22-Elec-A3 Signals and Communications · December 2018

Question 5 of 5: Square-Law Frequency Downconverter — IF Selection and Filter Bandwidth

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

Notes on this paper

Paper format. National Examinations, December 2018 — 16-Elec-A3 Signals and Communications. Three hours, closed book (an approved Casio or Sharp calculator is permitted). Answer all 5 questions; all 5 questions are of equal value (20 marks each, 100 marks total). Candidates are urged to state any assumption made where a question admits more than one reading.

Reference texts.

B. P. Lathi and R. Ding, Modern Digital and Analog Communication Systems, 4th ed. — Ch. 2–4 (Fourier series and transforms), Ch. 4 (amplitude modulation, DSB/SSB), Ch. 6 (sampling and PCM).
B. P. Lathi, Linear Systems and Signals, 2nd ed. — Ch. 6–7 (Fourier analysis of periodic and aperiodic signals), Ch. 5 (discrete-time systems).
A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. — Ch. 3 (Fourier series), Ch. 7 (sampling), Ch. 10 (z-transform).
J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. — Ch. 3 (z-transform), Ch. 5 (frequency-domain analysis of LTI systems), Ch. 9 (filter structures).
S. Haykin, Communication Systems, 5th ed. — Ch. 3 (amplitude modulation), Ch. 5 (pulse-code modulation and quantisation noise).

Note on the Question 4 figure. The block diagram shows both summing junctions drawn as plain adders — there are no minus signs at any input — so the printed arm gains are the multipliers exactly as shown: −3/4 on the first delay output into the input adder, +1/2 on the second delay output into the same adder, and +1/2 on the first delay output feeding forward to the output adder. The consequence, developed in Question 4, is that one pole falls outside the unit circle. That is a genuine property of the printed network, and the answer below treats it as such.

Question 5: Square-Law Frequency Downconverter — IF Selection and Filter Bandwidth (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. Received signal $A_c m(t)\cos(2\pi f_c t + \theta)$ summed with a local oscillator $A_l\cos(2\pi f_l t)$, the sum squared, then band-pass filtered at $f_i = |f_c - f_l|$. The message $m(t)$ is band-limited to $B$ Hz, and $f_c$, $f_l$ are both much larger than $B$ and than $f_i$.

Find. (a) the smallest usable IF; (b) the smallest BPF bandwidth; and (c) the largest BPF bandwidth, in each case such that the output is an undistorted copy of the modulated message.

Q5 - the square-law downconverter. Adding the local oscillator BEFORE the non-linearity is what creates the wanted cross-product; the squarer alone cannot translate frequency.

Approach. Expand the square of the sum, identify every term and the band it occupies, isolate the wanted cross-product at $f_i$, and then impose the requirement that the filter passband contain the whole wanted band and none of the others.

Square the sum. With $s(t) = A_c m(t)\cos(2\pi f_c t + \theta) + A_l\cos(2\pi f_l t)$, squaring gives three groups — two squares and a cross-product: $$s^{2}(t) = \underbrace{A_c^{2}m^{2}(t)\cos^{2}(2\pi f_c t + \theta)}_{\text{(i)}} + \underbrace{A_l^{2}\cos^{2}(2\pi f_l t)}_{\text{(ii)}} + \underbrace{2A_c A_l\, m(t)\cos(2\pi f_c t + \theta)\cos(2\pi f_l t)}_{\text{(iii)}} .$$ The cross-product is the only term that can translate the message to a new carrier, which is why the oscillator must be added before the non-linearity rather than after it.
Expand each group and list its band. Using $\cos^{2}\phi = \tfrac{1}{2}(1+\cos 2\phi)$ and $2\cos\alpha\cos\beta = \cos(\alpha-\beta)+\cos(\alpha+\beta)$: $$\text{(i)} = \frac{A_c^{2}}{2}m^{2}(t) + \frac{A_c^{2}}{2}m^{2}(t)\cos(4\pi f_c t + 2\theta), \qquad \text{(ii)} = \frac{A_l^{2}}{2} + \frac{A_l^{2}}{2}\cos(4\pi f_l t),$$ $$\text{(iii)} = A_c A_l\, m(t)\cos\big(2\pi (f_c - f_l)t + \theta\big) + A_c A_l\, m(t)\cos\big(2\pi (f_c + f_l)t + \theta\big).$$ Since $m(t)$ occupies $0$ to $B$, the product $m^{2}(t)$ occupies $0$ to $2B$ (squaring convolves the spectrum with itself and so doubles its width). The occupied bands are therefore:

Term	Centre frequency	Occupied band
$\tfrac{A_l^{2}}{2}$ (DC)	0	impulse at 0
$\tfrac{A_c^{2}}{2}m^{2}(t)$	baseband	$0$ to $2B$
wanted: $A_cA_l\,m(t)\cos(2\pi f_i t + \theta)$	$f_i = \|f_c-f_l\|$	$f_i - B$ to $f_i + B$
$A_cA_l\,m(t)\cos(2\pi (f_c+f_l)t + \theta)$	$f_c + f_l$	$f_c+f_l-B$ to $f_c+f_l+B$
$\tfrac{A_l^{2}}{2}\cos(4\pi f_l t)$	$2f_l$	impulse at $2f_l$
$\tfrac{A_c^{2}}{2}m^{2}(t)\cos(4\pi f_c t + 2\theta)$	$2f_c$	$2f_c-2B$ to $2f_c+2B$

Q5 - spectrum at the squarer output (schematic, drawn for fi = 6B). The wanted term sits at fi with width 2B; its nearest neighbour is the baseband m-squared term, which reaches up to 2B. Keeping the two apart is the whole design constraint - the far-off terms at fc + fl, 2fc and 2fl never bind.

Identify the binding neighbour. The wanted term is centred at $f_i$ and is $2B$ wide, running from $f_i - B$ to $f_i + B$. Of the unwanted terms, three ($f_c+f_l$, $2f_l$, $2f_c$) sit at roughly twice the carrier frequency and are enormously far away, since $f_c, f_l \gg f_i$. The only close neighbour is the baseband $m^{2}(t)$ term, which extends upward from DC to $2B$. The entire design problem is therefore the separation between $2B$ and $f_i - B$.
(a) Smallest usable IF. The wanted band must lie entirely above the $m^{2}$ tail, so its lower edge must clear $2B$: $$f_i - B \ge 2B \;\Longrightarrow\; f_i \ge 3B \;\Longrightarrow\; \boxed{f_{i,\min} = 3B}$$ Equivalently, $|f_c - f_l| \ge 3B$: the local oscillator must be offset from the carrier by at least three message bandwidths. Choosing $f_i$ any smaller lets the self-product $m^{2}(t)$ overlap the wanted sidebands, and because that overlap is additive and non-invertible, the resulting distortion cannot be removed by any later filtering.
(b) Smallest BPF bandwidth. The filter must pass the wanted term undistorted, and that term is $2B$ wide (a message of bandwidth $B$ modulated onto $f_i$ produces sidebands $B$ above and $B$ below). Any narrower window would truncate the outer sidebands and distort the recovered message. Hence $$\boxed{W_{\min} = 2B}$$ This is a lower limit set by the signal, independent of $f_i$.
(c) Largest BPF bandwidth. The filter may be widened only until its lower skirt reaches the $m^{2}$ term at $2B$. For a filter of bandwidth $W$ centred at $f_i$, the lower edge sits at $f_i - W/2$, so the constraint is $$f_i - \frac{W}{2} \ge 2B \;\Longrightarrow\; W \le 2\left(f_i - 2B\right)$$ $$\boxed{W_{\max} = 2\,(f_i - 2B)}$$ The upper skirt is unconstrained in practice, since the next term upward sits near $f_c + f_l \gg f_i$. Note the consistency check: at the minimum IF of part (a), $f_i = 3B$, this gives $W_{\max} = 2(3B - 2B) = 2B = W_{\min}$ — the two limits coincide exactly, which is another way of seeing that $3B$ is the smallest IF for which any valid filter exists at all.

Check: assumption on the relative size of $f_c$ and $f_l$. The conclusion that the baseband $m^{2}$ term is the only binding neighbour rests on $f_c + f_l - B \gg f_i + W/2$ and $2f_l \gg f_i + W/2$, which hold comfortably in any real downconverter (where $f_c$ and $f_l$ are close to each other and both far above the IF). If a design were attempted with $f_l$ comparable to $B$ — not a downconverter in any useful sense — the upper terms would also have to be checked, and $W_{\max}$ would become $\min\{2(f_i-2B),\; 2(f_c+f_l-B-f_i)\}$.

Quantity	Symbol	Result
Wanted output term	—	$A_cA_l\,m(t)\cos(2\pi f_i t + \theta)$, occupying $f_i \pm B$
Binding interference	—	$\tfrac{A_c^{2}}{2}m^{2}(t)$, occupying 0 to $2B$
(a) Smallest IF	$f_{i,\min}$	$3B$
(b) Smallest BPF bandwidth	$W_{\min}$	$2B$
(c) Largest BPF bandwidth	$W_{\max}$	$2(f_i - 2B)$

Topic. Non-linear (square-law) frequency conversion; spectral bookkeeping of intermodulation products; IF and filter-bandwidth selection.

Key relations. $\cos^{2}\phi = \tfrac{1}{2}(1+\cos 2\phi)$; $2\cos\alpha\cos\beta = \cos(\alpha-\beta)+\cos(\alpha+\beta)$; squaring a signal of bandwidth $B$ produces a self-product of bandwidth $2B$; a message of bandwidth $B$ on a carrier occupies $2B$; an ideal BPF of centre $f_i$ and bandwidth $W$ passes $[f_i - W/2,\, f_i + W/2]$.

Why this works. A square-law device is a practical stand-in for an ideal multiplier: a true four-quadrant multiplier is expensive at radio frequencies, whereas any device with a quadratic term in its transfer characteristic (a diode or FET biased into its curved region) will do. Squaring the sum of two signals produces the cross-term $2ab$, which is exactly the product a mixer needs; the price is the two square terms $a^{2}$ and $b^{2}$, which appear as unwanted products and set the design constraints. This is why the answers are entirely about spectral geometry: the physics is done once you have written $(a+b)^{2} = a^{2}+2ab+b^{2}$, and everything after that is deciding which of the resulting bands the filter can be made to separate.

Common pitfall. The most common error is forgetting that squaring doubles the message bandwidth, so the $m^{2}$ term is taken to extend only to $B$ instead of $2B$; that yields $f_{i,\min}=2B$ and $W_{\max}=2(f_i-B)$, both wrong. The second is confusing the filter's bandwidth with its half-width when writing the lower-edge condition, which gives $W_{\max}=f_i-2B$ — half the correct value. A useful guard is the consistency check performed in Step 6: at $f_i = f_{i,\min}$ the maximum and minimum bandwidths must coincide, and any pair of answers failing that test contains an error.

Source. Lathi & Ding, Modern Digital and Analog Communication Systems, 4th ed., §4.3 (non-linear modulators and frequency mixers); Haykin, Communication Systems, 5th ed., §3.7 (frequency translation and superheterodyne receivers).

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