WEBVTT
Kind: captions
Language: en
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so welcome to lecture series on mathematical
methods and its applications now now we will
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discuss fourier sine and cosine transforms
we have already seen what fourier transforms
00:00:30.550 --> 00:00:35.470
are we have we have seen that fourier transform
is given by fourier transform of function
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f f t is given by minus into plus infinity
f t e k power minus eta omega t dt which
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is f omega and its inverse its inverse which
is f omega inverse of f omega is f t which
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is given by one upon two pi integral minus
to plus infinity it is a f omega e k
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power eta omega t t omega so that is how we
can find out inverse fourier transform
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and fourier transform of ft
00:01:24.979 --> 00:01:33.630
now sine and cosine transforms before discussing
sine and cosine transform let us discuss shifting
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property of fourier transform also so suppose
fourier transform fts of omega and omega naught
00:01:39.750 --> 00:01:45.049
is belongs to r then fourier transform of
this is the very simple property e k power
00:01:45.049 --> 00:01:52.829
eta omega naught t into f t let us find fourier
transform of this so fourier transform of
00:01:52.829 --> 00:01:58.530
this is nothing but minus into plus infinity
you apply a definition ok this definition
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e k power eta omega naught t into f t into
e k power minus eta omega t dt so this is
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given by minus to plus infinity f t e k power
minus eta omega minus omega naught into t
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into dt
so in this expression instead of omega we
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have omega minus omega naught so this is but
f of omega minus omega naught so this is called
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a shifting property i mean i mean frequency
shifting property of a fourier transforms
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so this we can use some time while solving
some problems ok now next is fourier cosine
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transform the fourier cosine transform function
f t is defined as zero to infinity f t cos
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omega tdt now which is given as fourier cosine
transform of f t now it comes from basically
00:02:59.489 --> 00:03:03.519
from fourier cosine integral representation
of function f t what is the fourier cosine
00:03:03.519 --> 00:03:08.349
integral representation of function f t it
is f t is equals to one upon pi integral zero
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to infinity it is a omega cos omega d t omega
where a omega is given by this expression
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ok
now if you compare one and three so a omega
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is nothing but two times f c omega when you
compare these two ok so when you substitute
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it here so we obtained f t equal to this which
is the which is called inverse fourier cosine
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transform of f t so what is a fourier cosine
transform and what is this inverse so basically
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fourier cosine transform of f t it is given
by it is zero to infinity f t e k power minus
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eta omega t ok which is a fourier cosine transform
of ok because ok it is cosine so it contains
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cosine terms it is f t into cos omega t dt
which we are calling as fourier cosine transform
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of it is ag we are calling as fc omega ok
we are calling it as fc omega ok now the inverse
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of this is given by inverse of a inverse of
a fourier cosine transform is given by it
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is two upon pi integral zero to infinity
f t it is fc omega here
00:04:56.730 --> 00:05:06.861
it is fc omega cos omega t dt or d omega so
this will be the inverse fourier transform
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of inverse fourier cosine transform of
f t ok now similarly if we define fourier
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sine transform so fourier sine transform is
given by zero to infinity f t sin omega t
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dt which we denote as fx of omega and it comes
on fourier sine internal representation of
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f t because we know that fourier sine integral
representation of a function f t is given
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by this expression where b omega is this now
see if compare this with this expression so
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b omega is nothing but two times fx omega
ok so if you substitute this this expression
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b omega two time fx omega over here sorry
over here so it will be two upon pi times
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zero to infinity fx omega sine omega t dt
so what is a fourier sine integral representation
00:06:00.630 --> 00:06:05.960
of function fourier sine integral representation
of function f t it will be nothing but if
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it is zero to infinity f t sin omega t d t
which we are given as fourier sine integral
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representation zero to pi by this expression
and the inverse of a fourier inverse of a
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fourier sine transform of f t will be nothing
about two upon pi times integral zero to infinity
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fx omega sin omega t into d omega which is
nothing but f t ok
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so now let us find out fourier sine and cosine
transform of this function ok so suppose you
00:06:48.230 --> 00:06:59.390
find if fx is what fx is given as k when
zero less than x less than a and it is zero
00:06:59.390 --> 00:07:07.490
when x is greater than a so suppose is i want
to find out fourier cosine transform this
00:07:07.490 --> 00:07:18.910
function so this will be equal to zero to
infinity f t ft is ok f t cos omega t d t
00:07:18.910 --> 00:07:27.480
which is equals to zero to a it is k into
cos omega t d t which is equals to k now when
00:07:27.480 --> 00:07:36.810
we integrate this is sin omega t upon omega
from zero to a nothing but k upon omega sine
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omega here so this will be this is basically
fourier cosine transform of this function
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f t
now similarly if you want to find out fourier
00:07:49.390 --> 00:07:54.390
sine transform this function so fourier sine
transform this function f t will be nothing
00:07:54.390 --> 00:08:06.810
but again zero to infinity f t sin omega t
into dt which is equals to zero to a f t ft
00:08:06.810 --> 00:08:16.310
is k sin omega t into dt which is equal to
k can come out it is minus cos omega t
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upon minus upon omega from zero to a it is
minus k upon omega times it is cos omega a
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to minus one so it is k upon omega times one
minus cos a so this will be the fourier sine
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transform of this function f t
so in this way we can find out fourier cosine
00:08:42.479 --> 00:08:51.250
or fourier sine transform of any function
f t using these expressions ok now fourier
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transforms also satisfy some properties
what are they now first of all the existence
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condition if f x is absolutely integrable
on the positive side of x axis and piecewise
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continuous in every finite interval then the
fourier cosine and sine transform of f exists
00:09:11.529 --> 00:09:16.879
so this is the condition that function must
be absolutely integrable on the positive side
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of x axis and it must be piecewise continuous
then fourier sine and cosine transform of
00:09:22.740 --> 00:09:27.779
function will exist now the property is the
first property is fourier cosine and sine
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transform also satisfy linearity property
so this property all satisfy that we can easily
00:09:34.300 --> 00:09:44.129
is the c also that is easy to prove
basically what is what is fourier cosine
00:09:44.129 --> 00:09:50.860
transform of suppose a f plus bg it is nothing
but by a definition it is zero to infinity
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a f plus b g into cos omega t d t so that
will be nothing but a times zero to infinity
00:10:02.759 --> 00:10:10.790
f into cos omega t d t this has to be the
function of t these are function of t plus
00:10:10.790 --> 00:10:22.269
b times zero to infinity g cos omega t into
dt so this is nothing but a times fc of fc
00:10:22.269 --> 00:10:34.389
of f t and this is plus b times fc of gt
so hence fourier cosine transform satisfy
00:10:34.389 --> 00:10:40.420
linearity properties similarly we can we can
show that a fourier sine transform also satisfy
00:10:40.420 --> 00:10:46.689
linearity property ok the first thing the
second thing is cosine and sine transform
00:10:46.689 --> 00:10:50.640
derivatives
now if function is continuous and absolutely
00:10:50.640 --> 00:10:57.089
integrable on the positive side of x axis
and f dash is piecewise continuous on each
00:10:57.089 --> 00:11:04.589
finite interval ok and suppose fx tends to
zero fx tends to infinity then these two results
00:11:04.589 --> 00:11:15.180
hold now let us write these results so what
is fourier cosine transform of f dash x that
00:11:15.180 --> 00:11:22.050
will be nothing but a definition or power
f t ft or fx you can say anything that will
00:11:22.050 --> 00:11:33.860
be a zero to infinity f dash t into cos omega
t into dt this is by the definition of fourier
00:11:33.860 --> 00:11:41.560
cosine transform you simply replace f by f
dash in this expression ok so that will be
00:11:41.560 --> 00:11:47.689
given by is equal to now you apply integration
by parts you take it first function you take
00:11:47.689 --> 00:11:56.100
a second function first as it is now integration
of second it is f t from zero to infinity
00:11:56.100 --> 00:12:04.870
minus integration derivative of first is omega
sin omega t with negative sign the positive
00:12:04.870 --> 00:12:16.170
negative positive and it is f t dt
now when you take upper limit now this we
00:12:16.170 --> 00:12:23.529
are assumed that as s tending to infinity
or t tending to infinity fx tend to zero so
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as t tend to infinity this will tend to zero
so zero into some finite quantity will be
00:12:28.809 --> 00:12:37.079
zero and that will be it is nothing but
zero minus f zero because when t is zero
00:12:37.079 --> 00:12:49.250
it is f zero plus omega times integral zero
to infinity it is a sin omega t into f t dt
00:12:49.250 --> 00:12:56.339
ok so this is nothing but omega will come
here this omega and this is nothing but fourier
00:12:56.339 --> 00:13:03.330
sine transform of f t ok this is fourier sine
transform of f t so this is fourier sine transform
00:13:03.330 --> 00:13:12.639
of f t minus f zero so this is the derivation
for the first part that the fourier derivative
00:13:12.639 --> 00:13:18.850
of fourier cosine transform of f will be nothing
but omega times fourier sine transform of
00:13:18.850 --> 00:13:24.439
f t minus f zero ok
now second part fourier sine transform of
00:13:24.439 --> 00:13:35.499
f dash x so fourier sine transform of f dash
f for t is nothing but zero to infinity f
00:13:35.499 --> 00:13:46.600
dash t into sin omega t dt this is by a
definition of fourier sine transform you simply
00:13:46.600 --> 00:13:53.949
replace f by f dash t in this expression now
you again apply integration by parts it is
00:13:53.949 --> 00:14:03.379
one it is two now it is first as it is integration
of second from zero to infinity minus integral
00:14:03.379 --> 00:14:17.889
derivative of first is omega cos omega t into
f f t dt so when when t tend to zero when
00:14:17.889 --> 00:14:24.439
t tend to infinity f t tend to zero this is
this is by the assumption ok so this is tend
00:14:24.439 --> 00:14:31.449
to zero now when t is zero sin zero is zero
so we have zero here from both the limits
00:14:31.449 --> 00:14:40.269
minus omega will come out it is zero to infinity
cos omega t into f t dt
00:14:40.269 --> 00:14:46.540
so this is minus omega times this is nothing
but fourier cosine transform of f t it is
00:14:46.540 --> 00:14:55.369
fourier cosine transform of f t so this is
how we can obtain the fourier derivative of
00:14:55.369 --> 00:15:00.220
fourier sine transform which is nothing but
minus omega time fourier cosine transform
00:15:00.220 --> 00:15:06.749
of f t now suppose you want to find out fourier
cosine transform of second derivatives so
00:15:06.749 --> 00:15:13.360
that also we can find out in the same result
now what is fourier transform i have just
00:15:13.360 --> 00:15:18.790
fourier cosine transform f dash x fourier
cosine transform f dash x is nothing or f
00:15:18.790 --> 00:15:25.980
dash t is nothing but the this is we have
derived it is omega times fourier cosine transform
00:15:25.980 --> 00:15:32.519
of f dash fx minus f zero ok this we have
derived
00:15:32.519 --> 00:15:38.939
next is fourier sine transform of f dash
fourier sine transform of f dash t is nothing
00:15:38.939 --> 00:15:49.310
but minus omega fourier sine transform of
fourier cosine transform of f t so these expressions
00:15:49.310 --> 00:15:55.481
we have just derived suppose suppose now you
want to find out fourier cosine transform
00:15:55.481 --> 00:16:02.110
of double derivative of f t so this is we
want to find out ok it is fourier cosine that
00:16:02.110 --> 00:16:08.769
should must be signed ok now suppose i want
to find out fourier cosine transform of second
00:16:08.769 --> 00:16:16.579
derivative of f t so this will be nothing
but now you simply replace f by f dash
00:16:16.579 --> 00:16:21.110
in this expression so while you replace s
by f dash in this expression
00:16:21.110 --> 00:16:30.410
so fourier sine transform of f double t will
be nothing but omega times fourier sine transform
00:16:30.410 --> 00:16:41.680
of f double dash f dash t you are replacing
f by f dash t so it is f dash t minus f dash
00:16:41.680 --> 00:16:48.809
zero which is equals to omega times now fourier
sine transform of f dash t is we already know
00:16:48.809 --> 00:17:00.319
it is minus omega fourier minus omega fourier
cosine transform of f t minus f dash zero
00:17:00.319 --> 00:17:06.850
so when you simplify it is nothing but minus
omega square fourier cosine transform of a
00:17:06.850 --> 00:17:14.521
f t minus f dash zero so this will be the
fourier cosine transform of second derivative
00:17:14.521 --> 00:17:18.980
of f
now similarly if we want to find out fourier
00:17:18.980 --> 00:17:28.150
sine transform of second derivative of f so
that also you can find out now in this expression
00:17:28.150 --> 00:17:35.240
you simply replace f by f dash when you replace
f by f dash in this expression it is minus
00:17:35.240 --> 00:17:43.760
omega fourier cosine transform of f dash t
now this equals to minus omega now fourier
00:17:43.760 --> 00:17:49.890
cosine transform of f dash t from this expression
is nothing you can simply substitute it here
00:17:49.890 --> 00:17:59.560
it is omega fourier sine transform of f t
minus f zero so that will be nothing but minus
00:17:59.560 --> 00:18:15.930
omega square fourier sine transform of f t
minus omega f zero ok so ok its minus omega
00:18:15.930 --> 00:18:19.970
square into minus minus plus ok so it will
be plus
00:18:19.970 --> 00:18:25.880
so in this way we can obtain the fourier transform
of second derivative of f or the fourier sine
00:18:25.880 --> 00:18:31.880
transform second derivative of f so similarly
if we want to obtain the fourier cosine or
00:18:31.880 --> 00:18:36.480
sine transform of third derivatives or the
higher derivatives so that also we can obtain
00:18:36.480 --> 00:18:46.880
using the same concept ok so this is
how we can find out fourier sine or cosine
00:18:46.880 --> 00:18:52.500
transform of function if a function is
known to us and how we can find out fourier
00:18:52.500 --> 00:18:56.960
sine or cosine derivatives that also we have
seen or the higher derivatives that also we
00:18:56.960 --> 00:19:02.091
have seen so next in the next lecture we will
see some more properties of fourier transform
00:19:02.091 --> 00:19:08.070
and and how how the problems can be solved
on that that we will see in next class ok
00:19:08.070 --> 00:19:08.620
so
thank you very much