WEBVTT
Kind: captions
Language: en
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yeah hello every one so in the last lecture
we have seen that what is the existence theorem
00:00:24.550 --> 00:00:31.410
for laplace transforms we have seen that if
a function is a piece wise continuous and
00:00:31.410 --> 00:00:39.470
is of exponential order alpha then its laplace
transform always exists otherwise if function
00:00:39.470 --> 00:00:46.110
is not piece wise continuous or is not of
exponential order then then laplace may or
00:00:46.110 --> 00:00:55.400
may not exist that also we have seen so now
in this topic we will see what are the
00:00:55.400 --> 00:01:05.280
properties of laplace transforms so we will
start with the first property that we call
00:01:05.280 --> 00:01:11.100
as shifting property now shifting property
is first we have first translation and
00:01:11.100 --> 00:01:16.460
then we have second translation of the shifting
property what is first translation first translation
00:01:16.460 --> 00:01:31.219
means if laplace transform of f t is say f
p then if this holds then laplace transform
00:01:31.219 --> 00:01:42.799
of e ki power a t into f t is nothing but
f of p minus a that means if we know the laplace
00:01:42.799 --> 00:01:49.200
transform f t which is given as f p if you
know this then laplace transform of e ki a
00:01:49.200 --> 00:01:56.460
t into f t is nothing but you simply replace
p by p minus a so then we will get the laplace
00:01:56.460 --> 00:02:03.619
transform of this function e ki power at into
f t now what are the proof of this result
00:02:03.619 --> 00:02:08.190
how will obtain this result now what is
lapalce transform of e ki power a t into f
00:02:08.190 --> 00:02:18.260
t by the definition this is nothing but zero
to infinity e ki power a t f t into e ki power
00:02:18.260 --> 00:02:23.931
minus p t d t this is by the definition of
laplace transform
00:02:23.931 --> 00:02:37.470
so this is equal to zero to infinity e ki
power minus p minus a times t into f t d t
00:02:37.470 --> 00:02:46.340
so this is nothing but zero to infinity suppose
suppose p minus a is some xi of f t where
00:02:46.340 --> 00:03:02.950
xi is nothing but p minus a then this quantity
is nothing but f of xi because what is f p
00:03:02.950 --> 00:03:09.760
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
00:03:09.760 --> 00:03:16.069
p we have xi in the same expression so we
replace f as xi and what is xi xi is p minus
00:03:16.069 --> 00:03:25.900
a so it is p minus a so hence laplace of e
ki power a t into f t is nothing but f
00:03:25.900 --> 00:03:35.590
of p minus a so simply replace p by p minus
a now the now second translation or the
00:03:35.590 --> 00:03:46.280
shifting property is if laplace transform
of f t is f p and g t is given as f of t minus
00:03:46.280 --> 00:04:01.180
a when t greater than a and zero when t less
than a then laplace transform of g t is nothing
00:04:01.180 --> 00:04:17.489
but e ki power minus a p times f p again
again we can prove this using the definition
00:04:17.489 --> 00:04:23.960
of laplace transform laplace transform of
g t is nothing but we have to prove this result
00:04:23.960 --> 00:04:35.580
as so this is equal to zero to infinity e
ki power minus p t g t d t by the definition
00:04:35.580 --> 00:04:43.890
of laplace transforms
this is equal to zero to a e ki power minus
00:04:43.890 --> 00:05:01.770
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
00:05:01.770 --> 00:05:08.810
because g t is zero and t less than a so it
is zero so this quantity zero plus because
00:05:08.810 --> 00:05:16.710
g t is zero so integration value will be zero
plus so this is a to infinity e ki power minus
00:05:16.710 --> 00:05:24.560
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
00:05:24.560 --> 00:05:42.870
here d t now this expression so therefore
therefore what is laplace of g t therefore
00:05:42.870 --> 00:05:53.940
laplace of g t will be nothing but from here
we got a into infinity e ki power minus p
00:05:53.940 --> 00:06:07.990
t f of t minus a into d t now now you can
take t minus a equal to some new variable
00:06:07.990 --> 00:06:16.910
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
00:06:16.910 --> 00:06:24.190
tend into infinity z tend into infinity e
ki power minus p t from here a is a plus z
00:06:24.190 --> 00:06:34.461
so it is a plus z times f z into d z so this
is nothing but since e ki power minus p a
00:06:34.461 --> 00:06:40.949
is free from z we are integrating it with
respect to z so we can take it outside the
00:06:40.949 --> 00:06:52.449
integration e ki power minus a p zero to infinity
e ki power minus p z f z d z which is same
00:06:52.449 --> 00:07:00.919
as e ki power minus a p so this is nothing
but laplace transform of f t or f z so this
00:07:00.919 --> 00:07:11.389
is f p this is nothing but laplace transform
of f t so e ki power minus a p into f p so
00:07:11.389 --> 00:07:16.319
hence laplace of g t is nothing but e ki power
minus a p into f p so this is called second
00:07:16.319 --> 00:07:25.139
translation or shifting property
now to how we can solve problems based
00:07:25.139 --> 00:07:32.449
on this let us solve few examples on this
the laplace transform of t cube e ki power
00:07:32.449 --> 00:07:42.620
three t
so so we have just discussed that if laplace
00:07:42.620 --> 00:07:53.889
transform of f t is f p then laplace transform
of e ki power a t into f t is nothing but
00:07:53.889 --> 00:08:06.889
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
00:08:06.889 --> 00:08:18.699
so what is laplace transform of t cube it
is it is gamma four upon p ki power four
00:08:18.699 --> 00:08:24.110
this is nothing but factorial three upon p
ki power four which is nothing but six upon
00:08:24.110 --> 00:08:36.000
p ki four so six upon p ki power four so this
is f p now you have to compute laplace transform
00:08:36.000 --> 00:08:47.540
of e ki power three into t cube so simply
replace f p in f p p by p minus a and a here
00:08:47.540 --> 00:08:54.720
is three so that is nothing but f of p minus
three so that is nothing but six upon p minus
00:08:54.720 --> 00:09:03.880
three ki power four so that will be the laplace
transform of t cube e ki power three t
00:09:03.880 --> 00:09:16.710
next problem suppose suppose the next problem
next problem is e ki power minus four t cos
00:09:16.710 --> 00:09:26.529
hyperbolic two t so again here if you compare
with this so f t is cos hyperbolic two t so
00:09:26.529 --> 00:09:34.611
first find laplace of cos hyperbolic two t
let us put equal to f p and then replace p
00:09:34.611 --> 00:09:42.670
by p minus a here a is minus four so replace
p by p plus four so what is laplace transform
00:09:42.670 --> 00:09:50.260
of cos hyperbolic two t we already know that
laplace transform of cos hyperbolic a t is
00:09:50.260 --> 00:09:59.440
p upon p square minus a square so it is p
upon p square minus four so laplace transform
00:09:59.440 --> 00:10:06.870
of e ki power so this if f p this is f p so
laplace transform of e ki power minus four
00:10:06.870 --> 00:10:14.690
t into cos hyperbolic two t will be nothing
but f of p you replace p by p minus a a is
00:10:14.690 --> 00:10:24.240
minus four so p plus four so that is nothing
but p plus four upon p plus four whole square
00:10:24.240 --> 00:10:29.020
minus four
so that this will be the laplace transform
00:10:29.020 --> 00:10:35.880
of the e ki power minus four t cos hyperbolic
two t now the third problem suppose you want
00:10:35.880 --> 00:10:45.300
to solve the third problem it is e ki power
minus t sin square t so first find laplace
00:10:45.300 --> 00:10:53.760
transform of sin square t put it equal to
f p and replace p by p minus a and here is
00:10:53.760 --> 00:11:00.930
a minus one so replace p by p plus one so
first find lapalce transform of sin square
00:11:00.930 --> 00:11:13.640
t so sin square t is nothing but one minus
cos two t by two so this is nothing but one
00:11:13.640 --> 00:11:22.180
by two laplace transform of one minus one
by two laplace transform of cos two t so laplace
00:11:22.180 --> 00:11:29.670
transform of one is one by p minus one by
two laplace transform of cos two t is p upon
00:11:29.670 --> 00:11:40.420
p square plus a square so p upon p square
plus four so this equal to f p now laplace
00:11:40.420 --> 00:11:49.610
transform of e ki power minus t square t will
be nothing but f of you replace p by p minus
00:11:49.610 --> 00:11:59.620
a a is minus one that is p plus one so that
is nothing but one upon two p plus one minus
00:11:59.620 --> 00:12:08.329
p plus one upon two into p plus one whole
square plus four so that will be the laplace
00:12:08.329 --> 00:12:12.870
transform of e ki power minus t sin square
t
00:12:12.870 --> 00:12:27.699
again in the last problem we can use shifting
property because what is it is sin hyperbolic
00:12:27.699 --> 00:12:39.680
three t into cos square t it is nothing but
laplace of e ki power three t minus e ki power
00:12:39.680 --> 00:12:50.550
three t by two cos square t so it is one by
two laplace transform of e ki power three
00:12:50.550 --> 00:12:59.720
t cos square t minus one by two laplace transform
of e ki power minus three t into cos square
00:12:59.720 --> 00:13:11.430
t so we will find laplace transform of cos
square t put it equal to f p and then in the
00:13:11.430 --> 00:13:19.070
first part replace p by p minus three and
for the second part replace p by p plus three
00:13:19.070 --> 00:13:23.540
so what is laplace transform of cos square
t that again we can find out cos square t
00:13:23.540 --> 00:13:32.640
will be nothing but laplace transform of one
plus cos two t by two that is nothing but
00:13:32.640 --> 00:13:39.110
one by two laplace transform of one is one
by p so that is one by p plus one by two laplace
00:13:39.110 --> 00:13:49.500
transform of cos two t is p upon p square
plus a square so p upon p square plus four
00:13:49.500 --> 00:14:02.370
so this is f p now what is laplace transform
of this e ki power three t cos square t
00:14:02.370 --> 00:14:09.149
this is nothing but f of p minus a and what
is laplace transform of e ki power minus three
00:14:09.149 --> 00:14:18.870
t cos square t this is nothing but f of p
plus three so substituting these two values
00:14:18.870 --> 00:14:24.110
you replace p by p minus three for the first
part and you replace p by p plus three for
00:14:24.110 --> 00:14:29.051
the second part and you substitute these values
here so we will get the laplace transform
00:14:29.051 --> 00:14:32.250
of sin hyperbolic three t into cos square
t
00:14:32.250 --> 00:14:47.329
now let us find laplace transform of t sin
four t so sin four t doesnt involve as
00:14:47.329 --> 00:14:59.779
such doesnt involve e ki power a t but we
can break it so laplace of t sin four t so
00:14:59.779 --> 00:15:09.139
laplace of t now sin four t can be written
as e ki power four iota t minus e ki power
00:15:09.139 --> 00:15:19.940
minus four iota t upon two iota this is how
we can replace sin four t now this is nothing
00:15:19.940 --> 00:15:31.290
but one by two i laplace transform of t into
e ki power four iota t minus one by two i
00:15:31.290 --> 00:15:40.410
laplace transform of t into e ki power minus
four iota t so again here here a is four iota
00:15:40.410 --> 00:15:47.290
if you compare with this and here a is minus
four iota so you find laplace of t first let
00:15:47.290 --> 00:15:54.440
it put it equal to f p and for the first expression
replace p by p minus four iota and for second
00:15:54.440 --> 00:16:02.730
expression replace p by p plus four iota so
what is laplace transform of t what is laplace
00:16:02.730 --> 00:16:10.920
transform of t laplace of t is nothing but
one by p square it is equals to f p and for
00:16:10.920 --> 00:16:19.779
this part it is nothing but one upon two iota
f of p minus four iota minus one by two iota
00:16:19.779 --> 00:16:26.779
f of p plus four iota by by this shifting
property
00:16:26.779 --> 00:16:33.300
so this is nothing but one upon two iota what
is f of p minus four iota one upon p minus
00:16:33.300 --> 00:16:41.800
four iota whole square minus one upon p plus
four iota whole square so the simplify it
00:16:41.800 --> 00:16:47.449
one upon two iota into so this will go here
this will go here simplify it so this is nothing
00:16:47.449 --> 00:16:57.649
but two a b two into four iota into p two
a b into two again two one two again comes
00:16:57.649 --> 00:17:04.690
then it is nothing but p square plus sixteen
whole square this two iota will cancel with
00:17:04.690 --> 00:17:14.000
two iota and this is nothing but eight p upon
p square plus sixteen whole square so this
00:17:14.000 --> 00:17:24.830
will be the laplace transformation of t sin
four t now now this problem on shifting
00:17:24.830 --> 00:17:36.530
property lapalce transform of e ki power minus
two t sin under root t so again we first we
00:17:36.530 --> 00:17:42.500
find out laplace transform of sin under
root t and using shifting property we will
00:17:42.500 --> 00:17:50.110
replace p by p plus two because here a is
minus two so what is laplace transform of
00:17:50.110 --> 00:17:59.300
sin under root t let us see so what is what
is sin theta sin theta is theta minus theta
00:17:59.300 --> 00:18:06.300
cube upon factorial three plus theta ki five
upon factorial five and so on what is sin
00:18:06.300 --> 00:18:13.280
under root t sin under root will be t ki power
half minus t ki power three by two upon factorial
00:18:13.280 --> 00:18:21.080
thee plus t ki power five by two upon factorial
three factorial five and so on
00:18:21.080 --> 00:18:28.150
now it is laplace of sin under root t will
be nothing but will be nothing but laplace
00:18:28.150 --> 00:18:33.890
transform of this entire expression and by
the linearity property we can split in term
00:18:33.890 --> 00:18:42.820
wise so this is nothing but laplace transform
of laplace transform of t ki power half
00:18:42.820 --> 00:18:49.950
minus one by factorial three laplace transform
of t ki power three by two plus one by factorial
00:18:49.950 --> 00:18:59.050
five laplace transform f t ki power five by
two so this is t ki power half is gamma three
00:18:59.050 --> 00:19:09.500
by two upon p ki power three by two using
laplace transform of t ki power n minus
00:19:09.500 --> 00:19:18.370
one by factorial three it is gamma five by
two upon p ki power five two plus one by factorial
00:19:18.370 --> 00:19:27.930
five gamma seven by two upon p ki power seven
by two and so on so this is gamma three by
00:19:27.930 --> 00:19:35.390
two is because gamma n plus one is gamma n
so gamma three by two will be half gamma half
00:19:35.390 --> 00:19:44.030
and gamma half is under root pi so it is one
by two under root pi p ki power three by two
00:19:44.030 --> 00:19:52.150
minus one by factorial three it is nothing
but three by two one by two under root pi
00:19:52.150 --> 00:19:59.080
again the property of gamma function p ki
power five by two plus one by factorial five
00:19:59.080 --> 00:20:10.700
it is five by two three by two one by two
under root pi p ki power seven by two
00:20:10.700 --> 00:20:22.560
so what finally we get so laplace transform
of sin under root t this will be nothing but
00:20:22.560 --> 00:20:26.930
now you can take one by two under root pi
upon this quantity common form of all the
00:20:26.930 --> 00:20:36.700
terms so this is one by two p ki power three
by two into under root pi inside bracket we
00:20:36.700 --> 00:20:44.880
get one minus so this is factorial three is
three into two into one three three cancels
00:20:44.880 --> 00:20:51.280
out under root two this we have taken common
one so this is nothing but this is nothing
00:20:51.280 --> 00:21:05.470
but one by two square into p whole power
upon one factorial because one two comes from
00:21:05.470 --> 00:21:10.800
here one to from here one two is common so
it is two square from here five into five
00:21:10.800 --> 00:21:18.150
cancels out so it is nothing but plus one
by two square p whole square upon factorial
00:21:18.150 --> 00:21:28.690
two this comes when you simplify five five
cancels out three also cancels so we will
00:21:28.690 --> 00:21:35.040
get two square whole square upon factorial
two and this is under root two p ki power
00:21:35.040 --> 00:21:44.700
three by two and it is nothing but e ki power
minus one by four p so that is the laplace
00:21:44.700 --> 00:21:52.600
transform of sin under root t so this is f
p and for to find laplace transform of
00:21:52.600 --> 00:21:57.840
laplace transform of e ki power minus two
t you simply replace p by a p plus two in
00:21:57.840 --> 00:22:04.901
this expression this is f p you replace p
by p plus two you will get laplace transform
00:22:04.901 --> 00:22:12.840
of this quantity
again using second translation property
00:22:12.840 --> 00:22:18.460
we can easily find out laplace transform of
g t because if you compare with second shifting
00:22:18.460 --> 00:22:25.050
property second translation so laplace transform
of f t is f p then laplace transform g
00:22:25.050 --> 00:22:32.270
t is e ki power minus a p times f p by
second translation property so if we compare
00:22:32.270 --> 00:22:41.680
with this here here in the first problem
what is g t what is f t minus a now f of p
00:22:41.680 --> 00:22:55.540
minus a for the first problem f t will be
t cube then only it is t minus a a is one
00:22:55.540 --> 00:23:03.910
whole cube so what is laplace transform of
f t it is nothing but factorial three upon
00:23:03.910 --> 00:23:14.270
p ki power four that is six upon p ki power
four now the laplace transform this g t will
00:23:14.270 --> 00:23:26.410
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
00:23:26.410 --> 00:23:34.560
p is the this quantity so this is nothing
but e ki power minus p e ki power six into
00:23:34.560 --> 00:23:40.870
e ki power minus p upon p ki power four so
this will be the laplace transform of the
00:23:40.870 --> 00:23:48.160
first problem by the second second translation
property again for the for the second part
00:23:48.160 --> 00:24:01.220
you see that here here f t is this quantity
if you compare with the this property here
00:24:01.220 --> 00:24:11.170
f t is the f t given here in this term f t
here is cos t so the laplace of cos t is
00:24:11.170 --> 00:24:17.670
p upon p square plus one
so laplace of this f t will be nothing but
00:24:17.670 --> 00:24:23.990
using second translation property laplace
of ft will be nothing but e ki power minus
00:24:23.990 --> 00:24:32.590
a p a is two pi by three e ki power minus
two pi by three p into p into laplace of cos
00:24:32.590 --> 00:24:42.920
t laplace of cos t is nothing but p upon p
square plus one so this is how we can find
00:24:42.920 --> 00:24:50.840
out laplace transform of this problems
using second translation property now the
00:24:50.840 --> 00:24:59.060
same shifting property also hold for inverse
laplace transforms inverse laplace if inverse
00:24:59.060 --> 00:25:05.420
laplace of f t is f p then inverse laplace
transform of e minus a is e ki power a t into
00:25:05.420 --> 00:25:13.180
f t and the second shifting property states
that if laplace inverse of f p is f t then
00:25:13.180 --> 00:25:19.410
laplace inverse of e ki power minus a p into
f p f p is nothing but f of t minus a when
00:25:19.410 --> 00:25:28.460
t is greater than zero a t less than a this
is by the shifting property itself now
00:25:28.460 --> 00:25:37.170
based on this let us try to solve this problems
so the first problem is laplace of three p
00:25:37.170 --> 00:25:50.750
upon plus two upon p square minus two p plus
five so laplace of this is three times p upon
00:25:50.750 --> 00:25:59.590
p square minus two p plus five plus two times
laplace oh laplace inverse it is laplace inverse
00:25:59.590 --> 00:26:10.950
yeah so it is laplace inverse of one upon
p square minus two p plus five so it is three
00:26:10.950 --> 00:26:19.210
laplace inverse of p upon it is p minus one
whole square plus four it is two times laplace
00:26:19.210 --> 00:26:27.610
inverse of one upon p minus one whole square
plus four
00:26:27.610 --> 00:26:36.630
now here p minus so you subtract one and add
one here so this is three laplace inverse
00:26:36.630 --> 00:26:46.110
of p minus one upon p minus one whole square
plus four and three into one upon this quantity
00:26:46.110 --> 00:26:52.010
plus two into one upon this quantity will
be five time laplace inverse of one upon p
00:26:52.010 --> 00:27:06.600
minus one whole square plus four now now
now we know that laplace inverse of f p is
00:27:06.600 --> 00:27:19.900
f t if this happens then laplace inverse of
f p is minus a will be nothing but e ki power
00:27:19.900 --> 00:27:30.370
a t into f t by the inverse of property shifting
property now here if you use this definition
00:27:30.370 --> 00:27:40.660
f p minus a here a is one so it is nothing
but three e ki power a t a is one so three
00:27:40.660 --> 00:27:46.850
into e ki power t and f t and f t will be
laplace inverse of f p so it will be laplace
00:27:46.850 --> 00:27:58.930
inverse of p upon p square plus four you replace
you replace p plus one ok so it will be p
00:27:58.930 --> 00:28:05.920
upon p square plus four so that will be laplace
inverse of this quantity again plus you replace
00:28:05.920 --> 00:28:14.110
p by by this property is nothing but five
into e ki power t laplace inverse of one upon
00:28:14.110 --> 00:28:20.750
p square plus four so this is nothing but
three into e ki power t this is cos two t
00:28:20.750 --> 00:28:32.900
plus five into e ki power t you divide multiply
into two so it is two upon this sin two t
00:28:32.900 --> 00:28:38.241
so that will be the laplace inverse of this
expression that will be laplace transform
00:28:38.241 --> 00:28:48.860
of this expression similarly when you apply
the same thing in second problem laplace
00:28:48.860 --> 00:29:00.890
inverse of this so it is two p plus one ok
it is one upon one upon under root four p
00:29:00.890 --> 00:29:07.990
plus five second problem so you take four
common first so it is one by two laplace inverse
00:29:07.990 --> 00:29:19.490
of one upon under root p plus five by four
use this shifting property now here a is minus
00:29:19.490 --> 00:29:29.300
minus five by four so it will be one by
two e ki power minus five by four t laplace
00:29:29.300 --> 00:29:38.410
inverse of one by under root p by the shifting
property of for inverse laplace transforms
00:29:38.410 --> 00:29:45.320
and this is again is equal to one by two e
ki power minus five by four t we know that
00:29:45.320 --> 00:29:52.380
now for to find laplace inverse of this expression
we know that laplace transform of t ki power
00:29:52.380 --> 00:29:59.140
n is nothing but gamma n plus one upon p ki
power n plus one
00:29:59.140 --> 00:30:04.360
so laplace inverse of one upon p ki power
n plus one is nothing but t ki power n upon
00:30:04.360 --> 00:30:14.220
gamma n plus one so replace n by n by minus
half so it is t ki power minus half upon gamma
00:30:14.220 --> 00:30:21.800
half so that is nothing but one by two e ki
power minus five by four t five by four into
00:30:21.800 --> 00:30:31.440
t into one by under root pi into t so that
will be the laplace inverse of this expression
00:30:31.440 --> 00:30:38.260
now to solve the last problem we can use partial
fractions this we use partial fraction to
00:30:38.260 --> 00:30:45.440
simply this and then we apply the shifting
property to find out the laplace inverse of
00:30:45.440 --> 00:30:53.200
this expression the third problem we can solve
now to find out laplace inverse of this
00:30:53.200 --> 00:30:58.630
two problems again we will use second shifting
second translation property for inverse laplace
00:30:58.630 --> 00:31:05.690
transforms first we will find laplace inverse
of here we have problem e ki power e ki power
00:31:05.690 --> 00:31:14.400
minus p pi by three upon p square plus one
so first find laplace inverse of one upon
00:31:14.400 --> 00:31:24.460
p square plus one so that is nothing but sin
t now using second second translation property
00:31:24.460 --> 00:31:33.860
laplace inverse of e ki power minus p pi
by three upon p square plus one will be nothing
00:31:33.860 --> 00:31:51.010
but sin t minus pi by three when t greater
than pi by three and zero t when pi by three
00:31:51.010 --> 00:31:56.710
and this is because by the by the second
property by the second translation property
00:31:56.710 --> 00:32:01.010
we have seen that laplace inverse of this
quantity is nothing but f of t minus a when
00:32:01.010 --> 00:32:09.700
t greater than a zero t when less than a here
here a is minus pi by three so you
00:32:09.700 --> 00:32:23.330
replace t minus pi by three in this f t
this is f t this is f t ok now for second
00:32:23.330 --> 00:32:30.230
the last problem again we will first find
laplace inverse of p plus one upon one p square
00:32:30.230 --> 00:32:38.440
p plus one and then using second shifting
property we will find laplace inverse of
00:32:38.440 --> 00:32:48.040
entire expression so what is
so first we will find laplace inverse of this
00:32:48.040 --> 00:32:58.370
expression p plus one upon p square plus
p plus one we will call it f t and then for
00:32:58.370 --> 00:33:07.830
this entire expression we replace we replace
t by t minus pi the laplace inverse of this
00:33:07.830 --> 00:33:26.040
will be nothing but minus p pi this will be
nothing but f of f of t plus t minus pi
00:33:26.040 --> 00:33:31.010
when t greater than pi and zero when t less
than pi
00:33:31.010 --> 00:33:39.570
so this we can we can obtain by using second
property so laplace of inverse of this we
00:33:39.570 --> 00:33:49.080
can find making perfect square when the denominator
let us to make it this is p plus half whole
00:33:49.080 --> 00:33:59.440
square plus three by four and this is nothing
but laplace inverse of p plus half upon p
00:33:59.440 --> 00:34:12.320
plus half whole square plus three by four
plus one by two laplace inverse of one upon
00:34:12.320 --> 00:34:21.329
p plus half whole square plus three by four
and it is e ki power minus one by two t by
00:34:21.329 --> 00:34:33.450
using shifting property p plus half plus
half is one and it is one by two square plus
00:34:33.450 --> 00:34:42.250
yeah its ok e ki power e ki power a t a
is minus half minus half is laplace inverse
00:34:42.250 --> 00:34:51.450
of p upon p square plus under root three by
two whole square plus one by two e ki power
00:34:51.450 --> 00:34:58.140
minus one by two t laplace inverse of one
upon p square plus under root by three by
00:34:58.140 --> 00:35:05.260
two whole square now this we can find out
this is nothing but cos a t a is under
00:35:05.260 --> 00:35:09.690
root three by two you multiply and divide
by this quantity here this will be nothing
00:35:09.690 --> 00:35:18.730
but sin a t and a is under root three by
two so that will be f t and you replace
00:35:18.730 --> 00:35:23.620
you substitute f t here so you will get laplace
inverse of this expression so
00:35:23.620 --> 00:35:24.410
thank you very much