Question 5 of 6: Spectral Analysis of a Composite Message — Bandwidth Under DSB and SSB

Question 5: Spectral Analysis of a Composite Message — Bandwidth Under DSB and SSB (1/6 of paper)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $m(t)=\cos^3(2\pi f_mt)+\text{sinc}(f_mt)$ with $\text{sinc}(u)=\sin(\pi u)/(\pi u)$; carrier frequency $f_c=20f_m$.

Find. (a) message spectrum and bandwidth; (b) DSB spectrum and bandwidth; (c) LSB-SSB spectrum and bandwidth; (d) an exact DSB demodulator (general, any bandlimited message); (e) an up-converter from $20f_m$ to $25f_m$.

Approach. Expand $\cos^3\theta$ with the triple-angle identity to expose its two discrete spectral lines; recognize the second term as a sinc pulse whose Fourier transform is a baseband rectangle; combine the two for the total message bandwidth, then shift/mirror the result for DSB and trim to one side for SSB. The recovery and re-mixing systems are standard coherent-detection and frequency-translation stages.

1. Part (a) — message spectrum and bandwidth. The triple-angle identity $\cos^3\theta=\tfrac34\cos\theta+\tfrac14\cos3\theta$ turns the first term into two discrete tones: $$\Big(\cos2\pi f_mt\Big)^3=\tfrac34\cos(2\pi f_mt)+\tfrac14\cos(2\pi\cdot3f_mt)$$ giving spectral lines at $\pm f_m$ (amplitude $3/8$ each, i.e. half of $3/4$ from splitting the cosine into two exponentials) and $\pm3f_m$ (amplitude $1/8$ each). The second term is a sinc pulse; by the standard pair $\text{sinc}(at)\leftrightarrow\tfrac1a\,\text{rect}(f/a)$, $\text{sinc}(f_mt)$ transforms to a flat rectangle of height $1/f_m$ spanning $|f|<f_m/2$ — narrower than the $\pm3f_m$ lines, so it does not extend the bandwidth. The highest frequency present is therefore the $3f_m$ line: $$BW_{message}=\boxed{3f_m}$$
|M(f)|: discrete lines at ±f_m (3/8) and ±3f_m (1/8), plus a rect of half-width f_m/2 from the sinc term (all axes in units of f_m).
2. Part (b) — DSB spectrum and bandwidth. DSB shifts (and mirrors) the message spectrum to $\pm f_c$: $S_{DSB}(f)=\tfrac12[M(f-f_c)+M(f+f_c)]$. Since $f_c=20f_m\gg3f_m$, the two shifted copies never overlap, and the modulated bandwidth is simply double the message bandwidth: $$BW_{DSB}=2\times3f_m=\boxed{6f_m}$$ occupying $f_c\pm3f_m=[17f_m,23f_m]$ (and the mirror band at $-f_c\pm3f_m$).
|S_DSB(f)|: the message spectrum shifted to ±20f_m (=±f_c), full double-sideband width 6f_m on each side.
3. Part (c) — LSB-SSB spectrum and bandwidth. Lower-sideband SSB keeps only the half of each DSB image between $f_c-3f_m$ and $f_c$ (and its Hermitian mirror on the negative side), discarding the upper sideband entirely: $$BW_{SSB}=\boxed{3f_m}$$ — exactly the message bandwidth, half of the DSB figure, since only one sideband is transmitted.
|S_SSB(f)| (lower sideband only): occupies [17f_m,20f_m] and its mirror, half the DSB bandwidth.
4. Part (d) — exact DSB recovery (any bandlimited message). A coherent (synchronous) demodulator recovers $m(t)$ exactly for ANY message of bandwidth $W$: multiply the DSB signal by a LOCAL OSCILLATOR at exactly the carrier frequency and phase, $\cos(2\pi f_ct)$, which translates the sidebands back to baseband (and to $\pm2f_c$); an ideal low-pass filter with cutoff $W$ then rejects the $2f_c$ image and passes $m(t)$ unchanged, scaled by a constant: $$s(t)\cos(2\pi f_ct)=\tfrac12A_cm(t)+\tfrac12A_cm(t)\cos(4\pi f_ct) \ \xrightarrow{\text{LPF, cutoff }W}\ \tfrac12A_cm(t)$$ This works for ANY bandlimited $m(t)$ because the derivation never used the specific shape of $m(t)$, only that its spectrum is confined to $|f|<W\ll f_c$.
Coherent demodulator: multiply by cos(2πf_c t), then low-pass filter to cutoff W.
5. Part (e) — up-converting $20f_m\to25f_m$. Mix the $f_c=20f_m$ DSB signal with a local oscillator at $5f_m$ (since $20f_m+5f_m=25f_m$); the mixer output has images centred at $20f_m-5f_m=15f_m$ and $20f_m+5f_m=25f_m$, each carrying a full copy of the original modulated spectrum (bandwidth $6f_m$). A bandpass filter passing $22f_m$ to $28f_m$ keeps only the wanted $25f_m$-centred copy and rejects the $15f_m$ image: $$s(t)\cos(2\pi\cdot5f_mt)\ \xrightarrow{\text{BPF, }22f_m\text{-}28f_m}\ \text{DSB signal, carrier }25f_m$$
Up-converter: mix with cos(2π·5f_m t), bandpass-filter to keep the 25f_m-centred image.