--- title: "USB-SSB expression for a random band limited message" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:usb-ssb-random-message" source: https://www.bctengineer.com/q/semester_6/communication_systems/usb-ssb-random-message site: BCT Engineering --- # USB-SSB expression for a random band limited message Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2067 Mangsir (CS I), 10 marks - Video: [Understanding Single Sideband](https://www.youtube.com/watch?v=sv1xzlBut1I) — Rohde & Schwarz - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/usb-ssb-random-message - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question Derive the expression for the USB-SSB signal in terms of the carrier c(t) = Ac cos(ωct) and a random band limited signal m(t). ## Short answer The USB-SSB signal is s_USB(t) = (Ac/2)[m(t) cos(omega_c t) - m_hat(t) sin(omega_c t)], where m_hat(t) is the Hilbert transform of the band-limited message. It occupies half the bandwidth of DSB and requires coherent detection. ## Answer **SSB-SC** transmits only one sideband of a DSB-SC wave. For a random band-limited message $m(t)$ (bandwidth $W$), the upper sideband (USB) signal is expressed using the Hilbert transform $\hat m(t)$ of the message. #### 1. Starting point: DSB-SC $$ s_{DSB}(t) = A_cm(t)\cos\omega_ct \;\Leftrightarrow\; S_{DSB}(f) = \frac{A_c}{2}[M(f-f_c) + M(f+f_c)] $$ ```text M(f) S_DSB(f) /\ USB LSB LSB USB / \ __/|\__ __/|\__ ----/----\---- --/---|---\-----/---|---\--> f -W 0 W -fc fc ``` #### 2. Selecting the USB The USB is the part of $S_{DSB}(f)$ with $|f| > f_c$. Using the unit step in frequency, $u(f) = \frac{1}{2}[1 + \text{sgn}(f)]$: $$ S_{USB}(f) = \frac{A_c}{2}M(f-f_c)u(f-f_c) + \frac{A_c}{2}M(f+f_c)u(-(f+f_c)) $$ Substituting $u$: $$ \begin{aligned} S_{USB}(f) &= \frac{A_c}{4}[M(f-f_c) + M(f+f_c)] \\ &\quad + \frac{A_c}{4}[M(f-f_c)\,\text{sgn}(f-f_c) - M(f+f_c)\,\text{sgn}(f+f_c)] \end{aligned} $$ #### 3. Hilbert transform The Hilbert transform shifts every frequency component of $m(t)$ by −90°: $$ \hat m(t) = m(t) * \frac{1}{\pi t} \;\Leftrightarrow\; \hat M(f) = -j\,\text{sgn}(f)\,M(f) $$ So $M(f)\,\text{sgn}(f) = j\hat M(f)$, and $$ M(f-f_c)\text{sgn}(f-f_c) - M(f+f_c)\text{sgn}(f+f_c) = j[\hat M(f-f_c) - \hat M(f+f_c)] $$ #### 4. Inverse Fourier transform Using the modulation theorem: $$ \begin{aligned} m(t)\cos\omega_ct &\Leftrightarrow \frac{1}{2}[M(f-f_c) + M(f+f_c)] \\ \hat m(t)\sin\omega_ct &\Leftrightarrow \frac{1}{2j}[\hat M(f-f_c) - \hat M(f+f_c)] \end{aligned} $$ So the first bracket of $S_{USB}(f)$ gives $\frac{A_c}{2}m(t)\cos\omega_ct$, and the second gives $$ \frac{jA_c}{4}[\hat M(f-f_c) - \hat M(f+f_c)] = \frac{jA_c}{4}\cdot 2j\,\mathcal{F}\{\hat m(t)\sin\omega_ct\} = -\frac{A_c}{2}\mathcal{F}\{\hat m(t)\sin\omega_ct\} $$ Therefore $$ s_{USB}(t) = \frac{A_c}{2}\left[m(t)\cos\omega_ct - \hat m(t)\sin\omega_ct\right] $$ Similarly, $s_{LSB}(t) = \frac{A_c}{2}[m(t)\cos\omega_ct + \hat m(t)\sin\omega_ct]$. #### 5. Verification with a single tone Let $m(t) = A_m\cos\omega_mt$, so $\hat m(t) = A_m\sin\omega_mt$: $$ s_{USB}(t) = \frac{A_cA_m}{2}[\cos\omega_mt\cos\omega_ct - \sin\omega_mt\sin\omega_ct] = \frac{A_cA_m}{2}\cos(\omega_c + \omega_m)t $$ Only the upper side frequency is present, which confirms the result. #### 6. Interpretation - $m(t)\cos\omega_ct$ is the in-phase part; $\hat m(t)\sin\omega_ct$ is the quadrature part that cancels the lower sideband. - The bandwidth is $W$, half of DSB, and the power is that of one sideband only. - The envelope is $\frac{A_c}{2}\sqrt{m^2 + \hat m^2}$, not $m(t)$, so coherent detection is required. - The equation directly gives the **phase-shift method** of SSB generation: ```text m(t) -+---------->[ X ]--+ | ^cos | + | | +-->(+/-)--> s_SSB | [ Osc ]--+ | - | |-90 deg | +-[-90 deg]->[ X ]-+ (Hilbert) ^sin ``` (subtract for USB, add for LSB). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which is the USB-SSB signal for message m(t) and Hilbert transform m^(t)? A. (Ac/2)[m(t) cos ωc t + m^(t) sin ωc t] B. Ac m(t) cos ωc t C. (Ac/2)[m(t) cos ωc t − m^(t) sin ωc t] D. (Ac/2)[m(t) sin ωc t − m^(t) cos ωc t] Answer: C. The in-phase term minus the quadrature term keeps the upper sideband; adding gives the LSB. (Key point: Final USB expression) 2. What does the Hilbert transform do to m(t)? A. Doubles every frequency B. Shifts every frequency component by −90° (M^(f) = −j sgn(f) M(f)) C. Shifts every frequency component by +180° D. Removes negative frequencies and doubles the rest Answer: B. The Hilbert transform is a −90° phase shift of all frequency components with magnitude unchanged. (Key point: Hilbert transform) 3. How is the USB picked from the DSB-SC spectrum in the derivation? A. By multiplying by a unit step u(f − fc) (and its mirror), written using sgn(f) B. By squaring the DSB-SC signal C. By differentiating the DSB-SC signal D. By adding a carrier to the DSB-SC signal Answer: A. The USB is the part with |f| > fc; the step function u(f) = (1 + sgn f)/2 selects it. (Key point: Selecting the USB) 4. For m(t) = Am cos ωm t, what does the USB expression reduce to? A. (Ac Am/2) cos(ωc − ωm)t B. Ac Am cos ωc t C. (Ac Am/2) cos ωm t D. (Ac Am/2) cos(ωc + ωm)t Answer: D. cos ωm t cos ωc t − sin ωm t sin ωc t = cos(ωc + ωm)t, so only the upper side frequency remains. (Key point: Single-tone check) ## Related in Communication Systems - [SSB wave for a low pass message signal](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Assb-lowpass-message) — Communication Systems, chapter 2 - [SSB in terms of Hilbert transform; SSB generation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Assb-hilbert-derivation-generation) — Communication Systems, chapter 2, asked 3× - [Band pass and band limited signals; LTI properties](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Abandpass-bandlimited-lti-properties) — Communication Systems, chapter 1 - [Advantages of SSB over DSB-FC; SSB expression](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Assb-advantages-derive) — Communication Systems, chapter 2 - [SSB vs full carrier AM; ring modulator](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Assb-vs-am-ring-modulator) — Communication Systems, chapter 2 - [FM expression after change of modulating signal](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-new-waveform-500hz-2khz) — Communication Systems, chapter 3 ## Related in other subjects - [Aliasing in a band-limited signal](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aaliasing-bandlimited-w) — Signals and Systems, chapter 4 - [Band-limited signals; aliasing; Nyquist rate](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Abandlimited-aliasing-nyquist) — Signals and Systems, chapter 4 - [Small signal model, AC resistance of diode](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Apn-diode-small-signal-ac-resistance) — Basic Electronics Engineering, chapter 2, asked 2× - [Message transfer parts and signal unit addressing](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Amtp-signal-unit-addressing) — Telecommunications, chapter 5