Question 5 of 7: Superheterodyne receiver: local-oscillator choice, image frequency and band planning (20 marks)

Question 5 — Superheterodyne receiver: local-oscillator choice, image frequency and band planning (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. Intermediate-frequency centre $f_{IF} = 800\ \text{kHz}$; wanted carrier $f_c = 2\ \text{MHz}$; channel bandwidth $B_{ch} = 20\ \text{kHz}$; a fixed (non-tracking) RF front-end filter; the AM band is centred on 2 MHz.

Find. (a) the two possible local-oscillator frequencies and which is preferable for block tuning; (b) the image frequency for that choice; (c) the frequency range of the band and how many 20 kHz stations it holds.

Approach. The mixer produces the IF from the magnitude of the difference $|f_{LO} - f_c|$, which admits two solutions. Choosing between them is a tuning-ratio argument. The image is then the second input frequency that maps to the same IF, and part (c) turns the image requirement into a limit on how wide a fixed front-end filter can be.

1. Solve the mixing equation for both local-oscillator options. The IF is produced by the difference product, so $|f_{LO} - f_c| = f_{IF}$ and $$f_{LO} = f_c + f_{IF} = 2.0 + 0.8 = 2.8\ \text{MHz} \qquad\text{(high-side injection)},$$ $$f_{LO} = f_c - f_{IF} = 2.0 - 0.8 = 1.2\ \text{MHz} \qquad\text{(low-side injection)}.$$ $$\boxed{\;f_{LO} = 2.8\ \text{MHz}\ \text{or}\ 1.2\ \text{MHz}\;}$$
2. Choose between them on the tuning-ratio argument. The band is centred at 2 MHz and, as part (c) establishes, spans 1.2–2.8 MHz. Tuning across it requires the oscillator to sweep $$\text{high side: } 2.0\!-\!3.6\ \text{MHz},\ \text{ratio } 1.80; \qquad \text{low side: } 0.4\!-\!2.0\ \text{MHz},\ \text{ratio } 5.00 .$$ Since a variable capacitor’s frequency ratio goes as $\sqrt{C_{\max}/C_{\min}}$, a 1.8:1 sweep is easy and a 5:1 sweep is not. High-side injection also keeps the oscillator entirely above the received band, so its harmonics and leakage cannot fall on a wanted channel. Hence $$\boxed{\;f_{LO} = 2.8\ \text{MHz (high-side injection) is preferable}\;}$$
3. Locate the image frequency. The image is the other input that differs from the local oscillator by the IF, on the far side: $$f_{\text{image}} = f_{LO} + f_{IF} = f_c + 2f_{IF} = 2.0 + 1.6 \qquad\Longrightarrow\qquad \boxed{\;f_{\text{image}} = 3.6\ \text{MHz}\;}$$ Checking: $|2.8 - 3.6| = 0.8\ \text{MHz} = f_{IF}$, so a station at 3.6 MHz would land on top of the wanted one and no amount of IF selectivity could separate them — image rejection must be done before the mixer.
4. Turn image rejection into a limit on the band width. With a fixed front-end filter, whatever passes to the mixer is admitted for every channel the radio can tune. Let the band run from $f_L$ to $f_H$. The image of the lowest channel is $f_L + 2f_{IF}$; if that frequency lies inside the band, it is a legitimate station and the filter cannot possibly remove it. Safety therefore requires $$f_L + 2f_{IF} \ge f_H \qquad\Longrightarrow\qquad B_{\text{band}} = f_H - f_L \le 2f_{IF} = 1.6\ \text{MHz}.$$ This is the governing design rule for a fixed-filter superheterodyne: the receivable band can be no wider than twice the IF.
5. Centre that band on 2 MHz. Taking the widest permissible band and centring it as the question specifies, $$f_L = 2.0 - 0.8 = 1.2\ \text{MHz},\qquad f_H = 2.0 + 0.8 = 2.8\ \text{MHz}$$ $$\boxed{\;\text{AM band } = 1.2\ \text{MHz to } 2.8\ \text{MHz}\;}$$ At the limiting condition the image of the bottom channel lands exactly on the top edge, $1.2 + 1.6 = 2.8\ \text{MHz}$, so the front-end filter has just enough selectivity to do its job. (This is also why the low-side oscillator would have had to reach 0.4 MHz in step 2.)
6. Count the channels. Each AM station occupies its full 20 kHz bandwidth, so $$N = \frac{B_{\text{band}}}{B_{ch}} = \frac{1.6\times10^{6}}{20\times10^{3}} \qquad\Longrightarrow\qquad \boxed{\;N = 80\ \text{stations}\;}$$ with carriers spaced 20 kHz apart from $1.21\ \text{MHz}$ to $2.79\ \text{MHz}$. If guard bands between adjacent channels were required, the count would fall accordingly.