--- title: "Ring modulator for DSB-SC; characteristics of DSB-SC" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:ring-modulator-dsbsc-characteristics" source: https://www.bctengineer.com/q/semester_6/communication_systems/ring-modulator-dsbsc-characteristics site: BCT Engineering --- # Ring modulator for DSB-SC; characteristics of DSB-SC Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2064 Shrawan (CS I), 5+3 marks - Video: [Generation of DSB SC modulation by Ring Modulator](https://www.youtube.com/watch?v=0UZi2jTqnHQ) — HOD Electronics & Communication Engineering - 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-characteristics - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question Show how can a ring modulator be used to generate DSB-SC signal, explain with neat diagrams. What are the basic characteristics of DSB-SC? ## Short answer A ring modulator multiplies the message by a ±1 square wave from four diodes switched by a large carrier. After bandpass filter, output is (4/pi)m(t) cos(omega_c t), a DSB-SC signal with no carrier, bandwidth 2W, 100% power efficiency, and needing coherent detection. ## Answer A **ring modulator** is a double-balanced diode modulator in which four diodes form a ring and are switched by the carrier, so the output is the message multiplied by a square wave, which after filtering is DSB-SC. #### Circuit ```text T1 D1 T2 m(t) ))|((a-----|>|-----b))|(( --> BPF --> s(t) |(( \ / ))| |(( D4 D2 ))| |(( / \ ))| )|( d-----|<|----c ))| | D3 | +----( c(t) )-----+ carrier at centre taps ``` #### Operation 1. The carrier $c(t)$ is a large square wave (or large sinusoid), $|c(t)| \gg |m(t)|$, so it alone controls the diodes. 2. **Positive half cycle of carrier:** D1 and D3 are forward biased, D2 and D4 off. The message reaches the output with the same polarity: $v(t) = +m(t)$. 3. **Negative half cycle:** D2 and D4 conduct, D1 and D3 are off. The paths cross over, so $v(t) = -m(t)$. 4. Hence $v(t) = m(t)\,c(t)$, where $c(t)$ is a ±1 square wave: $$ c(t) = \frac{4}{\pi}\left[\cos\omega_ct - \frac{1}{3}\cos3\omega_ct + \frac{1}{5}\cos5\omega_ct - \dots\right] $$ $$ v(t) = \frac{4}{\pi}m(t)\cos\omega_ct - \frac{4}{3\pi}m(t)\cos3\omega_ct + \dots $$ 5. Because $c(t)$ has zero mean, there is no carrier term and no baseband term in the output. A band-pass filter at $f_c$ with bandwidth $2W$ gives $$ s(t) = \frac{4}{\pi}m(t)\cos\omega_ct \quad \text{(DSB-SC)} $$ ```text m(t) ___ / \___ c(t) |-| |-| |-| |_| |_| |_ v(t) m(t) chopped and flipped every half cycle ``` #### Basic characteristics of DSB-SC - Contains only USB and LSB; the carrier is suppressed. - Bandwidth $= 2W$, same as DSB-FC. - All transmitted power is in the sidebands: power efficiency 100%. For a tone, power $= P_c\mu^2/2$, a saving of 66.7% vs DSB-FC at $\mu = 1$. - The envelope is $|m(t)|$; the carrier phase reverses at each zero crossing of $m(t)$. - Cannot use an envelope detector; needs coherent detection with exact frequency and phase (Costas loop, squaring loop or pilot carrier). - Used in stereo FM (L−R signal), colour TV chroma, and QAM. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the ring modulator, what happens in the negative half cycle of the carrier? A. All diodes conduct B. D1 and D3 conduct, passing +m(t) C. No diode conducts and the output is zero D. D2 and D4 conduct and the paths cross, giving −m(t) Answer: D. The crossed path reverses polarity, so v(t) = −m(t) in the negative half cycle. (Key point: Diode action) 2. What is the ring modulator output before filtering? A. m(t) multiplied by a ±1 square wave c(t) B. m(t) plus a square wave C. m(t) multiplied by a unit step D. m(t) squared Answer: A. Output is +m(t) or −m(t) in alternate half cycles, i.e. m(t) c(t) with c(t) a ±1 square wave. (Key point: Output as a product) 3. Why does the output contain neither a carrier nor a baseband m(t) term? A. The BPF removes the carrier only B. The message has zero mean C. The square wave c(t) has zero mean, so it has no DC term D. The diodes absorb them Answer: C. A ±1 square wave has only odd harmonics of fc and no DC, so m(t) c(t) has only m(t) cos(nωc t) terms. (Key point: Why no carrier or baseband term) 4. Which statement about DSB-SC is correct? A. It can be detected by a simple envelope detector B. Its bandwidth is W, half that of DSB-FC C. It needs coherent detection with exact frequency and phase D. Its power efficiency is 33% Answer: C. Its envelope is |m(t)|, not m(t), so coherent detection is needed. Bandwidth is 2W and all power is in sidebands (100% efficient). 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