- 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.