22-Elec-A3 Signals and Communications · December 2015

Question 6 of 6: Spectral Inversion and Frequency Conversion

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

Notes on this paper

Paper format: 07-Elec-A3 Signals and Communications, December 2015 — 3 hours, closed book, a standard non-programmable calculator permitted. Six questions of equal value (20 marks each); any five constitute a complete paper, and only the first five presented are marked. All six are solved here, since the set is intended as a study resource.

Reference texts. B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. (AM/FM, PCM, frequency conversion); A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. (Fourier series, LTI systems); J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. (z-transform, difference equations, stability); S. Haykin, Communication Systems, 5th ed. (modulation and detection).

Question 6: Spectral Inversion and Frequency Conversion (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. Part (a): a real band-pass signal whose positive-frequency spectrum is asymmetric about its centre frequency $f_{0}$ — it rises steeply below $f_{0}$ and decays slowly above it. The required output has the same centre frequency but the mirrored shape. Part (b): a modulated signal at $f_{c} = 10$ MHz, to be moved to 12 MHz, using only a square-wave local oscillator whose fundamental can be set anywhere up to 1 MHz.

Find. Block diagrams for (a) the spectral inverter and (b) the frequency converter, with all oscillator frequencies specified.

Figure 6.1 — Input spectrum (top) and the required mirrored output (bottom), both about the same centre frequency $f_{0}$.

Approach. Mixing a signal with an oscillator produces sum and difference terms; the difference term $f_{\mathrm{LO}} - f$ is a reflection of the input spectrum about $f_{\mathrm{LO}}/2$, so choosing $f_{\mathrm{LO}} = 2f_{0}$ reflects the spectrum about $f_{0}$ itself while returning it to the same centre frequency. For part (b), a square wave contains odd harmonics, so a sub-1-MHz oscillator can still deliver the required 2 MHz shift through its third harmonic.

Multiply by an oscillator at twice the carrier. Take $v(t) = s(t) \cdot 2\cos\!\left(2\pi(2f_{0})t\right)$. In the frequency domain this produces two shifted copies, $$V(f) = S(f - 2f_{0}) + S(f + 2f_{0})$$
Identify the mirrored term. For a real signal $S(-f) = S^{*}(f)$, so the copy built from the negative-frequency half of $S$ appears at positive frequencies as $S^{*}(2f_{0} - f)$. A component originally at $f_{0} + \delta$ now sits at $2f_{0} - (f_{0} + \delta) = f_{0} - \delta$: energy above the carrier has moved below it by the same offset, and vice versa. That is exactly the required mirroring, and it is still centred on $f_{0}$.
Remove the unwanted sum term. The other copy occupies a band centred on $3f_{0}$, which is far from $f_{0}$ for any reasonable bandwidth, so a band-pass filter centred on $f_{0}$ and just wide enough to accept the signal isolates the mirrored component: $$\boxed{\ s(t) \rightarrow \times\,2\cos(2\pi(2f_{0})t) \rightarrow \text{BPF centred on } f_{0} \rightarrow \text{mirrored output}\ }$$

Figure 6.2 — Spectral inverter. Mixing with an oscillator at $2f_{0}$ folds the spectrum about $f_{0}$; the band-pass filter discards the image near $3f_{0}$.

Part (b) is constrained: the required shift is 2 MHz, but the oscillator cannot be set above 1 MHz. The key observation is that the oscillator produces a square wave, whose Fourier series contains odd harmonics at $f_{\mathrm{LO}}, 3f_{\mathrm{LO}}, 5f_{\mathrm{LO}}, \ldots$ with amplitudes falling as $1/k$. Any one of those harmonics can serve as the mixing tone provided the output filter selects the product we want.

Determine the required shift. $$\Delta f = 12\ \text{MHz} - 10\ \text{MHz} = 2\ \text{MHz}$$
Choose an odd harmonic within the oscillator's range. Using the third harmonic, $3f_{\mathrm{LO}} = 2$ MHz, so $$f_{\mathrm{LO}} = \frac{2\ \text{MHz}}{3} = 0.667\ \text{MHz} \le 1\ \text{MHz} \quad\checkmark$$
Confirm the wanted product and check the others. Mixing produces components at $10 \pm kf_{\mathrm{LO}}$ MHz for odd $k$: $9.333$ and $10.667$ ($k=1$), $8$ and $\boxed{12}$ ($k=3$), $6.667$ and $13.333$ ($k=5$). The wanted 12 MHz output is separated from its nearest neighbours (10.667 MHz and 13.333 MHz) by 1.33 MHz, so a band-pass filter centred on 12 MHz with a modest skirt selects it cleanly: $$\boxed{\,f_{\mathrm{LO}} = \tfrac{2}{3}\ \text{MHz},\ \text{use the 3rd harmonic, BPF at 12 MHz}\,}$$

Figure 6.3 — Frequency converter using the third harmonic of a 2/3 MHz square-wave oscillator. Selecting the sum product at 12 MHz preserves the modulation and only translates the carrier.

Two alternatives are equally acceptable. The fifth harmonic of a 0.4 MHz oscillator also lands on 2 MHz, at the cost of a weaker mixing tone (amplitude $\propto 1/5$). Alternatively, two cascaded conversion stages each using a 1 MHz oscillator fundamental move the carrier $10 \rightarrow 11 \rightarrow 12$ MHz, which relaxes the filter requirement at each stage but doubles the hardware. The single-stage third-harmonic solution is the most economical answer to the question as posed.

Quantity	Result
(a) Inverter oscillator	$2\cos(2\pi(2f_{0})t)$ — oscillator at $2f_{0}$
(a) Output filter	Band-pass centred on $f_{0}$ (rejects the image near $3f_{0}$)
(a) Mechanism	Difference term $2f_{0} - f$ mirrors the spectrum about $f_{0}$
(b) Required shift	2 MHz
(b) Oscillator setting	$f_{\mathrm{LO}} = 2/3$ MHz (0.667 MHz), third harmonic used
(b) Output filter	Band-pass centred on 12 MHz; nearest unwanted product 1.33 MHz away
(b) Alternatives	0.4 MHz oscillator on its 5th harmonic, or two 1 MHz stages in cascade

Topic: Mixing as frequency translation and spectral reflection; harmonic mixing with a square-wave local oscillator.

Key relations: $s(t)\cos(2\pi f_{L}t) \leftrightarrow \tfrac{1}{2}\left[S(f - f_{L}) + S(f + f_{L})\right]$; difference term $f_{L} - f$ reflects the spectrum about $f_{L}/2$; real signal symmetry $S(-f) = S^{*}(f)$; square wave $= \dfrac{4}{\pi}\sum_{k\ \text{odd}}\dfrac{1}{k}\sin(2\pi k f_{\mathrm{LO}}t)$.

Why this works: A mixer does not "shift" a spectrum so much as create two shifted copies, one of which is a reflected version drawn from the negative-frequency half of the input. Choosing the local oscillator at exactly twice the carrier makes that reflected copy land back on the original carrier frequency, so the output is simultaneously frequency-aligned and spectrally inverted — the same principle behind the speech scramblers that inverted voice bands in early secure telephony. In part (b), the square wave is effectively a bank of oscillators at $f_{\mathrm{LO}}, 3f_{\mathrm{LO}}, 5f_{\mathrm{LO}}, \ldots$; harmonic mixing exploits that bank to reach frequencies the fundamental cannot, which is standard practice in microwave receivers where a clean high-frequency oscillator is expensive.

Common pitfall: Attempting the inversion by mixing to baseband and back — a real oscillator at $f_{0}$ folds the upper and lower halves of the band onto each other and destroys the signal irrecoverably; the reflection must be done at $2f_{0}$, or else with a quadrature (Weaver/Hartley) structure. In part (b) the usual error is declaring the task impossible because 2 MHz exceeds the oscillator's 1 MHz limit, which overlooks the square wave's harmonic content. A related slip is proposing an even harmonic: an ideal square wave has none.

Source: Lathi & Ding, Modern Digital and Analog Communication Systems, 4th ed., Ch. 4 (frequency mixing, converters, and superheterodyne reception); Haykin, Communication Systems, 5th ed., Ch. 3.

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