--- title: "SSB vs full carrier AM; ring modulator" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:ssb-vs-am-ring-modulator" source: https://www.bctengineer.com/q/semester_6/communication_systems/ssb-vs-am-ring-modulator site: BCT Engineering --- # SSB vs full carrier AM; ring modulator Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2070 Bhadra (CS I), 2+5 marks - Video: [SSB Modulation in communication systems](https://www.youtube.com/watch?v=iwmwQeQT_Jk) — Signals Systems - 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-ring-modulator - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question How is SSB different from conventional full carrier AM? Describe how ring modulator can be used to generate DSB-SC. ## Short answer SSB transmits only one sideband with carrier suppressed, using half the bandwidth of conventional AM and saving 83% power. A ring modulator switches the message with a carrier square wave via diodes, producing DSB-SC which is filtered to retain the component at the carrier frequency. ## Answer #### SSB versus conventional AM SSB-SC transmits only one sideband with the carrier suppressed, while conventional AM (DSB-FC) sends the carrier and both sidebands. | Point | DSB-FC AM | SSB-SC | |---|---|---| | Spectrum sent | Carrier + USB + LSB | One sideband only | | Bandwidth | $2W$ | $W$ | | Power ($\mu=1$) | $1.5P_c$ | $0.25P_c$ (83.3% saving) | | Efficiency | ≤ 33.3% | 100% | | Demodulation | Envelope detector | Coherent detector needed | | Receiver | Simple, cheap | Complex (carrier recovery) | | Noise/fading | More affected | Less selective fading | | Equation | $A_c[1+\mu m(t)]\cos\omega_ct$ | $\frac{A_c}{2}[m\cos\omega_ct \mp \hat m\sin\omega_ct]$ | #### Ring modulator for DSB-SC A ring modulator uses four diodes connected in a ring between two centre-tapped transformers. The carrier is applied between the centre taps and acts as a switch; the message enters at the input transformer. ```text T1 D1 T2 m(t) ))|(( a----|>|----b ))|(( s(t) |(( \ / ))| |(( D4 D2 ))| |(( / \ ))| )|( d----|<|----c ))| | D3 | +----( c(t) )---+ (carrier between centre taps) ``` **Operation** 1. The carrier amplitude is much larger than $m(t)$, so the carrier alone decides which diodes conduct. 2. **Positive carrier half cycle:** D1 and D3 conduct, D2 and D4 are off. The message passes to the output with its original polarity. 3. **Negative carrier half cycle:** D2 and D4 conduct, D1 and D3 are off. The connections are crossed, so the message reaches the output **inverted**. 4. So the output is $m(t)$ multiplied by a bipolar square wave $c(t)$ of frequency $f_c$: $$ c(t) = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}\cos[2\pi f_c(2n-1)t] $$ $$ v(t) = m(t)c(t) = \frac{4}{\pi}m(t)\cos\omega_ct - \frac{4}{3\pi}m(t)\cos 3\omega_ct + \dots $$ 5. The square wave has no DC term, so no carrier and no baseband $m(t)$ appear at the output. A band-pass filter centred at $f_c$ (bandwidth $2W$) keeps only $$ s(t) = \frac{4}{\pi}m(t)\cos\omega_ct $$ which is DSB-SC. The circuit is balanced, so the carrier leakage is very small when the diodes are matched. The condition $f_c > W$ (in practice $f_c \gg W$) keeps the bands around $f_c$ and $3f_c$ separate. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Compared with DSB-FC at mu=1, SSB needs A. half the bandwidth and 0.25Pc instead of 1.5Pc B. twice the bandwidth C. half bandwidth and 1.5Pc D. the same bandwidth and 0.25Pc Answer: A. SSB bandwidth is W, and its power is 0.25Pc (83.3% saving). (Key point: Bandwidth and power) 2. How is SSB demodulated? A. No demodulation needed B. Square-law detector only C. Coherent detector D. Envelope detector Answer: C. The carrier is suppressed, so a local carrier is needed. (Key point: Detection) 3. In the ring modulator, what does the carrier do? A. Is amplified by the diodes B. Acts as a switch deciding which diodes conduct C. Is added to the message D. Filters the message Answer: B. A large carrier makes D1,D3 conduct on positive half and D2,D4 on negative half. (Key point: Ring modulator operation) 4. What does the ring modulator output multiply m(t) by? A. A constant B. A sine wave at 2fc C. A 0/1 pulse train D. A bipolar square wave at fc Answer: D. The message polarity flips every half cycle; the BPF leaves (4/pi) m(t) cos wc t. (Key point: Output) ## 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 - [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 - [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 - [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 - [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