I'm asked to prove using "duality property" the Fourier transform of
$$\frac{1}{\pi t} = -j sgn(f)$$
I have the proof steps but I'm quit not understanding it:
- multiply by $j = \frac{j}{j(\pi t)}$
- $j2\pi x(-w)$
- $2j\pi(-0.5+U(-w))$
multiplying by $2$ and factoring $- sgn$ gives:
$$-j( 1-2U(-w))= -j*sgn (f)$$
*correction: ignore the above steps as they are confusing, consider this answer:
prove $$\frac{1}{πt} ↔ -j sgn(f)$$
Using the duality theorem which state:
If $x(t) ↔ X(f)$
then $X(t) ↔ x(-f)$
Let $x(t) = \frac{1}{πt}$ and $X(t) = -j sgn(f)$
Then by duality theorem:
$-j sgn(t) ↔ – \frac{1}{πf}$
Calculating Fourier Transform of $–jsgn(t)$:
$\mathcal{F}[-j sgn(t) ]= -j \mathcal{F}[sgn(t)]$
Fourier transform of $sgn(t)$ is :
$\mathcal{F}[sgn(t)] = \frac{1}{πfj}$
Thus
$\frac{1}{πt} ↔ -j* \frac{1}{jπf} = – \frac{1}{πf}$
Therefore if we apply the duality again on this result we ge
$\frac{1}{πt} = -j sgn(f)$
Best Answer
The duality property is the statement that if
$$\mathcal{F}(h(t)) = H(\omega)$$
then
$$\mathcal{F}(H(t)) = h(-\omega)$$
To use the duality property to prove the statement you need to show that
$$\mathcal{F}\left(i~\text{sign}(t)\right) = -\frac{1}{\omega\pi}$$
This can be done by a direct computation of $\mathcal{F}\left(i~\text{sign}(t)\right)$. To do this note that
$$\frac{d}{dx} \text{sign} (x) = 2\delta(x)$$
and by using the property
$$ \mathcal{F}\left(\frac{dg(x)}{dx}\right) = 2\pi i\omega \mathcal{F}(g(x))$$
togeather with $\mathcal{F}(\delta) = 1$ the result follows (I have used the convention $\mathcal{F}(g) \equiv \int g(x)e^{-2\pi i \omega} dx$ above).