Question 3 of 5: DSB, Full AM and SSB of a Two-Component Message

Question 3: DSB, Full AM and SSB of a Two-Component Message (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. Message $m(t) = \cos(2\pi f_m t) + \sin(2\pi f_m t)$ with $f_m = 2$ kHz; carrier frequency $f_c = 100$ kHz; AM modulation index $\mu = 0.8$. A unit carrier amplitude $A_c = 1$ V is assumed throughout, since the paper does not specify one.

Find. Time-domain expressions and spectra for DSB-SC, full AM at $\mu = 0.8$, and upper-sideband SSB.

Approach. Collapse the two-component message into a single phase-shifted cosine, which makes its peak value and its spectrum immediate; then apply the standard modulation identities, keeping real and imaginary parts of every spectral line separate as the question's “plot it” instruction demands.

1. Rewrite the message as one cosine. Adding a cosine and a sine of the same frequency and equal amplitude gives $$m(t) = \cos(2\pi f_m t) + \sin(2\pi f_m t) = \sqrt{2}\,\cos\!\left(2\pi f_m t - \frac{\pi}{4}\right),$$ so the message is a single 2 kHz tone with $$\boxed{\,|m|_{\max} = \sqrt{2} = 1.414\ \text{V}, \qquad \overline{m^2} = 1\ \text{W}\,}$$ This one step supplies everything the later parts need: the normalising constant for AM and the message power for the efficiency calculation.
2. Write the message spectrum. Using $\cos \leftrightarrow \tfrac12[\delta(f-f_m)+\delta(f+f_m)]$ and $\sin \leftrightarrow \tfrac{1}{2j}[\delta(f-f_m)-\delta(f+f_m)]$, $$M(f) = \frac{1-j}{2}\,\delta(f - f_m) + \frac{1+j}{2}\,\delta(f + f_m),$$ a Hermitian pair of magnitude $|M| = \sqrt{2}/2 = 0.707$ at $\mp 45^\circ$.
3. Part (a) — DSB-SC in the time domain. Double-sideband suppressed-carrier modulation multiplies the message by the carrier: $$\boxed{\,s_{\text{DSB}}(t) = m(t)\cos(2\pi f_c t) = \sqrt{2}\cos\!\left(2\pi (2000)t - \tfrac{\pi}{4}\right)\cos\!\left(2\pi (10^5) t\right)\,}$$ Expanding the product gives the four-tone form $\tfrac{\sqrt2}{2}\left[\cos(2\pi(102\,\text{kHz})t - \tfrac{\pi}{4}) + \cos(2\pi(98\,\text{kHz})t + \tfrac{\pi}{4})\right]$. Its transmitted power is $\overline{m^2}/2 = 0.5$ W.

The envelope of the DSB-SC waveform is $|m(t)|$, and the carrier phase reverses by $180^\circ$ each time $m(t)$ crosses zero. That is precisely why an envelope detector cannot demodulate DSB-SC: the envelope follows $|m|$, losing the sign.

4. Part (b) — DSB-SC spectrum. Modulation shifts the message spectrum to $\pm f_c$ and halves it: $$S_{\text{DSB}}(f) = \tfrac12\left[M(f-f_c) + M(f+f_c)\right],$$ giving four lines at $\pm 98$ kHz and $\pm 102$ kHz: $$S(102\ \text{kHz}) = \frac{1-j}{4}, \quad S(98\ \text{kHz}) = \frac{1+j}{4}, \quad S(-98\ \text{kHz}) = \frac{1-j}{4}, \quad S(-102\ \text{kHz}) = \frac{1+j}{4}.$$ Each has magnitude $\sqrt{2}/4 = 0.354$. The real part is even in $f$ and the imaginary part odd, as any real signal requires. The occupied bandwidth is $2f_m = 4$ kHz.

5. Part (c) — full AM in the time domain. Full AM adds a carrier and scales the message by the modulation index relative to its peak: $$s_{\text{AM}}(t) = A_c\left[1 + \mu\,\frac{m(t)}{|m|_{\max}}\right]\cos(2\pi f_c t),$$ $$\boxed{\,s_{\text{AM}}(t) = \left[1 + 0.8\cos\!\left(2\pi(2000)t - \tfrac{\pi}{4}\right)\right]\cos\!\left(2\pi(10^5)t\right)\ \text{V}\,}$$ Note the normalisation by $|m|_{\max} = \sqrt2$, not by unity: dividing the raw $m(t)$ by $\sqrt2$ turns the sum of a cosine and a sine into the single unit-peak tone $\cos(2\pi f_m t - \pi/4)$. The envelope minimum is $A_c(1-\mu) = 0.2$ V, comfortably positive, so envelope detection is distortionless.

6. Part (d) — full AM spectrum. The dc term in the bracket produces carrier lines; the tone produces sidebands: $$S_{\text{AM}}(f) = \frac{A_c}{2}\left[\delta(f-f_c)+\delta(f+f_c)\right] + \frac{A_c\mu}{4}\left[e^{-j\pi/4}\delta(f-f_c-f_m) + e^{+j\pi/4}\delta(f-f_c+f_m) + \text{(conjugate pair at } -f)\right].$$ Numerically the carrier lines are purely real at $0.5$, and each of the four sidebands has magnitude $A_c\mu/4 = 0.2$ with real and imaginary parts $\pm 0.1414$. Power efficiency follows from the normalised message power $\overline{m_n^2} = \tfrac12$: $$\eta = \frac{\mu^2\overline{m_n^2}}{1+\mu^2\overline{m_n^2}} = \frac{0.64(0.5)}{1+0.64(0.5)} = 0.2424,$$ so only 24.2% of the transmitted power carries information — the price of the carrier that makes envelope detection possible.

7. Part (e) — upper-sideband SSB. The standard phasing (Hartley) form of USSB modulation is $$s_{\text{USB}}(t) = \tfrac12\left[m(t)\cos(2\pi f_c t) - \hat{m}(t)\sin(2\pi f_c t)\right],$$ where $\hat m$ is the Hilbert transform of $m$ ($\cos \to \sin$, $\sin \to -\cos$), so here $\hat m(t) = \sin(2\pi f_m t) - \cos(2\pi f_m t)$. Substituting and applying the angle-sum identities, every $f_c - f_m$ term cancels and $$s_{\text{USB}}(t) = \tfrac12\left[\cos(2\pi (f_c+f_m)t) + \sin(2\pi(f_c+f_m)t)\right],$$ $$\boxed{\,s_{\text{USB}}(t) = \frac{\sqrt{2}}{2}\cos\!\left(2\pi(102\,\text{kHz})\,t - \frac{\pi}{4}\right) = 0.707\cos(2\pi(102\times10^3)t - 45^\circ)\ \text{V}\,}$$ The spectrum retains only the DSB lines at $\pm 102$ kHz, with the same weights $\tfrac{1}{4}(1 \mp j)$; the occupied bandwidth halves to $f_m = 2$ kHz and the transmitted power halves to 0.25 W.