Question 2 of 6: Sampling a band-limited periodic signal (12 marks)

Question 2: Sampling a band-limited periodic signal (12 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. A periodic continuous-time signal whose Fourier series is band-limited to nineteen harmonics, sampled uniformly at six samples per period.

Find. (a) whether the sampled sequence is periodic and with what fundamental period; (b) whether $f_s$ clears the Nyquist rate; (c) the discrete-time Fourier series coefficients of $x[n]$ expressed through the given $a_k$.

Approach. Substitute $t=nT$ into the given series; the ratio $T/T_p$ then decides the period, the highest harmonic decides the Nyquist rate, and grouping the exponentials that become indistinguishable after sampling produces the discrete-time coefficients.

1. Sample the Fourier series. Putting $t=nT$ into the given expansion, $$x[n]=x_c(nT)=\sum_{k=-9}^{9}a_k\,e^{j2\pi k nT/10^{-3}} .$$ The whole problem hinges on the ratio of the two time scales, $$\frac{T}{T_p}=\frac{\tfrac16\times 10^{-3}}{10^{-3}}=\frac16 ,$$ so the exponent becomes $j2\pi k n/6$ and $$x[n]=\sum_{k=-9}^{9}a_k\,e^{j2\pi k n/6}.$$
2. Read off the period (part a). Every exponential $e^{j2\pi kn/6}$ repeats when $n$ advances by $6$, because $e^{j2\pi k(n+6)/6}=e^{j2\pi kn/6}e^{j2\pi k}=e^{j2\pi kn/6}$ for integer $k$. No smaller advance works for all $k$ present, since $e^{j2\pi k n/6}$ with $k=1$ needs the full six samples. The sampled signal is therefore periodic, with $$\boxed{N=6\ \text{samples}} .$$ This is the general rule that sampling a periodic signal yields a periodic sequence exactly when $T/T_p$ is rational; here $T/T_p=1/6$ and six samples span precisely one continuous-time period.
3. Compare the sampling rate with the Nyquist rate (part b). The highest frequency actually present is the ninth harmonic, $$f_{\max}=9f_0=9\times 1\ \text{kHz}=9\ \text{kHz},$$ so the Nyquist rate — twice the highest frequency — is $$f_{\text{Nyq}}=2f_{\max}=18\ \text{kHz}.$$ The actual sampling rate is $$f_s=\frac1T=\frac{1}{\tfrac16\times10^{-3}}=6\ \text{kHz}.$$ Since $6\ \text{kHz}$ is a factor of three below $18\ \text{kHz}$, $$\boxed{f_s = 6\ \text{kHz}\ \lt\ f_{\text{Nyq}} = 18\ \text{kHz}\ \Rightarrow\ T\ \text{is NOT small enough; the signal aliases}} $$ Only the harmonics with $|k|f_0 \lt f_s/2 = 3\ \text{kHz}$, that is $|k|\le 2$, survive unaliased; harmonics 3 through 9 all fold back into the base band.

Part (c) asks for the discrete-time Fourier series. Because the discrete exponential $e^{j2\pi kn/6}$ depends on $k$ only through $k \bmod 6$, harmonics whose indices differ by a multiple of six are indistinguishable once sampled — that indistinguishability is the aliasing found in part (b), now expressed algebraically.

4. Collect the aliased harmonics into six bins. Write $x[n]=\sum_{m=0}^{5}b_m e^{j2\pi mn/6}$ and gather every original coefficient whose index is congruent to $m$ modulo 6: $$\boxed{\,b_m=\sum_{\substack{|k|\le 9\\ k\equiv m\ (\mathrm{mod}\ 6)}} a_k ,\qquad m=0,1,\dots,5\,}$$
5. Write the six sums out explicitly. Sweeping $k=-9,\dots,9$ and sorting by $k \bmod 6$ gives $$\begin{aligned} b_0 &= a_{-6}+a_{0}+a_{6}, &\qquad b_3 &= a_{-9}+a_{-3}+a_{3}+a_{9},\\ b_1 &= a_{-5}+a_{1}+a_{7}, &\qquad b_4 &= a_{-8}+a_{-2}+a_{4},\\ b_2 &= a_{-4}+a_{2}+a_{8}, &\qquad b_5 &= a_{-7}+a_{-1}+a_{5}. \end{aligned}$$ The nineteen given coefficients are accounted for exactly once each ($3+3+3+4+3+3=19$), which is a useful arithmetic check on the bookkeeping. Note that $b_3$ — the bin at $\omega=\pi$, the folding frequency — collects four terms rather than three, because both extreme harmonics $a_{\pm 9}$ land there.

If instead the discrete Fourier series is written in the analysis form used in the appendix, $X[m]=\sum_{n=0}^{5}x[n]W_{6}^{mn}$, the two conventions differ only by the length factor, $X[m]=6\,b_m$.