PROFESSOR: Hi everyone.
Welcome back.
So today I'd like to look at a problem
on manipulating Fourier series.
Specifically, we're asked to find the Fourier
series of cosine 2t minus pi/4.
And then the second problem is given a square wave
function, which takes on the value of minus 1 and 1,
and it's 2pi periodic.
We're also told that the square wave function has the Fourier
series 4/pi, 1/n sine n*t.
And the question is to find the Fourier series of three related
functions.
So the first function takes on the value of 0 and 4
on the intervals minus pi to 0 and 0 to pi.
Note that this function is also 2pi periodic.
The second problem it is to find the Fourier series
of a function which is minus 1 from minus 1 to 0,
and 1 from 0 to 1.
So this is the square wave function,
but now has period two.
And then lastly we're asked to find
the Fourier series of a function f of t, which is absolute t.
And it's defined as absolute t on the interval
from minus pi to pi.
However this function f is going to be
extended on the entire domain with period 2pi.
So I'll let you think about this problem
and I'll be back in a moment.
Hi everyone.
Welcome back.
So let's take a look at the first problem,
finding the Fourier series of cosine 2t minus pi/4.
So in this class when we say find the Fourier series, what
we're really looking for are coefficients a_0, a_n and b_n,
such that, for example, we can write our function out
as a sum of sine and cosines with these values a_n and b_n.
Now, what we could do is we could take this function,
plug it into the integral formulas
and compute, using integrals, the value a_0, a_n, and b_n.
However, there's a bit of a trick.
And that's to realize that this function, cosine 2t minus pi/4,
is really just a sinusoidal function
with a single frequency.
And it's currently in an amplitude-phase form.
And we can always convert an amplitude-phase form
of a sinusoidal function into some amplitude times a sine
and some amplitude times a cosine and add the two up.
So specifically, this can be done
by recalling the cosine identity,
cosine A minus B is cosine A times cosine
B plus sine of A sine of B.
And if we identify A with 2t and B with pi/4,
we obtain that f of t is cosine 2t cosine
pi/4 plus sine 2t sine of pi/4.
And now sine pi/4 and cosine pi/4 are both 1 over root 2.
So f of t becomes 1 over root 2 cosine 2t
plus 1 over root 2 sine 2t.
And that's it.
So when we take a look at this expression for f of t,
we see that this is actually in exactly the same form
that we were seeking in the beginning, the Fourier series
representation for f.
So this is the end of the problem.
We don't have to actually compute any integrals.
So now let's take a look at part two.
So for part two, we're asked to find the Fourier series
of several different functions.
And I'll just take a look at part one.
So in this case, f of t takes on the value of 0
from minus pi to zero, and the value of 4 from 0 to pi.
And if I were to quickly sketch this function,
it would look like a square wave, 0, pi, minus pi.
This is f of t.
And we see that it's actually related-- I can draw in dashes
just to be clear-- what the original square wave
function might look like.
So this is square wave of t.
We see that if we were to take the square wave function
and shift it up, the square wave takes on values of plus 1
and minus 1.
But if we take the square wave function
and shift it up 1 unit, and then rescale it,
we'll then get the function we're looking for, f of t.
So specifically, we can write f of t in terms
of the square wave as well.
We have to shift the square wave function up 1,
and then we have to amplify it by a factor of 2.
So then when we seek the Fourier series for f, all we have to do
is substitute the Fourier series for the square wave function
in.
So let's do this.
So we have 1 plus 4/pi.
So I'm writing here this Fourier series for the square wave
function, n odd, 1/n sine n*t.
So by manipulating the Fourier series for the square wave
function, we can arrive at a Fourier series for f of t,
n odd, 1/n sine n*t.
So this is the first function.
The second function f of t-- I'll just rewrite
it again-- f of t takes on the value of negative 1
from negative 1 to 0, and takes on the value of plus 1
from 0 to 1.
So it's somewhat useful to draw, just to quickly sketch,
this function.
And I'll draw in my axes here.
So this is f of t.
This is t.
And it transitions.
It makes its jumps at the integers.
So for example, it jumps from plus 1 to minus 1 at 1,
and from plus 1 to minus 1 at minus 1 as well.
And if I were to just quickly sketch the square wave function
underneath of it, the square wave function
makes its transitions at minus pi and pi.
So this is the square wave function.
This is t.
So we see that we can take the square wave function
and squeeze it to obtain this function f of t.
And the question now is, what factor
do we have to squeeze the square wave function by to get f of t?
Well, usually what I like to do is
take one characteristic point in the sketch
for in this case the square wave function.
So a nice characteristic point is
when the square wave function has a 0,
so when it goes through the origin,
and compare it to the same point on this function f of t.
So we see that the point pi has to get
stretched to the point 1.
So we basically need to take the time axis in the square wave
function and squeeze it by a factor of pi.
So what this means is that f of t
is going to be the square wave.
And when we do what mathematicians sometimes
called a dilatation, a stretch or a squeeze.
If we're squeezing by some factor,
that factor always appears in the argument of the function
we're squeezing.
If it's a stretch, then it would be t over the stretch factor.
And we can quickly check here that if t equals 1,
so as t goes from 0 to 1, so when t is 1,
the square wave function is being evaluated at pi.
And I'll now write out the Fourier series.
So this is going to be 4/pi, sum of n odd, 1/n sine of n*pi*t.
And there's another quick check we can do.
And that's this function f we were told
is periodic with period 2.
And if we just take a look at this function here,
the Fourier series representation of it,
we see that each sine n*pi*t is in fact periodic with period 2.
So that's just another quick, back of the envelope check
that we've done our calculation correctly.
So this concludes part two.
And then lastly than for part three, f of t
is the absolute value of t from minus pi to pi.
And it's extended beyond minus pi to pi
to make it 2pi periodic.
And at first it might look like this function f of t
is not related to the square wave function.
But if we look at it a little more carefully,
and analyze the derivative of f of t,
we see that this function takes on the value plus 1
on the interval of 0 to pi and takes on the value of minus 1
on the interval of minus pi to 0.
And this is exactly the square wave.
So what we can do is we can write
f of t-- it's going to be an integral of the square wave
function.
So specifically, we have the integral of square wave,
which is going to be the integral of 4/pi 1/n sine n*t,
summation n odd.
And I'm going to leave this as an indefinite integral.
So when we integrate, we have a constant of integration
c plus 4/pi sum of 1 over now n squared.
And we have negative cosine n*t.
And this is n odd again.
And now we have one last question to answer.
What is the constant of integration?
Well, notice how the constant of integration
can be identified with the a_0 term of a Fourier series.
So if we want the constant of integration,
we can just calculate it directly
by the integral formula for the a_0 of this function.
So what this means is c is going to be 1/(2pi)--
essentially it's just the average of the function--
but it's pi to pi, the absolute value of t, which is 1/pi,
pi squared over 2.
So we get pi/2.
So just to quickly recap, in this problem
we were given several functions which were related
to the square wave function.
And we started off knowing what the Fourier
series for the square wave function was.
And by manipulating or rewriting these functions
in terms of the square wave function,
we were able to compute their Fourier series.
And we were able to do this essentially
without having to evaluate any integrals.
OK.
So I hope you enjoyed this problem,
and I'll see you next time.