inverse Fourier transform

[logistic_guy]

Determine the inverse Fourier transform of the frequency function \(\displaystyle G(f)\) defined by the amplitude and phase spectra shown in the figure.

 

[logistic_guy]

One way to define these spectrums is:

\(\displaystyle G(f) = \begin{cases} j, & -W < f < 0 \\-j, & 0 \leq f < W \\0, & \text{otherwise}\end{cases}\)

Or

\(\displaystyle G(f) = -j\text{sgn}(f) + ju(f - W) - ju(-f - W)\)

 

[logistic_guy]

Let \(\displaystyle h(t) = \text{sgn}(t)\).

\(\displaystyle \mathcal{F}\{h(t)\} = H(f) = \frac{1}{j\pi f}\)

By duality property,

\(\displaystyle \mathcal{F}^{-1}\{\text{sgn}(f)\} = H(-t) = -\frac{1}{j\pi t}\)

Let \(\displaystyle G_1(f) = -j\text{sgn}(f)\), then

\(\displaystyle \mathcal{F}^{-1}\{-j\text{sgn}(f)\} = -j\frac{-1}{j\pi t} = \textcolor{green}{\frac{1}{\pi t}}\)

 

[logistic_guy]

We will define \(\displaystyle u(t)\) as

\(\displaystyle u(t) = \frac{1}{2}\bigg(\text{sgn}(t) + 1\bigg)\)


\(\displaystyle \mathcal{F}\bigg\{u(t)\bigg\} = \frac{1}{2}\bigg(\frac{1}{j\pi f} + \delta(f)\bigg)\)


\(\displaystyle \mathcal{F}\bigg\{u(t - W)\bigg\} = \frac{1}{2}\bigg(\frac{1}{j\pi f} + \delta(f)\bigg)e^{-j2\pi Wf} \ \ \ \ \ \textcolor{red}{\bold{shifting \ property.}}\)


\(\displaystyle \mathcal{F}^{-1}\bigg\{u(f - W)\bigg\} = \frac{1}{2}\bigg(-\frac{1}{j\pi t} + \delta(t)\bigg)e^{j2\pi Wt} \ \ \ \ \ \textcolor{indigo}{\bold{duality \ property.}}\)

 

[logistic_guy]

Using the same reasoning above, we get:

\(\displaystyle \mathcal{F}^{-1}\bigg\{u\left(f + W\right)\bigg\} = \frac{1}{2}\bigg(-\frac{1}{j\pi t} + \delta(t)\bigg)e^{-j2\pi Wt}\)

 

[logistic_guy]

\(\displaystyle G(f) = -j\text{sgn}(f) + ju(f - W) - ju(-f - W)\)

Now we can find the inverse Fourier transform of the above.


\(\displaystyle g(t) = \mathcal{F}^{-1}\bigg\{G(f)\bigg\} = \frac{1}{\pi t} + j\frac{1}{2}\bigg(-\frac{1}{j\pi t} + \delta(t)\bigg)e^{j2\pi Wt} - j\frac{1}{2}\bigg(-\frac{1}{j\pi t} + \delta(t)\bigg)e^{-j2\pi Wt}\)



\(\displaystyle = \frac{1}{\pi t} - \frac{1}{2\pi t}e^{j2\pi Wt} + j\frac{\delta(t)}{2} + \frac{1}{2\pi t}e^{-j2\pi Wt} - j\frac{\delta(t)}{2}\)



\(\displaystyle = \frac{1}{\pi t} - \frac{1}{2\pi t}e^{j2\pi Wt} + \frac{1}{2\pi t}e^{-j2\pi Wt}\)

With a little simplification, we get:

\(\displaystyle g(t) = \textcolor{indigo}{\frac{1}{\pi t} - \frac{j\sin(2\pi Wt)}{\pi t}}\)