- 2074 Bhadra (CS I) · 8 marks
Ring modulator for DSB-SC
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.
With waveforms and necessary derivation, explain Ring Modulator method of generating DSB-SC signal.
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
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: .
- Negative half cycle of carrier: D3 and D4 conduct, D1 and D2 are off. The connections are crossed, so .
So the output is , where is a ±1 square wave of frequency .
Derivation
The ±1 square wave (odd harmonics only, zero DC):
There is no carrier term and no message term (the square wave has no DC). A band-pass filter centred at with bandwidth keeps only
which is DSB-SC. Condition: and in practice so that the bands at , do not overlap ().
Waveforms
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
|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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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