22-Elec-A3 Signals and Communications · May 2018

Question 5 of 5: Spectral Mirroring and Frequency Conversion with a Limited Oscillator

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

Question 5: Spectral Mirroring and Frequency Conversion with a Limited Oscillator (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.

Find. A block diagram that mirrors the spectrum about the carrier without moving it; and a block diagram that shifts a 10 MHz carrier to 12 MHz using only a sub-1 MHz square-wave oscillator.

Approach. For (a), recognise that mirroring about $f_0$ is conjugation of the complex envelope, and that mixing with a carrier at $2f_0$ achieves it in one step (with a quadrature demodulator / remodulator as an equivalent alternative). For (b), exploit the fact that a square wave is rich in odd harmonics, so a 2 MHz mixing tone can be synthesised from a 666.7 kHz fundamental.

1. See what mirroring means (part a). Write the input as $s(t) = x_I(t)\cos 2\pi f_0 t - x_Q(t)\sin 2\pi f_0 t$, i.e. $s(t) = \mathrm{Re}\{\tilde{x}(t)e^{j2\pi f_0 t}\}$ with complex envelope $\tilde{x} = x_I + jx_Q$. The positive-frequency content of $s$ is $\tilde{X}(f-f_0)/2$, so reflecting it about $f_0$ means replacing $\tilde{X}(f)$ by $\tilde{X}(-f)$ — equivalently replacing $\tilde{x}(t)$ by $\tilde{x}^{*}(t)$. Mirroring a band-pass spectrum about its own carrier is therefore exactly conjugating the complex envelope, and the wanted output is $$y(t) = x_I(t)\cos 2\pi f_0 t + x_Q(t)\sin 2\pi f_0 t.$$
2. Realise it with a single mixer at twice the carrier. Multiply the input by $2\cos(2\pi (2f_0) t)$. A component of $s$ at $f_0 + \delta$ (where $|\delta| < B/2$) generates products at $$\left|(f_0+\delta) - 2f_0\right| = f_0 - \delta \qquad\text{and}\qquad (f_0+\delta) + 2f_0 = 3f_0 + \delta .$$ The difference product lands at $f_0 - \delta$: every offset above the carrier has become the same offset below it, and vice versa — precisely the required mirror image, at the same centre frequency. The sum products cluster near $3f_0$ and are removed by a band-pass filter centred on $f_0$ with the signal's own bandwidth $B$. Hence $$\boxed{\,y(t) = \mathrm{BPF}_{f_0,\,B}\bigl\{\,2s(t)\cos\bigl(2\pi (2f_0)t\bigr)\,\bigr\}\,}$$ Formally, $2\cos(4\pi f_0 t)\,\mathrm{Re}\{\tilde{x}e^{j2\pi f_0t}\}$ contains the term $\mathrm{Re}\{\tilde{x}^{*}e^{j2\pi f_0 t}\}$, which is the conjugated-envelope signal derived in step 1.
3. State the quadrature alternative. If an oscillator at $2f_0$ is unavailable, the same result follows from a quadrature demodulator/remodulator pair: mix $s(t)$ with $2\cos 2\pi f_0 t$ and with $-2\sin 2\pi f_0 t$, low-pass each branch to recover $x_I$ and $x_Q$, then remodulate as $y = x_I\cos 2\pi f_0 t + x_Q \sin 2\pi f_0 t$ — that is, re-combine with the sign of the quadrature branch reversed. This is the Weaver-style implementation and is the one used in digital radios, where the conjugation is a single sign change on the imaginary part of the sample stream.

4. Identify the required shift (part b). Converting 10 MHz to 12 MHz needs a mixing tone at $$f_{shift} = f_{out} - f_{in} = 12 - 10 = 2\ \text{MHz},$$ but the only oscillator available tops out at 1 MHz. A sine-wave oscillator would make the task impossible; a square-wave oscillator does not, because its Fourier series $$v_{LO}(t) = \frac{4V}{\pi}\left[\cos 2\pi f_{LO} t - \tfrac13\cos 2\pi (3f_{LO}) t + \tfrac15\cos 2\pi (5f_{LO}) t - \cdots\right]$$ contains strong odd harmonics. We may therefore mix against the $3^{\text{rd}}$, $5^{\text{th}}$, … harmonic instead of the fundamental.
5. Choose the harmonic and the oscillator setting. The needed 2 MHz must be an odd multiple of $f_{LO}$ with $f_{LO} \le 1$ MHz. The fundamental ($f_{LO} = 2$ MHz) violates the limit; the third harmonic does not: $$3 f_{LO} = 2\ \text{MHz} \quad\Rightarrow\quad \boxed{\,f_{LO} = \tfrac{2}{3}\ \text{MHz} = 666.7\ \text{kHz}\,}$$ Note that an even harmonic is unavailable — a symmetric square wave has none — so 2 MHz cannot be obtained as $2\times 1$ MHz. The fifth harmonic would need $f_{LO} = 400$ kHz and is an equally valid, if weaker, alternative (its amplitude is $1/5$ of the fundamental against $1/3$).
6. Select the wanted product with a band-pass filter. Mixing the 10 MHz signal with the full square wave generates sum and difference products against every harmonic: $$9.333,\ 10.667\ \text{MHz (fundamental)};\quad 8,\ \mathbf{12}\ \text{MHz (3rd)};\quad 6.667,\ 13.333\ \text{MHz (5th)};\ \ldots$$ The wanted 12 MHz product is separated from its nearest neighbour (13.333 MHz) by 1.333 MHz, so a band-pass filter centred on 12 MHz with a bandwidth equal to the modulated signal's is entirely straightforward to build. The complete system is therefore $$\boxed{\,\text{mixer driven by a 666.7 kHz square wave} \to \text{BPF centred on 12 MHz}\,}$$ The mixer's conversion loss against the third harmonic is $20\log_{10}3 = 9.5$ dB worse than against the fundamental, so a post-filter amplifier is usual in practice.