yeah hello every one so in the last lecture we have seen that what is the existence theorem
for laplace transforms we have seen that if a function is a piece wise continuous and
is of exponential order alpha then its laplace transform always exists otherwise if function
is not piece wise continuous or is not of exponential order then then laplace may or
may not exist that also we have seen so now in this topic we will see what are the
properties of laplace transforms so we will start with the first property that we call
as shifting property now shifting property is first we have first translation and
then we have second translation of the shifting property what is first translation first translation
means if laplace transform of f t is say f p then if this holds then laplace transform
of e ki power a t into f t is nothing but f of p minus a that means if we know the laplace
transform f t which is given as f p if you know this then laplace transform of e ki a
t into f t is nothing but you simply replace p by p minus a so then we will get the laplace
transform of this function e ki power at into f t now what are the proof of this result
how will obtain this result now what is lapalce transform of e ki power a t into f
t by the definition this is nothing but zero to infinity e ki power a t f t into e ki power
minus p t d t this is by the definition of laplace transform
so this is equal to zero to infinity e ki power minus p minus a times t into f t d t
so this is nothing but zero to infinity suppose suppose p minus a is some xi of f t where
xi is nothing but p minus a then this quantity is nothing but f of xi because what is f p
f p is laplace of f t which is zero to infinity e ki power minus p t f t d t so instead of
p we have xi in the same expression so we replace f as xi and what is xi xi is p minus
a so it is p minus a so hence laplace of e ki power a t into f t is nothing but f
of p minus a so simply replace p by p minus a now the now second translation or the
shifting property is if laplace transform of f t is f p and g t is given as f of t minus
a when t greater than a and zero when t less than a then laplace transform of g t is nothing
but e ki power minus a p times f p again again we can prove this using the definition
of laplace transform laplace transform of g t is nothing but we have to prove this result
as so this is equal to zero to infinity e ki power minus p t g t d t by the definition
of laplace transforms this is equal to zero to a e ki power minus
p t g t d t plus a to infinity e ki power minus p t g t d t now zero to a g t is zero
because g t is zero and t less than a so it is zero so this quantity zero plus because
g t is zero so integration value will be zero plus so this is a to infinity e ki power minus
p t when t is greater than a so g t is f t minus a so you simply take f of t minus a
here d t now this expression so therefore therefore what is laplace of g t therefore
laplace of g t will be nothing but from here we got a into infinity e ki power minus p
t f of t minus a into d t now now you can take t minus a equal to some new variable
say z d t will be d z so this will be equals to when t is a so z will be zero when t is
tend into infinity z tend into infinity e ki power minus p t from here a is a plus z
so it is a plus z times f z into d z so this is nothing but since e ki power minus p a
is free from z we are integrating it with respect to z so we can take it outside the
integration e ki power minus a p zero to infinity e ki power minus p z f z d z which is same
as e ki power minus a p so this is nothing but laplace transform of f t or f z so this
is f p this is nothing but laplace transform of f t so e ki power minus a p into f p so
hence laplace of g t is nothing but e ki power minus a p into f p so this is called second
translation or shifting property now to how we can solve problems based
on this let us solve few examples on this the laplace transform of t cube e ki power
three t so so we have just discussed that if laplace
transform of f t is f p then laplace transform of e ki power a t into f t is nothing but
f of p minus a so here f t is t cube if you compare this with this so there f t is t cube
so what is laplace transform of t cube it is it is gamma four upon p ki power four
this is nothing but factorial three upon p ki power four which is nothing but six upon
p ki four so six upon p ki power four so this is f p now you have to compute laplace transform
of e ki power three into t cube so simply replace f p in f p p by p minus a and a here
is three so that is nothing but f of p minus three so that is nothing but six upon p minus
three ki power four so that will be the laplace transform of t cube e ki power three t
next problem suppose suppose the next problem next problem is e ki power minus four t cos
hyperbolic two t so again here if you compare with this so f t is cos hyperbolic two t so
first find laplace of cos hyperbolic two t let us put equal to f p and then replace p
by p minus a here a is minus four so replace p by p plus four so what is laplace transform
of cos hyperbolic two t we already know that laplace transform of cos hyperbolic a t is
p upon p square minus a square so it is p upon p square minus four so laplace transform
of e ki power so this if f p this is f p so laplace transform of e ki power minus four
t into cos hyperbolic two t will be nothing but f of p you replace p by p minus a a is
minus four so p plus four so that is nothing but p plus four upon p plus four whole square
minus four so that this will be the laplace transform
of the e ki power minus four t cos hyperbolic two t now the third problem suppose you want
to solve the third problem it is e ki power minus t sin square t so first find laplace
transform of sin square t put it equal to f p and replace p by p minus a and here is
a minus one so replace p by p plus one so first find lapalce transform of sin square
t so sin square t is nothing but one minus cos two t by two so this is nothing but one
by two laplace transform of one minus one by two laplace transform of cos two t so laplace
transform of one is one by p minus one by two laplace transform of cos two t is p upon
p square plus a square so p upon p square plus four so this equal to f p now laplace
transform of e ki power minus t square t will be nothing but f of you replace p by p minus
a a is minus one that is p plus one so that is nothing but one upon two p plus one minus
p plus one upon two into p plus one whole square plus four so that will be the laplace
transform of e ki power minus t sin square t
again in the last problem we can use shifting property because what is it is sin hyperbolic
three t into cos square t it is nothing but laplace of e ki power three t minus e ki power
three t by two cos square t so it is one by two laplace transform of e ki power three
t cos square t minus one by two laplace transform of e ki power minus three t into cos square
t so we will find laplace transform of cos square t put it equal to f p and then in the
first part replace p by p minus three and for the second part replace p by p plus three
so what is laplace transform of cos square t that again we can find out cos square t
will be nothing but laplace transform of one plus cos two t by two that is nothing but
one by two laplace transform of one is one by p so that is one by p plus one by two laplace
transform of cos two t is p upon p square plus a square so p upon p square plus four
so this is f p now what is laplace transform of this e ki power three t cos square t
this is nothing but f of p minus a and what is laplace transform of e ki power minus three
t cos square t this is nothing but f of p plus three so substituting these two values
you replace p by p minus three for the first part and you replace p by p plus three for
the second part and you substitute these values here so we will get the laplace transform
of sin hyperbolic three t into cos square t
now let us find laplace transform of t sin four t so sin four t doesnt involve as
such doesnt involve e ki power a t but we can break it so laplace of t sin four t so
laplace of t now sin four t can be written as e ki power four iota t minus e ki power
minus four iota t upon two iota this is how we can replace sin four t now this is nothing
but one by two i laplace transform of t into e ki power four iota t minus one by two i
laplace transform of t into e ki power minus four iota t so again here here a is four iota
if you compare with this and here a is minus four iota so you find laplace of t first let
it put it equal to f p and for the first expression replace p by p minus four iota and for second
expression replace p by p plus four iota so what is laplace transform of t what is laplace
transform of t laplace of t is nothing but one by p square it is equals to f p and for
this part it is nothing but one upon two iota f of p minus four iota minus one by two iota
f of p plus four iota by by this shifting property
so this is nothing but one upon two iota what is f of p minus four iota one upon p minus
four iota whole square minus one upon p plus four iota whole square so the simplify it
one upon two iota into so this will go here this will go here simplify it so this is nothing
but two a b two into four iota into p two a b into two again two one two again comes
then it is nothing but p square plus sixteen whole square this two iota will cancel with
two iota and this is nothing but eight p upon p square plus sixteen whole square so this
will be the laplace transformation of t sin four t now now this problem on shifting
property lapalce transform of e ki power minus two t sin under root t so again we first we
find out laplace transform of sin under root t and using shifting property we will
replace p by p plus two because here a is minus two so what is laplace transform of
sin under root t let us see so what is what is sin theta sin theta is theta minus theta
cube upon factorial three plus theta ki five upon factorial five and so on what is sin
under root t sin under root will be t ki power half minus t ki power three by two upon factorial
thee plus t ki power five by two upon factorial three factorial five and so on
now it is laplace of sin under root t will be nothing but will be nothing but laplace
transform of this entire expression and by the linearity property we can split in term
wise so this is nothing but laplace transform of laplace transform of t ki power half
minus one by factorial three laplace transform of t ki power three by two plus one by factorial
five laplace transform f t ki power five by two so this is t ki power half is gamma three
by two upon p ki power three by two using laplace transform of t ki power n minus
one by factorial three it is gamma five by two upon p ki power five two plus one by factorial
five gamma seven by two upon p ki power seven by two and so on so this is gamma three by
two is because gamma n plus one is gamma n so gamma three by two will be half gamma half
and gamma half is under root pi so it is one by two under root pi p ki power three by two
minus one by factorial three it is nothing but three by two one by two under root pi
again the property of gamma function p ki power five by two plus one by factorial five
it is five by two three by two one by two under root pi p ki power seven by two
so what finally we get so laplace transform of sin under root t this will be nothing but
now you can take one by two under root pi upon this quantity common form of all the
terms so this is one by two p ki power three by two into under root pi inside bracket we
get one minus so this is factorial three is three into two into one three three cancels
out under root two this we have taken common one so this is nothing but this is nothing
but one by two square into p whole power upon one factorial because one two comes from
here one to from here one two is common so it is two square from here five into five
cancels out so it is nothing but plus one by two square p whole square upon factorial
two this comes when you simplify five five cancels out three also cancels so we will
get two square whole square upon factorial two and this is under root two p ki power
three by two and it is nothing but e ki power minus one by four p so that is the laplace
transform of sin under root t so this is f p and for to find laplace transform of
laplace transform of e ki power minus two t you simply replace p by a p plus two in
this expression this is f p you replace p by p plus two you will get laplace transform
of this quantity again using second translation property
we can easily find out laplace transform of g t because if you compare with second shifting
property second translation so laplace transform of f t is f p then laplace transform g
t is e ki power minus a p times f p by second translation property so if we compare
with this here here in the first problem what is g t what is f t minus a now f of p
minus a for the first problem f t will be t cube then only it is t minus a a is one
whole cube so what is laplace transform of f t it is nothing but factorial three upon
p ki power four that is six upon p ki power four now the laplace transform this g t will
be nothing but e ki power minus a s e ki power minus a p a is one here a p into f p and f
p is the this quantity so this is nothing but e ki power minus p e ki power six into
e ki power minus p upon p ki power four so this will be the laplace transform of the
first problem by the second second translation property again for the for the second part
you see that here here f t is this quantity if you compare with the this property here
f t is the f t given here in this term f t here is cos t so the laplace of cos t is
p upon p square plus one so laplace of this f t will be nothing but
using second translation property laplace of ft will be nothing but e ki power minus
a p a is two pi by three e ki power minus two pi by three p into p into laplace of cos
t laplace of cos t is nothing but p upon p square plus one so this is how we can find
out laplace transform of this problems using second translation property now the
same shifting property also hold for inverse laplace transforms inverse laplace if inverse
laplace of f t is f p then inverse laplace transform of e minus a is e ki power a t into
f t and the second shifting property states that if laplace inverse of f p is f t then
laplace inverse of e ki power minus a p into f p f p is nothing but f of t minus a when
t is greater than zero a t less than a this is by the shifting property itself now
based on this let us try to solve this problems so the first problem is laplace of three p
upon plus two upon p square minus two p plus five so laplace of this is three times p upon
p square minus two p plus five plus two times laplace oh laplace inverse it is laplace inverse
yeah so it is laplace inverse of one upon p square minus two p plus five so it is three
laplace inverse of p upon it is p minus one whole square plus four it is two times laplace
inverse of one upon p minus one whole square plus four
now here p minus so you subtract one and add one here so this is three laplace inverse
of p minus one upon p minus one whole square plus four and three into one upon this quantity
plus two into one upon this quantity will be five time laplace inverse of one upon p
minus one whole square plus four now now now we know that laplace inverse of f p is
f t if this happens then laplace inverse of f p is minus a will be nothing but e ki power
a t into f t by the inverse of property shifting property now here if you use this definition
f p minus a here a is one so it is nothing but three e ki power a t a is one so three
into e ki power t and f t and f t will be laplace inverse of f p so it will be laplace
inverse of p upon p square plus four you replace you replace p plus one ok so it will be p
upon p square plus four so that will be laplace inverse of this quantity again plus you replace
p by by this property is nothing but five into e ki power t laplace inverse of one upon
p square plus four so this is nothing but three into e ki power t this is cos two t
plus five into e ki power t you divide multiply into two so it is two upon this sin two t
so that will be the laplace inverse of this expression that will be laplace transform
of this expression similarly when you apply the same thing in second problem laplace
inverse of this so it is two p plus one ok it is one upon one upon under root four p
plus five second problem so you take four common first so it is one by two laplace inverse
of one upon under root p plus five by four use this shifting property now here a is minus
minus five by four so it will be one by two e ki power minus five by four t laplace
inverse of one by under root p by the shifting property of for inverse laplace transforms
and this is again is equal to one by two e ki power minus five by four t we know that
now for to find laplace inverse of this expression we know that laplace transform of t ki power
n is nothing but gamma n plus one upon p ki power n plus one
so laplace inverse of one upon p ki power n plus one is nothing but t ki power n upon
gamma n plus one so replace n by n by minus half so it is t ki power minus half upon gamma
half so that is nothing but one by two e ki power minus five by four t five by four into
t into one by under root pi into t so that will be the laplace inverse of this expression
now to solve the last problem we can use partial fractions this we use partial fraction to
simply this and then we apply the shifting property to find out the laplace inverse of
this expression the third problem we can solve now to find out laplace inverse of this
two problems again we will use second shifting second translation property for inverse laplace
transforms first we will find laplace inverse of here we have problem e ki power e ki power
minus p pi by three upon p square plus one so first find laplace inverse of one upon
p square plus one so that is nothing but sin t now using second second translation property
laplace inverse of e ki power minus p pi by three upon p square plus one will be nothing
but sin t minus pi by three when t greater than pi by three and zero t when pi by three
and this is because by the by the second property by the second translation property
we have seen that laplace inverse of this quantity is nothing but f of t minus a when
t greater than a zero t when less than a here here a is minus pi by three so you
replace t minus pi by three in this f t this is f t this is f t ok now for second
the last problem again we will first find laplace inverse of p plus one upon one p square
p plus one and then using second shifting property we will find laplace inverse of
entire expression so what is so first we will find laplace inverse of this
expression p plus one upon p square plus p plus one we will call it f t and then for
this entire expression we replace we replace t by t minus pi the laplace inverse of this
will be nothing but minus p pi this will be nothing but f of f of t plus t minus pi
when t greater than pi and zero when t less than pi
so this we can we can obtain by using second property so laplace of inverse of this we
can find making perfect square when the denominator let us to make it this is p plus half whole
square plus three by four and this is nothing but laplace inverse of p plus half upon p
plus half whole square plus three by four plus one by two laplace inverse of one upon
p plus half whole square plus three by four and it is e ki power minus one by two t by
using shifting property p plus half plus half is one and it is one by two square plus
yeah its ok e ki power e ki power a t a is minus half minus half is laplace inverse
of p upon p square plus under root three by two whole square plus one by two e ki power
minus one by two t laplace inverse of one upon p square plus under root by three by
two whole square now this we can find out this is nothing but cos a t a is under
root three by two you multiply and divide by this quantity here this will be nothing
but sin a t and a is under root three by two so that will be f t and you replace
you substitute f t here so you will get laplace inverse of this expression so
thank you very much