WEBVTT
Kind: captions
Language: en
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welcome to the series of lectures on mathematical
methods and applications so we have seen so
00:00:24.820 --> 00:00:30.119
many thing about laplace transforms we have
seen what laplace transforms are what are
00:00:30.119 --> 00:00:34.739
the important properties of laplace transform
how can you find out laplace transform of
00:00:34.739 --> 00:00:41.129
some important functions that inverse etcetera
in the last lecture we have seen what unit
00:00:41.129 --> 00:00:46.640
step function is how to find out laplace transform
unit step of function and some problem based
00:00:46.640 --> 00:00:50.920
on it
now in this lecture we will see what are what
00:00:50.920 --> 00:00:56.210
is dirac delta functions and how to find
out laplace transform dirac delta function
00:00:56.210 --> 00:01:03.201
and some problems based on it so first
first thing is what is dirac delta function
00:01:03.201 --> 00:01:09.770
or we call it unit impulse function so it
is defined like this derived delta at t equal
00:01:09.770 --> 00:01:20.220
to a is nothing but one upon epsilon where
a is tending to t tending from a to a plus
00:01:20.220 --> 00:01:35.880
epsilon and zero otherwise
and where epsilon tend to zero so this
00:01:35.880 --> 00:01:42.840
is how we define dirac delta function so
if we roughly speaking if epsilon tend to
00:01:42.840 --> 00:01:48.590
zero so t equal to a and this will tends to
infinity
00:01:48.590 --> 00:01:55.490
so basically if we speak about a definition
of dirac delta function so we can say that
00:01:55.490 --> 00:02:02.460
at t equal to a that is at one point it is
having infinite value and it and all other
00:02:02.460 --> 00:02:10.250
values at for all other points it is zero
ok so whenever we would solve such type
00:02:10.250 --> 00:02:16.880
of problems or based on dirac delta we always
use this definition that is it is one by epsilon
00:02:16.880 --> 00:02:23.019
when t is varying from a to a plus epsilon
zero otherwise and where epsilon tend to zero
00:02:23.019 --> 00:02:29.360
so what what does this definition means
this means when a is varying from a to a to
00:02:29.360 --> 00:02:37.040
a plus epsilon where epsilon tend to zero
sent then this value is one upon epsilon so
00:02:37.040 --> 00:02:45.819
this value is one so a to a plus epsilon this
value is one and otherwise this value is zero
00:02:45.819 --> 00:02:53.040
so this is what the graph of dirac delta
function is this is t this is dirac delta
00:02:53.040 --> 00:03:00.010
function s t equal to t minus a so this is
how we define we can define dirac delta function
00:03:00.010 --> 00:03:08.720
so now the important property of delta function
is that zero to infinity the integration of
00:03:08.720 --> 00:03:18.330
dirac delta function is always one so that
we can easily show zero to infinity this
00:03:18.330 --> 00:03:28.129
integral is equal to now upon the definition
of dirac delta function it is it is one
00:03:28.129 --> 00:03:33.689
by epsilon when t varying from a to a plus
epsilon so and zero otherwise so this integral
00:03:33.689 --> 00:03:39.700
has a value only when t varying from a to
a plus epsilon and it is one by of seven d
00:03:39.700 --> 00:03:48.120
epsilon d t and limit epsilon tend to zero
because we epsilon tend to zero so this is
00:03:48.120 --> 00:03:55.660
nothing but limit epsilon tend to zero one
by epsilon integration a to a plus epsilon
00:03:55.660 --> 00:04:07.980
d t and this is further equal to limit delta
epsilon tend to zero one by epsilon t a to
00:04:07.980 --> 00:04:16.440
a plus epsilon and it is equals to limit epsilon
tend to zero one by epsilon a plus epsilon
00:04:16.440 --> 00:04:26.430
minus a which is nothing but one so that means
the total area covered by this function the
00:04:26.430 --> 00:04:33.770
dirac delta function from zero to infinity
the total area is always one so we can always
00:04:33.770 --> 00:04:40.530
define dirac delta function like this that
this value is one by epsilon when t varying
00:04:40.530 --> 00:04:46.090
a to a plus epsilon zero otherwise and that
we are epsilon tend to zero and the total
00:04:46.090 --> 00:04:51.740
area covered by that dirac delta function
is always one ok
00:04:51.740 --> 00:04:57.310
the next property is a we call it filtering
property of dirac delta function it states
00:04:57.310 --> 00:05:03.270
that if f t be is a continuous and integrable
in zero to infinity then this integral is
00:05:03.270 --> 00:05:11.330
always equal to f a so this also so we can
prove easily let us try to prove this result
00:05:11.330 --> 00:05:20.889
so what this property is it is zero
to infinity f t into dirac delta of t minus
00:05:20.889 --> 00:05:28.040
a d t is nothing but f t so this property
is call it filtering property of dirac
00:05:28.040 --> 00:05:33.770
delta function so how we will prove it now
we will apply the definition of delta function
00:05:33.770 --> 00:05:43.000
so the left hand side is zero to infinity
f t into dirac delta of t at a into d t which
00:05:43.000 --> 00:05:52.860
is equals to now limit epsilon tend to zero
a to a plus epsilon it is one by epsilon f
00:05:52.860 --> 00:06:02.639
t into d t this is by a definition of delta
function because dirac delta function attains
00:06:02.639 --> 00:06:08.430
a value one by epsilon when t varying from
a to a plus epsilon and epsilon tend to zero
00:06:08.430 --> 00:06:21.349
ok now this is equal to limit this tend to
zero a to a plus epsilon into f t d t upon
00:06:21.349 --> 00:06:30.300
epsilon ok
now when epsilon tend to zero this is a to
00:06:30.300 --> 00:06:37.539
a that is zero and this is zero this is zero
by zero form and function is continuous ok
00:06:37.539 --> 00:06:43.229
and integrable so we can apply l hospital
rule to find out the value because it is
00:06:43.229 --> 00:06:49.730
zero by zero form so we can apply l hospital
rule so it is equal to limit epsilon tend
00:06:49.730 --> 00:07:03.910
to zero d by s epsilon of a to a plus epsilon
f t d t upon one we differentiate respect
00:07:03.910 --> 00:07:11.160
of epsilon the numerator and the denominator
ok now this is equal to limit epsilon tend
00:07:11.160 --> 00:07:16.740
to zero now when you take d by d epsilon of
this function with respect of epsilon so this
00:07:16.740 --> 00:07:28.750
is nothing but f t a plus epsilon into one
minus zero upon one ok and when epsilon tend
00:07:28.750 --> 00:07:37.039
to zero it is nothing but f a so this is called
filtering property of dirac delta function
00:07:37.039 --> 00:07:44.169
that integration zero to infinity f t into
dirac delta t minus a d t is nothing but f
00:07:44.169 --> 00:07:49.319
a ok
so this how we can prove this the next property
00:07:49.319 --> 00:07:56.620
of dirac delta is the laplace transform of
dirac delta is nothing but e to the power
00:07:56.620 --> 00:08:08.509
minus a t so laplace transform of dirac delta
function so it is nothing but zero to infinity
00:08:08.509 --> 00:08:17.460
e to the power minus p t dirac delta of t
minus a into d t by definition of laplace
00:08:17.460 --> 00:08:23.551
transforms now by the definition dirac delta
function it is equal to limit epsilon tend
00:08:23.551 --> 00:08:33.979
to zero a to a plus epsilon into e to the
power minus p t into one by epsilon into d
00:08:33.979 --> 00:08:45.750
t and which is which is further equal to limit
epsilon tend to zero a to a plus epsilon e
00:08:45.750 --> 00:08:55.850
to the power minus p t d t upon epsilon ok
now it is again zero by zero form because
00:08:55.850 --> 00:09:00.220
there are epsilon tend to zero the numerator
is zero and the denominator is zero so zero
00:09:00.220 --> 00:09:05.721
by zero form so again you will apply l hospital
rule to simplify this so this is [no/nothing]
00:09:05.721 --> 00:09:12.330
nothing but equal to limit epsilon tend to
zero again d by d epsilon of the numerator
00:09:12.330 --> 00:09:23.429
quantity upon denominator is one the derivative
denominator respect to epsilon is one and
00:09:23.429 --> 00:09:30.750
it is nothing but limit epsilon tend to zero
e to the power minus p in place of t we have
00:09:30.750 --> 00:09:38.900
the upper limit which is a plus epsilon and
when into one minus zero of course upon one
00:09:38.900 --> 00:09:46.450
and when seven tend to zero it is nothing
but e to the power minus a p so that is how
00:09:46.450 --> 00:09:51.770
we can find out laplace transform of dirac
delta function and it is nothing but e to
00:09:51.770 --> 00:10:00.300
the power minus a p and of course when a is
zero the laplace transform of delta t is nothing
00:10:00.300 --> 00:10:06.620
but one because when you substitute a equal
to one in this expression so we will get
00:10:06.620 --> 00:10:13.699
laplace transform of dirac delta function
at t equal to zero is nothing but one ok
00:10:13.699 --> 00:10:19.140
now let us solve some problems based on dirac
delta function now whenever we want to find
00:10:19.140 --> 00:10:27.390
out laplace transform of some f t into dirac
delta at t equal to a suppose so this will
00:10:27.390 --> 00:10:35.049
be equal to zero to infinity e to the power
minus p t f t into dirac delta t minus at
00:10:35.049 --> 00:10:41.830
t equal to a t and when you apply a definition
of definition of dirac delta function it
00:10:41.830 --> 00:10:51.910
is nothing but limit delta tend to zero a
to a plus epsilon into one by epsilon e to
00:10:51.910 --> 00:11:00.530
the power minus p t into f t d t and which
is again equal to ok epsilon tend to zero
00:11:00.530 --> 00:11:09.049
limit epsilon tend to zero integration a to
plus epsilon e to the power minus p t f t
00:11:09.049 --> 00:11:17.270
d t upon epsilon so it will become it will
it will always become zero by zero form so
00:11:17.270 --> 00:11:22.310
you will apply l hospital rule to simplify
this expression and then you will you can
00:11:22.310 --> 00:11:28.179
get the laplace transform of f t into dirac
delta or t equal to a ok so this is how we
00:11:28.179 --> 00:11:32.569
can find out laplace transform of f t into
dirac delta function
00:11:32.569 --> 00:11:39.731
so now let us try to find out laplace transform
of these problems first is laplace transform
00:11:39.731 --> 00:11:51.150
of unit step function at t equal to one into
dirac delta t equal to one ok this so this
00:11:51.150 --> 00:12:00.680
is nothing but zero to infinity u one t
that is f t f t is a entire expression into
00:12:00.680 --> 00:12:07.870
e to the power minus p t into d t which is
further equal to first apply definition of
00:12:07.870 --> 00:12:15.040
dirac delta function ok by definition of dirac
delta function is nothing but limit epsilon
00:12:15.040 --> 00:12:26.240
tend to zero one by epsilon a to a plus epsilon
u ok here a is one so this will be one only
00:12:26.240 --> 00:12:37.580
and this is one ok and u one t into e to
the power minus p t into d t now apply definition
00:12:37.580 --> 00:12:45.280
of u one t so u one so u one t at t equal
to t more than one by definition of unit
00:12:45.280 --> 00:12:48.890
step function is one
so it is nothing but limit epsilon tend to
00:12:48.890 --> 00:12:59.809
zero one by epsilon one to one plus epsilon
into e to the power minus p t d t because
00:12:59.809 --> 00:13:05.559
u one t is one when t is greater than one
otherwise it is zero so now it is nothing
00:13:05.559 --> 00:13:14.640
but limit epsilon tend to zero one by epsilon
ok it it is now zero by zero form because
00:13:14.640 --> 00:13:21.650
you can always write it like this so it
is zero by zero form ok so you have to apply
00:13:21.650 --> 00:13:27.820
l hospital rule to simplify this so it is
nothing but limit epsilon tend to zero d by
00:13:27.820 --> 00:13:35.939
d epsilon of integral one to one plus epsilon
e to the power minus p t d t upon one and
00:13:35.939 --> 00:13:42.209
which is nothing but limit epsilon tend to
zero e to the power minus p and t is nothing
00:13:42.209 --> 00:13:49.560
but one plus epsilon and it is nothing but
e k power minus p so that will be the laplace
00:13:49.560 --> 00:13:56.100
transform of the first problem ok
now let us compute laplace transform of second
00:13:56.100 --> 00:14:04.530
problem it is again on dirac delta function
so the second problem is basically power t
00:14:04.530 --> 00:14:13.290
involved in this problem so it is laplace
transform of e k power t into dirac delta
00:14:13.290 --> 00:14:21.939
at t equal at a equal to three at a equal
to three so it is nothing but so i can write
00:14:21.939 --> 00:14:29.670
directly it is limit epsilon tend to zero
instead of a i have three here so it is three
00:14:29.670 --> 00:14:36.290
to three plus epsilon three to three plus
epsilon f t is e to the power minus e to the
00:14:36.290 --> 00:14:46.959
power t and e to the power minus p t is
is here also upon epsilon because it is a
00:14:46.959 --> 00:14:52.601
laplace transform and i directly applied definition
of dirac delta function so it is epsilon tend
00:14:52.601 --> 00:14:59.970
to zero three to three plus epsilon ok it
is limit limit epsilon tend to zero three
00:14:59.970 --> 00:15:04.880
to three plus epsilon e to the power three
into e to the power minus p t d t upon epsilon
00:15:04.880 --> 00:15:11.310
so it is zero by zero form so you will [app/again]again
apply l hospital rule to simplify this so
00:15:11.310 --> 00:15:18.130
it is or you can first integrate because
it is integrable it is one minus p into three
00:15:18.130 --> 00:15:25.630
plus epsilon upon one limit epsilon tends
to zero and it is nothing but e to the power
00:15:25.630 --> 00:15:35.390
three one minus p is it so so this is the
final expression which we can obtain using
00:15:35.390 --> 00:15:42.830
this by applying l hospital rule ok now the
next problem next problem is laplace transform
00:15:42.830 --> 00:15:56.020
of t square into u one t plus t into dirac
delta at t at t equal to two so this is nothing
00:15:56.020 --> 00:16:04.560
but laplace transform of t square into u one
t plus laplace transform of t into dirac delta
00:16:04.560 --> 00:16:10.439
at t equal to two
now to compute the first expression we will
00:16:10.439 --> 00:16:17.870
use second shifting property of unit step
function which states that laplace of f t
00:16:17.870 --> 00:16:27.650
minus into u t minus a is nothing but e
to the power minus a p f p where this f p
00:16:27.650 --> 00:16:36.930
is nothing but laplace of f t ok so we have
to break this t square in the powers of t
00:16:36.930 --> 00:16:43.720
minus one because it is f t minus a and a
is one so first we find first find the laplace
00:16:43.720 --> 00:16:48.350
of this then laplace of this the addition
of this two will give the laplace of the entire
00:16:48.350 --> 00:16:57.520
expression so laplace of t square into u one
t this can be find out laplace of i can write
00:16:57.520 --> 00:17:03.679
it like this t minus one plus one whole square
u one t
00:17:03.679 --> 00:17:17.789
this is nothing but laplace of a square plus
two a b plus b square
00:17:17.789 --> 00:17:27.500
and this is nothing but laplace of t minus
one whole square u one t plus two times laplace
00:17:27.500 --> 00:17:41.900
of t minus one u one t plus laplace of u one
t now using that that property it is nothing
00:17:41.900 --> 00:17:51.880
but e to the power minus a p a is one so it
is minus p into into laplace of f t so laplace
00:17:51.880 --> 00:18:03.390
of t square plus two into e to the power minus
p again and laplace of t plus and this is
00:18:03.390 --> 00:18:11.580
nothing but e to the power minus p into laplace
of one so this is nothing but e to the power
00:18:11.580 --> 00:18:20.640
minus p into t square is nothing but two upon
p cube plus two e to the power minus p into
00:18:20.640 --> 00:18:28.410
t is nothing but one p square plus e to the
power minus p into one by p so that is the
00:18:28.410 --> 00:18:33.710
laplace of the first expression laplace of
t square u one t
00:18:33.710 --> 00:18:38.850
now we will compute laplace of second expression
how we compute laplace of second expression
00:18:38.850 --> 00:18:54.880
it is laplace of t into dirac delta t minus
two ok so this can be written as zero to infinity
00:18:54.880 --> 00:19:02.930
t dirac delta t minus two e to the power minus
p t into d t and applying the definition of
00:19:02.930 --> 00:19:12.690
dirac delta function it is nothing but
limit epsilon tend to zero two to two plus
00:19:12.690 --> 00:19:27.350
epsilon t e to the power minus p t d t upon
epsilon so it is it is zero by zero form
00:19:27.350 --> 00:19:33.900
again we can apply l hospital rule to simplify
this expression so we will obtain this
00:19:33.900 --> 00:19:41.310
will be equal to limit epsilon tend to zero
two plus epsilon into e to the power minus
00:19:41.310 --> 00:19:49.191
two p into two plus epsilon into one minus
zero upon one so when you take epsilon tend
00:19:49.191 --> 00:19:58.400
to zero here it is nothing but two into e
to the power minus two p ok so the sum of
00:19:58.400 --> 00:20:04.560
these two expressions will give will give
us the epsilon the laplace transform of this
00:20:04.560 --> 00:20:14.250
expression ok
now the last problem of this slide
00:20:14.250 --> 00:20:22.700
laplace transform of laplace transform
of t square into dirac delta at t equal
00:20:22.700 --> 00:20:31.590
to t equal to one so this problem also can
be solved easily it is zero to infinity t
00:20:31.590 --> 00:20:40.920
square dirac delta minus one into e to the
power minus p t d t and when you apply definition
00:20:40.920 --> 00:20:50.860
of dirac delta it is nothing but limit epsilon
tend to zero one to one plus epsilon t square
00:20:50.860 --> 00:20:59.090
e to the power minus p t d t upon epsilon
now it is zero by zero form so you can directly
00:20:59.090 --> 00:21:04.800
apply l hospital rule to simplify this so
this is nothing but limit epsilon tend to
00:21:04.800 --> 00:21:13.890
zero one plus epsilon whole square e to the
power minus p into one plus epsilon minus
00:21:13.890 --> 00:21:20.830
zero upon one and when you take epsilon tend
to zero it is nothing but e to the power minus
00:21:20.830 --> 00:21:30.810
p so that will be the laplace transform of
t square into dirac delta at t equal to one
00:21:30.810 --> 00:21:38.630
ok so that is how we can solve the problems
based on dirac delta function i mean laplace
00:21:38.630 --> 00:21:46.860
to find out the function with dirac
delta function ok
00:21:46.860 --> 00:21:52.620
now let us evaluate the following integrals
zero to infinity which are having the involvement
00:21:52.620 --> 00:22:00.790
of dirac delta so we can directly use filtering
property of dirac delta function what are
00:22:00.790 --> 00:22:05.450
filtering property that we have already
proved so this is a filtering property of
00:22:05.450 --> 00:22:15.650
dirac delta function that is zero to infinity
f t into dirac delta of t minus a d t is nothing
00:22:15.650 --> 00:22:26.900
but f a ok so suppose want to solve first
problem the first problem is zero to infinity
00:22:26.900 --> 00:22:33.640
sin square t dirac delta at pi by four so
if we compare with this expression so f t
00:22:33.640 --> 00:22:40.550
is nothing but sin square t and a is nothing
but pi by four so directly using this filtering
00:22:40.550 --> 00:22:51.200
property this expression will be nothing but
sin square pi by four now sin pi by four is
00:22:51.200 --> 00:22:57.440
one by under root two it is whole square which
is nothing but one by two so directly using
00:22:57.440 --> 00:23:02.970
filtering property we can find out the value
of this problem
00:23:02.970 --> 00:23:11.300
now similarly if we take the second problem
the second problem is nothing but it is zero
00:23:11.300 --> 00:23:27.940
to infinity t square cos two t into dirac
delta at t equal to pi by two again if we
00:23:27.940 --> 00:23:35.090
compare this with this expression so f t is
nothing but t square cos two t and a is pi
00:23:35.090 --> 00:23:45.440
by two so by using this property this expression
will have the value equal to t square cos
00:23:45.440 --> 00:23:56.650
two t at t equal to pi by two so this nothing
but pi by two whole square cos pi and cos
00:23:56.650 --> 00:24:07.180
pi is minus one so it is minus pi square by
four it is minus pi square by four
00:24:07.180 --> 00:24:14.721
now let us see the third problem of the slide
that is zero to infinity it is zero to
00:24:14.721 --> 00:24:26.470
infinity f t into derived delta at t pi by
two d t and f t is defined like this f t is
00:24:26.470 --> 00:24:37.900
given to us it is it is zero when t lying
between zero and pi by two and it is sin t
00:24:37.900 --> 00:24:46.590
when t is greater than equal to pi by two
so basically what this function is it is zero
00:24:46.590 --> 00:25:00.450
into u zero t minus u pi by two t plus
sin t into u pi by two t so i express this
00:25:00.450 --> 00:25:10.250
function in terms of unit step function ok
so it is nothing but sin t into u at pi by
00:25:10.250 --> 00:25:20.710
two into t so now this is equal to by filtering
property this will be nothing but the value
00:25:20.710 --> 00:25:27.059
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
00:25:27.059 --> 00:25:35.650
f that is sin t into u pi by two into t at
t equal to pi by two
00:25:35.650 --> 00:25:39.720
now at t equal to pi by two this is one it
is a unit step function the value will be
00:25:39.720 --> 00:25:47.920
one so it is nothing but sin pi by two into
one which is nothing but one so hence we can
00:25:47.920 --> 00:25:54.620
find out the value of this expression so
we have seen that what is dirac delta function
00:25:54.620 --> 00:26:01.210
how we can solve some some problems
based on the based on based on this
00:26:01.210 --> 00:26:06.850
how we can solve these type of integral using
filtering property or delta function so
00:26:06.850 --> 00:26:12.200
in the next lecture we will see that what
are the various applications of the laplace
00:26:12.200 --> 00:26:13.380
transforms so
thank you