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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: vo=+m(t)v_o = +m(t).
  • Negative half cycle of carrier: D3 and D4 conduct, D1 and D2 are off. The connections are crossed, so vo=−m(t)v_o = -m(t).

So the output is v(t)=m(t) cs(t)v(t) = m(t)\,c_s(t), where cs(t)c_s(t) is a ±1 square wave of frequency fcf_c.

Derivation

The ±1 square wave (odd harmonics only, zero DC):

cs(t)=4π∑n=1∞(−1)n+12n−1cos⁡[2πfc(2n−1)t]c_s(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)cs(t)=4πm(t)cos⁡ωct−43πm(t)cos⁡3ωct+45πm(t)cos⁡5ωct−…\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 fcf_c with bandwidth 2W2W keeps only

s(t)=4πm(t)cos⁡ωcts(t) = \frac{4}{\pi}m(t)\cos\omega_ct

which is DSB-SC. Condition: fc>Wf_c > W and in practice fc≫Wf_c \gg W so that the bands at fcf_c, 3fc3f_c do not overlap (3fc−W>fc+W3f_c - W > f_c + W).

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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