Short notes

DSB-SC Modulation Short Notes

In a balanced modulator circuit built around an MC1496 IC, feeding a 1 kHz audio tone and a 1 MHz carrier produces an output that contains components at 999 kHz and 1001 kHz — but not at 1 MHz. The carrier is suppressed by the balanced configuration, leaving only the two sidebands. This is DSB-SC: Double Sideband Suppressed Carrier. The MC1496 achieves better than 50 dB carrier suppression with good component matching, and this principle underpins AM stereo, QPSK, and quadrature mixers in modern SDR hardware.

ECE, EI

How it works

DSB-SC modulation multiplies the message signal m(t) directly by the carrier cos(2πf_c t): s(t) = m(t)·cos(2πf_c t). In the frequency domain, this convolution shifts M(f) to ±f_c: S(f) = ½[M(f−f_c) + M(f+f_c)]. For a single-tone message A_m·cos(2πf_m t), s(t) = (A_m/2)·[cos(2π(f_c−f_m)t) + cos(2π(f_c+f_m)t)] — two sidebands, no carrier. Bandwidth is 2W where W is the message bandwidth. Demodulation requires a coherent (synchronous) local carrier at exactly f_c in phase — multiplying the received signal by the local carrier and low-pass filtering recovers m(t), but a phase error θ reduces output by cos(θ) and a frequency error causes a beating distortion.

Key points to remember

DSB-SC power efficiency is 100% — all transmitted power carries information. Compared to full AM where the carrier wastes 67% of total power at 100% modulation index, DSB-SC is twice as power-efficient for the same information content. Bandwidth of DSB-SC equals 2 × message bandwidth — identical to full AM, but with no carrier component. Coherent detection requires a pilot carrier or a Costas loop to extract carrier phase at the receiver. Single Sideband (SSB) reduces bandwidth by half by transmitting only one sideband, at the cost of more complex receiver design. The Hilbert transform of m(t) is used in SSB generation via the phase method.

Exam tip

Anna University communication systems papers always ask you to draw the DSB-SC spectrum for a single-tone message and explain why a coherent detector is mandatory — make sure you draw both positive and negative frequency axes showing the four impulses at ±(fc±fm) with amplitude Am/2.

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