. So, in the previous lecture we finished
the the ideas behind time frequency representation.
So, in this lecture we will talk about Fourier
series various notions of convergence, properties
and so on. So, we will go slightly into some
details concerning the Fourier series ok.
Now, just to recall 
most of you would have done this over the
interval minus pi to pi you can expand f of
x as a 0 plus summation k equals one to infinity
ak cos kx plus b k sin k x ok.
So, this is a general representation for m
series a naught is your dc term,which is one
upon 2 pi integral minus pi to pi f of x dx
and these aks are basically, the Fourier coefficients
a n is given by 1 upon pi integral minus pi
to plus pi f of x cos n x d x and b n is basically
the inner product of f of x with sin nx right
and a is bis are the 
Fourier coefficients .
Now, let us delve a little carefully 
consider the function f of x equals x on minus
pi to pi . So, this function is clearly an
order odd function right, it is straightforward
it is an odd function therefore, you do not
have these ais and you can compute these bis.
So, let us say the b k coefficient is given
by this formula . 
And if you did carefully integration by parts
and compute at b k you can say that F of x
is summation k equals 1 to infinity 2 times
minus 1 power k plus 1, upon k sin k x this
is what you get as an expansion right. Which
most of you would have done in your undergraduate
in electrical sciences, which of our electronics
communication electric in whichever stream
electrical sciences you would have done this.
Now, for this example F of x is not 2 pi periodical
. So, let us form a function f tilde, which
is a periodic extension of f ok. So, if you
just sketch this function .
You, get these get the functions for 0 or
then extended by 2 pi either side, if you
can just write this as right . Now, observe
that there are these jump points like plus
minus k pi are points of discontinuities or
jump jump form of discontinuity. So, F of
x converges to f tilde x at points, where
f tilde is continuous .
However, we have 
discontinuity points at plus minus k pi, where
k is some integer right, and and at those
points at these will have to prove this result
anyway we will go into the details and depth
through the subsequent lectures. At these
points F of x will converge to f of k pi plus
plus f of k pi minus; that means, you take
the left hand limit right hand limit and you
take the mean value .
Of course, when we observe this figure the
discontinuity points are at odd multiples
of k here. So, therefore, I would just ah
say that this is going to be at plus minus
2 k plus 1 times pi and k belongs to set of
integers positive integers and you have to
include 0 as well . And we would have basically
studied this in our undergraduate curriculum
ok.
So, now let us look at the finite sum case
.
So, N instead of being infinite capital N
being infinite is now finite ; that means,
I take the first capital N terms in the Fourier
series expansion and then I compute .
Now, there is a weird thing that one would
notice 
as a function of x if you plot the function
right, ideally it has to be like this . Now,
at this point there is a sharp jump right.
So, if you plot this and graph this function
S N of x is capital N of x for different values
of N you will see that there will be some
blips, at this point and then the signal just
jumps through this point I mean from at this
point pi there is some some positive value
it has and from some positive value it has
suddenly jumped to a negative value the thing
is really sharp right.
That means, the graph of this function must
travel from if if you just look at what this
value is right I mean f of x equals x. So,
therefore, this point is pi this point is
minus pi on the ordinates right. So, from
pi to minus pi it has to just jump in a very
small interval and this you can sort of imagine
like a waterfall or it is called the waterfall
effect.
So, imagine you have a river it is going straight
and then as you approach a waterfall I do
not think anybody would want to just go through
the process of going through the boat on the
on the river with leading to a waterfall,
because you can see the rapids right as the
rapid increase in exactly the same fluctuation
phenomena, which is happening here there is
rapids and then there is a huge fall right.
So, we would like to investigate to so, this
is because of a truncation truncation and
this point of discontinuity. So, at this point
of discontinuity you are trying to represent
this summation and this is not accurate limit
is not defined at this point there is a sharp
jump right. So, therefore, at this point of
discontinuity the accuracy of these blips,
it really gets worse right I mean as; that
means, if you sketch 20 100 1000 so on on
for N, you can see that there is a variation
in this height and and then the oscillations
and this phenomenon is called blips effect
.
So, I call this as blips 
accuracy of the blips around the discontinuity
points 
gets 'worse' . So, what it means the graph
of S N of x for small finite N capital 'N'
say N equals I just take 10 terms right, must
travel from pi to minus pi in a very short
interval .
So, this is something like the you know like
imagine you have a very small orifice and
you just have to just go through that point
of discontinuity right. So, the blips occurring
just before and after 
the points of 
discontinuity is called Gibbs effect .
This is a very very important phenomena I
do not know I mean you have to really exercises
you have to write a program compute this.
And then observe, what is happening right
and an artist's way of visualization of the
Gibbs phenomenon is imagining, that you are
going through a river rapid and then there
is and if you are finding the oscillations
are becoming larger you better beware that
there is a steep fall expected right it is
it is like that.
Now, a few things to note the height of the
blip is approximately the same for large N
and the width gets smaller as N gets larger
. So, this is a through observations actually
I will give you a homework when you will actually
prove that these results are indeed correct.
So, what I would like to do is like you to
do is plot S N of x for N equals 10 say 100
1000 dot dot and sketch this observe 
Gibbs effect, then I would also want you to
investigate 
for a saw tooth wave, which is given by S
of t equals t 0 less than or equal to t less
than or equal to pi upon 2 and pi minus t
from pi upon 2 less than or equal to t less
than or equal to pi right. I give you this
function observe what do you observe and what
do you what can you conclude from this. So,
this is just a homework exercise for you .
So, it is just not computing the Fourier series
is what we have done? I mean we have to understand
many subtle details in this series expansion
right can we approximate all functions are
there any conditions, what does the function
take at the point of continuity? What does
it take at the point of discontinuity right?
So, on and so forth and these subtle issues
are what we will be dealing with through the
next set of lectures ok.
So, we will first start with a theorem suppose
f 
is a piecewise continuous function on the
interval a less than or equal to x less than
or equal to b, then limit k going to infinity
integral a to b f of x cos kx dx is limit
k going to infinity integral a to b f of x
sin kx dx and this is equal to 0, f is a piecewise
continuous function.
So, what does this mean? So, a piecewise continuous
function has only finite number of discontinuities
right, at those piecewise points. And therefore,
they are only a finite number of Fourier coefficients
larger than a certain absolute value given
a certain positive number right, I give you
certain positive number only a finite number
of Fourier coefficients, which are bigger
than this number and basically this implies
you can do data compression for these type
of functions or signals I use them interchangeably
right.
So, let us see the proof of this result . So,
you can imagine that you have a high frequency
wave this is almost like, my sin wave right
and I have some function that I am sketching
this is y equals f of x and this is like my
y equals sin kx ok.
Now, as the frequency increases right f of
x is nearly constant between 2 adjacent periods
of sin kx and and and cos kx right. So, what
is happening is if I take 
imagine this really very high frequency and
I have a function here some function I increase
this frequency a lot right you can say that
this between these 2 peaks it is a nearly
constant right I mean this is really not a
very high frequency, but I imagine terahertz
if tera is not sufficient for you just bring
in some huge number for frequency; that means,
it is almost you really cannot see it, it
is as if it is piecewise flat I mean this
is this is even high frequency probably right,
if you just put a function around that you
probably see it is it is constant right.
Now, this is the idea. Now, we take that high
frequency and then we you know you take the
inner product of that with this function and
let the frequency go to infinity you have
to show that this result is 0. I mean you
can it can imagine why this is true, because
over a cycle the value is constant and the
swing basically gives you a positive value
and a negative value it should hopefully sort
of cancel out, but they will actually prove
this result, well as the frequency 
k increases f of x is nearly constant on 2
adjacent periods of sin kx and cos kx .
The integral over each such small period is
0 over piecewise intervals . Now, in the case
of a differentiable 
function f let us assume that this function
is differentiable. Now integral a to b f of
x cos kx dx I am considering this integral
this inner product here.
So, this is sin kx I am integrating this by
parts first function integral of the second
function minus integral of the second function
derivative of the first function since, I
need to take the derivative of the first function
I am introducing that it is a differentiable
function ok. So, this also gives you a hint
sort of how mathematicians work is also playing
around with with these quantities right, I
mean integrating the second function from
a to b minus integral from a to b of this
this form .
Now, let us look at each of these terms and
call it 1 I call it 2 . So, now, turn 1 is
basically sin k be f of b minus sin k a f
of a divided by k . Now, f of x is a and nearly
constant and therefore, the sign is bounded
function. So, therefore, this this basically
heads to 0 heads to 0. And in the second part
right so, you have to use the fact that when
k goes large f of b is f of a right, and and
then apply this this rule here it is straight
forward. And then in this part the derivative
of this function is going to be 0, because
it is a constant function. So, this vanishes.
So, 2 is also heads to 0, because f of x is
constant when when this is really large, because
of this observation ok .
So, with this we are ready to discuss the
notion of convergence at a point of continuity
ok.
So, this is a just going back, we this concludes
I mean you can take the sin you can take the
cos it is the same thing it does not really
matter. So, therefore, this concludes the
proof. So, let us stop here if you have any
questions we will we will take it and then
we will get on to the next.
