Question 6 of 6: Baseband, DSB-SC, USB and LSB Spectra

Question 6: Baseband, DSB-SC, USB and LSB Spectra (20 marks)

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. Three baseband messages and a carrier $\cos\omega_c t$ with $\omega_c=10{,}000$ rad/s ($f_c=1591.5$ Hz). All frequencies below are in rad/s.

Find. The baseband and DSB-SC line spectra, the USB and LSB line sets, and an explanation of the shifting mechanism.

Approach. Every message here is a sum of cosines, so each spectrum is a set of impulse pairs. Use $\cos\omega_m t\leftrightarrow\pi\left[\delta(\omega-\omega_m)+\delta(\omega+\omega_m)\right]$ and the modulation theorem, which shifts a copy of $M(\omega)$ to $\pm\omega_c$ at half height. Case (iii) must first be expanded with a product-to-sum identity.

1. State the modulation theorem. $$m(t)\cos\omega_c t\ \longleftrightarrow\ \tfrac12 M(\omega-\omega_c)+\tfrac12 M(\omega+\omega_c).$$ A baseband line at $\omega_m$ therefore reappears at $\omega_c+\omega_m$ (upper sideband) and $\omega_c-\omega_m$ (lower sideband) with half the original weight, and there is no line at $\omega_c$ itself — the carrier is suppressed.
2. (a), (b) Case (i): $m=\cos 2000t$. Baseband lines at $\omega=\pm2000$ of weight $\pi$. Modulating gives lines at $$\boxed{\ \omega=\pm 8000\ (\text{LSB}),\quad \pm 12{,}000\ (\text{USB})\ }$$ each of weight $\pi/2$. The transmission bandwidth is $12{,}000-8000=4000$ rad/s, twice the message bandwidth — the defining penalty of DSB.
3. (a), (b) Case (ii): $m=3\cos 2000t+\cos 4000t$. Baseband lines at $\pm2000$ (weight $3\pi$) and $\pm4000$ (weight $\pi$). Each is shifted independently by superposition, giving LSB lines at $8000$ and $6000$ and USB lines at $12{,}000$ and $14{,}000$ rad/s. Note the ordering inversion: the higher baseband tone (4000) maps to the lower LSB line (6000), because the LSB is a frequency-reversed image.
4. (a), (b) Case (iii) — expand the product first. $\cos 2000t\cos 6000t=\tfrac12\left[\cos 4000t+\cos 8000t\right]$, so the baseband already contains two tones at $4000$ and $8000$ rad/s, each of weight $\pi/2$. Modulating gives LSB lines at $10{,}000-4000=6000$ and $10{,}000-8000=2000$, and USB lines at $14{,}000$ and $18{,}000$ rad/s. Because the highest baseband component (8000) is still below $\omega_c$, no sideband folds through zero and the signal remains recoverable.
5. (c) Identify the sidebands. For each case the USB is the set of lines above $\omega_c=10{,}000$ rad/s and the LSB the set below it; each carries the complete message information, which is what makes single-sideband transmission possible at half the bandwidth.
6. (d) Explain the shifting. Multiplication in time is convolution in frequency, and convolving with the carrier pair $\pi[\delta(\omega-\omega_c)+\delta(\omega+\omega_c)]$ simply translates the whole baseband spectrum to sit around $\pm\omega_c$. The USB is a faithful (upright) translated copy: $\omega_m\mapsto\omega_c+\omega_m$ preserves ordering. The LSB is an inverted copy: $\omega_m\mapsto\omega_c-\omega_m$ reverses ordering, so the top of the message sits at the bottom of the LSB. Both copies are at half amplitude, and their spectral content is redundant, which is why coherent detection (multiplying by $\cos\omega_c t$ again and low-pass filtering) recovers $\tfrac12 m(t)$ exactly.