welcome to the series of lectures on mathematical methods and applications so we have seen so
many thing about laplace transforms we have seen what laplace transforms are what are
the important properties of laplace transform how can you find out laplace transform of
some important functions that inverse etcetera in the last lecture we have seen what unit
step function is how to find out laplace transform unit step of function and some problem based
on it now in this lecture we will see what are what
is dirac delta functions and how to find out laplace transform dirac delta function
and some problems based on it so first first thing is what is dirac delta function
or we call it unit impulse function so it is defined like this derived delta at t equal
to a is nothing but one upon epsilon where a is tending to t tending from a to a plus
epsilon and zero otherwise and where epsilon tend to zero so this
is how we define dirac delta function so if we roughly speaking if epsilon tend to
zero so t equal to a and this will tends to infinity
so basically if we speak about a definition of dirac delta function so we can say that
at t equal to a that is at one point it is having infinite value and it and all other
values at for all other points it is zero ok so whenever we would solve such type
of problems or based on dirac delta we always use this definition that is it is one by epsilon
when t is varying from a to a plus epsilon zero otherwise and where epsilon tend to zero
so what what does this definition means this means when a is varying from a to a to
a plus epsilon where epsilon tend to zero sent then this value is one upon epsilon so
this value is one so a to a plus epsilon this value is one and otherwise this value is zero
so this is what the graph of dirac delta function is this is t this is dirac delta
function s t equal to t minus a so this is how we define we can define dirac delta function
so now the important property of delta function is that zero to infinity the integration of
dirac delta function is always one so that we can easily show zero to infinity this
integral is equal to now upon the definition of dirac delta function it is it is one
by epsilon when t varying from a to a plus epsilon so and zero otherwise so this integral
has a value only when t varying from a to a plus epsilon and it is one by of seven d
epsilon d t and limit epsilon tend to zero because we epsilon tend to zero so this is
nothing but limit epsilon tend to zero one by epsilon integration a to a plus epsilon
d t and this is further equal to limit delta epsilon tend to zero one by epsilon t a to
a plus epsilon and it is equals to limit epsilon tend to zero one by epsilon a plus epsilon
minus a which is nothing but one so that means the total area covered by this function the
dirac delta function from zero to infinity the total area is always one so we can always
define dirac delta function like this that this value is one by epsilon when t varying
a to a plus epsilon zero otherwise and that we are epsilon tend to zero and the total
area covered by that dirac delta function is always one ok
the next property is a we call it filtering property of dirac delta function it states
that if f t be is a continuous and integrable in zero to infinity then this integral is
always equal to f a so this also so we can prove easily let us try to prove this result
so what this property is it is zero to infinity f t into dirac delta of t minus
a d t is nothing but f t so this property is call it filtering property of dirac
delta function so how we will prove it now we will apply the definition of delta function
so the left hand side is zero to infinity f t into dirac delta of t at a into d t which
is equals to now limit epsilon tend to zero a to a plus epsilon it is one by epsilon f
t into d t this is by a definition of delta function because dirac delta function attains
a value one by epsilon when t varying from a to a plus epsilon and epsilon tend to zero
ok now this is equal to limit this tend to zero a to a plus epsilon into f t d t upon
epsilon ok now when epsilon tend to zero this is a to
a that is zero and this is zero this is zero by zero form and function is continuous ok
and integrable so we can apply l hospital rule to find out the value because it is
zero by zero form so we can apply l hospital rule so it is equal to limit epsilon tend
to zero d by s epsilon of a to a plus epsilon f t d t upon one we differentiate respect
of epsilon the numerator and the denominator ok now this is equal to limit epsilon tend
to zero now when you take d by d epsilon of this function with respect of epsilon so this
is nothing but f t a plus epsilon into one minus zero upon one ok and when epsilon tend
to zero it is nothing but f a so this is called filtering property of dirac delta function
that integration zero to infinity f t into dirac delta t minus a d t is nothing but f
a ok so this how we can prove this the next property
of dirac delta is the laplace transform of dirac delta is nothing but e to the power
minus a t so laplace transform of dirac delta function so it is nothing but zero to infinity
e to the power minus p t dirac delta of t minus a into d t by definition of laplace
transforms now by the definition dirac delta function it is equal to limit epsilon tend
to zero a to a plus epsilon into e to the power minus p t into one by epsilon into d
t and which is which is further equal to limit epsilon tend to zero a to a plus epsilon e
to the power minus p t d t upon epsilon ok now it is again zero by zero form because
there are epsilon tend to zero the numerator is zero and the denominator is zero so zero
by zero form so again you will apply l hospital rule to simplify this so this is [no/nothing]
nothing but equal to limit epsilon tend to zero again d by d epsilon of the numerator
quantity upon denominator is one the derivative denominator respect to epsilon is one and
it is nothing but limit epsilon tend to zero e to the power minus p in place of t we have
the upper limit which is a plus epsilon and when into one minus zero of course upon one
and when seven tend to zero it is nothing but e to the power minus a p so that is how
we can find out laplace transform of dirac delta function and it is nothing but e to
the power minus a p and of course when a is zero the laplace transform of delta t is nothing
but one because when you substitute a equal to one in this expression so we will get
laplace transform of dirac delta function at t equal to zero is nothing but one ok
now let us solve some problems based on dirac delta function now whenever we want to find
out laplace transform of some f t into dirac delta at t equal to a suppose so this will
be equal to zero to infinity e to the power minus p t f t into dirac delta t minus at
t equal to a t and when you apply a definition of definition of dirac delta function it
is nothing but limit delta tend to zero a to a plus epsilon into one by epsilon e to
the power minus p t into f t d t and which is again equal to ok epsilon tend to zero
limit epsilon tend to zero integration a to plus epsilon e to the power minus p t f t
d t upon epsilon so it will become it will it will always become zero by zero form so
you will apply l hospital rule to simplify this expression and then you will you can
get the laplace transform of f t into dirac delta or t equal to a ok so this is how we
can find out laplace transform of f t into dirac delta function
so now let us try to find out laplace transform of these problems first is laplace transform
of unit step function at t equal to one into dirac delta t equal to one ok this so this
is nothing but zero to infinity u one t that is f t f t is a entire expression into
e to the power minus p t into d t which is further equal to first apply definition of
dirac delta function ok by definition of dirac delta function is nothing but limit epsilon
tend to zero one by epsilon a to a plus epsilon u ok here a is one so this will be one only
and this is one ok and u one t into e to the power minus p t into d t now apply definition
of u one t so u one so u one t at t equal to t more than one by definition of unit
step function is one so it is nothing but limit epsilon tend to
zero one by epsilon one to one plus epsilon into e to the power minus p t d t because
u one t is one when t is greater than one otherwise it is zero so now it is nothing
but limit epsilon tend to zero one by epsilon ok it it is now zero by zero form because
you can always write it like this so it is zero by zero form ok so you have to apply
l hospital rule to simplify this so it is nothing but limit epsilon tend to zero d by
d epsilon of integral one to one plus epsilon e to the power minus p t d t upon one and
which is nothing but limit epsilon tend to zero e to the power minus p and t is nothing
but one plus epsilon and it is nothing but e k power minus p so that will be the laplace
transform of the first problem ok now let us compute laplace transform of second
problem it is again on dirac delta function so the second problem is basically power t
involved in this problem so it is laplace transform of e k power t into dirac delta
at t equal at a equal to three at a equal to three so it is nothing but so i can write
directly it is limit epsilon tend to zero instead of a i have three here so it is three
to three plus epsilon three to three plus epsilon f t is e to the power minus e to the
power t and e to the power minus p t is is here also upon epsilon because it is a
laplace transform and i directly applied definition of dirac delta function so it is epsilon tend
to zero three to three plus epsilon ok it is limit limit epsilon tend to zero three
to three plus epsilon e to the power three into e to the power minus p t d t upon epsilon
so it is zero by zero form so you will [app/again]again apply l hospital rule to simplify this so
it is or you can first integrate because it is integrable it is one minus p into three
plus epsilon upon one limit epsilon tends to zero and it is nothing but e to the power
three one minus p is it so so this is the final expression which we can obtain using
this by applying l hospital rule ok now the next problem next problem is laplace transform
of t square into u one t plus t into dirac delta at t at t equal to two so this is nothing
but laplace transform of t square into u one t plus laplace transform of t into dirac delta
at t equal to two now to compute the first expression we will
use second shifting property of unit step function which states that laplace of f t
minus into u t minus a is nothing but e to the power minus a p f p where this f p
is nothing but laplace of f t ok so we have to break this t square in the powers of t
minus one because it is f t minus a and a is one so first we find first find the laplace
of this then laplace of this the addition of this two will give the laplace of the entire
expression so laplace of t square into u one t this can be find out laplace of i can write
it like this t minus one plus one whole square u one t
this is nothing but laplace of a square plus two a b plus b square
and this is nothing but laplace of t minus one whole square u one t plus two times laplace
of t minus one u one t plus laplace of u one t now using that that property it is nothing
but e to the power minus a p a is one so it is minus p into into laplace of f t so laplace
of t square plus two into e to the power minus p again and laplace of t plus and this is
nothing but e to the power minus p into laplace of one so this is nothing but e to the power
minus p into t square is nothing but two upon p cube plus two e to the power minus p into
t is nothing but one p square plus e to the power minus p into one by p so that is the
laplace of the first expression laplace of t square u one t
now we will compute laplace of second expression how we compute laplace of second expression
it is laplace of t into dirac delta t minus two ok so this can be written as zero to infinity
t dirac delta t minus two e to the power minus p t into d t and applying the definition of
dirac delta function it is nothing but limit epsilon tend to zero two to two plus
epsilon t e to the power minus p t d t upon epsilon so it is it is zero by zero form
again we can apply l hospital rule to simplify this expression so we will obtain this
will be equal to limit epsilon tend to zero two plus epsilon into e to the power minus
two p into two plus epsilon into one minus zero upon one so when you take epsilon tend
to zero here it is nothing but two into e to the power minus two p ok so the sum of
these two expressions will give will give us the epsilon the laplace transform of this
expression ok now the last problem of this slide
laplace transform of laplace transform of t square into dirac delta at t equal
to t equal to one so this problem also can be solved easily it is zero to infinity t
square dirac delta minus one into e to the power minus p t d t and when you apply definition
of dirac delta it is nothing but limit epsilon tend to zero one to one plus epsilon t square
e to the power minus p t d t upon epsilon now it is zero by zero form so you can directly
apply l hospital rule to simplify this so this is nothing but limit epsilon tend to
zero one plus epsilon whole square e to the power minus p into one plus epsilon minus
zero upon one and when you take epsilon tend to zero it is nothing but e to the power minus
p so that will be the laplace transform of t square into dirac delta at t equal to one
ok so that is how we can solve the problems based on dirac delta function i mean laplace
to find out the function with dirac delta function ok
now let us evaluate the following integrals zero to infinity which are having the involvement
of dirac delta so we can directly use filtering property of dirac delta function what are
filtering property that we have already proved so this is a filtering property of
dirac delta function that is zero to infinity f t into dirac delta of t minus a d t is nothing
but f a ok so suppose want to solve first problem the first problem is zero to infinity
sin square t dirac delta at pi by four so if we compare with this expression so f t
is nothing but sin square t and a is nothing but pi by four so directly using this filtering
property this expression will be nothing but sin square pi by four now sin pi by four is
one by under root two it is whole square which is nothing but one by two so directly using
filtering property we can find out the value of this problem
now similarly if we take the second problem the second problem is nothing but it is zero
to infinity t square cos two t into dirac delta at t equal to pi by two again if we
compare this with this expression so f t is nothing but t square cos two t and a is pi
by two so by using this property this expression will have the value equal to t square cos
two t at t equal to pi by two so this nothing but pi by two whole square cos pi and cos
pi is minus one so it is minus pi square by four it is minus pi square by four
now let us see the third problem of the slide that is zero to infinity it is zero to
infinity f t into derived delta at t pi by two d t and f t is defined like this f t is
given to us it is it is zero when t lying between zero and pi by two and it is sin t
when t is greater than equal to pi by two so basically what this function is it is zero
into u zero t minus u pi by two t plus sin t into u pi by two t so i express this
function in terms of unit step function ok so it is nothing but sin t into u at pi by
two into t so now this is equal to by filtering property this will be nothing but the value
of f t equal to a and here a is pi by two so the value of f at pi by two so value of
f that is sin t into u pi by two into t at t equal to pi by two
now at t equal to pi by two this is one it is a unit step function the value will be
one so it is nothing but sin pi by two into one which is nothing but one so hence we can
find out the value of this expression so we have seen that what is dirac delta function
how we can solve some some problems based on the based on based on this
how we can solve these type of integral using filtering property or delta function so
in the next lecture we will see that what are the various applications of the laplace
transforms so thank you