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SSB vs AM and DSB-SC; balanced modulator

Short answer

SSB uses half the bandwidth of conventional AM or DSB-SC by transmitting only one sideband, with no carrier. A balanced modulator feeds +m(t) and -m(t) to two identical AM modulators; subtracting outputs cancels the carrier, producing DSB-SC: s(t) = 2Ac ka m(t)cos ωct.

How does SSB differ from conventional AM and DSB-SC? Describe the process of generation of DSB-SC AM wave using Balance modulator.

SSB vs conventional AM and DSB-SC

SSB transmits only one sideband, without the carrier. Conventional AM sends the carrier and both sidebands, and DSB-SC sends both sidebands without the carrier.

PointConventional AMDSB-SCSSB
ComponentsCarrier + USB + LSBUSB + LSBUSB or LSB
Bandwidth2W2W2W2WWW
Power at μ=1\mu=11.5Pc1.5P_c0.5Pc0.5P_c0.25Pc0.25P_c
Efficiency≤ 33.3%100%100%
DetectionEnvelopeCoherentCoherent
ComplexityLowMediumHigh

DSB-SC generation by balanced modulator

A balanced modulator uses two identical AM modulators in a balanced configuration, so that the carrier cancels and only the sidebands remain.

 m(t) ---> AM modulator 1 --> s1(t) --+
                ^                     | (+)
                |                    (S)--> s(t) = s1 - s2
          Ac cos wc t                 | (-)
                |                     |
                v                     |
 -m(t) --> AM modulator 2 --> s2(t) --+

Modulator 1 gets +m(t)+m(t) and modulator 2 gets −m(t)-m(t); the same carrier goes to both.

s1(t)=Ac[1+kam(t)]cos⁡ωcts2(t)=Ac[1−kam(t)]cos⁡ωcts(t)=s1(t)−s2(t)=2Acka m(t)cos⁡ωct\begin{aligned} s_1(t) &= A_c[1 + k_am(t)]\cos\omega_ct \\ s_2(t) &= A_c[1 - k_am(t)]\cos\omega_ct \\ s(t) &= s_1(t) - s_2(t) = 2A_ck_a\,m(t)\cos\omega_ct \end{aligned}

The output is DSB-SC: the carrier terms cancel exactly.

Practical circuit: two diodes (or transistors) with a centre-tapped transformer. The carrier is fed so that its currents flow in opposite directions through the two halves of the output transformer and cancel, while the message drives the diodes in opposite senses so that the sideband currents add.

Spectrum: for a single tone m(t)=Amcos⁡ωmtm(t) = A_m\cos\omega_mt:

s(t)=AckaAm[cos⁡(ωc+ωm)t+cos⁡(ωc−ωm)t]s(t) = A_ck_aA_m[\cos(\omega_c+\omega_m)t + \cos(\omega_c-\omega_m)t]
        |    LSB   (no carrier)   USB
        |     |                    |
 -------+-----+---------+----------+----> f
             fc-fm      fc       fc+fm

Requirement: the two modulators must be perfectly matched; any imbalance lets some carrier through (carrier leakage).

✓ Answer checked by Claude Opus 5.5 (high effort)

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