22-Elec-A3 Signals and Communications · May 2015

Question 5 of 7: Square-Law Frequency Downconverter — Choice of IF and Filter Bandwidth

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

Notes on this paper

Paper format. 07-Elec-A3 Signals and Communications, May 2015 — 3 hours, closed book, non-programmable calculator permitted. Seven questions, all of equal value (20 marks each); any five constitute a complete paper and only the first five in the answer book are marked. All seven are solved here, since the set is a study resource.

Reference texts. B. P. Lathi & Z. Ding, Modern Digital and Analog Communication Systems, 5th ed.; B. P. Lathi, Linear Systems and Signals, 2nd ed.; A. V. Oppenheim & A. S. Willsky, Signals and Systems, 2nd ed.; J. G. Proakis & D. G. Manolakis, Digital Signal Processing, 4th ed.; S. Haykin & M. Moher, Communication Systems, 5th ed.

Check: pulse definition. The pulse is p(t) = t/a — a ramp, not a constant — and every part of Question 1 is solved against that reading. A constant-height pulse would make parts (a)–(c) trivial and would not produce the odd-harmonic structure the question is clearly aiming at.

Question 5: Square-Law Frequency Downconverter — Choice of IF 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. Input $u(t) = A_c m(t)\cos(2\pi f_c t + \theta) + A_l\cos(2\pi f_l t)$ applied to a squarer, followed by a BPF centred on $f_i = |f_c - f_l|$; the message $m(t)$ is band-limited to $B$ Hz.

Find. The minimum usable IF, and the minimum and maximum BPF bandwidths that pass the wanted product undistorted while rejecting every other product.

Approach. Expand the square into its three groups of terms, place each group on the frequency axis with its own bandwidth, and impose the requirement that the passband contain the whole wanted band and nothing else.

Expand the square. With $u = v + w$ where $v = A_c m(t)\cos(2\pi f_c t + \theta)$ and $w = A_l\cos(2\pi f_l t)$, $$u^2 = v^2 + 2vw + w^2$$ The wanted term is the cross-product, which after the product-to-sum identity is $$2vw = A_cA_l\,m(t)\Big[\cos\big(2\pi(f_c-f_l)t + \theta\big) + \cos\big(2\pi(f_c+f_l)t + \theta\big)\Big]$$ The first cosine is the message translated to the IF: it occupies $f_i \pm B$, a band of width $2B$.
Locate the unwanted products. The self-terms give $$v^2 = \frac{A_c^2m^2(t)}{2}\Big[1 + \cos(4\pi f_ct + 2\theta)\Big], \qquad w^2 = \frac{A_l^2}{2}\Big[1 + \cos(4\pi f_lt)\Big]$$ Since squaring a signal doubles its bandwidth, $m^2(t)$ occupies $0$ to $2B$. The complete inventory of output components is therefore: a baseband block from 0 to $2B$ (plus a DC line), the wanted band at $f_i \pm B$, the sum-frequency band at $(f_c+f_l) \pm B$, a line at $2f_l$, and a block at $2f_c \pm 2B$.
Figure Q5.1 — spectral inventory at the squarer output (not to scale). The m²(t) block reaching 2B is the constraint that sets both the minimum IF and the maximum filter bandwidth.
Smallest usable IF (part a). The nearest interferer below the wanted band is the baseband $m^2(t)$ block, whose upper edge is at $2B$. The lower edge of the wanted band is $f_i - B$, so the two must not overlap: $$f_i - B \gt 2B \;\Longrightarrow\; \boxed{\,f_i \gt 3B\ \ (\text{smallest IF} = 3B)\,}$$ Below this the squared-message spectrum folds into the IF band and no filter can separate them, because the two overlap in frequency rather than merely sitting close together.
Smallest BPF bandwidth (part b). The filter must pass the entire wanted band without truncating either sideband. That band runs from $f_i - B$ to $f_i + B$, so $$\boxed{\,B_{\text{BPF,min}} = 2B\,}$$ Anything narrower clips the outer message frequencies and produces linear (amplitude) distortion of $m(t)$.
Largest BPF bandwidth (part c). Widening the symmetric passband about $f_i$ eventually admits an unwanted product. The binding constraint is again the baseband block: the lower passband edge $f_i - B_{\text{BPF}}/2$ must stay above $2B$, giving $$\frac{B_{\text{BPF}}}{2} \lt f_i - 2B \;\Longrightarrow\; \boxed{\,B_{\text{BPF,max}} = 2\,(f_i - 2B)\,}$$ The upper edge must simultaneously stay below the sum-frequency band, $B_{\text{BPF}} \lt 2\big[(f_c+f_l) - B - f_i\big]$, but in any practical downconverter $f_c + f_l \gg f_i$ so the low-side limit is the one that binds. Note that both bounds collapse together, $2(f_i - 2B) \to 2B$, exactly at the minimum IF $f_i = 3B$ found in part (a) — a satisfying consistency check: at the smallest usable IF there is precisely one admissible filter bandwidth.
Numerical illustration. For a 5 kHz message with $f_c = 90$ MHz and $f_l = 79.3$ MHz, the IF is 10.7 MHz, the smallest usable IF would be only 15 kHz, and the filter bandwidth may lie anywhere between 10 kHz and $2(10.7\ \text{MHz} - 10\ \text{kHz}) = 21.38$ MHz — a very wide window, which is why real receivers choose the BPF bandwidth from selectivity and adjacent-channel requirements rather than from these distortion limits.

Quantity	Result
Wanted product	$A_cA_l\,m(t)\cos(2\pi f_i t + \theta)$, occupying $f_i \pm B$
Unwanted products	DC and $0$–$2B$ (from $m^2$); $2f_l$; $(f_c+f_l)\pm B$; $2f_c \pm 2B$
Smallest IF	$f_i = 3B$
Smallest BPF bandwidth	$2B$
Largest BPF bandwidth	$2(f_i - 2B)$, subject to $2[(f_c+f_l)-B-f_i]$

Topic. Non-linear (square-law) mixing for frequency translation, and the spectral-planning constraints that set the IF and filter bandwidth.

Key relations. $(v+w)^2 = v^2 + 2vw + w^2$; $\cos A\cos B = \frac{1}{2}[\cos(A-B)+\cos(A+B)]$; squaring doubles bandwidth, so $m(t)$ band-limited to $B$ gives $m^2(t)$ band-limited to $2B$; an ideal BPF passes $[f_i - B_{\text{BPF}}/2,\ f_i + B_{\text{BPF}}/2]$.

Why this works. A squarer is a non-linear device, so it generates the sum and difference frequencies that a purely linear circuit never could — this is exactly the mechanism a mixer exploits. The price is that it also generates the self-products $v^2$ and $w^2$, which are useless but real, so the design problem is entirely one of spectral bookkeeping: list every product with its bandwidth, then choose $f_i$ and the filter so the wanted band is isolated. The doubling of bandwidth on squaring is the single fact that drives the $3B$ answer.

Common pitfall. Treating the $m^2(t)$ term as occupying only $B$ Hz, which gives the wrong answer $f_i \gt 2B$. A second is forgetting that the filter is specified by its total bandwidth while the constraints are on the distance from the centre, so the factor of two must appear in both the minimum and the maximum.

Source. Lathi & Ding, Modern Digital and Analog Communication Systems, §4.2 and §4.8 (non-linear modulators, frequency conversion and the IF); Haykin & Moher, Communication Systems, §3.6 (superheterodyne receiver).

← Question 4 Next: Question 6 →

Quantity	Result
Wanted product	\(A_cA_l\,m(t)\cos(2\pi f_i t + \theta)\), occupying \(f_i \pm B\)
Unwanted products	DC and \(0\)–\(2B\) (from \(m^2\)); \(2f_l\); \((f_c+f_l)\pm B\); \(2f_c \pm 2B\)
Smallest IF	\(f_i = 3B\)
Smallest BPF bandwidth	\(2B\)
Largest BPF bandwidth	\(2(f_i - 2B)\), subject to \(2[(f_c+f_l)-B-f_i]\)