22-Elec-A3 Signals and Communications · December 2013

Question 3 of 6: Impulse Sampling of a Squared-Sinc Signal

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

Notes on this paper

Paper format. National Exams, December 2013 — 07-Elec-A3, Signals and Communications. Three hours, closed book (one two-sided aid sheet and an approved calculator permitted). Six questions, all of equal value at 20 marks; five constitute a complete paper.

All six questions are solved here. The examination marks only the first five answers submitted, but this document is a study resource, so every question — and every sub-part — is worked in full.

Reference texts.

B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. — Ch. 3 (signal transmission), Ch. 4 (amplitude modulation), Ch. 5 (angle modulation), Ch. 6 (sampling).
B. P. Lathi, Linear Systems and Signals, 2nd ed. — Ch. 6 (Fourier series), Ch. 7 (Fourier transform).
A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. — Ch. 4 (Fourier transform properties), Ch. 7 (sampling), Ch. 10 (z-transform).
S. Haykin, Communication Systems, 5th ed. — Ch. 3 (amplitude modulation), Ch. 4 (angle modulation).
J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. — Ch. 3 (z-transform and inversion).

Question 3: Impulse Sampling of a Squared-Sinc Signal (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. $x(t) = \operatorname{sinc}^2(10\pi t)$ with $\operatorname{sinc}(u) = \sin u / u$; impulse sampling at $f_s = $ 10, 20 and 30 Hz; reconstruction filter an ideal LPF of bandwidth 10 Hz.

Find. For each rate: the sampled time signal, its spectrum, whether $x(t)$ is recoverable, and the spectrum at the output of the 10 Hz low-pass filter.

Approach. Find the bandwidth of $x(t)$ from the transform pair for $\operatorname{sinc}$, square it (which convolves the spectrum with itself and doubles the bandwidth), compare each sampling rate with the Nyquist rate, and apply the impulse-sampling spectrum formula.

Find the bandwidth of the signal. The standard pair is $\operatorname{sinc}(2\pi B t) \leftrightarrow \frac{1}{2B}\operatorname{rect}\!\left(\frac{f}{2B}\right)$. Matching $2\pi B = 10\pi$ gives $B = 5$ Hz, so $\operatorname{sinc}(10\pi t)$ occupies $|f| \lt 5$ Hz. Squaring in time convolves in frequency: $$X(f) = \tfrac{1}{10}\operatorname{rect}\!\left(\tfrac{f}{10}\right) * \tfrac{1}{10}\operatorname{rect}\!\left(\tfrac{f}{10}\right) = \tfrac{1}{10}\,\Delta\!\left(\tfrac{f}{10}\right)$$ a triangle of half-width 10 Hz and peak $X(0) = 0.1$. Hence $$\boxed{\,B_x = 10\ \text{Hz}, \qquad f_{\text{Nyquist}} = 2B_x = 20\ \text{Hz}\,}$$ The peak checks against $X(0) = \int x(t)\,dt = 0.1$.

The message x(t) = sinc²(10πt) (left) and its triangular spectrum X(f) (right), which is strictly band-limited to |f| ≤ 10 Hz.

Sketch the sampled signals (part a). Impulse sampling gives $x_s(t) = \sum_n x(nT_s)\,\delta(t-nT_s)$, so each impulse has area equal to the sample value. At $f_s = 10$ Hz the sample instants $t = n/10$ fall on every zero crossing of $\operatorname{sinc}^2(10\pi t)$, so only the sample at $t=0$ survives and $x_s(t) = \delta(t)$ — a dramatic illustration of information loss. At 20 Hz the odd-indexed samples are $\left(\tfrac{2}{n\pi}\right)^2$ and even-indexed ones vanish; at 30 Hz the main lobe is sampled three times and the sidelobes are resolved.

Part (a) — the impulse-sampled signal at each rate. Impulse areas equal the sample values x(nT_s); the dashed curve is the underlying x(t).

Sketch the sampled spectra (part b). The spectrum of an impulse-sampled signal is the periodic repetition $$X_s(f) = f_s\sum_{k=-\infty}^{\infty} X(f - k f_s)$$ so identical triangles of peak $f_s X(0)$ are centred on every multiple of $f_s$. At 10 Hz the triangles (each 20 Hz wide) overlap heavily; at 20 Hz adjacent triangles just touch at $f = \pm 10$ Hz, where each is already zero; at 30 Hz they are separated by a 10 Hz guard band on each side.

Part (b) — the sampled spectrum X_s(f) = f_s∑X(f − kf_s). Grey dashed triangles are the individual replicas; the solid curve is their sum.

Decide recoverability at each rate (part c). The sampling theorem requires $f_s \gt 2B_x = 20$ Hz for guaranteed recovery, with equality the critical case.
- (i) $f_s = 10$ Hz — not recoverable. The rate is below Nyquist, so replicas overlap and alias. In fact the two neighbouring triangles sum to a perfectly flat spectrum: for $0 \le f \lt 10$, $f_s[X(f)+X(f-10)] = 10\left[0.1\left(1-\tfrac{f}{10}\right)+0.1\tfrac{f}{10}\right] = 1$ for all $f$, consistent with $x_s(t) = \delta(t)$, whose transform is indeed constant. All spectral shape has been destroyed.
- (ii) $f_s = 20$ Hz — recoverable (critical case). The replicas touch at $f = \pm 10$ Hz but the triangle is already zero there, so no energy overlaps and an ideal LPF of bandwidth exactly 10 Hz returns $20X(f)$. Recovery is exact in theory, though it demands a brick-wall filter and is not robust in practice.
- (iii) $f_s = 30$ Hz — recoverable with margin. The baseband triangle is isolated between $\pm 10$ Hz with the nearest replica edge at 20 Hz, leaving a 10 Hz guard band for a realisable filter roll-off.
Sketch the low-pass filter outputs (part d). The ideal LPF of bandwidth 10 Hz passes $|f| \le 10$ Hz unchanged and rejects everything else, so $Y(f) = X_s(f)\,\operatorname{rect}(f/20)$: $$Y(f) = \begin{cases} \text{flat at } 1, & f_s = 10\ \text{Hz (aliased)}\\ 20\,X(f) = 2\Delta(f/10), & f_s = 20\ \text{Hz}\\ 30\,X(f) = 3\Delta(f/10), & f_s = 30\ \text{Hz}\end{cases}$$ The 20 Hz and 30 Hz outputs are exact scaled copies of $X(f)$ — divide by $f_s$ to recover $x(t)$ exactly. The 10 Hz output is a rectangle, whose inverse transform $y(t) = 20\operatorname{sinc}(20\pi t)$ bears no resemblance to the original signal.

Part (d) — output of the ideal low-pass filter of bandwidth 10 Hz. The red dashed triangle is the wanted f_sX(f); only the lowest rate departs from it.

Quantity	Result
Signal bandwidth $B_x$	10 Hz
Spectrum $X(f)$	Triangle, peak $X(0)=0.1$, zero at $\|f\|=10$ Hz
Nyquist rate	20 Hz
(i) $f_s = 10$ Hz	Aliased; $x_s(t)=\delta(t)$; $X_s$ flat at 1; not recoverable
(ii) $f_s = 20$ Hz	Critical sampling; replicas just touch; recoverable, LPF output $20X(f)$, peak 2
(iii) $f_s = 30$ Hz	Oversampled, 10 Hz guard band; recoverable, LPF output $30X(f)$, peak 3

Topic. Uniform impulse sampling, spectral replication, aliasing and ideal reconstruction.

Key relations. $x_s(t) = x(t)\!\sum_n\delta(t-nT_s) \leftrightarrow X_s(f) = f_s\sum_k X(f-kf_s)$; sampling theorem $f_s \gt 2B$; $\operatorname{sinc}(2\pi Bt) \leftrightarrow \tfrac{1}{2B}\operatorname{rect}(f/2B)$; multiplication in time $\leftrightarrow$ convolution in frequency.

Why this works. Multiplying by an impulse train is multiplying by a periodic signal whose Fourier series coefficients are all equal to $f_s$; each of those coefficients shifts a complete copy of $X(f)$ to a multiple of $f_s$. Recovery is possible precisely when those copies do not overlap, which is the sampling theorem in one line. The squared-sinc is a good exam vehicle because squaring in the time domain doubles the bandwidth — a student who reads $B=5$ Hz off the sinc alone concludes that 10 Hz sampling suffices, and the degenerate $x_s(t)=\delta(t)$ result at that rate shows spectacularly how wrong that is.

Common pitfall. Forgetting the bandwidth doubling under squaring. A second one is calling the 20 Hz case unrecoverable because the replicas touch: touching is harmless when the spectrum has already decayed to zero at the fold frequency, though it would be fatal for a spectrum with a non-zero edge value (such as a raw sinc, whose rectangular spectrum has a discontinuity there).

Source. Lathi & Ding, Modern Digital and Analog Communication Systems, 4th ed., §6.1; Oppenheim & Willsky, Signals and Systems, 2nd ed., §7.1–7.2.

← Question 2 Next: Question 4 →

Quantity	Result
Signal bandwidth \(B_x\)	10 Hz
Spectrum \(X(f)\)	Triangle, peak \(X(0)=0.1\), zero at \(\|f\|=10\) Hz
Nyquist rate	20 Hz
(i) \(f_s = 10\) Hz	Aliased; \(x_s(t)=\delta(t)\); \(X_s\) flat at 1; not recoverable
(ii) \(f_s = 20\) Hz	Critical sampling; replicas just touch; recoverable, LPF output \(20X(f)\), peak 2
(iii) \(f_s = 30\) Hz	Oversampled, 10 Hz guard band; recoverable, LPF output \(30X(f)\), peak 3