Question 5 of 7: Superheterodyne Receiver — LO Tuning, Image Frequency and Front-End Filter

Question 5: Superheterodyne Receiver — LO Tuning, Image Frequency and Front-End Filter (1/5 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.

Find. (a) the two LO tuning frequencies and which is preferable for a tunable channel block; (b) the image frequency for the preferred choice; (c) the AM band edges and the number of stations the fixed front-end filter admits.

Approach. A mixer produces $f_c\pm f_{LO}$ terms, so either $f_{LO}=f_c+f_{IF}$ (high-side injection) or $f_{LO}=f_c-f_{IF}$ (low-side injection) puts one of them at $f_{IF}$; compare the LO TUNING RATIO each requires across a channel block to find the preferable one. The image frequency is the other input frequency that also mixes to $f_{IF}$ with the same LO. The fixed front-end filter must pass the whole tuned channel block while keeping every channel's image OUTSIDE that same passband.

1. Part (a) — two LO frequencies and the preferable one. Either sideband of the mixer product can land at $f_{IF}$: $$f_{LO}=f_c+f_{IF}=2.8\ \text{MHz (high-side)}\qquad\text{or}\qquad f_{LO}=f_c-f_{IF}=1.2\ \text{MHz (low-side)}$$ For a receiver that must tune across a BLOCK of channels (LO frequency varies with $f_c$), compare the required LO tuning RATIO $f_{LO,\max}/f_{LO,\min}$ for each choice: adding the constant $f_{IF}$ (high-side) compresses the ratio toward 1 relative to the RF's own ratio, while subtracting it (low-side) expands the ratio. A smaller LO tuning ratio is easier to realize with a single tracking variable capacitor/inductor, which is exactly why real AM receivers use high-side injection almost universally. So $$\boxed{f_{LO}=2.8\ \text{MHz (high-side injection) is preferable}}$$
2. Part (b) — image frequency. With high-side injection, the image is the OTHER input frequency that also mixes to $f_{IF}$ with the same LO, i.e. the one $f_{IF}$ further beyond the LO from $f_c$: $$f_{image}=f_{LO}+f_{IF}=f_c+2f_{IF}=2\ \text{MHz}+1.6\ \text{MHz}=\boxed{3.6\ \text{MHz}}$$
3. Part (c) — fixed front-end filter: AM band and station count. A FIXED (non-tracking) front-end filter must pass every channel in the tuned block, from $f_{c,\min}$ to $f_{c,\max}$, while each channel's own image ($f_c+2f_{IF}$, high-side) must stay outside that same passband, so the lowest channel's image cannot fall below the highest channel: $f_{c,\max}\le f_{c,\min}+2f_{IF}$, i.e. the whole band the front-end can unambiguously cover is capped at $2f_{IF}=1.6$ MHz, centered on 2 MHz: $$f_{lo,edge}=2-0.8=1.2\ \text{MHz}\qquad f_{hi,edge}=2+0.8=2.8\ \text{MHz}$$ $$\boxed{\text{AM band: }1.2\ \text{MHz to }2.8\ \text{MHz (1.6 MHz wide)}}$$ With 20 kHz channels, the number of stations that fit in this span is $$N=\dfrac{1.6\ \text{MHz}}{20\ \text{kHz}}=\boxed{80\ \text{stations}}$$