FT of Step Function

The Fourier transform of the unit step function u(t) contains both a Dirac delta at zero frequency and a frequency-dependent term, reflecting the step's DC component and its slow roll-off with frequency. Filter designers use this result to understand how an ideal integrator responds across the spectrum.

Core Concept

The unit step u(t) equals 1 for t >= 0 and 0 for t < 0. Like the signum function, it is not absolutely integrable because it never decays to zero. Its Fourier transform therefore cannot be found by directly evaluating the standard integral, and the result involves a Dirac delta.

The derivation uses the identity u(t) = 1/2 + (1/2)*sgn(t). The FT of the constant 1/2 is (1/2)*delta(f) by the duality property of the impulse. The FT of (1/2)*sgn(t) is (1/2) * 1/(j*pi*f) = 1/(j2*pi*f). Adding these two contributions gives the full transform.

The delta(f)/2 term represents the DC (zero-frequency) content of the step function, which has an average value of 1/2 when integrated symmetrically about t = 0. The 1/(j2*pi*f) term describes how the step's spectrum rolls off at higher frequencies, falling as 1/f.

Key Formula

FT pair: u(t) <-> (1/2)*delta(f) + 1/(j2*pi*f). In angular frequency: u(t) <-> pi*delta(omega) + 1/(j*omega). The magnitude of the continuous part is 1/(2*pi|f|), which decays as 1/f. The phase of the continuous part is -90 degrees for f > 0 and +90 degrees for f < 0 because of the j in the denominator.

Derivation using u(t) = 1/2 + (1/2)*sgn(t)

Step 1: Express u(t) in terms of known signals
  u(t) = 1/2 + (1/2)*sgn(t)

Step 2: FT of the constant 1/2
  FT{1} = delta(f)  [by duality: FT of delta(t) = 1, so FT of 1 = delta(f)]
  FT{1/2} = (1/2)*delta(f)

Step 3: FT of (1/2)*sgn(t)
  FT{sgn(t)} = 1/(j*pi*f)
  FT{(1/2)*sgn(t)} = 1/(2*j*pi*f) = 1/(j2*pi*f)

Step 4: Add by linearity
  U(f) = (1/2)*delta(f) + 1/(j2*pi*f)

Final Answer: U(f) = (1/2)*delta(f) + 1/(j2*pi*f)

In omega convention:
  U(omega) = pi*delta(omega) + 1/(j*omega)

Exam Tip: The FT of u(t) is NOT simply 1/(j2*pi*f). That result misses the delta(f)/2 term representing the DC content. A common GATE question gives a system response involving 1/(j*omega) and asks about the time-domain signal; always check whether a pi*delta(omega) term should accompany it. Forgetting this term means your inverse FT will miss the step component.

Properties Summary

• FT pair in f: u(t) <-> (1/2)*delta(f) + 1/(j2*pi*f).
• FT pair in omega: u(t) <-> pi*delta(omega) + 1/(j*omega).
• The delta term captures DC; the 1/f term captures the spectral roll-off.
• Relation to sgn: U(f) = (1/2)*delta(f) + (1/2)*FT{sgn(t)}.
• Derivative relation: d/dt[u(t)] = delta(t), so FT{u(t)} * j2*pi*f = 1, confirming the 1/(j2*pi*f) continuous part.
• Integration property: if Y(f) = X(f)/(j2*pi*f) + (1/2)*X(0)*delta(f), the inverse FT is the running integral of x(t).

Quick Revision

• u(t) = 1/2 + (1/2)*sgn(t).
• FT{u(t)} = (1/2)*delta(f) + 1/(j2*pi*f).
• In omega: pi*delta(omega) + 1/(j*omega).
• The delta(f)/2 term accounts for the non-zero average value of u(t).
• d/dt u(t) = delta(t) and j2*pi*f * U(f) = 1 are consistent.
• u(t) + u(-t) = 1 for all t, so FT{u(t)} + FT{u(-t)} = delta(f), which confirms U(f) + U(-f) = delta(f).
• Exam trap: writing FT{u(t)} = 1/(j2*pi*f) and omitting the (1/2)*delta(f). This is the most common error for this topic and directly loses marks in GATE.