Shifted Fourier transform

Question

Please can some one help and give me a direction to evaluate the following shifted Fourier transform: \begin{alignat}{2} s(x_c) =&\frac{1}{\Delta x_0} \int_{x_c-\Delta x_0}^{x_c+\Delta x_0}f_{b}(x-x_c)exp(jx)dx \end{alignat} When $f_{b}(x-x_c)=1$ then I am able to derive the previous integration which is : \begin{alignat}{2} s(x_c) =&\frac{1}{\Delta x_0} \int_{x_c-\Delta x_0}^{x_c+\Delta x_0}exp(jx)dx\\ =& sinc(\frac{\Delta x_0}{2})exp(jx_c) \end{alignat} For a generale case where the Fourier transform of $f_{b}$ equal to $F_{b}$ what we obtained? Thank you.

Answer 1

You can evaluate the integral by changing variables. Let say $u=x-x_c$, then $du=dx$ and the lower and upper bounds of integral will be $-\Delta x_0$ and $+\Delta x_0$, so: $$s(x_0) = \frac{1}{\Delta x_0} \int _{-\Delta x_0} ^{+\Delta x_0} f_b (u) e^{j(u+x_c)} du$$

$$ = \frac{e^{jx_c}}{\Delta x_0} \int _{-\Delta x_0} ^{+\Delta x_0} f_b (u) e^{ju} du$$

So if you define $\int _{-\Delta x_0} ^{+\Delta x_0} f_b (u) e^{ju} du$ as the Fourier transform of $f_b(x)$ then: $$ s(x_0)=\frac{e^{jx_c}}{\Delta x_0} F_b(\Delta x_0).$$

NOTE: Actually it is not correct to call $s(x_0)$ shifted Fourier transform. As in Fourier transform the limits should be from $-\infty$ to $+\infty$ and also the output is not a function of frequency variable.