ASN6 a,b- Prove given Fourier Transform property

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Problem Statement

6(a) Show ${ℱ}[\underset{-\mathrm{\infty }}{\overset{t}{\int }}s(\lambda )d\lambda ]=\frac{S(f)}{j2\pi f}\text{ if }S(f_0)=0$.

6(b) If $S(f_0)\ne 0$ can you find ${ℱ}[\underset{-\mathrm{\infty }}{\overset{t}{\int }}s(\lambda )d\lambda ]$ in terms of $S(f_0)$?

Answer

a)Remember that dummy variable $\lambda$ was used in substitution such that $\lambda =t-t_0$

Then $s(\lambda )=s(t-t_0)={ℱ}[S(f)-S(f_0)]$

and $\underset{-\mathrm{\infty }}{\overset{t}{\int }}s(\lambda )d\lambda =\underset{-\mathrm{\infty }}{\overset{t}{\int }}{ℱ}[S(f)-S(f_0)]d\lambda$

The problem statement says to make $S(f_0)=0$ that makes the above equation simplify to

$\underset{-\mathrm{\infty }}{\overset{t}{\int }}s(\lambda )d\lambda =\underset{-\mathrm{\infty }}{\overset{t}{\int }}{ℱ}[S(f)]dt$

Taking the inverse Fourier Transform and changing the order of intgration

$\underset{-\mathrm{\infty }}{\overset{t}{\int }}s(\lambda )d\lambda =\underset{-\mathrm{\infty }}{\overset{t}{\int }}{e}^{j2\pi ft}dt\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}S(f)df=\frac{e^{j2\pi ft}}{j2\pi f}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}S(f)df$

Then

$\underset{-\mathrm{\infty }}{\overset{t}{\int }}s(\lambda )d\lambda =\underset{\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}S(f)\frac{e^{j2\pi ft}}{j2\pi f}df={{ℱ}}^{-1}\left[\frac{S(f)}{j2\pi f}\right]$

Therefore it is demonstrated that ${ℱ}[\underset{-\mathrm{\infty }}{\overset{t}{\int }}s(\lambda )d\lambda ]=\frac{S(f)}{j2\pi f}$

b) If $S(f_0)\ne 0$

Then

$\underset{-\mathrm{\infty }}{\overset{t}{\int }}s(\lambda )d\lambda =\underset{-\mathrm{\infty }}{\overset{t}{\int }}{{ℱ}}^{-1}[S(f)-S(f_0)]d\lambda =\underset{-\mathrm{\infty }}{\overset{t}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi ft}[S(f)-S(f_0)]d\lambda d\lambda$

$\underset{-\mathrm{\infty }}{\overset{t}{\int }}s(\lambda )d\lambda =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi ft}S(f)d\lambda d\lambda -\underset{-\mathrm{\infty }}{\overset{t}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi ft}S(f_0)d\lambda d\lambda$

$\underset{-\mathrm{\infty }}{\overset{t}{\int }}s(\lambda )d\lambda =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{e^{j2\pi ft}}{j2\pi f}S(f)d\lambda -\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{e^{j2\pi ft}}{j2\pi f}S(f_0)d\lambda$

Using $dt=d\lambda$ and taking the Fourier transform of the equation

answer is ${ℱ}[\underset{-\mathrm{\infty }}{\overset{t}{\int }}s(\lambda )d\lambda ]=\frac{e^{j2\pi ft}}{j2\pi f}S(f)-\frac{e^{j2\pi ft}}{j2\pi f}S(f_0)$