--- title: "Hilbert transform in frequency domain" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:hilbert-define-frequency-domain" source: https://www.bctengineer.com/q/semester_6/communication_systems/hilbert-define-frequency-domain site: BCT Engineering --- # Hilbert transform in frequency domain Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2080 Chaitra (CS I), 2+3 marks - Video: [Hilber Transform](https://www.youtube.com/watch?v=wYrNwRuQFB4) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/hilbert-define-frequency-domain - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question Define and describe Hilbert Transform with the help of mathematical expression in frequency domain. ## Short answer The Hilbert transform in the frequency domain is Ĥ(f) = −j sgn(f) X(f). Its magnitude response is 1 (for f≠0) and phase is −90° for positive frequencies, +90° for negative frequencies; in time domain it is the convolution x(t)*1/(πt). ## Answer The **Hilbert transform (HT)** of a real signal $x(t)$ is a new signal $\hat{x}(t)$ in which every frequency component of $x(t)$ is phase shifted by $-90^\circ$ (for positive frequencies) and $+90^\circ$ (for negative frequencies), with amplitude unchanged. #### Time-domain definition $$ \hat{x}(t) = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{x(\tau)}{t-\tau}\,d\tau = x(t) * \frac{1}{\pi t} $$ So HT is the output of a linear time-invariant filter with impulse response $h(t) = \dfrac{1}{\pi t}$. #### Frequency-domain expression The Fourier transform of $\dfrac{1}{\pi t}$ is $-j\,\text{sgn}(f)$. Using the convolution property: $$ \hat{X}(f) = H(f)X(f) = -j\,\text{sgn}(f)\,X(f) $$ where $$ H(f) = -j\,\text{sgn}(f) = \begin{cases} -j = e^{-j\pi/2}, & f > 0 \\ 0, & f = 0 \\ +j = e^{+j\pi/2}, & f < 0 \end{cases} $$ Therefore: - $|H(f)| = 1$ for all $f \neq 0$: amplitude spectrum is unchanged. - $\angle H(f) = -90^\circ$ for $f>0$ and $+90^\circ$ for $f<0$. ```text |H(f)| angle H(f) 1 ------+------ +90 ------+ | | ----------+-------> f ------------+-------> f +------ -90 ``` The Hilbert transformer is thus an **ideal wideband $90^\circ$ phase shifter**. **Example:** $x(t) = \cos 2\pi f_0 t$ has $X(f) = \tfrac12[\delta(f-f_0) + \delta(f+f_0)]$. $$ \hat{X}(f) = \tfrac{1}{2}[-j\delta(f-f_0) + j\delta(f+f_0)] \Rightarrow \hat{x}(t) = \sin 2\pi f_0 t $$ **Uses:** generation of SSB (phase-shift method), analytic signal $x(t) + j\hat{x}(t)$ and complex envelope in bandpass signal analysis. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The Hilbert transform is the output of an LTI filter with which impulse response? A. h(t) = sgn(t) B. h(t) = 1/(pi t) C. h(t) = delta(t) D. h(t) = pi t Answer: B. x^(t) = x(t) * 1/(pi t). (Key point: HT filter impulse response) 2. What is H(f) of the Hilbert transformer? A. j sgn(f) B. sgn(f) C. -j for all f D. -j sgn(f) Answer: D. H(f) = -j for f > 0 and +j for f < 0, i.e. -j sgn(f). (Key point: Frequency response) 3. Which statement about |H(f)| and angle H(f) is correct? A. |H(f)| = 1 and the phase is -90° for f > 0 and +90° for f < 0 B. |H(f)| = 0 for f < 0 and the phase is 0 C. |H(f)| = 1 and the phase is +90° for f > 0 D. |H(f)| = sgn(f) with zero phase Answer: A. It is an all-pass wideband 90° phase shifter. (Key point: Magnitude and phase) 4. What is the Hilbert transform of cos(2 pi f0 t)? A. -sin(2 pi f0 t) B. cos(2 pi f0 t) C. sin(2 pi f0 t) D. -cos(2 pi f0 t) Answer: C. The spectral lines are multiplied by -j at +f0 and +j at -f0, which gives a sine. 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