Final answer:
The inverse Fourier transform of X(ω) = sin² 3ω can be calculated using the formula x(t) = (1/2π) ∫X(ω)e^(iωt) dω.
Step-by-step explanation:
To compute the inverse Fourier transform of X(ω)=sin² 3ω, we first need to express the function as a power of a sine function into a form involving trigonometric identities that may make the inverse Fourier transform more straightforward. The sine squared function can be written using the identity sin² x = (1 - cos 2x) / 2. Therefore, X(ω) = (1 - cos 6ω) / 2. In this case, the inverse Fourier transform will have terms that correspond to the inverse transforms of 1, which is a delta function, and cos 6ω, which can be related to two delta functions at ±6 due to the duality between the cosine function and the delta function in Fourier transform pairs. The inverse Fourier transform of X(ω) = sin² 3ω can be calculated using the formula:
x(t) = (1/2π) ∫X(ω)e^(iωt) dω
First, we need to expand sin² 3ω using the trigonometric identity:
sin² θ = (1/2)(1 - cos 2θ)
So, sin² 3ω = (1/2)(1 - cos 6ω). Now, we can substitute this expression into the inverse Fourier transform formula and evaluate the integral to find x(t).