22-Elec-A3 Signals and Communications · December 2016

Question 5 of 6: Spectra of DSB and SSB Modulation, Coherent Detection and Frequency Translation

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

Question 5: Spectra of DSB and SSB Modulation, Coherent Detection and Frequency Translation (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. $m(t) = \cos^{3}(2\pi f_m t) + \operatorname{sinc}(f_m t)$, where the second term is written in the paper as $\sin(\pi f_m t)/(\pi f_m t)$. Carrier $A\cos(2\pi f_c t)$ with $f_c = 20 f_m$.

Find. The message spectrum and bandwidth; the DSB and lower-sideband SSB spectra and their bandwidths; a coherent DSB demodulator; and a frequency translator that moves the DSB signal to a carrier of $25 f_m$.

Approach. Expand the cubed cosine with the triple-angle identity to expose its discrete lines, transform the sinc term to a rectangle, take the larger of the two as the message bandwidth, and then apply the modulation and frequency-shifting properties of the Fourier transform.

1. Expand the cubed cosine. Using $\cos^{3}\theta = \tfrac34\cos\theta + \tfrac14\cos 3\theta$, $$\cos^{3}(2\pi f_m t) = \tfrac34\cos(2\pi f_m t) + \tfrac14\cos(2\pi\,3f_m t),$$ which contributes impulses of weight $\tfrac38$ at $f = \pm f_m$ and $\tfrac18$ at $f = \pm 3f_m$. (A numerical projection of $\cos^{3}$ onto the two cosines returns exactly $0.75$ and $0.25$, confirming the identity.)
2. Transform the sinc term. With $\operatorname{sinc}(f_m t) = \dfrac{\sin \pi f_m t}{\pi f_m t}$, the standard pair gives $$\operatorname{sinc}(f_m t) \;\longleftrightarrow\; \frac{1}{f_m}\,\Pi\!\left(\frac{f}{f_m}\right),$$ a rectangle of height $1/f_m$ extending over $|f| \lt f_m/2$. This term is therefore band-limited to $f_m/2$ — half the fundamental, and well inside the discrete lines.
3. Message spectrum and bandwidth (part a). The complete spectrum is the flat rectangle on $|f| \lt f_m/2$ plus four impulses at $\pm f_m$ and $\pm 3f_m$. The highest frequency present is the third harmonic, so $$\boxed{B_m = 3f_m}$$ The sinc term does not set the bandwidth here; the cubed cosine does.

4. DSB spectrum (part b). Multiplying by the carrier translates the message spectrum to $\pm f_c$ and halves it: $$s_{\text{DSB}}(t) = A\,m(t)\cos(2\pi f_c t) \;\longleftrightarrow\; \frac{A}{2}\left[M(f - f_c) + M(f + f_c)\right].$$ Each copy occupies $f_c \pm B_m$, i.e. from $17f_m$ to $23f_m$ about the positive carrier, and the transmission bandwidth is $$\boxed{B_{\text{DSB}} = 2B_m = 6f_m}$$ The impulses land at $f_c \pm f_m$ ($19f_m$ and $21f_m$) and $f_c \pm 3f_m$ ($17f_m$ and $23f_m$); the flat rectangle occupies $f_c \pm f_m/2$.

5. Lower-sideband SSB (part c). Retaining only the sideband below the carrier deletes everything above $f_c$, leaving the band from $f_c - B_m = 17f_m$ up to $f_c = 20f_m$ (and its mirror image at negative frequencies). Hence $$\boxed{B_{\text{SSB}} = B_m = 3f_m}$$ which is exactly half the DSB requirement — the spectral-efficiency argument for SSB. Note the spectrum is now asymmetric about the carrier, which is why an envelope detector cannot demodulate it.

6. Exact recovery from DSB (part d). Because DSB-SC carries no carrier line, the receiver must supply a local oscillator that is coherent in both frequency and phase. Multiplying the received signal by $2\cos(2\pi f_c t)$, $$2\,s_{\text{DSB}}(t)\cos(2\pi f_c t) = A\,m(t)\left[1 + \cos(2\pi\,2f_c t)\right] = A\,m(t) + A\,m(t)\cos(4\pi f_c t).$$ The first term is the message at baseband; the second is centred on $2f_c = 40f_m$. An ideal low-pass filter of cut-off $3f_m$ (anywhere between $3f_m$ and $37f_m$ would do) removes the high-frequency term and leaves $A\,m(t)$ exactly. Because the argument never used any property of $m(t)$ beyond its bandwidth, the receiver works for arbitrary messages of bandwidth $3f_m$, as the question requires.

7. Frequency translation to $25 f_m$ (part e). Mixing the DSB signal with a local oscillator at $f_{\text{LO}}$ produces sum and difference bands centred at $f_c + f_{\text{LO}}$ and $f_c - f_{\text{LO}}$. Choosing $$f_{\text{LO}} = 25f_m - f_c = 25f_m - 20f_m = \boxed{5f_m}$$ places the wanted band at $25f_m$ (spanning $22f_m$ to $28f_m$) and the unwanted image at $15f_m$ (spanning $12f_m$ to $18f_m$). The two are separated by $4f_m$ of clear spectrum, so an ordinary band-pass filter centred at $25f_m$ with a passband $6f_m$ wide isolates the wanted output. The result is a genuine DSB signal on a $25f_m$ carrier, since mixing preserves the two-sideband structure.