What is the Fourier transform of a unit step input u(t), and what is notable about it?

• A. FT{u(t)} = π δ(ω) + 1/(jω) (distribution sense); indicates a constant spectral component plus a singular term at zero frequency.
• B. FT{u(t)} = δ(ω) only.
• C. FT{u(t)} = 1/(jω) only.
• D. FT{u(t)} = 0 for all ω.

Answer: (A)

The key idea is that the unit step does not have a regular Fourier transform as a usual function, because it does not decay in time. In the sense of distributions, its transform consists of two pieces: a delta function at zero frequency and a principal-value term 1/(jω). This combination, π δ(ω) + 1/(jω), captures both how the step carries a constant component over time and how its spectrum distributes across nonzero frequencies. The delta at ω = 0 reflects the nonzero average of the step for t > 0, while the 1/(jω) term describes the spread of energy across frequencies with a 1/ω decay and the corresponding 90-degree phase behavior.

This aligns with the derivative property: the derivative of the unit step is the delta function in time, whose transform is 1, and applying the relation jω U(ω) = F{du/dt} gives U(ω) = 1/(jω) plus the necessary δ(ω) term to satisfy the distributional identity. Hence the correct transform is a combination of a constant spectral component and a singular term at zero frequency, not just one of those pieces alone.