22-Elec-A3 Signals and Communications · December 2018

Question 3 of 5: Spectra of a Composite Message under DSB and SSB Modulation

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 3: Spectra of a Composite Message under DSB and SSB Modulation (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 $m(t) = \cos^{3}(2\pi f_m t) + \dfrac{\sin(\pi f_m t)}{\pi f_m t}$, carrier $A\cos(2\pi f_c t)$ with $f_c = 30 f_m$. The second message term is a sinc function; using the normalised form $\operatorname{sinc}(x) = \sin(\pi x)/(\pi x)$, it is $\operatorname{sinc}(f_m t)$.

Find. (a) $M(f)$ and the message bandwidth; (b) the DSB spectrum and its bandwidth; (c) the lower-sideband SSB spectrum and its bandwidth; (d) a receiver that recovers any message of this bandwidth exactly; and (e) a system that retunes the DSB signal from $30f_m$ to $35f_m$.

Approach. Expand the cubed cosine with the triple-angle identity to turn it into two ordinary tones, transform the sinc into a rectangle, and then apply the modulation theorem — multiplication by a carrier simply copies $M(f)$ to $\pm f_c$ and halves it.

1. Expand the cubed cosine. The power-reduction identity $\cos^{3}u = \tfrac{3}{4}\cos u + \tfrac{1}{4}\cos 3u$ gives $$\cos^{3}(2\pi f_m t) = \tfrac{3}{4}\cos(2\pi f_m t) + \tfrac{1}{4}\cos(2\pi (3f_m) t).$$ So the first message term is not one tone but two: a strong one at $f_m$ and a weaker third harmonic at $3f_m$. Recognising this is the crux of part (a) — the cubing operation has already tripled the bandwidth before any modulation takes place.
2. Transform the sinc term. With $\dfrac{\sin(\pi f_m t)}{\pi f_m t} = \operatorname{sinc}(f_m t)$ and the standard pair $\operatorname{sinc}(at) \leftrightarrow \dfrac{1}{|a|}\operatorname{rect}\!\left(\dfrac{f}{a}\right)$, $$\operatorname{sinc}(f_m t) \;\longleftrightarrow\; \frac{1}{f_m}\operatorname{rect}\!\left(\frac{f}{f_m}\right),$$ a flat rectangle of height $1/f_m$ occupying $|f| \lt f_m/2$. This term is strictly band-limited and, crucially, narrower than the tones it accompanies.
3. (a) Assemble $M(f)$ and read the bandwidth. Using $\cos(2\pi f_1 t) \leftrightarrow \tfrac{1}{2}[\delta(f-f_1)+\delta(f+f_1)]$, $$M(f) = \tfrac{3}{8}\big[\delta(f-f_m)+\delta(f+f_m)\big] + \tfrac{1}{8}\big[\delta(f-3f_m)+\delta(f+3f_m)\big] + \frac{1}{f_m}\operatorname{rect}\!\left(\frac{f}{f_m}\right).$$ The highest occupied frequency is the impulse pair at $3f_m$ (the rectangle stops at $f_m/2$), so $$\boxed{B_m = 3f_m}$$

4. (b) Apply the modulation theorem. The DSB (suppressed-carrier) signal is $s(t) = A\,m(t)\cos(2\pi f_c t)$, whose transform is $$S(f) = \frac{A}{2}\big[M(f-f_c) + M(f+f_c)\big].$$ Each copy of $M(f)$ is halved in height and shifted to $\pm f_c = \pm 30f_m$, so the positive- frequency copy runs from $30f_m - 3f_m$ to $30f_m + 3f_m$, i.e. $27f_m$ to $33f_m$. Its internal structure is the mirror image of $M(f)$ about the carrier: impulses at $27f_m,\,29f_m,\,31f_m,\,33f_m$ plus a rectangle across $29.5f_m$ to $30.5f_m$. Hence $$\boxed{B_{\text{DSB}} = 2B_m = 6f_m}$$

5. (c) Retain only the lower sideband. Lower-sideband SSB keeps, from each DSB copy, only the part lying below the carrier in absolute frequency. The positive-frequency content therefore runs from $27f_m$ up to $30f_m$: the impulses at $27f_m$ and $29f_m$ survive, those at $31f_m$ and $33f_m$ are removed, and only the lower half of the rectangle (from $29.5f_m$ to $30f_m$) remains. The bandwidth collapses to $$\boxed{B_{\text{SSB}} = B_m = 3f_m}$$ exactly half that of DSB, which is the entire motivation for single-sideband transmission: the two sidebands of a real message are conjugate-symmetric and therefore redundant.

6. (d) Recover the message: coherent detection. Multiplying the received DSB signal by a local carrier of the same frequency and phase gives $$A\,m(t)\cos(2\pi f_c t)\cdot 2\cos(2\pi f_c t) = A\,m(t)\big[1 + \cos(2\pi (2f_c) t)\big] = A\,m(t) + A\,m(t)\cos(2\pi (2f_c) t),$$ using $2\cos^{2}\phi = 1 + \cos 2\phi$. The first term is the message at baseband (occupying up to $3f_m$) and the second is a copy centred at $2f_c = 60f_m$ (occupying $57f_m$ to $63f_m$). A low-pass filter of cut-off $3f_m$ and gain $1/A$ removes the second and returns $m(t)$ exactly. Because this detector is linear in the message, it works for an arbitrary message of bandwidth $3f_m$ — unlike an envelope detector, which would fail here since the DSB signal has no carrier and $m(t)$ changes sign.

7. (e) Retune the carrier from $30f_m$ to $35f_m$. The required shift is $35f_m - 30f_m = 5f_m$, so mix the DSB signal with a local oscillator at $5f_m$: $$s(t)\cdot 2\cos\big(2\pi (5f_m) t\big) \;\longrightarrow\; \text{copies centred at } 30f_m - 5f_m = 25f_m \ \text{and}\ 30f_m + 5f_m = 35f_m .$$ A band-pass filter centred at $35f_m$ with bandwidth $6f_m$ selects the wanted copy. The separation is generous: the unwanted copy occupies $22f_m$ to $28f_m$ while the wanted one occupies $32f_m$ to $38f_m$, leaving a $4f_m$ guard band, so a modest filter suffices. Note that this stage does not require phase coherence — a phase error in the local oscillator merely adds a constant phase to the retuned carrier, which the eventual coherent detector will track.