--- title: "SSB vs AM and DSB-SC; balanced modulator" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:ssb-vs-am-balanced-modulator" source: https://www.bctengineer.com/q/semester_6/communication_systems/ssb-vs-am-balanced-modulator site: BCT Engineering --- # SSB vs AM and DSB-SC; balanced modulator Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2075 Bhadra (CS I), 3+5 marks - Video: [Introduction to Amplitude Modulation | Double Side Band Suppressed (DSB-SC) Carrier Explained](https://www.youtube.com/watch?v=3wBlB-TFwkQ) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/ssb-vs-am-balanced-modulator - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question How does SSB differ from conventional AM and DSB-SC? Describe the process of generation of DSB-SC AM wave using Balance modulator. ## Short answer SSB uses half the bandwidth of conventional AM or DSB-SC by transmitting only one sideband, with no carrier. A balanced modulator feeds +m(t) and -m(t) to two identical AM modulators; subtracting outputs cancels the carrier, producing DSB-SC: s(t) = 2Ac ka m(t)cos ωct. ## Answer #### SSB vs conventional AM and DSB-SC SSB transmits only one sideband, without the carrier. Conventional AM sends the carrier and both sidebands, and DSB-SC sends both sidebands without the carrier. | Point | Conventional AM | DSB-SC | SSB | |---|---|---|---| | Components | Carrier + USB + LSB | USB + LSB | USB or LSB | | Bandwidth | $2W$ | $2W$ | $W$ | | Power at $\mu=1$ | $1.5P_c$ | $0.5P_c$ | $0.25P_c$ | | Efficiency | ≤ 33.3% | 100% | 100% | | Detection | Envelope | Coherent | Coherent | | Complexity | Low | Medium | High | #### DSB-SC generation by balanced modulator A balanced modulator uses **two identical AM modulators** in a balanced configuration, so that the carrier cancels and only the sidebands remain. ```text m(t) ---> AM modulator 1 --> s1(t) --+ ^ | (+) | (S)--> s(t) = s1 - s2 Ac cos wc t | (-) | | v | -m(t) --> AM modulator 2 --> s2(t) --+ ``` Modulator 1 gets $+m(t)$ and modulator 2 gets $-m(t)$; the same carrier goes to both. $$ \begin{aligned} s_1(t) &= A_c[1 + k_am(t)]\cos\omega_ct \\ s_2(t) &= A_c[1 - k_am(t)]\cos\omega_ct \\ s(t) &= s_1(t) - s_2(t) = 2A_ck_a\,m(t)\cos\omega_ct \end{aligned} $$ The output is DSB-SC: the carrier terms cancel exactly. **Practical circuit:** two diodes (or transistors) with a centre-tapped transformer. The carrier is fed so that its currents flow in opposite directions through the two halves of the output transformer and cancel, while the message drives the diodes in opposite senses so that the sideband currents add. **Spectrum:** for a single tone $m(t) = A_m\cos\omega_mt$: $$ s(t) = A_ck_aA_m[\cos(\omega_c+\omega_m)t + \cos(\omega_c-\omega_m)t] $$ ```text | LSB (no carrier) USB | | | -------+-----+---------+----------+----> f fc-fm fc fc+fm ``` **Requirement:** the two modulators must be perfectly matched; any imbalance lets some carrier through (carrier leakage). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In terms of Pc, what are total powers of AM, DSB-SC and SSB at mu = 1? A. 1.5Pc, 1.5Pc and 0.5Pc B. 1.5Pc, 0.5Pc and 0.25Pc C. 0.5Pc, 1.5Pc and 0.25Pc D. 1.5Pc, 0.25Pc and 0.5Pc Answer: B. The sidebands carry mu^2 Pc/4 each, 0.25Pc each. (Key point: Power at mu = 1) 2. What do the two modulators receive and how are the outputs combined? A. m(t) only into both; added B. Different carriers; multiplied C. m(t) and -m(t); added D. +m(t) and -m(t) with the same carrier; outputs are subtracted Answer: D. s1 - s2 = 2 Ac ka m(t) cos wc t. (Key point: Balanced modulator principle) 3. What happens to the carrier term in the output? A. It cancels exactly B. It doubles C. It is shifted by 90° D. It is unchanged Answer: A. Both s1 and s2 have Ac cos wc t, which cancels on subtraction. (Key point: Carrier cancellation) 4. What causes carrier leakage? A. Too much message power B. A very high carrier frequency C. Mismatch between the two modulators D. Narrow bandwidth Answer: C. Perfect cancellation needs matched modulators. (Key point: Practical limitation) ## Related in Communication Systems - [DSB-AM vs SSB; ring modulator for DSB-SC](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adsbam-ssb-compare-ring-modulator) — Communication Systems, chapter 2 - [Compare DSB-AM, DSB-SC, SSB, VSB; SSB by phase discrimination](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Acompare-am-variants-ssb-phase) — Communication Systems, chapter 2 - [Balanced modulator output is DSB-SC](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Abalanced-modulator-dsbsc) — 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 - [DSB-AM, DSB-SC and SSB spectra; synchronous detection](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adsbam-dsbsc-ssb-spectrum-synchronous) — Communication Systems, chapter 2 - [Compare DSB-AM, DSB-SC and SSB](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Acompare-dsbam-dsbsc-ssb) — Communication Systems, chapter 2 ## Related in other subjects - [Conventional vs renewable energy for generation](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aconventional-vs-renewable-generation) — Power System Analysis I, chapter 1, asked 2×