6 - Fourier Transform 2

(a) Show ${\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]={\frac {S(f)}{j2\pi f}}{\mbox{ if }}S(0)=0$. Hint: $S(0)=S(f)|_{_{f=0}}=\int _{-\infty }^{\infty }s(t)e^{-j2\pi (f\rightarrow 0)t}\,dt=\int _{-\infty }^{\infty }s(t)\,dt$

(b) If $S(0)\neq 0$ can you find ${\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]$ in terms of $\displaystyle S(0)$?

(c) Do another property on the Wiki and review a second property

Find ${\mathcal {F}}\left[e^{j2\pi f_{0}t}s(t)\right]$
First ${\mathcal {F}}\left[e^{j2\pi f_{0}t}s(t)\right]=\int _{-\infty }^{\infty }e^{j2\pi f_{0}t}s(t)e^{-j2\pi ft}$
or rearranging we get $\int _{-\infty }^{\infty }e^{j2\pi f_{0}t}s(t)e^{-j2\pi ft}dt=\int _{-\infty }^{\infty }s(t)e^{j2\pi t(f_{0}-f)}dt$
Which leads to $\int _{-\infty }^{\infty }s(t)e^{j2\pi t(f_{0}-f)}dt=S(f-f_{0})$
So ${\mathcal {F}}\left[e^{j2\pi f_{0}t}s(t)\right]=S(f-f_{0})$

Reviewed Nicks 2nd Fourier transform made comment about one possible error other than that looked good