so welcome to lecture series on mathematical methods and its applications now now we will
discuss fourier sine and cosine transforms we have already seen what fourier transforms
are we have we have seen that fourier transform is given by fourier transform of function
f f t is given by minus into plus infinity f t e k power minus eta omega t dt which
is f omega and its inverse its inverse which is f omega inverse of f omega is f t which
is given by one upon two pi integral minus to plus infinity it is a f omega e k
power eta omega t t omega so that is how we can find out inverse fourier transform
and fourier transform of ft
now sine and cosine transforms before discussing sine and cosine transform let us discuss shifting
property of fourier transform also so suppose fourier transform fts of omega and omega naught
is belongs to r then fourier transform of this is the very simple property e k power
eta omega naught t into f t let us find fourier transform of this so fourier transform of
this is nothing but minus into plus infinity you apply a definition ok this definition
e k power eta omega naught t into f t into e k power minus eta omega t dt so this is
given by minus to plus infinity f t e k power minus eta omega minus omega naught into t
into dt so in this expression instead of omega we
have omega minus omega naught so this is but f of omega minus omega naught so this is called
a shifting property i mean i mean frequency shifting property of a fourier transforms
so this we can use some time while solving some problems ok now next is fourier cosine
transform the fourier cosine transform function f t is defined as zero to infinity f t cos
omega tdt now which is given as fourier cosine transform of f t now it comes from basically
from fourier cosine integral representation of function f t what is the fourier cosine
integral representation of function f t it is f t is equals to one upon pi integral zero
to infinity it is a omega cos omega d t omega where a omega is given by this expression
ok now if you compare one and three so a omega
is nothing but two times f c omega when you compare these two ok so when you substitute
it here so we obtained f t equal to this which is the which is called inverse fourier cosine
transform of f t so what is a fourier cosine transform and what is this inverse so basically
fourier cosine transform of f t it is given by it is zero to infinity f t e k power minus
eta omega t ok which is a fourier cosine transform of ok because ok it is cosine so it contains
cosine terms it is f t into cos omega t dt which we are calling as fourier cosine transform
of it is ag we are calling as fc omega ok we are calling it as fc omega ok now the inverse
of this is given by inverse of a inverse of a fourier cosine transform is given by it
is two upon pi integral zero to infinity f t it is fc omega here
it is fc omega cos omega t dt or d omega so this will be the inverse fourier transform
of inverse fourier cosine transform of f t ok now similarly if we define fourier
sine transform so fourier sine transform is given by zero to infinity f t sin omega t
dt which we denote as fx of omega and it comes on fourier sine internal representation of
f t because we know that fourier sine integral representation of a function f t is given
by this expression where b omega is this now see if compare this with this expression so
b omega is nothing but two times fx omega ok so if you substitute this this expression
b omega two time fx omega over here sorry over here so it will be two upon pi times
zero to infinity fx omega sine omega t dt so what is a fourier sine integral representation
of function fourier sine integral representation of function f t it will be nothing but if
it is zero to infinity f t sin omega t d t which we are given as fourier sine integral
representation zero to pi by this expression and the inverse of a fourier inverse of a
fourier sine transform of f t will be nothing about two upon pi times integral zero to infinity
fx omega sin omega t into d omega which is nothing but f t ok
so now let us find out fourier sine and cosine transform of this function ok so suppose you
find if fx is what fx is given as k when zero less than x less than a and it is zero
when x is greater than a so suppose is i want to find out fourier cosine transform this
function so this will be equal to zero to infinity f t ft is ok f t cos omega t d t
which is equals to zero to a it is k into cos omega t d t which is equals to k now when
we integrate this is sin omega t upon omega from zero to a nothing but k upon omega sine
omega here so this will be this is basically fourier cosine transform of this function
f t now similarly if you want to find out fourier
sine transform this function so fourier sine transform this function f t will be nothing
but again zero to infinity f t sin omega t into dt which is equals to zero to a f t ft
is k sin omega t into dt which is equal to k can come out it is minus cos omega t
upon minus upon omega from zero to a it is minus k upon omega times it is cos omega a
to minus one so it is k upon omega times one minus cos a so this will be the fourier sine
transform of this function f t so in this way we can find out fourier cosine
or fourier sine transform of any function f t using these expressions ok now fourier
transforms also satisfy some properties what are they now first of all the existence
condition if f x is absolutely integrable on the positive side of x axis and piecewise
continuous in every finite interval then the fourier cosine and sine transform of f exists
so this is the condition that function must be absolutely integrable on the positive side
of x axis and it must be piecewise continuous then fourier sine and cosine transform of
function will exist now the property is the first property is fourier cosine and sine
transform also satisfy linearity property so this property all satisfy that we can easily
is the c also that is easy to prove basically what is what is fourier cosine
transform of suppose a f plus bg it is nothing but by a definition it is zero to infinity
a f plus b g into cos omega t d t so that will be nothing but a times zero to infinity
f into cos omega t d t this has to be the function of t these are function of t plus
b times zero to infinity g cos omega t into dt so this is nothing but a times fc of fc
of f t and this is plus b times fc of gt so hence fourier cosine transform satisfy
linearity properties similarly we can we can show that a fourier sine transform also satisfy
linearity property ok the first thing the second thing is cosine and sine transform
derivatives now if function is continuous and absolutely
integrable on the positive side of x axis and f dash is piecewise continuous on each
finite interval ok and suppose fx tends to zero fx tends to infinity then these two results
hold now let us write these results so what is fourier cosine transform of f dash x that
will be nothing but a definition or power f t ft or fx you can say anything that will
be a zero to infinity f dash t into cos omega t into dt this is by the definition of fourier
cosine transform you simply replace f by f dash in this expression ok so that will be
given by is equal to now you apply integration by parts you take it first function you take
a second function first as it is now integration of second it is f t from zero to infinity
minus integration derivative of first is omega sin omega t with negative sign the positive
negative positive and it is f t dt now when you take upper limit now this we
are assumed that as s tending to infinity or t tending to infinity fx tend to zero so
as t tend to infinity this will tend to zero so zero into some finite quantity will be
zero and that will be it is nothing but zero minus f zero because when t is zero
it is f zero plus omega times integral zero to infinity it is a sin omega t into f t dt
ok so this is nothing but omega will come here this omega and this is nothing but fourier
sine transform of f t ok this is fourier sine transform of f t so this is fourier sine transform
of f t minus f zero so this is the derivation for the first part that the fourier derivative
of fourier cosine transform of f will be nothing but omega times fourier sine transform of
f t minus f zero ok now second part fourier sine transform of
f dash x so fourier sine transform of f dash f for t is nothing but zero to infinity f
dash t into sin omega t dt this is by a definition of fourier sine transform you simply
replace f by f dash t in this expression now you again apply integration by parts it is
one it is two now it is first as it is integration of second from zero to infinity minus integral
derivative of first is omega cos omega t into f f t dt so when when t tend to zero when
t tend to infinity f t tend to zero this is this is by the assumption ok so this is tend
to zero now when t is zero sin zero is zero so we have zero here from both the limits
minus omega will come out it is zero to infinity cos omega t into f t dt
so this is minus omega times this is nothing but fourier cosine transform of f t it is
fourier cosine transform of f t so this is how we can obtain the fourier derivative of
fourier sine transform which is nothing but minus omega time fourier cosine transform
of f t now suppose you want to find out fourier cosine transform of second derivatives so
that also we can find out in the same result now what is fourier transform i have just
fourier cosine transform f dash x fourier cosine transform f dash x is nothing or f
dash t is nothing but the this is we have derived it is omega times fourier cosine transform
of f dash fx minus f zero ok this we have derived
next is fourier sine transform of f dash fourier sine transform of f dash t is nothing
but minus omega fourier sine transform of fourier cosine transform of f t so these expressions
we have just derived suppose suppose now you want to find out fourier cosine transform
of double derivative of f t so this is we want to find out ok it is fourier cosine that
should must be signed ok now suppose i want to find out fourier cosine transform of second
derivative of f t so this will be nothing but now you simply replace f by f dash
in this expression so while you replace s by f dash in this expression
so fourier sine transform of f double t will be nothing but omega times fourier sine transform
of f double dash f dash t you are replacing f by f dash t so it is f dash t minus f dash
zero which is equals to omega times now fourier sine transform of f dash t is we already know
it is minus omega fourier minus omega fourier cosine transform of f t minus f dash zero
so when you simplify it is nothing but minus omega square fourier cosine transform of a
f t minus f dash zero so this will be the fourier cosine transform of second derivative
of f now similarly if we want to find out fourier
sine transform of second derivative of f so that also you can find out now in this expression
you simply replace f by f dash when you replace f by f dash in this expression it is minus
omega fourier cosine transform of f dash t now this equals to minus omega now fourier
cosine transform of f dash t from this expression is nothing you can simply substitute it here
it is omega fourier sine transform of f t minus f zero so that will be nothing but minus
omega square fourier sine transform of f t minus omega f zero ok so ok its minus omega
square into minus minus plus ok so it will be plus
so in this way we can obtain the fourier transform of second derivative of f or the fourier sine
transform second derivative of f so similarly if we want to obtain the fourier cosine or
sine transform of third derivatives or the higher derivatives so that also we can obtain
using the same concept ok so this is how we can find out fourier sine or cosine
transform of function if a function is known to us and how we can find out fourier
sine or cosine derivatives that also we have seen or the higher derivatives that also we
have seen so next in the next lecture we will see some more properties of fourier transform
and and how how the problems can be solved on that that we will see in next class ok
so thank you very much