WEBVTT
Kind: captions
Language: en
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so welcome to lecture series on mathematical
methods and its applications so we have seen
00:00:23.840 --> 00:00:29.890
what fourier transforms are and also we have
seen fourier sin and cosine transforms so
00:00:29.890 --> 00:00:35.690
how can write fourier transforms function
f t fourier transform of a function f t
00:00:35.690 --> 00:00:46.730
is given by minus infinity to plus infinity
f t e k power minus eta omega t into
00:00:46.730 --> 00:00:52.870
d omega into d t ok and we are calling it
f omega function of omega
00:00:52.870 --> 00:01:01.260
so and the inverse fourier transform of f
omega is f f omega is given by it is one
00:01:01.260 --> 00:01:09.860
upon two pi integral minus infinity to plus
infinity that is f omega e k power eta omega
00:01:09.860 --> 00:01:17.120
t and it is d omega so which is nothing but
f t so this we have already discussed and
00:01:17.120 --> 00:01:22.770
this come from complex form of fourier integrals
this we have already seen that fourier
00:01:22.770 --> 00:01:26.640
transforms f t is nothing but given by this
expression and the inverse fourier transforms
00:01:26.640 --> 00:01:31.080
is given by this expression
now let us come to convolution theorem
00:01:31.080 --> 00:01:35.810
for fourier transforms so before discussing
convolution theorem let us discuss a more
00:01:35.810 --> 00:01:42.730
popular of fourier transforms so first of
all fourier transform of derivates ok so let
00:01:42.730 --> 00:01:50.099
f t be continuous on the x axis and f t tend
into zero as mod t tend into infinity now
00:01:50.099 --> 00:01:57.980
further suppose f dash t is absolutely integrable
on the x axis then fourier transform of f
00:01:57.980 --> 00:02:02.500
dash t is given by eta omega fourier transforms
of f t
00:02:02.500 --> 00:02:10.179
so this thing is very easy to proof fourier
transform of f dash t so by the definition
00:02:10.179 --> 00:02:17.570
of fourier transforms it is nothing but minus
infinity to plus infinity f dash t into
00:02:17.570 --> 00:02:24.880
e k power minus eta omega t into d t so
this will be nothing but let us suppose it
00:02:24.880 --> 00:02:32.310
is first function it is second function so
first as it is integral of second function
00:02:32.310 --> 00:02:41.549
from minus infinity to plus infinity and minus
integral derivate of first is minus eta omega
00:02:41.549 --> 00:02:49.920
into e k power minus eta omega and integral
of second which is f t into d t
00:02:49.920 --> 00:02:56.780
so this is by integration by parts now
as we have assumed that as f t as mod t tend
00:02:56.780 --> 00:03:02.720
to infinity f t is tend to zero that is when
t is tend to plus or minus infinity f t is
00:03:02.720 --> 00:03:07.480
tend to zero so from the upper and the lower
limit this is tend in to zero so this is nothing
00:03:07.480 --> 00:03:16.379
but zero map plus eta omega times minus infinity
to plus infinity e k power minus eta omega
00:03:16.379 --> 00:03:25.019
t into f t d t now what it is it is nothing
but fourier transform of f t so it is eta
00:03:25.019 --> 00:03:35.769
omega times fourier transform of f t
so so hence the fourier transform f dash
00:03:35.769 --> 00:03:41.499
t is nothing but eta omega times fourier transform
of f dash t now suppose you want to find out
00:03:41.499 --> 00:03:49.620
fourier transform of f double dash t phase
into the phase f pi f dash in this expression
00:03:49.620 --> 00:03:58.459
so that will be nothing but eta omega fourier
transform of f dash t which is equal to fourier
00:03:58.459 --> 00:04:01.870
transform of f dash t is the already derived
is equal to this expression so this is nothing
00:04:01.870 --> 00:04:10.889
but eta omega whole square fourier transform
of f t so in the similar way if we proceed
00:04:10.889 --> 00:04:17.420
for the nth derivative for the same expression
so we get fourier transform of nth derivative
00:04:17.420 --> 00:04:27.970
of t will be nothing but eta omega k to the
power n fourier transform of f t
00:04:27.970 --> 00:04:33.430
so nth derivative of fourier transform of
f will be nothing but eta omega k to the power
00:04:33.430 --> 00:04:39.110
n fourier transform of f t so this is how
we can obtain fourier transform of derivatives
00:04:39.110 --> 00:04:46.840
now let us solve these two problems find fourier
transform of f t where f t equal to this ok
00:04:46.840 --> 00:04:54.310
find fourier transform of f t actually so
we have to find fourier transform of t e k
00:04:54.310 --> 00:05:06.100
power minus t square ok now this is nothing
but fourier transform of e k power minus
00:05:06.100 --> 00:05:13.650
t square is derivative is derivative is nothing
but minus two t times this thing you multiply
00:05:13.650 --> 00:05:18.919
and divide by this expression so this will
be nothing but when you simplify this so you
00:05:18.919 --> 00:05:24.199
get back this expression
now by the fourier transform derivative we
00:05:24.199 --> 00:05:28.639
know that fourier transfer of f dash t is
nothing but eta omega times fourier transform
00:05:28.639 --> 00:05:36.650
of f t so this is nothing but minus half of
eta omega times fourier transform of e k power
00:05:36.650 --> 00:05:44.090
minus t square because here here f t is
e k power minus t square is derivative and
00:05:44.090 --> 00:05:50.050
is derivative is given by eta omega fourier
transform of f t ok so this is this expression
00:05:50.050 --> 00:05:59.720
now let us find fourier transform of e t square
so fourier transform of e t square fourier
00:05:59.720 --> 00:06:05.379
transform of e k power minus t square will
be given by the minus and plus infinity f
00:06:05.379 --> 00:06:15.009
t into e k power minus eta omega t d t which
is equal to minus and plus infinity e k power
00:06:15.009 --> 00:06:23.750
minus so when you make perfect square it is
t plus eta omega by two whole square a square
00:06:23.750 --> 00:06:32.470
plus two a b and minus eta omega by two
whole square into d t
00:06:32.470 --> 00:06:38.949
so which is equal to it is minus minus plus
it is e k power minus omega square by four
00:06:38.949 --> 00:06:45.440
can come out because it is free from t and
it is integral minus or plus infinity e k
00:06:45.440 --> 00:06:54.090
power minus t plus eta omega by two whole
square into d t now you can take t plus
00:06:54.090 --> 00:07:09.960
eta omega by two as as suppose under
root p ok i or you can take it as suppose
00:07:09.960 --> 00:07:16.729
you can take it as z first ok so then d t
will be nothing but d z so it will be e k
00:07:16.729 --> 00:07:23.449
power minus omega square by four minus and
plus infinity e k power minus z square into
00:07:23.449 --> 00:07:30.189
d z limit will remain the same ok
now its an even function so this can be written
00:07:30.189 --> 00:07:37.120
as two times e k power minus omega square
by four zero to infinity e k power minus z
00:07:37.120 --> 00:07:47.689
square d z ok now we can let z equal to suppose
under root p so this will give this integral
00:07:47.689 --> 00:07:57.960
will give this will give fourier two
e k power minus omega square by four integral
00:07:57.960 --> 00:08:06.909
zero to infinity one by two under root p e
k power minus p into d p ok which is nothing
00:08:06.909 --> 00:08:12.360
but two two cancel out and it is e k power
minus omega square by four and it is gamma
00:08:12.360 --> 00:08:19.009
half by that definition gamma function this
is nothing but under root pi e k power minus
00:08:19.009 --> 00:08:23.810
omega square by four
so therefore the value of the value of this
00:08:23.810 --> 00:08:30.759
is nothing but minus one by two eta omega
under root pi e k power minus omega square
00:08:30.759 --> 00:08:42.300
for ok so here we use the here we use the
concept of derivative of fourier transform
00:08:42.300 --> 00:08:47.430
and to find out fourier transform of e
k power minus t square we go back to the main
00:08:47.430 --> 00:08:53.380
definition of fourier transforms they are
the simplify and we get the fourier transform
00:08:53.380 --> 00:09:00.890
of e k power minus t square ok now suppose
you want to solve this problem ok the next
00:09:00.890 --> 00:09:11.890
problem it is y dash minus two y will be equal
to e k power minus two t u naught t ok take
00:09:11.890 --> 00:09:16.570
fourier transform both the sides what is the
fourier transform y dash by the formula it
00:09:16.570 --> 00:09:27.480
is eta omega f omega here f omega is a fourier
transform of y t minus two f omega and is
00:09:27.480 --> 00:09:32.080
equal to you we know we already that fourier
transform of e k power minus a t u naught
00:09:32.080 --> 00:09:37.320
t u naught t is a unit step function at t
equal to zero we already know that the fourier
00:09:37.320 --> 00:09:44.500
transform of this is nothing but one upon
two plus eta omega because fourier transform
00:09:44.500 --> 00:09:49.410
of e k power minus a t u naught t is one upon
a plus eta omega ok
00:09:49.410 --> 00:09:55.420
so what we obtain from here we obtain that
f omega is nothing but one upon eta omega
00:09:55.420 --> 00:10:03.210
minus two into two plus eta omega when they
simplify this so this is nothing but a square
00:10:03.210 --> 00:10:12.120
minus b square and which is equals to minus
of one upon omega square plus four now we
00:10:12.120 --> 00:10:21.080
have also seen that fourier transform e
k power minus a mod t is nothing but two
00:10:21.080 --> 00:10:28.000
a upon a square plus omega square that we
have already derived ok so it is something
00:10:28.000 --> 00:10:35.420
like one upon omega square plus two square
so you can multiply and divide by four so
00:10:35.420 --> 00:10:40.570
minus four into four upon omega square plus
four
00:10:40.570 --> 00:10:49.660
so instead of a we have two so this is nothing
but so so if we take inverse fourier
00:10:49.660 --> 00:10:56.330
transform both the sides so y t will be nothing
but minus one by four and inverse of this
00:10:56.330 --> 00:11:02.630
will be given by that it is e k power minus
two into mod t so that will be the solution
00:11:02.630 --> 00:11:10.540
of this expression this i mean differential
equation ok so we take fourier transform both
00:11:10.540 --> 00:11:16.660
the sides simplify and the inverse of fourier
transform will give the solution of this differential
00:11:16.660 --> 00:11:24.460
equation ok
now differentiation respect to frequency
00:11:24.460 --> 00:11:30.062
omega let us suppose functions is piece wise
continuous on the x axis and let t k to the
00:11:30.062 --> 00:11:38.300
power n f t be absolutely integrable on the
x axis then then we have a result for fourier
00:11:38.300 --> 00:11:50.550
transforms then then fourier transform
of t k to the power n f t that is a multiplication
00:11:50.550 --> 00:12:02.070
of t is nothing but eta k power n and nth
derivative respect to omega of f omega ok
00:12:02.070 --> 00:12:15.010
nth derivative of omega with f omega ok
now now to proof this the proof is simple
00:12:15.010 --> 00:12:21.170
you take d by d omega d omega of f omega now
f omega is a fourier transform of f t what
00:12:21.170 --> 00:12:29.940
is nothing but d by d omega of minus to
plus infinity f t e k power minus eta omega
00:12:29.940 --> 00:12:37.520
t into d t because fourier transform f t is
given by this expression and f omega is nothing
00:12:37.520 --> 00:12:44.310
but fourier transform of f t so this will
be given by by the lagrange theorem it is
00:12:44.310 --> 00:12:52.790
nothing but del by del omega of f t into
e k power minus eta omega t into d t which
00:12:52.790 --> 00:13:00.860
is nothing but when you take derivative
respect to omega it is nothing but minus eta
00:13:00.860 --> 00:13:11.940
t into f t e k power minus eta omega t into
d t now minus eta will come out and the
00:13:11.940 --> 00:13:17.370
and this it is nothing but t f t keep our
minus eta omega t that will be nothing but
00:13:17.370 --> 00:13:26.010
fourier transform of t f t
so here so therefore f dash omega will be
00:13:26.010 --> 00:13:31.970
nothing but this expression now when you multiply
with eta both the sides so this is nothing
00:13:31.970 --> 00:13:42.640
but eta f omega will be equals to fourier
transform of t f t so hence fourier transform
00:13:42.640 --> 00:13:49.970
of this is nothing but this expression for
n equal to one so similarly if you take
00:13:49.970 --> 00:13:58.630
second derivate suppose you take second derivate
d two upon d w t of f f w so that will
00:13:58.630 --> 00:14:07.220
be nothing but by the same by the same
concept it is nothing but d by d w of integral
00:14:07.220 --> 00:14:15.610
minus into plus infinity first we take
first derivate by this expression del by
00:14:15.610 --> 00:14:26.080
del w of f t e k power minus eta omega t into
d t so that will be given by d by d w of minus
00:14:26.080 --> 00:14:34.370
infinity to plus infinity eta t f t e k power
minus eta omega t
00:14:34.370 --> 00:14:43.060
again when you take this d by d w inside and
you differentiate with respect to omega so
00:14:43.060 --> 00:14:50.160
that will be nothing but that will be nothing
but that will nothing but minus infinity
00:14:50.160 --> 00:15:01.100
to plus infinity it is minus eta square
t square f t e k power minus eta omega t into
00:15:01.100 --> 00:15:10.540
d t which is eta square times integral
minus into plus infinity t square f t e k
00:15:10.540 --> 00:15:18.540
power minus eta omega t d t and this is nothing
but minus fourier transform of t square f
00:15:18.540 --> 00:15:27.110
t ok this is nothing but fourier transform
of this ok
00:15:27.110 --> 00:15:40.370
now ok this implies fourier transform of t
square f t will be nothing but eta square
00:15:40.370 --> 00:15:48.700
times double derivative of f here we have
a single derivative ok here we have a single
00:15:48.700 --> 00:15:56.040
derivative and here we have a double derivative
so so that means similarly if you repeat
00:15:56.040 --> 00:16:01.680
this process n times so we will get back
this expression that is a fourier transform
00:16:01.680 --> 00:16:12.080
of t k power n f t will be nothing but
eta k power n f f omega nth derivative
00:16:12.080 --> 00:16:17.760
of f omega ok so in this way we will get by
this expression
00:16:17.760 --> 00:16:25.380
now using this suppose you want to find
out fourier transform of t e k power minus
00:16:25.380 --> 00:16:34.380
two mod t so that will be nothing but using
this expression it is nothing but eta k
00:16:34.380 --> 00:16:44.170
power one because n is one into fourier transform
of into derivative into d by d omega of f
00:16:44.170 --> 00:16:54.090
omega here f omega is the fourier transform
of this f t it is e k power minus two mod
00:16:54.090 --> 00:17:02.110
t and what is the fourier transform of this
it is two a upon four plus omega square
00:17:02.110 --> 00:17:10.089
so this will be nothing but eta d by d omega
of four upon four plus omega square which
00:17:10.089 --> 00:17:17.470
is nothing but four eta into minus one upon
four plus omega square the whole square into
00:17:17.470 --> 00:17:23.130
two omega
so that will be nothing but minus eight omega
00:17:23.130 --> 00:17:29.991
eta upon four plus omega square the whole
square so this will be the fourier transform
00:17:29.991 --> 00:17:40.559
of this expression now dirac delta function
we already dirac delta function in laplace
00:17:40.559 --> 00:17:45.549
transform that it is given by one by epsilon
twenty varying from a to a plus epsilon and
00:17:45.549 --> 00:17:50.769
zero otherwise and we are epsilon ten to zero
so roughly speaking dirac delta functions
00:17:50.769 --> 00:17:58.890
is infinity at a point i say t equal to a
and zero otherwise such that the total integral
00:17:58.890 --> 00:18:03.890
from minus infinity to plus infinity of dirac
delta function is one that we have already
00:18:03.890 --> 00:18:09.059
discussed in the laplace transforms
now the fourier transform of dirac delta function
00:18:09.059 --> 00:18:15.190
is given by this so this can be derived using
the definition of dirac delta function i mean
00:18:15.190 --> 00:18:23.679
dirac delta and fourier a transform so to
find fourier transform of dirac delta function
00:18:23.679 --> 00:18:33.600
it is nothing but minus to plus infinity f
t f t is the delta into e k power minus eta
00:18:33.600 --> 00:18:41.360
omega t into d t now dirac delta is given
by one by epsilon when t varying from a to
00:18:41.360 --> 00:18:46.690
a plus epsilon here epsilon trying to zero
so this we can define like this limit epsilon
00:18:46.690 --> 00:18:54.710
tend to zero a to a plus epsilon it is one
by epsilon into e k power minus eta omega
00:18:54.710 --> 00:18:59.820
t into d t
now it is limit epsilon trying to zero one
00:18:59.820 --> 00:19:06.519
by epsilon can come out and the integration
of this will be nothing but this term from
00:19:06.519 --> 00:19:16.010
a to a plus epsilon now it is nothing but
limit epsilon tend to zero ok one upon minus
00:19:16.010 --> 00:19:23.970
eta omega can come here and it is e k power
minus eta omega a plus epsilon minus e k power
00:19:23.970 --> 00:19:30.470
minus eta omega a by applying upper limit
minus lower limit now when epsilon tend to
00:19:30.470 --> 00:19:36.740
zero it is zero by zero form so you apply
a rule to simply this so you will take the
00:19:36.740 --> 00:19:42.450
derivate of the numerator and the denominator
so limit epsilon tend to zero the derivate
00:19:42.450 --> 00:19:52.409
of numerator will be nothing but it is
minus eta omega e k power minus eta omega
00:19:52.409 --> 00:20:01.789
a plus omega minus zero upon minus eta omega
derivate is respect to omega i mean epsilon
00:20:01.789 --> 00:20:04.800
ok
now these two terms cancels out and when you
00:20:04.800 --> 00:20:11.080
take epsilon tend to zero it is nothing but
e k power minus eta omega so which is the
00:20:11.080 --> 00:20:16.940
same as this expression now when you take
a equal to zero it is clear that when you
00:20:16.940 --> 00:20:23.059
take a equal to zero fourier transform of
dirac delta t at a equal to zero is nothing
00:20:23.059 --> 00:20:33.200
but one so this is very clear from this expression
when you substitute a equal to c
00:20:33.200 --> 00:20:38.559
now come back to convolution theorem now
the convolution of two function define in
00:20:38.559 --> 00:20:43.779
a same way as we did in a laplace transform
now here integral is from minus to plus
00:20:43.779 --> 00:20:50.860
infinity so convolution of two functions in
fourier transform is given by this so convolution
00:20:50.860 --> 00:21:00.940
of two functions in fourier transform is given
by minus to plus infinity f tau g t minus
00:21:00.940 --> 00:21:10.980
tau d tau or can be written as minus into
plus infinity f t minus tau g tau d tau
00:21:10.980 --> 00:21:18.100
because this convolution of two functions
is commutative ok
00:21:18.100 --> 00:21:23.140
now come come to convolution theorem now
what is states convolution theorem fot fourier
00:21:23.140 --> 00:21:29.860
transform suppose that f t and g t are piecewise
continuous bound it and absolutely integrable
00:21:29.860 --> 00:21:36.520
on the x axis then fourier transform of convolution
of two functions f and g will be given by
00:21:36.520 --> 00:21:43.320
fourier transform of f into fourier transform
of g which is nothing but f omega into g omega
00:21:43.320 --> 00:21:48.840
so we are we are fourier transform of f
t is f omega and fourier transform of g t
00:21:48.840 --> 00:21:53.279
is g omega
so now we will see the proof of convolution
00:21:53.279 --> 00:22:03.970
theorem so basically want to find out fourier
transform of f star g ok so by definition
00:22:03.970 --> 00:22:12.090
it is nothing but minus to plus infinity
convolution of this this function into e k
00:22:12.090 --> 00:22:18.450
power minus eta omega t d t this is by the
definition of fourier transform fourier transform
00:22:18.450 --> 00:22:25.870
of function f t is given by minus to plus
infinity f t k power minus eta omega t d t
00:22:25.870 --> 00:22:32.960
now convolution of two function this this
term is given by you replace this term as
00:22:32.960 --> 00:22:42.440
minus into plus infinity suppose f tau g t
minus tau d tau and whole multiplied by e
00:22:42.440 --> 00:22:49.990
k power minus eta omega t into d t
so this can be written as minus to plus infinity
00:22:49.990 --> 00:23:02.879
minus to plus infinity f tau g t minus tau
e k power minus eta omega t d tau into d t
00:23:02.879 --> 00:23:11.129
ok now now change the order of integration
both the limits are constant minus and plus
00:23:11.129 --> 00:23:15.279
infinity so change the order of integration
so when you change the order of integration
00:23:15.279 --> 00:23:24.299
is nothing but minus plus infinity minus into
plus infinity f tau g t minus tau e k power
00:23:24.299 --> 00:23:35.179
minus eta omega t d t into d tau
now suppose now suppose t minus tau as a new
00:23:35.179 --> 00:23:42.899
variable suppose z or d t will be d z ok so
when you substitute this variable over here
00:23:42.899 --> 00:23:50.560
in this expression so what we get we get so
basically we are simplifying fourier transform
00:23:50.560 --> 00:23:56.889
of convolution of two function so that will
be equal to this expression and this will
00:23:56.889 --> 00:24:03.740
further be equal to minus infinity to plus
infinity minus infinity to plus infinity this
00:24:03.740 --> 00:24:14.539
is f tau and this is g z e k power minus eta
w and t is nothing but tau plus z from this
00:24:14.539 --> 00:24:22.340
expression and d t is d z and d tau is d tau
so this is nothing but now you can write it
00:24:22.340 --> 00:24:30.490
like this minus to plus infinity f tau e k
power minus eta omega tau t tau because now
00:24:30.490 --> 00:24:36.480
you can separate two variables two via are
separating tau we can separately and z we
00:24:36.480 --> 00:24:49.200
can write separately so we can always do this
and this is nothing but fourier transform
00:24:49.200 --> 00:24:59.499
of f t and this is nothing but fourier transform
of g t so this is nothing but f f omega
00:24:59.499 --> 00:25:06.909
into g omega so hence hence fourier transform
of convolution of two functions f and g is
00:25:06.909 --> 00:25:16.190
nothing but f omega into g omega ok so this
is convolution theorem and here is a proof
00:25:16.190 --> 00:25:23.510
now from here it is also cleared at when the
fourier transform of f star g is f omega g
00:25:23.510 --> 00:25:32.000
omega so from here we can write that f star
g into t is nothing but fourier inverse
00:25:32.000 --> 00:25:42.120
of f omega into g omega now inverse fourier
transform is given by one upon two pi integral
00:25:42.120 --> 00:25:48.940
minus infinity to plus infinity the omega
function which is f omega into g omega e k
00:25:48.940 --> 00:25:54.700
power eta omega t into d omega
so that we will already know by definition
00:25:54.700 --> 00:26:01.940
of inverse fourier transforms ok now let us
find fourier inverse fourier transform
00:26:01.940 --> 00:26:10.210
of this f omega using convolution theorem
ok so how you can do that so we will recall
00:26:10.210 --> 00:26:15.661
convolution theorem again and using convolution
theorem we will try to find out the fourier
00:26:15.661 --> 00:26:23.299
inverse fourier transform of this function
so what is a function now what what we
00:26:23.299 --> 00:26:34.669
are find out fourier inverse of one upon six
plus five eta omega minus omega square so
00:26:34.669 --> 00:26:43.159
so this is f omega so this is f omega
so f omega is nothing but one upon six plus
00:26:43.159 --> 00:26:54.350
five eta omega minus omega square ok so
in our to apply convolution theorem we have
00:26:54.350 --> 00:27:01.899
to write as a product of two omega functions
and the convolution and the in i mean fourier
00:27:01.899 --> 00:27:07.840
inverse of both the omega functions we
should know so that convolution theorem we
00:27:07.840 --> 00:27:15.669
can apply ok so let us try to write this function
so we can easily write it as one upon six
00:27:15.669 --> 00:27:24.559
plus five eta omega plus eta square omega
square and this is nothing but we can simplify
00:27:24.559 --> 00:27:32.320
when we simplify so this is a square plus
five a plus six so this is eta omega plus
00:27:32.320 --> 00:27:41.739
two into eta omega plus three
so this is nothing but one upon eta omega
00:27:41.739 --> 00:27:50.519
plus two into one upon eta omega plus three
so this is something like g omega and this
00:27:50.519 --> 00:27:57.139
is something like h omega so we have write
this f omega as a product of two omega
00:27:57.139 --> 00:28:06.820
functions now to find out its inverse using
convolution theorem so f inverse of f omega
00:28:06.820 --> 00:28:16.110
which is equals to f inverse of g omega into
h omega will be equal to so we will apply
00:28:16.110 --> 00:28:22.299
convolution theorem it is nothing but convolution
of this and this so what is what is fourier
00:28:22.299 --> 00:28:31.700
inverse of g omega so fourier inverse of
g omega we know that fourier inverse of
00:28:31.700 --> 00:28:39.700
this is nothing but e k power minus two t
u naught t this we already know and fourier
00:28:39.700 --> 00:28:49.980
inverse of this expression one upon
fourier inverse of h omega h omega is this
00:28:49.980 --> 00:28:57.149
is nothing but e k power minus three t u naught
t ok this is by the this you already know
00:28:57.149 --> 00:29:03.659
that fourier transform of e k power minus
a t u naught t is one upon a plus eta omega
00:29:03.659 --> 00:29:08.930
ok
fourier transform of f omega which is equal
00:29:08.930 --> 00:29:16.610
to fourier inverse of fourier inverse transform
of f omega into g omega here using convolution
00:29:16.610 --> 00:29:22.029
theorem will be nothing but convolution of
these two function e k power minus two t u
00:29:22.029 --> 00:29:32.669
naught t e star with u e e k power minus
three t into u naught t the convolution of
00:29:32.669 --> 00:29:39.559
these two functions ok and that will be nothing
but minus infinity to plus infinity f tau
00:29:39.559 --> 00:29:50.399
that is e k power minus two tau u naught tau
into g t minus tau that is e minus three t
00:29:50.399 --> 00:30:01.649
minus tau u t minus tau and it is d t ok
this is by the convolution theorem because
00:30:01.649 --> 00:30:13.480
f tau into g t minus tau ok now let us simplify
this let us simplify this now here we have
00:30:13.480 --> 00:30:20.070
two unique step functions here and here so
let us find what it is so it is u naught tau
00:30:20.070 --> 00:30:29.510
into u t minus tau so now we have to see d
tau tau is a variable basically so we have
00:30:29.510 --> 00:30:37.720
to see respect to tau so when tau is less
than when tau is less than t ok and suppose
00:30:37.720 --> 00:30:45.200
greater than zero so when tau is less than
t this quantity is positive because t tau
00:30:45.200 --> 00:30:52.239
is less than t ok and this is also positive
one into one is one and when tau is greater
00:30:52.239 --> 00:30:57.919
than t so this is zero because this is
negative so this will be zero
00:30:57.919 --> 00:31:05.320
so when we apply this over here so this
will be equal to now e k power minus three
00:31:05.320 --> 00:31:15.269
t is free from tau can be come out and it
is e k power minus minus plus three tau into
00:31:15.269 --> 00:31:26.549
e k power minus two tau is e k power e k power
tau and this product is one when tau varying
00:31:26.549 --> 00:31:34.440
from zero to t otherwise it is zero so it
is from zero to tau from zero to t only and
00:31:34.440 --> 00:31:45.289
it is d tau ok so this will be equal to e
k power minus three t e k power tau zero to
00:31:45.289 --> 00:31:52.610
t which is equal to e k power minus three
t e k power t minus one so that will be the
00:31:52.610 --> 00:32:01.109
inverse fourier transform of this function
this f omega using convolution theorem
00:32:01.109 --> 00:32:09.210
so hence hence using convolution theorem
we can find out we can find out inverse
00:32:09.210 --> 00:32:17.009
fourier transforms also ok so the main application
of convolution theorem is to find out the
00:32:17.009 --> 00:32:22.259
fourier inverse so sometimes we have to find
out fourier inverse so it is important to
00:32:22.259 --> 00:32:27.639
use convolution theorem so so thats all
so
00:32:27.639 --> 00:32:28.320
thank you very much