Start with the impulse, the anchor of the whole table.

\(\mathcal{F}\{\delta(t)\}=\int_{-\infty}^{\infty}\delta(t)e^{-j\omega t}\,dt=1\)

by the sifting property. So c → ii: an impulse in time is flat in frequency, containing every frequency equally.

Now use duality for the constant. Since δ(t) transforms to 1, the dual statement is that 1 transforms to an impulse in frequency, scaled by 2π:

\(\mathcal{F}\{1\}=2\pi\delta(\omega)\)

So a → iv. Physically a DC signal has all of its energy at ω = 0 and nowhere else. Notice the pleasing symmetry: entries a and c are duals of each other, and this pair alone eliminates options 2, 3 and 4.

The unit step. Splitting u(t) into a DC part and an odd part, \(u(t)=\tfrac{1}{2}+\tfrac{1}{2}\mathrm{sgn}(t)\), and using \(\mathcal{F}\{\mathrm{sgn}(t)\}=2/(j\omega)\),

\(\mathcal{F}\{u(t)\}=\pi\delta(\omega)+\dfrac{1}{j\omega}\)

So b → i. The impulse term is the average value of the step (½, scaled by 2π) and the 1/jω term carries the discontinuity at t = 0. Forgetting the πδ(ω) term is the standard error here.

The remaining entry, |t|. By elimination d → iii, and it can be confirmed: \(|t|=t\,\mathrm{sgn}(t)\), and multiplication by t in the time domain corresponds to \(j\dfrac{d}{d\omega}\) in frequency, so

\(\mathcal{F}\{|t|\}=j\dfrac{d}{d\omega}\left(\dfrac{2}{j\omega}\right)=-\dfrac{2}{\omega^{2}}\)

(strictly in the generalised-function sense, ignoring an impulse term at the origin).

Assemble. a-iv, b-i, c-ii, d-iii — option 1.

The pattern behind the table. The narrower a signal is in time, the wider its transform: an impulse spreads over all ω, a constant collapses to a single ω. That inverse relationship is the essence of the uncertainty principle for signals.

Hence, the correct matching is a-iv, b-i, c-ii, d-iii.