--- title: "Ring modulator for DSB-SC" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:ring-modulator-dsbsc" source: https://www.bctengineer.com/q/semester_6/communication_systems/ring-modulator-dsbsc site: BCT Engineering --- # Ring modulator for DSB-SC Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2074 Bhadra (CS I), 8 marks - Video: [CT#12 Ring modulator / DSBSC generator in communication system || EC Academy](https://www.youtube.com/watch?v=swGseODNscA) — EC Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/ring-modulator-dsbsc - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question With waveforms and necessary derivation, explain Ring Modulator method of generating DSB-SC signal. ## Short answer Ring modulator uses four diodes switched by a square-wave carrier to multiply message by ±1, yielding v(t)=m(t)cs(t). The square wave's Fourier series has no DC, so carrier and message are suppressed. A BPF at fc selects the fundamental, giving DSB-SC s(t)=(4/π)m(t)cosωct. ## Answer A **ring modulator** (double-balanced modulator) generates DSB-SC using four diodes connected in a ring, switched ON and OFF by a high-amplitude square-wave carrier. It effectively multiplies the message by a ±1 square wave. #### Circuit ```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) ``` (D1 joins a–b and D2 joins c–d: the straight paths. D3 joins a–d and D4 joins b–c: the crossed paths. All four point the same way round the ring.) #### Working - **Positive half cycle of carrier:** D1 and D2 conduct, D3 and D4 are off. The message passes straight to the output: $v_o = +m(t)$. - **Negative half cycle of carrier:** D3 and D4 conduct, D1 and D2 are off. The connections are crossed, so $v_o = -m(t)$. So the output is $v(t) = m(t)\,c_s(t)$, where $c_s(t)$ is a ±1 square wave of frequency $f_c$. #### Derivation The ±1 square wave (odd harmonics only, zero DC): $$ c_s(t) = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}\cos[2\pi f_c(2n-1)t] $$ $$ \begin{aligned} v(t) &= m(t)c_s(t) \\ &= \frac{4}{\pi}m(t)\cos\omega_ct - \frac{4}{3\pi}m(t)\cos3\omega_ct + \frac{4}{5\pi}m(t)\cos5\omega_ct - \dots \end{aligned} $$ There is **no carrier term and no message term** (the square wave has no DC). A band-pass filter centred at $f_c$ with bandwidth $2W$ keeps only $$ s(t) = \frac{4}{\pi}m(t)\cos\omega_ct $$ which is DSB-SC. Condition: $f_c > W$ and in practice $f_c \gg W$ so that the bands at $f_c$, $3f_c$ do not overlap ($3f_c - W > f_c + W$). #### Waveforms ```text m(t) ____ ____ / \ / ---------/------\------------/------> t \____ __/ \__/ c_s(t) +-+ +-+ +-+ +-+ +-+ +-+ (+-1) | | | | | | | | | | | | -+ +-+ +-+ +-+ +-+ +-+ +- v(t) m(t) chopped and inverted every half carrier cycle s(t) DSB-SC: envelope |m(t)|, (BPF) carrier phase reverses at each zero crossing of m(t) ``` #### Spectrum ```text |V(f)| /\ /\ /\ / \ / \ / \ --+--+--+---+--+--+-----+--+--+--> f (none at 0) fc 3fc (kept) (removed) ``` **Advantages:** carrier and message are both suppressed at the output (double-balanced); good carrier suppression; stable. **Requirement:** matched diodes and centre-tapped transformers. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the ring modulator, what happens during the negative half cycle of the carrier? A. D1 and D2 conduct and output is +m(t) B. All four diodes conduct C. D3 and D4 conduct, connections are crossed and output is -m(t) D. No diode conducts and output is zero Answer: C. D3 and D4 turn on and cross the connections, so the output is -m(t); this makes the output m(t) times a +/-1 square wave. (Key point: Diode switching action) 2. The ring modulator output before filtering is v(t) = m(t) c_s(t), where c_s(t) is: A. A sinusoid at fc B. A 0/1 pulse train C. A +/-1 square wave of frequency fc D. A DC level Answer: C. The circuit alternately passes and inverts m(t), equivalent to multiplication by a +/-1 square wave. (Key point: Square wave multiplication) 3. Why does the output contain neither a carrier term nor a message term? A. The square wave has only odd harmonics and no DC B. The transformers block them C. The BPF is a notch filter D. The diodes are reverse biased Answer: A. The +/-1 square wave has zero DC (no m(t) term) and no component at fc multiplied by 1 alone, so only m(t) times harmonics remain. (Key point: Why no carrier or message appears) 4. A band-pass filter at fc with bandwidth 2W leaves which output? A. m(t) cos(3 wc t) B. (4/pi) m(t) cos(wc t), a DSB-SC signal C. m(t) + cos(wc t) D. (4/pi) cos(wc t) only Answer: B. Only the fundamental term (4/pi) m(t) cos(wc t) lies in the pass band; higher harmonic bands are removed. (Key point: BPF and final result) ## Related in Communication Systems - [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 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 - [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 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 - [DSB-FC vs DSB-SC; phase shift method for SSB](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adsbfc-dsbsc-ssb-phase-shift) — 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