00:01
In this question, we need to find out the fourier transform of the given functions.
00:05
Okay, so first let me write down the formula for the fourier transform.
00:08
So fourier transform of a function is given by f of omega equal to integral from minus infinity to infinity, p -raise to i omega -t, f of t, t, t, okay? so this is the fourier transform.
00:24
This formula we are going to use in every case.
00:27
So now you see our function is defined.
00:31
Okay, so in this case, the fourier transform will be given by, now you see minus infinity to infinity can be broken in three intervals here.
00:39
So it will be minus infinity to minus one, okay? of e raise to i omega t, f of t, t, t plus minus one to one, e raised to i omega t f of t d t plus one to infinity e raise to i omega t f of e d t okay so this will give us f of omega is equal now for minus less than minus one you see from this it will be zero and for greater than one it will be zero okay so it will be zero plus minus one to one and e -rae -rae and in this case the function is just one okay plus it will be zero again so we are getting f of omega to be equal to integral of just this so this integral we know is how much so it will be just two sine omega divided by omega okay if you integrate it so you will get this correct now let us see the second part so in that second part if you try to find out it is from minus 3 to 3 so we will again break the interval minus infinity to infinity into 3 parts so it will be minus infinity to minus 3 e -rase 2 i omega t f of d t okay and then from minus 3 to 3 e raise to i omega t f of t d t plus the last interval, which is from 3 to infinity, e raised to i omega t, f of t, d, d, t.
02:30
Okay? so in minus three to three, it is one by four, and rest other it is zero.
02:36
So it will be zero plus integral minus three to three, one divided by four, e raised to i omega t, d, d, t, plus again zero.
02:47
Okay...