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Current time:0:00Total duration:6:49

- Several videos ago,
we introduced the idea of a Fourier series. That I could take a periodic function, we started with the example of this square wave, and that I could represent it as the sum of weighted sines and cosines. And then we took a little bit of an interlude of building up some of our mathematical foundations, just establishing a bunch of properties of taking the definite integral over the period of that periodic function of sine and cosines and we established all of these properties. And now we're going to get the benefit from establishing all those because we're going to start actually finding at least a formula for
Fourier coefficients and then we can apply it to our original square wave to see that, hey, this could actually be a pretty straightforward thing. So right over here, I have rewritten a Fourier series expression, or I've rewritten a Fourier series for a periodic function f(t), let's say its period is 2π, and I'm going to use this in some of the properties that we have established to start solving for
these actual coefficients. And what I'm gonna do in this video, I'm gonna first try to solve for a-sub-0 and then in the next video, we're gonna solve for
an arbitrary a-sub-n. And either in that one or the next one, we'll also solve for an arbitrary b-sub-n. So to solve for a-sub-0, what we're going to do is take the definite integral of both sides from 0 to 2π. So 0 to 2π, dt of f(t), well that's gonna be the same thing as going from 0 to 2π of all of this stuff. And remember, this is an infinite series right over here, we have an infinite number of terms. And then we would have a dt out there. But we know from our
integration properties, taking the definite integral of a sum, even an infinite sum, is the same thing as the sum of the definite integrals. So that's going to be the same thing as taking this integral, dt plus this integral. And I could take the scaler out. Actually, let me just not do that, let me just write it like this. 0 to 2π, dt. 0 to 2π, dt. 0 to 2π, dt. This is getting a little monotonous, but it'll be worth it. 0 to 2π, dt. 0 to 2π, dt. 0 to 2π, dt. And we'll do it for every single one of the terms. And now what's nice is, we can look at our integration properties. This right over here, we could take these coefficients out. We could take this a-sub-1, put it in front of the integral sign. The a-sub-2, put it in front of the integral sign. The b-sub-1, put it in front of the integral sign. And then all you're left with is an integral from 0 to 2π of cosine of some integer multiple of t, dt. Well, we established
a couple of videos ago well, that's always
going to be equal to 0. The integral from 0 to 2π of cosine of some non-zero integer multiple of t, dt, that is equal to 0. And then the same thing is true for a sine of mt. So, this is gonna be fun. This is gonna be 0, based on what we just saw. If you just take that factor out of that integral, take that a-sub-1 out of the integral, it's gonna be a-sub-1 times 0. This is gonna be a-sub-2 times 0, that's gonna be 0. That's gonna be 0, that's gonna be 0, that's gonna be 0. Every term's gonna be 0, except for this one involving a-sub-0. And so, what is this going to be equal to? Well, let me write it this way. Let's take the integral of, the definite integral. Let me see where I have some space. So we're gonna take the definite integral from 0 to 2π of a-sub-0 dt, well that's the same thing. Once again, we can take
the coefficient out. And I can just put the dt like this and so that's going to be equal to a-sub-0 times t. Let me do that in magenta. Times t, evaluated at 2π and 0, which is going to be equal to a-sub-0 times 2π minus 0. Times 2π minus 0. Well, that's just 2π a-sub-0. So I could just write this a-sub-0 times 2π. So this expression right here is a-sub-0 times 2π. So let me scroll down a little bit. So I can rewrite this thing up here, the integral from 0 to 2π of f(t) dt which is equal to this integral. Well, we've just figured out that the integral from 0 to 2π of a-sub-0 dt is the same thing as a-sub-0 times 2π. Is equal to a-sub-0 times 2π. Times 2π. And so now, it's actually pretty straightforward to solve for a-sub-0. A-sub-0 is going to be equal to a-sub-0 is going to be equal to 1 over 2π, 1 over 2π times the definite integral from 0 to 2π, I'll just write the dt, of, let me write it a little bit, dt of f(t), oh I'll just write it like this. f(t) dt. And this is pretty cool because think about what this is. This over here, this is the average value of our function, this is the average value of f over the interval 0 to 2π. Average value of f over, over the interval, over we could say the interval from 0, the interval from 0 to 2π. And hopefully that actually makes intuitive sense. Because if I am, if you just think of
it from an engineering point of view, if you were just trying to engineer this, trying to just play around with these numbers, you know all of these cosines and sines, they oscillate between
positive 1 and negative 1. So in order to actually represent this function, you're gonna have to shift that oscillation. Sum of a bunch of oscillation is still gonna be an oscillation that's going to vary between positive 1 and our negative 1. And in order to shift it, well, that's what our a-sub-0 is going to do. And so what you, it makes sense that you would want to shift the oscillation so it oscillates around the average value of the function. Or you could say the average value of the function over, over an interval that's representative of a period of that function. And so that is what a-sub-0 is doing. A-sub-0 is just going to be that average value of the function.