Question 6 of 6: Bandpass Spectral Mirroring and Frequency Conversion with a Limited Local Oscillator

Question 6: Bandpass Spectral Mirroring and Frequency Conversion with a Limited Local Oscillator (1/6 of paper)

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. (a) A real bandpass signal $x(t)$ centred at $f_0$ with an asymmetric spectral shape. (b) A signal at 10 MHz to be converted to 12 MHz using only a square-wave LO whose fundamental is tunable up to 1 MHz.

Find. (a) a system producing the mirror-image spectrum about $f_0$; (b) a frequency-conversion system reaching 12 MHz using only the given ≤1 MHz LO.

Approach. For (a), use a local oscillator at TWICE the carrier frequency: mixing shifts each spectral component symmetrically about $2f_0$, which is exactly a reflection about $f_0$ once the unwanted image is filtered away. For (b), note the required shift is $12-10=2$ MHz $=2\times(1\text{ MHz})$, so the single available 1 MHz LO can be reused twice in cascade (no "other" oscillator needed, only the same one applied at two mixer stages).

1. Part (a) — spectral mirroring. Multiplying $x(t)$ by $\cos(2\pi(2f_0)t)$ shifts every component at frequency $f_0+\delta$ to two images, at $2f_0+(f_0+\delta)=3f_0+\delta$ and at $2f_0-(f_0+\delta)=f_0-\delta$. The second image is exactly the ORIGINAL component reflected about $f_0$ (a component that was $\delta$ above $f_0$ now sits $\delta$ below $f_0$, with the same magnitude). A bandpass filter centred on $f_0$ (same bandwidth as $x(t)$) rejects the unwanted image near $3f_0$ and keeps only the mirrored copy: $$\boxed{y(t)=\Big[x(t)\cos\big(2\pi(2f_0)t\big)\Big]_{\text{BPF at }f_0}}$$ This is verified algebraically for a single tone: a component $\cos(2\pi(f_0+\delta)t)$ times $\cos(2\pi\cdot2f_0t)$ produces $\tfrac12\cos(2\pi(3f_0+\delta)t)+\tfrac12\cos(2\pi(f_0-\delta)t)$ — the second term is precisely the mirror image sitting $\delta$ below $f_0$.
2. Part (b) — two-stage up-conversion with a single 1 MHz LO. The required 2 MHz shift is exactly twice the LO's maximum (1 MHz), so cascading two ordinary up-converting mixer stages — each driven by the SAME 1 MHz LO, each followed by a bandpass filter selecting the sum (upper) sideband — reaches the target with "no other oscillators": $$10\text{ MHz}\ \xrightarrow{\times1\text{ MHz LO, BPF sum}}\ 11\text{ MHz}\ \xrightarrow{\times\text{ same }1\text{ MHz LO, BPF sum}}\ \boxed{12\text{ MHz}}$$