original signal everywhere else the Fourier transform aka the area is zero so this would be the final Fourier transform of cosine PI X after accounting for that even symmetry this is the power of Fortier analysis it kind of scans your original signal for sinusoidal functions when it was on a wrong one the output was zero but once it found the right frequency it gave us a brief infinite output saying hey your original signal has a cosine of PI X within it so really all we're doing is looking for the omegas that output an infinite area because those tell us what sinusoids are in our signal if we had a more complicated function and applied the Fourier transform well we can think of this as four separate integrals with each term multiplied by the cosine Omega T when there are no matches then we get out zero area from everything but once our scanner finds a match we get that brief infinite area and this will happen for different times telling us we have four sinusoids in our function now of course it's easy to see what sinusoids make this up I mean they're right there but if I gave you a square wave for example then it's not remotely as obvious however with our scanner we'll figure it out I'll just call the square wave F of T and to find what sign you sit up we just sweep the Omega term and look for infinite areas right now Tomei equals zero we're multiplying by 1 which gives us an area of zero equal parts blue and green so that isn't of interest as I increase Omega this continues to be the case even here if I were to scroll sideways you see the blue and green regions all cancel but at Omega equals pi we get a sudden infinite area which means we have found a sinusoid that makes up our square wave that was hidden from us at first glance but the Fourier transform found it however this isn't the only sinusoid in our function we have to keep going now I'm going to skip around so if we jump to omega equals 2pi then the area will be zero if you look closely the blue and green regions do cancel out but when we go to 3pi there's a continuous pattern of more blue than green so we get negative infinite area meaning there's a cosine 3 pi T also making up our function with that coefficient will be negative for the negative area as I keep going it turns out only the odd integers times pi give us an infinite area meaning that our square wave can be created by summing infinitely many cosine curves together with this pattern which is the basic idea behind the Fourier series now the last few minutes have been all about either having zero area or infinite area which happens for periodic functions but can we relate this to the earlier parts when we were getting nonzero finite areas like is there anything deeper to the fact that we get this continuous change in area for the rectangular function from before well we could say that here at Omega equals 3 for example our scanner found a cosine of 3t that makes up the rectangular function the only problem is that we should be getting an infinite area right now if that were the case because when we multiply cosine 3t by itself we get that function with an infinite area but if we let that amplitude get infinitely close to 0 then we could get a finite area at least when we treat this kind of like a limit so there is in fact a cosine 3t making up our rectangular function but it's infinitely small if we increase I'll make it a three point one we still get a nonzero finite area which means there's also a cosine of three point one T in our function with an infinitely small amplitude this will be the case for pretty much every real number though meaning we have to sum up an infinite number a continuous spectrum of infinitely small sinusoids to make the rectangular function and when it comes to the visual intuition instead of spikes we show an actual continuous spectrum of values representing all the sinusoids that make up our signal and this animation I've shown before highlights how that sum can create a real finite function now I've done an entire video on the Laplace transform but real quick since it's so similar the biggest difference between Fourier and Laplace is the s and the exponent but that really represents alpha plus I Omega which I'll put into the equation and then I'll split up the exponential as shown now look besides just having a zero as one of the limits this is the exact same as the Fourier transform with one extra term so well Fourier acting as a scanner for sinusoids Laplace scans for sinusoids and Exponential's in the same way using areas here let's say this is our function starting at t equals zero Laplace says x cosine and sine Omega T just like before but also including e to the minus alpha T then find the area of both of those to make room up with these over here I'm only going to plot the cosine Omega T one now we need two axes just for our Laplace transform inputs one for Omega aka the sinusoids and another for alpha or the Exponential's if we set both constants to zero these are the areas and write that magnitude above the point zero comma zero now first I'm just gonna change Omega which will move the point up the imaginary axis and watch how the plot changes once we get to Omega equals three and again snaps above the x-axis so the scanner kind of has found the cosine of 3t in the original function but not really because we need an infinite area for that there's no indication of anything special here from our output but now I'm gonna change alpha or the exponential and I'll just remove the sine equation since the area is negligible from here on but again watch how the plot changes once we get to alpha equals negative 0.5 there's no longer any decay and we do get an infinite area that infinite area now means our scanners have found not only the sinusoid but the exponential in the original f of T just ignore the first negative and that exponential F by the way graphically these infinite areas are one of the only things we include on our plots and they're represented with an X so for those need to understand pole-zero plots really just find your poles and look at the Associated y coordinate and x coordinate because those tell you which sinusoids and Exponential's are in your equation pretty much everything you need to know we also plot values that give us zero area but those aren't as important for our purposes here the reason the plot is so useful is because many systems like RLC circuits masses on a spring and just general control systems yield sinusoidal and exponential outputs so we need something more powerful than the Fourier transform in order to analyze these again I have an entire video on laplace including the 3d intuition which I'll link below but if you want to dive more into the applications what we've seen here I highly recommend checking out brilliant differential equation series their first course starts at the basics but the second course is where you'll get into Fourier Laplace and much more that you likely didn't learn in an introductory differential equations course really includes visual explanations interactive exercises and constant practice problems to ensure you understand even the complex topics fundamentally it can successfully put them into practice this is an educational platform that hosts a variety of courses differential equations to vector calculus to relativity and more those looking to learn something new or just relearn old material plus the first 200 people to sign up will get 20% off their annual premium subscription by going to brilliant org slash major CREP or clicking the link below and with that I'm going to end that video there if you guys enjoyed be sure to LIKE and subscribe social media links are down below and I'll see you guys in the next video

Channel: Zach Star

Views: 389,641

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Id: 3gjJDuCAEQQ

Length: 17min 59sec (1079 seconds)

Published: Fri Dec 06 2019