Hilbert Transform

The Hilbert transform produces the analytic signal by shifting each frequency component of a real signal by 90 degrees. It is used in single-sideband modulation, envelope detection, and instantaneous frequency estimation in communications.

Core Concept

The Hilbert transform of a signal x(t) is written x̂(t) and is obtained by convolving x(t) with 1/(πt). In the frequency domain, this is equivalent to multiplying the Fourier transform X(f) by -j·sgn(f), where sgn(f) is +1 for positive frequencies and -1 for negative frequencies. The result is a 90-degree phase lag at every frequency component.

The analytic signal is formed as z(t) = x(t) + jx̂(t). Its Fourier transform is zero for all negative frequencies: Z(f) = 2X(f) for f > 0, and Z(0) = X(0). This one-sided spectrum is exactly what makes single-sideband (SSB) modulation possible — transmitting only positive frequencies halves the bandwidth compared to AM.

The instantaneous amplitude A(t) = |z(t)| gives the signal envelope and the instantaneous phase φ(t) = angle(z(t)) gives the phase at every moment. Differentiating φ(t) gives the instantaneous frequency, which tracks how the local oscillation rate of the signal changes over time.

Key Formula

Time domain: x̂(t) = H{x(t)} = (1/π)∫ x(τ)/(t-τ) dτ — a Cauchy principal value integral. Frequency domain: if X(f) = F{x(t)}, then F{x̂(t)} = -j·sgn(f)·X(f). The signum function sgn(f) = +1 for f > 0, -1 for f < 0. Analytic signal: z(t) = x(t)+jx̂(t) has transform Z(f) = 2u(f)X(f) where u(f) is the unit step in frequency. Key pairs: H{cos(2πf₀t)} = sin(2πf₀t) and H{sin(2πf₀t)} = -cos(2πf₀t).

Problem: Find the Hilbert transform of x(t) = cos(2πf₀t) + (1/3)cos(6πf₀t).

Step 1 — Apply Hilbert to each term separately (linearity):
  H{cos(2πf₀t)} = sin(2πf₀t)          [fundamental]
  H{(1/3)cos(6πf₀t)} = (1/3)sin(6πf₀t) [third harmonic, same rule]

Step 2 — Sum the results:
  x̂(t) = sin(2πf₀t) + (1/3)sin(6πf₀t)

Step 3 — Form the analytic signal:
  z(t) = x(t) + j x̂(t)
       = [cos(2πf₀t) + (1/3)cos(6πf₀t)] + j[sin(2πf₀t) + (1/3)sin(6πf₀t)]

Step 4 — Instantaneous envelope (approximate, assuming f₀ >> 0):
  |z(t)| approximated numerically — depends on phase relationship.

Final Answer:
  x̂(t) = sin(2πf₀t) + (1/3)sin(6πf₀t)
  Each cosine term becomes the corresponding sine term.

Exam Tip: The Hilbert transform is linear, so handle multi-component signals term by term. Remember H{H{x(t)}} = -x(t) — applying it twice gives the negative of the original. Also, H{x̂(t)} = -x(t). These identities appear in GATE as true/false questions. The analytic signal has no negative-frequency content — that is the defining property used in SSB derivations.

Properties Summary

• Definition: x̂(t) = (1/π)P.V.∫ x(τ)/(t-τ)dτ — Cauchy principal value convolution with 1/(πt).
• Frequency domain: F{x̂(t)} = -j·sgn(f)·X(f) — 90° phase shift at all frequencies.
• Linearity: H{ax(t)+by(t)} = ax̂(t)+bŷ(t).
• Double application: H{H{x(t)}} = -x(t) — two Hilbert transforms invert sign.
• Analytic signal: z(t) = x(t)+jx̂(t) has one-sided spectrum Z(f) = 2X(f)u(f).
• Key pairs: H{cos(2πf₀t)} = sin(2πf₀t); H{sin(2πf₀t)} = -cos(2πf₀t).
• Energy: the Hilbert transform preserves signal energy — x̂(t) and x(t) have equal energy.

Quick Revision

• Hilbert transform = convolution with 1/(πt) = multiply spectrum by -j·sgn(f).
• Phase shift is exactly -90° (π/2 lag) at every frequency.
• Applying Hilbert twice yields the negative original: H²{x}= -x.
• Analytic signal has spectrum only for f > 0.
• Envelope = |z(t)|, instantaneous phase = angle(z(t)).
• SSB upper sideband signal = Re{z(t)e^(j2πfct)}.
• Hilbert and original are orthogonal: ∫x(t)x̂(t)dt = 0 for finite-energy signals.
• Exam trap: writing H{sin(2πf₀t)} = cos(2πf₀t) instead of -cos(2πf₀t) — the sign is negative.