so welcome to lecture series on mathematical methods and its applications so we have seen
what fourier transforms are and also we have seen fourier sin and cosine transforms so
how can write fourier transforms function f t fourier transform of a function f t
is given by minus infinity to plus infinity f t e k power minus eta omega t into
d omega into d t ok and we are calling it f omega function of omega
so and the inverse fourier transform of f omega is f f omega is given by it is one
upon two pi integral minus infinity to plus infinity that is f omega e k power eta omega
t and it is d omega so which is nothing but f t so this we have already discussed and
this come from complex form of fourier integrals this we have already seen that fourier
transforms f t is nothing but given by this expression and the inverse fourier transforms
is given by this expression now let us come to convolution theorem
for fourier transforms so before discussing convolution theorem let us discuss a more
popular of fourier transforms so first of all fourier transform of derivates ok so let
f t be continuous on the x axis and f t tend into zero as mod t tend into infinity now
further suppose f dash t is absolutely integrable on the x axis then fourier transform of f
dash t is given by eta omega fourier transforms of f t
so this thing is very easy to proof fourier transform of f dash t so by the definition
of fourier transforms it is nothing but minus infinity to plus infinity f dash t into
e k power minus eta omega t into d t so this will be nothing but let us suppose it
is first function it is second function so first as it is integral of second function
from minus infinity to plus infinity and minus integral derivate of first is minus eta omega
into e k power minus eta omega and integral of second which is f t into d t
so this is by integration by parts now as we have assumed that as f t as mod t tend
to infinity f t is tend to zero that is when t is tend to plus or minus infinity f t is
tend to zero so from the upper and the lower limit this is tend in to zero so this is nothing
but zero map plus eta omega times minus infinity to plus infinity e k power minus eta omega
t into f t d t now what it is it is nothing but fourier transform of f t so it is eta
omega times fourier transform of f t so so hence the fourier transform f dash
t is nothing but eta omega times fourier transform of f dash t now suppose you want to find out
fourier transform of f double dash t phase into the phase f pi f dash in this expression
so that will be nothing but eta omega fourier transform of f dash t which is equal to fourier
transform of f dash t is the already derived is equal to this expression so this is nothing
but eta omega whole square fourier transform of f t so in the similar way if we proceed
for the nth derivative for the same expression so we get fourier transform of nth derivative
of t will be nothing but eta omega k to the power n fourier transform of f t
so nth derivative of fourier transform of f will be nothing but eta omega k to the power
n fourier transform of f t so this is how we can obtain fourier transform of derivatives
now let us solve these two problems find fourier transform of f t where f t equal to this ok
find fourier transform of f t actually so we have to find fourier transform of t e k
power minus t square ok now this is nothing but fourier transform of e k power minus
t square is derivative is derivative is nothing but minus two t times this thing you multiply
and divide by this expression so this will be nothing but when you simplify this so you
get back this expression now by the fourier transform derivative we
know that fourier transfer of f dash t is nothing but eta omega times fourier transform
of f t so this is nothing but minus half of eta omega times fourier transform of e k power
minus t square because here here f t is e k power minus t square is derivative and
is derivative is given by eta omega fourier transform of f t ok so this is this expression
now let us find fourier transform of e t square so fourier transform of e t square fourier
transform of e k power minus t square will be given by the minus and plus infinity f
t into e k power minus eta omega t d t which is equal to minus and plus infinity e k power
minus so when you make perfect square it is t plus eta omega by two whole square a square
plus two a b and minus eta omega by two whole square into d t
so which is equal to it is minus minus plus it is e k power minus omega square by four
can come out because it is free from t and it is integral minus or plus infinity e k
power minus t plus eta omega by two whole square into d t now you can take t plus
eta omega by two as as suppose under root p ok i or you can take it as suppose
you can take it as z first ok so then d t will be nothing but d z so it will be e k
power minus omega square by four minus and plus infinity e k power minus z square into
d z limit will remain the same ok now its an even function so this can be written
as two times e k power minus omega square by four zero to infinity e k power minus z
square d z ok now we can let z equal to suppose under root p so this will give this integral
will give this will give fourier two e k power minus omega square by four integral
zero to infinity one by two under root p e k power minus p into d p ok which is nothing
but two two cancel out and it is e k power minus omega square by four and it is gamma
half by that definition gamma function this is nothing but under root pi e k power minus
omega square by four so therefore the value of the value of this
is nothing but minus one by two eta omega under root pi e k power minus omega square
for ok so here we use the here we use the concept of derivative of fourier transform
and to find out fourier transform of e k power minus t square we go back to the main
definition of fourier transforms they are the simplify and we get the fourier transform
of e k power minus t square ok now suppose you want to solve this problem ok the next
problem it is y dash minus two y will be equal to e k power minus two t u naught t ok take
fourier transform both the sides what is the fourier transform y dash by the formula it
is eta omega f omega here f omega is a fourier transform of y t minus two f omega and is
equal to you we know we already that fourier transform of e k power minus a t u naught
t u naught t is a unit step function at t equal to zero we already know that the fourier
transform of this is nothing but one upon two plus eta omega because fourier transform
of e k power minus a t u naught t is one upon a plus eta omega ok
so what we obtain from here we obtain that f omega is nothing but one upon eta omega
minus two into two plus eta omega when they simplify this so this is nothing but a square
minus b square and which is equals to minus of one upon omega square plus four now we
have also seen that fourier transform e k power minus a mod t is nothing but two
a upon a square plus omega square that we have already derived ok so it is something
like one upon omega square plus two square so you can multiply and divide by four so
minus four into four upon omega square plus four
so instead of a we have two so this is nothing but so so if we take inverse fourier
transform both the sides so y t will be nothing but minus one by four and inverse of this
will be given by that it is e k power minus two into mod t so that will be the solution
of this expression this i mean differential equation ok so we take fourier transform both
the sides simplify and the inverse of fourier transform will give the solution of this differential
equation ok now differentiation respect to frequency
omega let us suppose functions is piece wise continuous on the x axis and let t k to the
power n f t be absolutely integrable on the x axis then then we have a result for fourier
transforms then then fourier transform of t k to the power n f t that is a multiplication
of t is nothing but eta k power n and nth derivative respect to omega of f omega ok
nth derivative of omega with f omega ok now now to proof this the proof is simple
you take d by d omega d omega of f omega now f omega is a fourier transform of f t what
is nothing but d by d omega of minus to plus infinity f t e k power minus eta omega
t into d t because fourier transform f t is given by this expression and f omega is nothing
but fourier transform of f t so this will be given by by the lagrange theorem it is
nothing but del by del omega of f t into e k power minus eta omega t into d t which
is nothing but when you take derivative respect to omega it is nothing but minus eta
t into f t e k power minus eta omega t into d t now minus eta will come out and the
and this it is nothing but t f t keep our minus eta omega t that will be nothing but
fourier transform of t f t so here so therefore f dash omega will be
nothing but this expression now when you multiply with eta both the sides so this is nothing
but eta f omega will be equals to fourier transform of t f t so hence fourier transform
of this is nothing but this expression for n equal to one so similarly if you take
second derivate suppose you take second derivate d two upon d w t of f f w so that will
be nothing but by the same by the same concept it is nothing but d by d w of integral
minus into plus infinity first we take first derivate by this expression del by
del w of f t e k power minus eta omega t into d t so that will be given by d by d w of minus
infinity to plus infinity eta t f t e k power minus eta omega t
again when you take this d by d w inside and you differentiate with respect to omega so
that will be nothing but that will be nothing but that will nothing but minus infinity
to plus infinity it is minus eta square t square f t e k power minus eta omega t into
d t which is eta square times integral minus into plus infinity t square f t e k
power minus eta omega t d t and this is nothing but minus fourier transform of t square f
t ok this is nothing but fourier transform of this ok
now ok this implies fourier transform of t square f t will be nothing but eta square
times double derivative of f here we have a single derivative ok here we have a single
derivative and here we have a double derivative so so that means similarly if you repeat
this process n times so we will get back this expression that is a fourier transform
of t k power n f t will be nothing but eta k power n f f omega nth derivative
of f omega ok so in this way we will get by this expression
now using this suppose you want to find out fourier transform of t e k power minus
two mod t so that will be nothing but using this expression it is nothing but eta k
power one because n is one into fourier transform of into derivative into d by d omega of f
omega here f omega is the fourier transform of this f t it is e k power minus two mod
t and what is the fourier transform of this it is two a upon four plus omega square
so this will be nothing but eta d by d omega of four upon four plus omega square which
is nothing but four eta into minus one upon four plus omega square the whole square into
two omega so that will be nothing but minus eight omega
eta upon four plus omega square the whole square so this will be the fourier transform
of this expression now dirac delta function we already dirac delta function in laplace
transform that it is given by one by epsilon twenty varying from a to a plus epsilon and
zero otherwise and we are epsilon ten to zero so roughly speaking dirac delta functions
is infinity at a point i say t equal to a and zero otherwise such that the total integral
from minus infinity to plus infinity of dirac delta function is one that we have already
discussed in the laplace transforms now the fourier transform of dirac delta function
is given by this so this can be derived using the definition of dirac delta function i mean
dirac delta and fourier a transform so to find fourier transform of dirac delta function
it is nothing but minus to plus infinity f t f t is the delta into e k power minus eta
omega t into d t now dirac delta is given by one by epsilon when t varying from a to
a plus epsilon here epsilon trying to zero so this we can define like this limit epsilon
tend to zero a to a plus epsilon it is one by epsilon into e k power minus eta omega
t into d t now it is limit epsilon trying to zero one
by epsilon can come out and the integration of this will be nothing but this term from
a to a plus epsilon now it is nothing but limit epsilon tend to zero ok one upon minus
eta omega can come here and it is e k power minus eta omega a plus epsilon minus e k power
minus eta omega a by applying upper limit minus lower limit now when epsilon tend to
zero it is zero by zero form so you apply a rule to simply this so you will take the
derivate of the numerator and the denominator so limit epsilon tend to zero the derivate
of numerator will be nothing but it is minus eta omega e k power minus eta omega
a plus omega minus zero upon minus eta omega derivate is respect to omega i mean epsilon
ok now these two terms cancels out and when you
take epsilon tend to zero it is nothing but e k power minus eta omega so which is the
same as this expression now when you take a equal to zero it is clear that when you
take a equal to zero fourier transform of dirac delta t at a equal to zero is nothing
but one so this is very clear from this expression when you substitute a equal to c
now come back to convolution theorem now the convolution of two function define in
a same way as we did in a laplace transform now here integral is from minus to plus
infinity so convolution of two functions in fourier transform is given by this so convolution
of two functions in fourier transform is given by minus to plus infinity f tau g t minus
tau d tau or can be written as minus into plus infinity f t minus tau g tau d tau
because this convolution of two functions is commutative ok
now come come to convolution theorem now what is states convolution theorem fot fourier
transform suppose that f t and g t are piecewise continuous bound it and absolutely integrable
on the x axis then fourier transform of convolution of two functions f and g will be given by
fourier transform of f into fourier transform of g which is nothing but f omega into g omega
so we are we are fourier transform of f t is f omega and fourier transform of g t
is g omega so now we will see the proof of convolution
theorem so basically want to find out fourier transform of f star g ok so by definition
it is nothing but minus to plus infinity convolution of this this function into e k
power minus eta omega t d t this is by the definition of fourier transform fourier transform
of function f t is given by minus to plus infinity f t k power minus eta omega t d t
now convolution of two function this this term is given by you replace this term as
minus into plus infinity suppose f tau g t minus tau d tau and whole multiplied by e
k power minus eta omega t into d t so this can be written as minus to plus infinity
minus to plus infinity f tau g t minus tau e k power minus eta omega t d tau into d t
ok now now change the order of integration both the limits are constant minus and plus
infinity so change the order of integration so when you change the order of integration
is nothing but minus plus infinity minus into plus infinity f tau g t minus tau e k power
minus eta omega t d t into d tau now suppose now suppose t minus tau as a new
variable suppose z or d t will be d z ok so when you substitute this variable over here
in this expression so what we get we get so basically we are simplifying fourier transform
of convolution of two function so that will be equal to this expression and this will
further be equal to minus infinity to plus infinity minus infinity to plus infinity this
is f tau and this is g z e k power minus eta w and t is nothing but tau plus z from this
expression and d t is d z and d tau is d tau so this is nothing but now you can write it
like this minus to plus infinity f tau e k power minus eta omega tau t tau because now
you can separate two variables two via are separating tau we can separately and z we
can write separately so we can always do this and this is nothing but fourier transform
of f t and this is nothing but fourier transform of g t so this is nothing but f f omega
into g omega so hence hence fourier transform of convolution of two functions f and g is
nothing but f omega into g omega ok so this is convolution theorem and here is a proof
now from here it is also cleared at when the fourier transform of f star g is f omega g
omega so from here we can write that f star g into t is nothing but fourier inverse
of f omega into g omega now inverse fourier transform is given by one upon two pi integral
minus infinity to plus infinity the omega function which is f omega into g omega e k
power eta omega t into d omega so that we will already know by definition
of inverse fourier transforms ok now let us find fourier inverse fourier transform
of this f omega using convolution theorem ok so how you can do that so we will recall
convolution theorem again and using convolution theorem we will try to find out the fourier
inverse fourier transform of this function so what is a function now what what we
are find out fourier inverse of one upon six plus five eta omega minus omega square so
so this is f omega so this is f omega so f omega is nothing but one upon six plus
five eta omega minus omega square ok so in our to apply convolution theorem we have
to write as a product of two omega functions and the convolution and the in i mean fourier
inverse of both the omega functions we should know so that convolution theorem we
can apply ok so let us try to write this function so we can easily write it as one upon six
plus five eta omega plus eta square omega square and this is nothing but we can simplify
when we simplify so this is a square plus five a plus six so this is eta omega plus
two into eta omega plus three so this is nothing but one upon eta omega
plus two into one upon eta omega plus three so this is something like g omega and this
is something like h omega so we have write this f omega as a product of two omega
functions now to find out its inverse using convolution theorem so f inverse of f omega
which is equals to f inverse of g omega into h omega will be equal to so we will apply
convolution theorem it is nothing but convolution of this and this so what is what is fourier
inverse of g omega so fourier inverse of g omega we know that fourier inverse of
this is nothing but e k power minus two t u naught t this we already know and fourier
inverse of this expression one upon fourier inverse of h omega h omega is this
is nothing but e k power minus three t u naught t ok this is by the this you already know
that fourier transform of e k power minus a t u naught t is one upon a plus eta omega
ok fourier transform of f omega which is equal
to fourier inverse of fourier inverse transform of f omega into g omega here using convolution
theorem will be nothing but convolution of these two function e k power minus two t u
naught t e star with u e e k power minus three t into u naught t the convolution of
these two functions ok and that will be nothing but minus infinity to plus infinity f tau
that is e k power minus two tau u naught tau into g t minus tau that is e minus three t
minus tau u t minus tau and it is d t ok this is by the convolution theorem because
f tau into g t minus tau ok now let us simplify this let us simplify this now here we have
two unique step functions here and here so let us find what it is so it is u naught tau
into u t minus tau so now we have to see d tau tau is a variable basically so we have
to see respect to tau so when tau is less than when tau is less than t ok and suppose
greater than zero so when tau is less than t this quantity is positive because t tau
is less than t ok and this is also positive one into one is one and when tau is greater
than t so this is zero because this is negative so this will be zero
so when we apply this over here so this will be equal to now e k power minus three
t is free from tau can be come out and it is e k power minus minus plus three tau into
e k power minus two tau is e k power e k power tau and this product is one when tau varying
from zero to t otherwise it is zero so it is from zero to tau from zero to t only and
it is d tau ok so this will be equal to e k power minus three t e k power tau zero to
t which is equal to e k power minus three t e k power t minus one so that will be the
inverse fourier transform of this function this f omega using convolution theorem
so hence hence using convolution theorem we can find out we can find out inverse
fourier transforms also ok so the main application of convolution theorem is to find out the
fourier inverse so sometimes we have to find out fourier inverse so it is important to
use convolution theorem so so thats all so
thank you very much