Question

find inverse fouriour transfrom​

Answer 1

Step-by-step explanation:

The inverse Fourier transform of δ(f−2) is

F−1[δ](t)=∫δ(f−2)ei2πftdf=ei2π2t=ei4πt

The 2nd equality holds by definition of the delta function.

Two commonly used definitions of the Fourier transform of x(t) are

X(ω)X^(f)=∫∞−∞x(t)e−iωtdt,=∫∞−∞x(t)e−i2πftdt,x(t)=12π∫∞−∞X(ω)eiωtdωx(t)=∫∞−∞X(f)ei2πftdf.

The two functions are related as X^(f)=X(2πf) and X(ω)=X^(f/2π).

I think your question essentially is: if you have a table that tells you the inverse Fourier transform of X(ω)=δ(ω−ω0) is 12πeiω0t from which it is easy to deduce that the inverse Fourier transform of δ(ω−2) is 12πei2t, how do you deduce from this that the inverse Fourier transform of X^(f−f0)=δ(f−f0) is ei2πf0t in general, and that the inverse Fourier transform of δ(f−2) is ei4πt? As williamdemeo showed you, and Willie Wong emphasized to you, just computing the Fourier integral

x(t)=∫∞−∞X^(f)ei2πftdf=∫∞−∞δ(f−2)ei2πftdf=ei2π2t=ei4πt

far easier than fussing with tables. But if you are dead set on using tables only, then note that if

x(t)=12π∫∞−∞X(ω)eiωtdω

is known to you, then since it does not matter what we call the variable of integration

x(t)=12π∫∞−∞X(ω)eiωtdω=12π∫∞−∞X(f)eiftdf=12π∫∞−∞X(f)ei2πf(t/2π)df

Let y(t) denote the rightmost integral. Then we have that y(t) is the inverse Fourier transform of X(f) evaluated at t/2π, and it happens to equal 2πx(t). So,

given that

x(t)=12π∫∞−∞X(ω)eiωtdω

is the inverse Fourier transform of X(ω), the inverse Fourier transform of X(f) is

∫∞−∞X(f)ei2πftdf=2π⋅x(2πt).

In particular, given that the inverse the inverse Fourier transform of δ(ω−2) is 12πei2t, the inverse Fourier transform of δ(f−2) is 2π12πei2⋅2πt=ei4πt.

Hope this helps you to gain Knowledge in Inverse Fourier transform

Answer 2

Answer:

I don't know the answer