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Short note: Hilbert transform

Short answer

Hilbert transform shifts all positive frequency components by -90° and negative ones by +90°. It is defined as x̂(t)=x(t)*1/(πt) and has transfer function -j sgn(f). Applications include generating SSB signals and forming the analytic signal.

Write a short note on Hilbert transform.

The Hilbert transform of a signal x(t)x(t), written x^(t)\hat x(t), is the signal obtained by shifting the phase of all its positive-frequency components by −90°-90° and all negative-frequency components by +90°+90°, while keeping amplitudes unchanged.

Time domain:

x^(t)=x(t)∗1πt=1π∫−∞∞x(τ)t−τdτ\hat x(t)=x(t)*\frac{1}{\pi t}=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{x(\tau)}{t-\tau}d\tau

Frequency domain: the Hilbert transformer is a filter with

H(f)=−j sgn(f),X^(f)=−j sgn(f) X(f)H(f)=-j\,\text{sgn}(f),\qquad \hat X(f)=-j\,\text{sgn}(f)\,X(f)

so ∣H(f)∣=1\lvert H(f)\rvert=1 and ∠H(f)=−90°\angle H(f)=-90° for f>0f>0, +90°+90° for f<0f<0. It is an ideal wideband 90°90° phase shifter.

Examples: cos⁡ω0t^=sin⁡ω0t\widehat{\cos\omega_0t}=\sin\omega_0t, sin⁡ω0t^=−cos⁡ω0t\widehat{\sin\omega_0t}=-\cos\omega_0t.

Properties:

  1. x(t)x(t) and x^(t)\hat x(t) have the same amplitude spectrum, energy and power.
  2. x^^(t)=−x(t)\hat{\hat x}(t)=-x(t).
  3. x(t)x(t) and x^(t)\hat x(t) are orthogonal: ∫x(t)x^(t)dt=0\int x(t)\hat x(t)dt=0.
  4. For low-pass m(t)m(t) and fcf_c above its bandwidth, m(t)cos⁡ωct^=m(t)sin⁡ωct\widehat{m(t)\cos\omega_ct}=m(t)\sin\omega_ct.

Applications:

  • SSB generation: s(t)=m(t)cos⁡ωct∓m^(t)sin⁡ωcts(t)=m(t)\cos\omega_ct\mp\hat m(t)\sin\omega_ct (phase-shift method).
  • Pre-envelope (analytic signal): x+(t)=x(t)+jx^(t)x_+(t)=x(t)+j\hat x(t), whose spectrum is zero for f<0f<0.
  • Band-pass signal representation in terms of in-phase and quadrature components, and complex envelope analysis.

✓ Answer checked by Claude Opus 5.5 (high effort)

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