WEBVTT
Kind: captions
Language: en
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so in the last class we have seen what are
laplace transforms and how to find laplace
00:00:26.250 --> 00:00:33.730
transform of some simple functions like
one t or e k power a t so what is laplace
00:00:33.730 --> 00:00:42.750
transform any function f t laplace transform
any function f t is nothing but zero to infinity
00:00:42.750 --> 00:00:56.980
f t e k power minus p t d t and we are calling
it f p function of p so also we have seen
00:00:56.980 --> 00:01:07.940
that laplace transform satisfy linearity property
so now in this lecture we will find laplace
00:01:07.940 --> 00:01:14.780
transform of some standard functions so first
what is laplace transform of t k power n so
00:01:14.780 --> 00:01:21.880
laplace transform of t k power n
let us try to find it laplace or t k power
00:01:21.880 --> 00:01:30.489
n so using the definition of laplace transform
laplace t k power n nothing but zero to infinity
00:01:30.489 --> 00:01:42.880
t k power n e k power minus p t d t now let
p t is equals to some variable say z so it
00:01:42.880 --> 00:01:55.440
will be nothing but zero to infinity z by
p whole power n e k power minus z and d t
00:01:55.440 --> 00:02:06.560
is nothing but d z upon p so it is nothing
but one upon p t k power n plus one integral
00:02:06.560 --> 00:02:18.650
zero to infinity z k power n e k power minus
z into d z and this function is nothing but
00:02:18.650 --> 00:02:25.840
gamma function so it is one upon p n plus
one gamma of n plus one
00:02:25.840 --> 00:02:38.850
now as gamma n plus one is factorial n
when n is a positive integer therefore laplace
00:02:38.850 --> 00:02:45.250
transform of t k power n will be factorial
n upon p k power n plus one when n is a positive
00:02:45.250 --> 00:02:53.390
integer now let us try to solve the two
problems given below laplace transform of
00:02:53.390 --> 00:03:09.210
three t k power three by two minus plus
five into t minus four e k power two t now
00:03:09.210 --> 00:03:14.760
again by the linearity property of laplace
transform this is nothing but three times
00:03:14.760 --> 00:03:25.340
laplace of three by two t k power three by
two plus five laplace of t minus four laplace
00:03:25.340 --> 00:03:37.361
of e k power two t so this is three into now
laplace of t k power n as in the first expression
00:03:37.361 --> 00:03:42.190
we are having it is gamma n plus one upon
p k power n plus one
00:03:42.190 --> 00:03:50.050
so here n is not a integer here n is positive
integer so we have to use the first definition
00:03:50.050 --> 00:03:56.720
of t k power n it is gamma plus one upon t
k power n plus one so it is gamma here n is
00:03:56.720 --> 00:04:07.740
three by two so it is five by two upon p k
power five by two plus five into laplace of
00:04:07.740 --> 00:04:13.520
t we have already seen that it is one by p
square also we can obtain here from here also
00:04:13.520 --> 00:04:21.489
when we put n equal to one in the sorry in
the second expression so it is one by p square
00:04:21.489 --> 00:04:29.190
minus four into laplace of e k power a t is
one upon p minus a so it will be nothing
00:04:29.190 --> 00:04:42.020
but one upon p minus two
so now now now gamma n plus one we already
00:04:42.020 --> 00:04:51.930
know it is n gamma n gamma n plus one nothing
but n gamma n and gamma is half under root
00:04:51.930 --> 00:05:04.539
pi this we already know so three into now
gamma five by two so here it is three by two
00:05:04.539 --> 00:05:17.419
plus one so it is three by two gamma three
by two upon p k power five by two plus five
00:05:17.419 --> 00:05:30.029
p square five by p square four upon p minus
two is equal to three into three by two again
00:05:30.029 --> 00:05:36.589
we will use a same expression instead of
n we have one by two because one by two plus
00:05:36.589 --> 00:05:42.771
is c by two so one by two plus one is c by
two it is one by two gamma one by two so one
00:05:42.771 --> 00:05:54.559
by two gamma one by two p k power five by
two plus five upon p square minus four upon
00:05:54.559 --> 00:06:04.480
p minus two so it is nothing but nine upon
four and gamma half is under root pi so it
00:06:04.480 --> 00:06:16.240
is under root pi upon p k power to five by
two plus five upon p square minus four upon
00:06:16.240 --> 00:06:21.189
p minus two so this is the laplace transform
of this function
00:06:21.189 --> 00:06:31.999
now now how to find laplace inverse of
the this expression p q plus two upon p
00:06:31.999 --> 00:06:46.539
k power five so laplace inverse of p cube
plus two upon p k power five so it is laplace
00:06:46.539 --> 00:07:01.660
inverse of one by p square plus two into laplace
inverse of one by p k power five now recall
00:07:01.660 --> 00:07:12.180
the definition laplace transform t k power
n is given as gamma n plus one upon p k power
00:07:12.180 --> 00:07:19.430
n plus one so what will be the laplace inverse
of laplace inverse of one upon p k power n
00:07:19.430 --> 00:07:26.379
plus one from here it is nothing but gamma
n plus one is a constant quantity so it will
00:07:26.379 --> 00:07:34.770
put it will go to the right hand side so it
will be noting but one upon gamma n plus one
00:07:34.770 --> 00:07:44.649
into t k power n by that definition
so laplace inverse of one by p square so
00:07:44.649 --> 00:07:53.110
one by p square means n is one when n is one
so it is nothing but t k power one upon gamma
00:07:53.110 --> 00:08:02.909
two plus two into now for this expression
n is four when n is four so it will be nothing
00:08:02.909 --> 00:08:14.939
but t k power four upon gamma five it is t
now gamma n plus one is factorial n when n
00:08:14.939 --> 00:08:22.219
is the positive integer so it is gamma two
to this factorial one plus two t k power four
00:08:22.219 --> 00:08:30.719
upon factorial four so it is t plus two t
k power four upon four into three into two
00:08:30.719 --> 00:08:38.510
into one so two two cancels out so it nothing
but twelve t plus t k power four divided by
00:08:38.510 --> 00:08:48.190
twelve so this will be the value of laplace
inverse of this function
00:08:48.190 --> 00:08:58.050
now how to find laplace of sin a t so now
we will see how to find laplace of sin a t
00:08:58.050 --> 00:09:08.470
so sin a t as we already know is nothing but
e k power iota a t minus e k power minus iota
00:09:08.470 --> 00:09:25.509
a t upon two iota this is sin a t now laplace
of sin a t using linearity property of laplace
00:09:25.509 --> 00:09:34.920
transform will be nothing but one upon two
iota laplace of e k power iota a t minus one
00:09:34.920 --> 00:09:45.449
upon two iota laplace of e k power minus iota
a t so it is nothing but one upon two iota
00:09:45.449 --> 00:09:53.480
as we already know let laplace of e k power
a t is one upon p minus a ok here instead
00:09:53.480 --> 00:10:02.350
of a we have iota a so we will replace a by
iota a one upon p minus iota a minus one upon
00:10:02.350 --> 00:10:07.050
two iota
again here instead of a we are having minus
00:10:07.050 --> 00:10:17.860
iota a so we will replace a by minus iota
a so we will get one upon p plus iota a so
00:10:17.860 --> 00:10:24.560
we will take one upon two iota a is common
so and the numerator will get we will get
00:10:24.560 --> 00:10:32.769
two a upon p square minus a square two iota
a sorry so two iota will cancels out so it
00:10:32.769 --> 00:10:39.660
is nothing but it is p square plus a square
when we multiply this two so it is a upon
00:10:39.660 --> 00:10:47.500
p square plus a square so this is laplace
transform of sin a t it is a upon p square
00:10:47.500 --> 00:10:53.269
plus a square
now so therefore if you want to find out laplace
00:10:53.269 --> 00:11:00.529
inverse of this function this f p this will
be nothing but sin at so what will be laplace
00:11:00.529 --> 00:11:07.660
inverse of one upon p square plus a square
it will be nothing but one by a sin a t now
00:11:07.660 --> 00:11:18.850
now now laplace transform of cos a t
in the similar way we can find out laplace
00:11:18.850 --> 00:11:28.899
transform of cos a t also so what is called
cos a t in terms of e so cos a t will be nothing
00:11:28.899 --> 00:11:41.089
but e k power iota a t plus e k power minus
iota a t upon two this we already known so
00:11:41.089 --> 00:11:46.780
to find laplace transform of cos a t again
we will apply linearity property of laplace
00:11:46.780 --> 00:11:53.459
transform laplace transforms of cos a t will
be nothing but one upon two laplace transform
00:11:53.459 --> 00:12:02.670
of e k power iota a t plus one by two laplace
transform of e k power minus iota a t
00:12:02.670 --> 00:12:09.279
so this is nothing but one upon two laplace
transform of e k power a t is one upon p minus
00:12:09.279 --> 00:12:19.370
a here instead of a we have iota a it is one
upon p minus iota a plus one by two one upon
00:12:19.370 --> 00:12:29.209
p plus iota a so when we simplify it so we
will get one by two into two p upon p square
00:12:29.209 --> 00:12:36.930
plus a square so two two cancels out so laplace
transforms of cos a t is nothing but p upon
00:12:36.930 --> 00:12:45.009
p square plus a square similarly laplace inverse
of this f p p upon p square plus a square
00:12:45.009 --> 00:12:54.069
is nothing but cos a t now now what is laplace
transform of sin hyperbolic a t now we will
00:12:54.069 --> 00:13:01.460
try to find out this expression to prove this
expression in fact so what is cos hyperbolic
00:13:01.460 --> 00:13:10.620
at sorry sin hyperbolic a t
sin hyperbolic a t is nothing but e k power
00:13:10.620 --> 00:13:21.070
a t minus e k power minus a t upon two this
we already know so how to find laplace transform
00:13:21.070 --> 00:13:27.439
of sin hyperbolic a t again we will use
linearity property of laplace transforms so
00:13:27.439 --> 00:13:35.250
laplace transform of sin hyperbolic a t will
be nothing but one by two laplace transform
00:13:35.250 --> 00:13:44.529
of e k power a t minus one by one by two laplace
form of e k power minus a t so it is one by
00:13:44.529 --> 00:13:54.570
two laplace transform of e k power a t is
one upon p minus a minus one upon two one
00:13:54.570 --> 00:14:05.519
upon p plus a so when we simplify this so
we will get one by two it is p plus a minus
00:14:05.519 --> 00:14:14.300
p plus a upon p square minus a square this
p p cancels out so it is nothing but a upon
00:14:14.300 --> 00:14:22.230
p square minus a square
so this would be the laplace of sin hyperbolic
00:14:22.230 --> 00:14:29.870
a t similarly if you want to find out laplace
inverse of this expression this f p so it
00:14:29.870 --> 00:14:38.680
will be nothing but sin hyperbolic a t
now what is laplace transform of cos hyperbolic
00:14:38.680 --> 00:14:45.439
a t in the similar way on the same lines we
can obtain the second expression the this
00:14:45.439 --> 00:14:52.790
expression for cos hyperbolic a t so what
is cos hyperbolic a t cos hyperbolic a t is
00:14:52.790 --> 00:15:02.839
nothing but e k power a t plus e k power minus
a t upon two this we already know
00:15:02.839 --> 00:15:11.249
now laplace transform of cos hyperbolic a
t will be one by two laplace transform of
00:15:11.249 --> 00:15:20.529
e k power a t plus one by two laplace transform
of e k power minus a t so it is nothing but
00:15:20.529 --> 00:15:30.370
one by two one upon p minus a laplace of
e k power a t we already know it is one upon
00:15:30.370 --> 00:15:41.130
p minus a plus one by two it is one upon p
plus a so it is one by two numerator we
00:15:41.130 --> 00:15:49.040
get two p upon p square minus a square two
two cancels out so this will be the laplace
00:15:49.040 --> 00:15:56.350
of cos hyperbolic a t now laplace inverse
of p upon p square minus a square is similarly
00:15:56.350 --> 00:16:03.240
cos hyperbolic a t because this laplace of
f t is p f then laplace inverse of f p will
00:16:03.240 --> 00:16:09.519
be nothing but f t that we already know so
this we already seen laplace inverse of
00:16:09.519 --> 00:16:14.959
these three expressions we already discussed
now how to find laplace of this two function
00:16:14.959 --> 00:16:22.279
this two simple function let us let us see
how to find laplace of these functions laplace
00:16:22.279 --> 00:16:34.889
of cos two t minus three so we know the laplace
of laplace transformation of cos a t laplace
00:16:34.889 --> 00:16:43.170
transform of cos a t is p upon p square
plus a square but how to find laplace transform
00:16:43.170 --> 00:16:51.649
of cos two t minus three that is a t plus
b types how to find out the laplace of this
00:16:51.649 --> 00:17:06.350
so it is cos a minus b so cos a minus b is
nothing but cos a cos b plus sin a sin b
00:17:06.350 --> 00:17:11.980
now cos three and sin three are the constants
free from t so can be taken out because of
00:17:11.980 --> 00:17:18.551
the linearity property of laplace transforms
so it is nothing but cos three laplace transforms
00:17:18.551 --> 00:17:28.280
of cos two t plus sin three laplace transform
of sin three t sin two t so this is nothing
00:17:28.280 --> 00:17:39.970
but cos three and laplace transform of cos
two t is p upon p square plus four and plus
00:17:39.970 --> 00:17:48.100
sin three laplace transform of sin two t is
two upon p square plus four p square plus
00:17:48.100 --> 00:17:58.090
a square a is two so this is nothing but p
into cos three plus two into sin three upon
00:17:58.090 --> 00:18:06.020
p square plus four so this will be the laplace
transformation of cos two t minus three
00:18:06.020 --> 00:18:10.840
now the second problem and now the next
problem laplace transform of sin three t into
00:18:10.840 --> 00:18:27.990
cos square t so laplace transform of sin three
t into cos square t
00:18:27.990 --> 00:18:36.040
now we know how to find laplace transform
of sin three t but we dont know how to find
00:18:36.040 --> 00:18:43.550
out laplace transform of product of two functions
when both involves t so first we try to break
00:18:43.550 --> 00:18:53.750
this function now sin three t what is cos
square t so cos square t can be written as
00:18:53.750 --> 00:19:07.000
one plus cos two t by two now this is nothing
but laplace transform of sin three t upon
00:19:07.000 --> 00:19:19.150
two plus sin three t into cos two t upon two
this can be written as one by two laplace
00:19:19.150 --> 00:19:35.630
transform of sin three t plus one by two laplace
transform of sin three t into cos two t so
00:19:35.630 --> 00:19:42.140
laplace of transform of sin three t we already
know this is this is three upon p square
00:19:42.140 --> 00:19:50.211
plus three square plus but again we have a
problem here how to find laplace transform
00:19:50.211 --> 00:20:03.560
of this so we multiply and divide by two in
this expression
00:20:03.560 --> 00:20:13.530
now two sin a cos b we know that it is equal
to sin a plus b plus sin a minus b so it is
00:20:13.530 --> 00:20:23.750
three by two p square plus nine plus one by
four laplace of it is sin five t a plus b
00:20:23.750 --> 00:20:32.150
two t plus three t five t plus this three
t minus two t is t
00:20:32.150 --> 00:20:39.810
so now we can find out the laplace of this
expression so this is three by two p square
00:20:39.810 --> 00:20:49.750
plus nine plus one by four laplace of sin
five t is a upon p square plus five square
00:20:49.750 --> 00:21:01.890
plus sin t is one upon p square plus one square
so this will be the laplace transform of this
00:21:01.890 --> 00:21:18.630
function sin three t into cos square t so
now
00:21:18.630 --> 00:21:23.940
now how to find laplace inverse of this simple
f p the simple function so laplace inverse
00:21:23.940 --> 00:21:36.250
of it is p minus three upon p square plus
four so this is simple to find out you
00:21:36.250 --> 00:21:49.430
just split this into two parts p into p square
plus four minus three upon p square plus four
00:21:49.430 --> 00:21:55.420
now by the linearity property of inverse laplace
transforms it is this into p upon p square
00:21:55.420 --> 00:22:07.500
plus four minus three times laplace inverse
of one upon p square plus four so this laplace
00:22:07.500 --> 00:22:14.250
inverse we already know this laplace inverse
of this expression is cos two t because it
00:22:14.250 --> 00:22:23.420
is four two square so it is cos two t because
laplace of cos two t is p upon p square plus
00:22:23.420 --> 00:22:32.110
two square that is p upon p square plus four
now it is minus now now laplace inverse
00:22:32.110 --> 00:22:41.130
of one upon p square plus a square is one
upon sin a t so here a is two so it is minus
00:22:41.130 --> 00:22:55.240
three by two sin two t so this will be the
laplace inverse of this f p
00:22:55.240 --> 00:23:05.650
now how to find laplace inverse of this
problem now now here f p is what f p is
00:23:05.650 --> 00:23:21.120
p plus two upon p minus one into p square
plus four so this is a linear part and this
00:23:21.120 --> 00:23:29.120
is a quadratic part so how to find out
laplace inverse of this f p so this can be
00:23:29.120 --> 00:23:36.920
written as now we will make use of partial
fractions first we will try to reduce into
00:23:36.920 --> 00:23:43.440
partial fractions and then we will try
to find out its laplace inverse so it can
00:23:43.440 --> 00:23:51.370
written as a upon p minus one plus
now since it is a quadratic part so in the
00:23:51.370 --> 00:23:59.520
numerator we will having a linear term linear
in p so it will be something b p plus c upon
00:23:59.520 --> 00:24:09.460
p square plus four so compare the coefficients
from the both side we will this implies p
00:24:09.460 --> 00:24:18.440
plus two will equals to this into this that
is a into p square plus four and plus b p
00:24:18.440 --> 00:24:28.550
plus c into p minus one now we have to compute
the values of a b and c first so we can make
00:24:28.550 --> 00:24:35.270
use of compare the coefficients of the both
sides here the coefficient of p square is
00:24:35.270 --> 00:24:48.260
a from here it is b and here there is no term
involving p square so a plus b will be zero
00:24:48.260 --> 00:24:57.150
and the coefficient of p here is zero and
the coefficient of p here is minus b plus
00:24:57.150 --> 00:25:06.670
c and here it is one so it will be one and
the constant here is four a minus c that here
00:25:06.670 --> 00:25:10.830
constant is two
so we are having three equations with three
00:25:10.830 --> 00:25:18.170
unknowns so we can find the values of a
b and c so a here we have already obtained
00:25:18.170 --> 00:25:29.820
actually so a is three by five b is minus
three by five and c is two by five this can
00:25:29.820 --> 00:25:37.660
easily compute by solving these three equations
ok so once we obtained the values of a b and
00:25:37.660 --> 00:25:51.100
c then it is nothing but sorry it is nothing
but three by five times this it is minus
00:25:51.100 --> 00:26:06.060
three plus two upon five now the laplace inverse
of this f p will be nothing but three by five
00:26:06.060 --> 00:26:13.870
times laplace inverse of one upon p minus
one minus three by five times laplace inverse
00:26:13.870 --> 00:26:23.960
of p upon p square plus four plus two by five
times laplace inverse of one upon p square
00:26:23.960 --> 00:26:30.070
plus four
so this is nothing but three upon five now
00:26:30.070 --> 00:26:36.200
laplace inverse of one upon t minus a e k
power a t here a is one so laplace inverse
00:26:36.200 --> 00:26:43.440
of one upon p minus a t is e k power t minus
three by five laplace inverse of this is cos
00:26:43.440 --> 00:26:51.750
two t ok so it will be cos two t a is two
and here the for this it is plus two by five
00:26:51.750 --> 00:26:59.180
laplace inverse of this expression a is two
so it will be nothing but one by two times
00:26:59.180 --> 00:27:09.960
sin two t so the final answer is e k power
three t e into three e k power t minus three
00:27:09.960 --> 00:27:21.020
cos two t this two cancels out plus sign two
t upon five
00:27:21.020 --> 00:27:27.230
so we have seen that how we can laplace transform
of some standard functions like sin a t cos
00:27:27.230 --> 00:27:35.130
a t sin hyperbolic a t or cos hyperbolic a
t and using these laplace transforms how
00:27:35.130 --> 00:27:40.400
we can find out laplace transform of some
functions or inverse laplace transforms
00:27:40.400 --> 00:27:40.850
thanks