22-Elec-A3 Signals and Communications · December 2019

Question 2 of 5: Two-Tone Message — Exact Spectra, DSB, SSB and Frequency Translation

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

Question 2: Two-Tone Message — Exact Spectra, DSB, SSB 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. Message $m(t) = \cos(2\pi f_m t) + \tfrac{1}{2}\sin(2\pi (2f_m) t)$ — a fundamental cosine of unit amplitude plus a second-harmonic sine of amplitude $\tfrac12$. Carrier $c(t) = A\cos(2\pi f_c t)$ with $f_c = 30 f_m$. Modulation is double-sideband suppressed-carrier (DSB-SC) unless stated otherwise.

Find. The time plot of $m(t)$; the exact (real and imaginary) spectra of $m(t)$, of the DSB signal and of the lower-sideband SSB signal, with the bandwidth in each case; and block diagrams for coherent recovery and for translation of the carrier to $40 f_m$.

Approach. Expand each tone with Euler's identity so that the cosine contributes purely real impulse pairs and the sine contributes purely imaginary ones, then apply the modulation theorem, which simply replicates $M(f)$ at $\pm f_c$ with weight $A/2$.

(a) The message in the time domain

The waveform repeats every $1/f_m$ because the second component is an exact harmonic of the first. Its asymmetry about the peak is the visible consequence of mixing a cosine (even) with a sine (odd), and this is precisely the asymmetry that shows up as a mixed real/imaginary spectrum in part (b).

(b) Exact spectrum of the message and its bandwidth

1. Expand each tone as a pair of impulses. Using $\cos\theta = \tfrac12(e^{j\theta}+e^{-j\theta})$ and $\sin\theta = \tfrac{1}{2j}(e^{j\theta}-e^{-j\theta})$, $$\cos(2\pi f_m t) \;\longleftrightarrow\; \tfrac{1}{2}\delta(f-f_m) + \tfrac{1}{2}\delta(f+f_m)$$ $$\tfrac{1}{2}\sin\!\left(2\pi (2f_m) t\right) \;\longleftrightarrow\; -j\tfrac{1}{4}\delta(f-2f_m) + j\tfrac{1}{4}\delta(f+2f_m)$$
2. Assemble $M(f)$ and separate real and imaginary parts. $$\boxed{\;M(f) = \tfrac{1}{2}\left[\delta(f-f_m)+\delta(f+f_m)\right] + \tfrac{j}{4}\left[\delta(f+2f_m)-\delta(f-2f_m)\right]\;}$$ The cosine therefore appears entirely in the real part, with weight $\tfrac12$ at $f = \pm f_m$, while the sine appears entirely in the imaginary part, with weight $-\tfrac14$ at $f = +2f_m$ and $+\tfrac14$ at $f = -2f_m$. The odd symmetry of the imaginary part and the even symmetry of the real part together confirm $M(-f) = M^*(f)$, as required for a real signal.
3. Read off the bandwidth. The highest frequency present is the second harmonic, so $$\boxed{W = 2f_m}$$ (measured one-sided, from DC to the highest spectral line).

(c) DSB spectrum and bandwidth

1. Apply the modulation theorem. For $s(t) = A\,m(t)\cos(2\pi f_c t)$, $$S(f) = \frac{A}{2}\left[M(f-f_c) + M(f+f_c)\right]$$ so each impulse of $M$ is copied to $f_c$ and $-f_c$ with its weight halved and scaled by $A$.
2. Place the lines and carry the signs. About the positive carrier the four lines sit at $f_c \pm f_m$ and $f_c \pm 2f_m$, i.e. at $\;29f_m,\;31f_m,\;28f_m,\;32f_m$. Their weights are $$\text{Re}: \ \frac{A}{4}\ \text{at } 29f_m \text{ and } 31f_m;\qquad \text{Im}: \ +\frac{A}{8}\ \text{at } 28f_m,\ \ -\frac{A}{8}\ \text{at } 32f_m$$ with the conjugate-symmetric mirror image about $f = 0$. Note that the carrier itself at $30 f_m$ is absent — this is suppressed-carrier DSB.
3. State the transmission bandwidth. The occupied band runs from $28f_m$ to $32f_m$, so $$\boxed{B_{\text{DSB}} = 2W = 4f_m}$$

(d) Lower-sideband SSB spectrum and bandwidth

Lower-sideband SSB retains only those components that lie below the carrier on the positive-frequency side (and their conjugate mirror images on the negative side). Of the four positive-frequency lines, $29f_m$ and $28f_m$ are retained and $31f_m$ and $32f_m$ are removed by the sideband filter. The retained content is $$S_{\text{LSB}}(f)\Big|_{f>0} = \frac{A}{4}\delta(f-29f_m) + j\frac{A}{8}\delta(f-28f_m)$$ and the occupied band now runs from $28f_m$ to $29f_m$ — but the bandwidth quoted for an SSB system is measured from the carrier reference to the outermost retained line, giving $$\boxed{B_{\text{SSB}} = W = 2f_m}$$ exactly half the DSB requirement, which is the entire commercial motivation for SSB.

(e) Exact recovery of $m(t)$ from the DSB signal

Because DSB-SC carries no discrete carrier line, the receiver must regenerate the carrier in both frequency and phase; a Costas loop or a squaring loop followed by a divide-by-two does this from the received signal itself. Multiplying by the recovered carrier gives $$s(t)\cdot 2\cos(2\pi f_c t) = A\,m(t)\left[1 + \cos(4\pi f_c t)\right]$$ so the message reappears at baseband alongside an image centred at $2f_c = 60f_m$. An ideal low-pass filter of cut-off $2f_m$ (the message bandwidth) removes the image, and a gain of $2/A$ restores the original scale. Because the detector is linear, this recovers any message of bandwidth $2f_m$ exactly, as the question demands — an envelope detector would not, since $m(t)$ takes negative values and DSB-SC has no carrier pedestal.

(f) Translating the carrier from $30f_m$ to $40f_m$

A second mixing stage shifts the whole DSB band without disturbing its internal structure. Multiplying by $2\cos(2\pi (10f_m)t)$ produces a sum band centred at $30f_m + 10f_m = 40f_m$ and a difference band centred at $30f_m - 10f_m = 20f_m$. Each band is $4f_m$ wide, so they occupy $38f_m$–$42f_m$ and $18f_m$–$22f_m$ respectively — separated by $16f_m$ of clear spectrum. A band-pass filter centred at $40f_m$ with bandwidth $4f_m$ therefore selects the wanted output cleanly, and the result is a DSB signal with exactly the same sideband structure translated to the new carrier.