--- title: "DSB-AM vs SSB; ring modulator for DSB-SC" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:dsbam-ssb-compare-ring-modulator" source: https://www.bctengineer.com/q/semester_6/communication_systems/dsbam-ssb-compare-ring-modulator site: BCT Engineering --- # DSB-AM vs SSB; ring modulator for DSB-SC Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2073 Magh (CS I), 4+6 marks - Video: [Ring Modulator or Chopper Modulator for Generation of DSB-SC signal || Lec-19](https://www.youtube.com/watch?v=kG2zX1RcmZE) — Engg. & Math Insights - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/dsbam-ssb-compare-ring-modulator - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question Compare DSB-AM and SSB wave in terms of transmission power and bandwidth. Describe how ring modulator can be used to generate DSB-SC. ## Short answer DSB-AM bandwidth 2fm, total power Pc(1+μ^2/2), useful power ≤33%; SSB bandwidth fm, power Pcμ^2/4, 100% useful. At μ=1, SSB saves 83% power and half bandwidth. Ring modulator uses four diodes switched by square-wave carrier to multiply message by ±1, then BPF extracts DSB-SC. ## Answer #### DSB-AM vs SSB: power and bandwidth Single tone, carrier amplitude $A_c$, index $\mu$, load $R$; $P_c = A_c^2/2R$. | Item | DSB-AM (DSB-FC) | SSB-SC | |---|---|---| | Components | Carrier + USB + LSB | One sideband only | | Bandwidth | $2f_m$ (or $2W$) | $f_m$ (or $W$) | | Total power | $P_c(1 + \mu^2/2)$ | $P_c\mu^2/4$ | | At $\mu = 1$ | $1.5P_c$ | $0.25P_c$ | | Useful fraction | ≤ 33.3% | 100% | **Power saving** of SSB over DSB-FC at $\mu = 1$: $$ \frac{1.5P_c - 0.25P_c}{1.5P_c} = \frac{1.25}{1.5} = 83.3\% $$ So SSB saves half the bandwidth and about 83% of the power. #### Ring modulator for DSB-SC The ring (double-balanced) modulator has four diodes in a ring between two centre-tapped transformers; a square-wave carrier, much larger than the message, is fed between the centre taps and acts as a switch. ```text T1 D1 T2 m(t) o--+ )||( a---->|----b )||( +--o | )||( |\ /| )||( | to | )||( | D3 D4 | )||( | BPF o-------+ )||( | \ / | )||( +--o | | X | | | | / \ | | | c---->|----d | | D2 | +--- c(t) ------+ (square-wave carrier between centre taps of T1 and T2) ``` **Working:** - Carrier positive: D1, D2 ON; D3, D4 OFF → output $= +m(t)$. - Carrier negative: D3, D4 ON; D1, D2 OFF → transformer connections reversed → output $= -m(t)$. So $v(t) = m(t)c_s(t)$ with $c_s(t)$ a ±1 square wave: $$ v(t) = m(t)\cdot\frac{4}{\pi}\left[\cos\omega_ct - \frac13\cos3\omega_ct + \frac15\cos5\omega_ct - \dots\right] $$ The square wave has no DC, so no message term appears, and no carrier term appears either. A BPF at $f_c$ (bandwidth $2W$) gives $$ s(t) = \frac{4}{\pi}m(t)\cos\omega_ct \quad \text{(DSB-SC)} $$ ```text m(t) ~~~~/\~~~~~/\~~~ c_s(t) +-+ +-+ +-+ +-+ v(t) m(t) sign-flipped each half cycle s(t) DSB-SC with phase reversal at m(t)=0 ``` Requirement: $f_c > W$ (in practice $f_c \gg W$) so that the bands around $f_c$ and $3f_c$ do not overlap; the diodes must be matched for good carrier suppression. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The bandwidths of DSB-AM and SSB are respectively: A. W and 2W B. 2W and W C. 2W and 2W D. W and W Answer: B. DSB-AM sends both sidebands (2W); SSB sends only one (W). (Key point: Bandwidth comparison) 2. At mu = 1, the total power of DSB-FC and SSB (single tone) are: A. 1.5 Pc and 0.25 Pc B. 1.5 Pc and 0.5 Pc C. 0.25 Pc and 1.5 Pc D. 2 Pc and 0.25 Pc Answer: A. DSB-FC: Pc(1 + 1/2) = 1.5 Pc; SSB: Pc mu^2/4 = 0.25 Pc, a saving of about 83.3%. (Key point: Power comparison) 3. How does the ring modulator act on the message? A. It adds the carrier to m(t) B. It squares m(t) C. It filters m(t) with a notch D. A large square-wave carrier switches diodes so the output is m(t) multiplied by +/-1 Answer: D. The carrier acts as a switch, passing m(t) straight or inverted in each half cycle. (Key point: Ring modulator working) 4. After the ring modulator, a BPF at fc yields: A. (4/pi) m(t) cos(wc t), with no carrier term B. m(t) + cos(wc t) C. cos(wc t) only D. m(t) cos(3 wc t) Answer: A. The square wave has no DC, so no carrier or message appears, and the BPF keeps the fundamental term. (Key point: Filtering to get DSB-SC) ## Related in Communication Systems - [SSB vs AM and DSB-SC; balanced modulator](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Assb-vs-am-balanced-modulator) — 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 - [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 - [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 - [Ring modulator for DSB-SC; characteristics of DSB-SC](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aring-modulator-dsbsc-characteristics) — 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 ## Related in other subjects - [AM wave power and bandwidth numerical](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aam-numeric-40-carrier-power) — Data Communication, chapter 5