Signum Function

The signum function extracts the algebraic sign of a signal, returning +1 for positive values and -1 for negative values. It models ideal limiters in comparator circuits and serves as the building block for the unit step function.

Core Concept

The signum function sgn(t) is defined as +1 when t > 0, -1 when t < 0, and 0 when t = 0. It is an odd function: sgn(-t) = -sgn(t). Unlike the unit step, it is antisymmetric about the origin. The step function and signum are directly related: u(t) = [1 + sgn(t)] / 2, which you can verify by checking both sides at positive and negative t.

The Fourier transform of sgn(t) requires care because sgn(t) is not absolutely integrable — its energy is infinite and it does not decay. The standard approach is to compute it as the limit of a decaying exponential pair: sgn(t) = lim(a->0) [e^(-at)u(t) - e^(at)u(-t)]. Taking transforms of both parts and letting a go to zero yields the result.

In the frequency domain, the Fourier transform of sgn(t) is 2/(j omega) = 1/(j pi f). This is a purely imaginary, odd function of frequency. The relationship between sgn(t) and differentiation is also useful: d/dt[sgn(t)] = 2 delta(t), because the function jumps by 2 at the origin.

Key Formula

sgn(t) = +1, t > 0; -1, t < 0; 0, t = 0. Fourier transform pair: sgn(t) <-> 2/(j omega) = 1/(j pi f). Relation to unit step: u(t) = [1 + sgn(t)] / 2, equivalently sgn(t) = 2u(t) - 1. Derivative: d/dt[sgn(t)] = 2 delta(t).

Problem: Verify the relation u(t) = [1 + sgn(t)] / 2 using Fourier transforms.

Known pairs:
  F{1}    = delta(f)  [DC component]
  F{sgn(t)} = 1/(j pi f)
  F{u(t)} = (1/2) delta(f) + 1/(j 2 pi f)

Step 1: Take Fourier transform of right side
  F{[1 + sgn(t)] / 2} = (1/2) F{1} + (1/2) F{sgn(t)}

Step 2: Substitute known transforms
  = (1/2) delta(f)  +  (1/2) * 1/(j pi f)
  = (1/2) delta(f)  +  1/(j 2 pi f)

Step 3: Compare with F{u(t)}
  F{u(t)} = (1/2) delta(f) + 1/(j 2 pi f)  [standard pair]

Final Answer:
  Both sides match exactly.
  This confirms u(t) = [1 + sgn(t)] / 2 is correct.

Exam Tip: Anna University and VTU questions on signum often ask you to derive F{u(t)} using F{sgn(t)}. Start from sgn(t) = 2u(t) - 1, rearrange to u(t) = [1 + sgn(t)]/2, then take the Fourier transform of both sides. The delta(f) term comes from the constant 1, and the 1/(j 2 pi f) term comes from (1/2) times the signum transform. Do not try to integrate u(t) directly — the integral does not converge without the delta term.

Properties Summary

• Definition: sgn(t) = +1 (t>0), -1 (t<0), 0 (t=0).
• Odd function: sgn(-t) = -sgn(t); Fourier transform is purely imaginary.
• Fourier transform: F{sgn(t)} = 2/(j omega) = 1/(j pi f).
• Relation to step: u(t) = [1 + sgn(t)]/2; sgn(t) = 2u(t) - 1.
• Derivative: d/dt[sgn(t)] = 2 delta(t); the jump of 2 at origin becomes an impulse.
• Power signal: P = 1 because |sgn(t)|^2 = 1 for all t != 0.

Quick Revision

• sgn(t) = +1 for t > 0, -1 for t < 0.
• Odd function; always zero at t = 0.
• Fourier transform: 2/(j omega) or equivalently 1/(j pi f).
• Relates to unit step: u(t) = [1 + sgn(t)] / 2.
• Derivative of sgn(t) is 2 delta(t).
• Power = 1; energy = infinity; it is a power signal.
• Exam trap: writing F{sgn(t)} = 1/(j omega) instead of 2/(j omega). The factor of 2 comes from the total jump of 2 across the discontinuity, not 1. Check against the unit step derivation to catch this error.