# Applying duality property to fourier transform of unit step function

For Continuous time aperiodic signals, the duality property of Continuous Time Fourier Transform (CTFT) is following

$$\mathscr{F}\Big\{x(t)\Big\} = X(f), \qquad\text{then} \quad \mathscr{F}\Big\{X(t)\Big\} = x(-f)$$

Now we know while Dirichlet conditions are not satisfied for unit step function $$u(t)$$, so its CTFT analysis and synthesis cannot be done. However, we can still do it provided we are willing to accept occurrence of singularity functions like Dirac delta impulse in its Fourier transform equation.

i.e.

\begin{align} \mathscr{F}\Big\{u(t)\Big\} &= \mathscr{F}\Big\{\tfrac{1}{2} + \tfrac{1}{2}\operatorname{sgn}(t) \Big\} \\ &= \frac{\delta(f)}{2} + \frac{1}{j2\pi f} \\ \end{align}

where the signum function,

$$\operatorname{sgn}(t) \triangleq \begin{cases} -1 \qquad & t<0 \\ 0 \qquad & t=0 \\ +1 \qquad & t>0 \\ \end{cases}$$

However if I apply duality property to the above result, then I should get following:

\begin{align} \mathscr{F}\Big\{\frac{\delta(t)}{2} + \frac{1}{j2\pi t}\Big\} &= u(-f) \\ &= \tfrac{1}{2} + \tfrac{1}{2}\operatorname{sgn}(-f) \\ &= \tfrac{1}{2} - \tfrac{1}{2}\operatorname{sgn}(f) \\ \end{align}

However when I read at least some books on Fourier transform, I find that result is $$u(f)$$ and not $$u(-f)$$. Question is why ?

• You get $u(-t)$. Just do the transform. $F[1/t]=-j\text{sgn}(f)$. Dual Property holds. – Max Apr 2 '19 at 7:07

## 1 Answer

The result you got is correct and it is also expected according to the two first formulas in your question. If $$X(f)$$ is the Fourier transform of $$x(t)$$, then the Fourier transform of $$X(t)$$ equals $$x(-f)$$. If $$x(t)=u(t)$$ is the unit step function then

$$X(f)=\frac12 \delta(f)+\frac{1}{j2\pi f}\tag{1}$$

and the Fourier transform of $$X(t)$$ is given by

$$\mathcal{F}\left\{\frac12 \delta(t)+\frac{1}{j2\pi t}\right\}=\frac12+\frac{1}{2 j}\mathcal{F}\left\{\frac{1}{\pi t}\right\}\tag{2}$$

where we recognize $$1/\pi t$$ as the impulse response of an ideal Hilbert transformer, the Fourier transform of which is given by

$$\mathcal{F}\left\{\frac{1}{\pi t}\right\}=-j\;\textrm{sgn}(f)\tag{3}$$

Combining $$(2)$$ and $$(3)$$ gives

$$\mathcal{F}\left\{X(t)\right\}=\frac12\big(1-\textrm{sgn}(f)\big)=u(-f)\tag{4}$$

just as expected.

Maybe you can clarify which books say otherwise.