The waveform is a step, and the Fourier transform of a step has two parts — an impulse at the origin and a term falling away as 1/ω — which is the shape drawn in option 1.

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

Where each term comes from. Write the unit step as a constant plus a signum function:

\(u(t)=\dfrac{1}{2}+\dfrac{1}{2}\text{sgn}(t)\)

Each half transforms separately.

The constant 1/2. A DC level exists at exactly one frequency, zero, so its transform is concentrated entirely at the origin:

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

This is the impulse marked \(\pi\) in every option — it is what makes the step's average value non-zero, and its presence in all four sketches means it cannot discriminate between them.

The signum part. This carries the transition and gives

\(\mathcal{F}\left\{\tfrac{1}{2}\text{sgn}(t)\right\}=\dfrac{1}{j\omega}\)

whose magnitude is \(1/|\omega|\) — large near the origin and decaying towards zero on both sides. That hyperbolic decay is the discriminating feature, and only option 1 shows it: curves rising steeply as \(\omega\to0\) from either side and falling away symmetrically.

Why the other sketches are wrong in principle. Options 2 and 3 show a magnitude that rises away from the origin and stays high, which would mean the step contains more energy at high frequency than at low — impossible for a waveform that is constant almost everywhere. Option 4 shows a curve dipping to a minimum near the origin, the exact inverse of the true behaviour.

A sanity check that needs no algebra. A step is a slowly varying signal: it changes once and then never again. Its spectrum must therefore be dominated by low frequencies, and the only high-frequency content comes from the single discontinuity — which is why the tail falls as \(1/\omega\) rather than vanishing altogether. A discontinuity always produces a \(1/\omega\) tail; a corner in the waveform would give \(1/\omega^{2}\).

Hence, the correct sketch is option 1.