22-Elec-A3 Signals and Communications · December 2017

Question 5 of 5: Superheterodyne Receiver — Image Frequency and Front-End Design

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

Notes on this paper

Paper format. Engineers Canada national examination 16-Elec-A3 — Signals and Communications, December 2017. Closed book (approved Casio or Sharp calculator permitted), 3 hours, 4 pages. Five questions, all compulsory, all of equal value (20 marks each). All five are solved in full below, every sub-part answered.

Reference texts. B. P. Lathi & Z. Ding, Modern Digital and Analog Communication Systems, 5th ed. (AM modulation and detection, PCM quantisation, superheterodyne receivers); B. P. Lathi, Linear Systems and Signals, 2nd ed. (Fourier series, ideal filtering, signal power); A. V. Oppenheim & A. S. Willsky, Signals and Systems, 2nd ed. (LTI systems, z-transform, BIBO stability); J. G. Proakis & D. G. Manolakis, Digital Signal Processing, 4th ed. (difference equations, direct-form structures); S. Haykin, Communication Systems, 5th ed. (image rejection, envelope detection, companding).

Question 5: Superheterodyne Receiver — Image Frequency and Front-End Design (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.

Quantity	Symbol	Value
Intermediate frequency	$f_{IF}$	820 kHz
Wanted carrier	$f_c$	2.2 MHz
Channel bandwidth (AM)	$B$	20 kHz
Tuning convention	—	upper-side (high-side) injection
Top of tuning range	$f_U$	2.2 MHz

Find. (a) the image frequency; (b) the local-oscillator frequency; (c) the argument for upper-side tuning; (d) a receiver block diagram, the fixed RF filter's centre frequency and bandwidth, the lowest tunable station $f_L$, and the LO tuning range.

Figure 5.2 — Both the wanted carrier and the image sit 820 kHz from the LO, so the mixer maps them onto the same IF; only the RF filter can separate them.

Approach. Use the mixer relations $f_{LO} = f_c + f_{IF}$ (upper-side) and $f_{\text{image}} = f_{LO} + f_{IF}$, then impose the condition that with a fixed front-end filter the image of the lowest wanted station must fall outside the filter's passband; this single inequality fixes $f_L$ and everything else follows.

(b) Local-oscillator frequency (needed first). Upper-side tuning places the oscillator above the wanted carrier by exactly the intermediate frequency: $$f_{LO} = f_c + f_{IF} = 2.200 + 0.820 = \boxed{3.020\ \text{MHz}}$$
(a) Image frequency. A mixer responds to any input that differs from the LO by the IF, so besides the wanted carrier at $f_{LO} - f_{IF}$ there is a second, unwanted response at $f_{LO} + f_{IF}$: $$f_{\text{image}} = f_{LO} + f_{IF} = f_c + 2f_{IF} = 2.200 + 2(0.820) = \boxed{3.840\ \text{MHz}}$$ A station at 3.84 MHz would be translated onto the same 820 kHz IF as the wanted 2.2 MHz station and would pass straight through the rest of the receiver. Since the two are separated by $2f_{IF} = 1.64$ MHz, the only place they can be distinguished is before the mixer — hence the RF front-end filter.
(c) Why upper-side tuning. The advantage is the far smaller tuning ratio demanded of the local oscillator. Tuning a band from $f_L$ to $f_U$ requires $$\text{upper side:}\ \frac{f_U + f_{IF}}{f_L + f_{IF}}, \qquad\qquad \text{lower side:}\ \frac{f_U - f_{IF}}{f_L - f_{IF}}$$ Adding the IF to both ends compresses the ratio; subtracting it stretches the ratio dramatically. For a standard 0.55–1.6 MHz broadcast band with this 820 kHz IF, upper-side tuning needs a ratio of only $(1.6+0.82)/(0.55+0.82) = 1.77$, whereas lower-side tuning is not realisable at all — the low end would demand a negative oscillator frequency, since $f_L = 0.55\ \text{MHz} \lt f_{IF}$. For this receiver's own band the upper-side ratio is $(2.2+0.82)/(0.56+0.82) = 2.19$.

This matters because a variable capacitor's capacitance ratio must equal the square of the frequency ratio ($f \propto 1/\sqrt{C}$): a 2.19:1 frequency swing needs a 4.8:1 capacitance swing, which is readily built, while the low-side alternative would need an impossible one. Upper-side tuning also guarantees $f_{LO} \gt f_{IF}$ everywhere, so the oscillator can never itself land on the IF and cause a whistle.
(d) Fix the lowest tunable station. With a fixed RF filter passing the whole band, image rejection can no longer be obtained by retuning the front end. The image of any station at $f$ sits at $f + 2f_{IF}$; for every image to lie above the band, it suffices that the image of the lowest station clear the top of the band: $$f_L + 2f_{IF} \ge f_U \;\Longrightarrow\; f_L \ge f_U - 2f_{IF} = 2.200 - 1.640 = \boxed{0.560\ \text{MHz}}$$ Taking the equality gives the widest possible tuning range, $f_L = 0.56$ MHz to $f_U = 2.2$ MHz. Any lower $f_L$ would place its image inside the passband, where nothing downstream could remove it.
Specify the RF front-end filter. The filter must pass every wanted channel in full, i.e. each carrier plus its $\pm 10$ kHz sidebands: $$f_{\text{low}} = f_L - \frac{B}{2} = 0.560 - 0.010 = 0.550\ \text{MHz}, \qquad f_{\text{high}} = f_U + \frac{B}{2} = 2.200 + 0.010 = 2.210\ \text{MHz}$$ so that $$\boxed{B_{RF} = 2.210 - 0.550 = 1.660\ \text{MHz}, \qquad f_{c,RF} = \frac{2.210 + 0.550}{2} = 1.380\ \text{MHz}}$$ The required selectivity is modest, $Q = f_{c,RF}/B_{RF} = 0.83$ — this is a broad preselector whose job is image rejection, not channel selection. Channel selection is done entirely by the 20 kHz IF filter.
Specify the LO tuning range. The oscillator tracks 820 kHz above each carrier across the whole band: $$f_{LO} \in \bigl[f_L + f_{IF},\ f_U + f_{IF}\bigr] = \boxed{1.380\ \text{MHz to } 3.020\ \text{MHz}}$$ a 2.19:1 ratio, achievable with a single ganged variable capacitor. The complete receiver is shown below: fixed RF band-pass, mixer driven by the tunable LO, fixed 820 kHz IF amplifier of 20 kHz bandwidth, envelope detector and audio amplifier.

Figure 5.1 — Superheterodyne receiver with a fixed RF front end. The LO tracks the wanted carrier 820 kHz above it (upper-side tuning).

Check: the image-rejection condition is applied at carrier level, which is the standard exam convention and yields the clean result $f_L = f_U - 2f_{IF} = 0.56$ MHz. Strictly, the image sidebands of the lowest station extend to $f_L + 2f_{IF} + B/2 = 2.21$ MHz and so graze the top edge of the passband. A design that also excludes the image sidebands raises the lowest station to $f_L = f_U + B/2 - 2f_{IF} = 0.57$ MHz, moving the LO range to 1.39–3.02 MHz; the practical filter skirt provides this 10 kHz of margin in any case.

Result	Value
(a) Image frequency	3.840 MHz
(b) Local-oscillator frequency	3.020 MHz
(c) Upper-side LO tuning ratio (this band)	2.19 : 1 (lower-side not realisable)
(d) Lowest tunable station $f_L$	0.560 MHz
(d) RF filter passband	0.550–2.210 MHz
(d) RF filter bandwidth / centre	1.660 MHz / 1.380 MHz
(d) LO tuning range	1.380–3.020 MHz
(d) IF filter	820 kHz, 20 kHz bandwidth

Topic. The superheterodyne receiver: frequency conversion, image response and the trade-off that sets the intermediate frequency.

Key relations. High-side injection $f_{LO} = f_c + f_{IF}$; image $f_{\text{image}} = f_c + 2f_{IF}$; low-side injection $f_{LO} = f_c - f_{IF}$ with image $f_c - 2f_{IF}$; fixed-preselector condition $f_L \ge f_U - 2f_{IF}$; capacitance ratio $C_{\max}/C_{\min} = (f_{\max}/f_{\min})^{2}$.

Why this works. A mixer forms sum and difference frequencies, and the IF filter accepts only $|f - f_{LO}| = f_{IF}$. That absolute value is the whole story: two input frequencies, one below the LO and one above it, satisfy it equally well, and no amount of filtering after the mixer can separate them because they have already been superimposed. The image is therefore an architectural consequence, not a component imperfection, and the only cure is selectivity ahead of the mixer. Raising $f_{IF}$ pushes the image further away and makes the preselector's job easier, but a higher IF makes the fixed IF filter's fractional bandwidth larger and channel selectivity harder — the classic tension that double-conversion receivers resolve with a high first IF and a low second IF.

Common pitfall. Quoting the image as $f_c + f_{IF}$ (that is the LO) rather than $f_c + 2f_{IF}$. In part (d), the other frequent error is forgetting that the RF filter must accommodate the channel sidebands, so its edges sit at $f_L - B/2$ and $f_U + B/2$ rather than at the carrier limits themselves; and quoting the geometric rather than the arithmetic centre when the question asks for the centre frequency of a band-pass specification.

Source. Haykin, Communication Systems, 5th ed., Ch. 3 (superheterodyne receiver and image rejection); Lathi & Ding, Modern Digital and Analog Communication Systems, 5th ed., Ch. 4 (frequency conversion and receiver architecture).

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