--- title: "Hilbert transform properties; distortionless transmission" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:hilbert-properties-distortionless" source: https://www.bctengineer.com/q/semester_6/communication_systems/hilbert-properties-distortionless site: BCT Engineering --- # Hilbert transform properties; distortionless transmission Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2076 Baisakh (CS I), 2+2+4 marks; 2076 Bhadra (CS I), 2+2+3 marks; 2075 Baisakh (CS I), 3+2+3 marks (asked 3 times) - Video: [Hilbert Transform (Basics, Properties and Applications), Cross Correlation and Autocorrelation, PSD](https://www.youtube.com/watch?v=vsHcyhLDZ2U) — Engineering Made Easy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/hilbert-properties-distortionless - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question What is Hilbert transform; describe it with mathematical expression and frequency response. Mention the properties of Hilbert Transform. Explain distortionless transmission line with its frequency response. ## Short answer The Hilbert transform m̂(t) = (1/π)∫ m(τ)/(t-τ)dτ, with H(f) = -j sgn(f). Properties: same amplitude spectrum, m̂̂ = -m, orthogonality. A distortionless line requires H(f) = K e^{-j2π f t_d}, i.e. constant amplitude and linear phase (constant group delay). ## Answer #### Hilbert transform The **Hilbert transform** $\hat x(t)$ of a signal $x(t)$ is obtained by shifting the phase of every positive-frequency component by $-90°$ and every negative-frequency component by $+90°$, without changing amplitudes. **Time domain:** it is the convolution of $x(t)$ with $\frac{1}{\pi t}$: $$ \hat x(t)=x(t)*\frac{1}{\pi t}=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{x(\tau)}{t-\tau}d\tau $$ **Frequency response:** since $\frac{1}{\pi t}\leftrightarrow-j\,\text{sgn}(f)$, $$ H(f)=-j\,\text{sgn}(f)=\begin{cases}-j=1\angle-90°, & f>0\\ +j=1\angle+90°, & f<0\end{cases},\qquad \hat X(f)=-j\,\text{sgn}(f)X(f) $$ ```text |H(f)| angle H(f) 1 +-------+------ +90 ----+ | | | --+-------+------> f -------+-------> f 0 | +---- -90 ``` **Inverse:** $x(t)=-\hat{\hat x}(t)$, i.e. $x(t)=-\frac{1}{\pi}\int\frac{\hat x(\tau)}{t-\tau}d\tau$. #### Properties of the Hilbert transform 1. $x(t)$ and $\hat x(t)$ have the **same amplitude spectrum**, hence the same ESD/PSD, energy and power. 2. Applying it twice gives the negative: $\hat{\hat x}(t)=-x(t)$. 3. $x(t)$ and $\hat x(t)$ are **orthogonal**: $\int_{-\infty}^{\infty}x(t)\hat x(t)dt=0$. 4. $\widehat{\cos\omega t}=\sin\omega t$ and $\widehat{\sin\omega t}=-\cos\omega t$. 5. For a low-pass $m(t)$ and carrier $f_c$ greater than its bandwidth: $\widehat{m(t)\cos\omega_ct}=m(t)\sin\omega_ct$. 6. It is a linear operation, and an even signal gives an odd Hilbert transform (and vice versa). Use: SSB generation $s(t)=m\cos\omega_ct\mp\hat m\sin\omega_ct$, pre-envelope and band-pass signal representation. #### Distortionless transmission (line) A transmission system or line is **distortionless** if the output is an exact replica of the input except for a constant scale factor and a constant delay: $$ y(t)=K\,x(t-t_d)\ \Rightarrow\ H(f)=K\,e^{-j2\pi ft_d} $$ **Frequency response requirements:** - **Amplitude:** $\lvert H(f)\rvert=K$ constant for all frequencies in the signal band; otherwise amplitude distortion. - **Phase:** $\theta(f)=-2\pi ft_d$, a straight line through the origin with slope $-2\pi t_d$; i.e. constant group delay $\tau_g=-\frac{1}{2\pi}\frac{d\theta}{df}=t_d$. Otherwise phase (delay) distortion. ```text |H(f)| theta(f) K +-------------- |\ | -----+-\------> f +-------------> f | \ slope -2*pi*td ``` **For a physical transmission line** with parameters $R, L, G, C$ per unit length, this is achieved when the **Heaviside condition** $\frac{R}{L}=\frac{G}{C}$ holds. Then the attenuation constant $\alpha=\sqrt{RG}$ and the velocity $v=\frac{1}{\sqrt{LC}}$ are both independent of frequency, so all frequency components are attenuated and delayed equally. Telephone lines achieve this approximately by loading coils that increase $L$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the Hilbert transform do to signal components? A. Shifts positive frequencies by +90 and negative by -90, changing amplitude B. Shifts positive frequencies by -90 and negative by +90, keeping amplitude C. Doubles the amplitude of positive frequencies only D. Delays all components by the same time Answer: B. H(f) = -j sgn(f) has magnitude 1, angle -90 for f>0 and +90 for f<0. (Key point: Hilbert phase action) 2. What is the Hilbert transform of the Hilbert transform of x(t)? A. x(t) B. -x(t) C. 0 D. 2x(t) Answer: B. Applying -j sgn(f) twice gives (-j)^2 sgn^2(f) = -1, so the result is -x(t). (Key point: Double transform) 3. A system is distortionless if its transfer function is: A. H(f) = K, with any phase B. H(f) = K exp(-j2 pi f td) C. H(f) = K f D. H(f) = K exp(-j2 pi f^2 td) Answer: B. Output must be K x(t - td): constant amplitude and linear phase through the origin (constant group delay). (Key point: Distortionless condition) 4. Which condition makes a physical line distortionless? A. R = G B. R/L = G/C C. L = C D. RL = GC^2 Answer: B. When R/L = G/C, attenuation sqrt(RG) and velocity 1/sqrt(LC) are independent of frequency. 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