Question 2 of 10: Sampling a Two-Tone Message, With and Without Aliasing

Question 2: Sampling a Two-Tone Message, With and Without Aliasing (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. $x(t)=2\cos(2\pi\,1000\,t)\cos(2\pi\,2000\,t)$; ideal impulse sampling at $f_s$; recovery filter ideal lowpass, unity gain, zero phase, cut-off $f_s/2$.

Find. $X(f)$, the Nyquist rate, and the sampled and recovered spectra and time waveforms at $f_s = 2f_{\text{Nyq}}$ and at $f_s = \tfrac{5}{6}f_{\text{Nyq}}$, with a comparison.

Approach. Expand the product of cosines into a sum of tones using the supplied identity, read the line spectrum straight off, then replicate at multiples of $f_s$ and see which lines land inside the recovery filter passband.

1. Convert the product into a sum of tones. The paper supplies $\cos\alpha\cos\beta = \tfrac12\cos(\alpha+\beta)+\tfrac12\cos(\alpha-\beta)$. With $\alpha = 2\pi(1000)t$ and $\beta = 2\pi(2000)t$ the factor 2 cancels the halves: $$\boxed{x(t)=\cos(2\pi\,1000\,t)+\cos(2\pi\,3000\,t)}$$ so the message is simply a 1 kHz tone plus a 3 kHz tone, each of unit amplitude.
2. Write down the line spectrum. Using $\cos 2\pi f_0 t \leftrightarrow \tfrac12[\delta(f-f_0)+\delta(f+f_0)]$, $$X(f)=\tfrac12\big[\delta(f\mp 1000)\big]+\tfrac12\big[\delta(f\mp 3000)\big]$$ i.e. four impulses of weight $1/2$ at $f = \pm 1$ kHz and $\pm 3$ kHz.

The highest frequency actually present is 3 kHz — note that this is the sum tone created by the multiplication, not either of the two frequencies written in the problem statement.

3. Nyquist rate. $$f_{\text{Nyq}} = 2f_{\max} = 2\times 3\ \text{kHz} = \boxed{6\ \text{kHz}}$$
4. Part (c): sample at twice the Nyquist rate. Here $f_s = 2f_{\text{Nyq}} = 12$ kHz, and $X_s(f) = f_s\sum_n X(f-nf_s)$ places copies of the four-line spectrum at every multiple of 12 kHz. Within $|f|\le 10$ kHz the visible lines are the originals at $\pm 1$ and $\pm 3$ kHz plus the lower skirt of the first replica at $12-3 = 9$ kHz (and its mirror at $-9$ kHz). Every impulse has weight $f_s/2 = 6000$.
5. Recover: the filter passband is empty of intruders. The recovery filter cuts off at $f_s/2 = 6$ kHz, which admits the $\pm 1$ and $\pm 3$ kHz lines and excludes the 9 kHz replica line. Since the filter has unity gain, the output carries the sampling scale factor: $$\boxed{y(t)=f_s\,x(t)=12000\big[\cos(2\pi\,1000\,t)+\cos(2\pi\,3000\,t)\big]}$$ The waveform shape is an exact replica of $x(t)$; only the amplitude is scaled, because the question specifies unity filter gain rather than the $1/f_s$ gain that would give literal equality.

Part (d) drops the rate below the Nyquist rate, and the behaviour changes qualitatively rather than just quantitatively.

6. Part (d): the undersampled replica pattern. Now $f_s = \tfrac56 f_{\text{Nyq}} = \tfrac56(6\ \text{kHz}) = 5$ kHz, which is below the 6 kHz Nyquist rate, so overlap is inevitable. Replicating the $\pm1$ and $\pm3$ kHz lines at multiples of 5 kHz puts positive-frequency lines at 1, 2, 3, 4, 6, 7, 8 and 9 kHz, each of weight $f_s/2 = 2500$. The line at 2 kHz is new: it is $|5-3| = 2$ kHz, the first replica of the 3 kHz tone folded down.
7. Recover: an impostor enters the passband. The filter now cuts off at $f_s/2 = 2.5$ kHz, admitting the genuine 1 kHz line and the aliased 2 kHz line while rejecting the true 3 kHz line, which has been pushed outside its own recovery band. Hence $$\boxed{y(t)=5000\big[\cos(2\pi\,1000\,t)+\cos(2\pi\,2000\,t)\big]}$$

Part (e) — comparison. In part (c) the recovered signal is a faithful copy of the message: both tones return at their correct frequencies and with their correct relative amplitudes, and the only difference from $x(t)$ is the harmless constant scale factor $f_s = 12000$ imposed by the unity-gain filter, which a single attenuator removes. In part (d) the recovered signal is irreversibly wrong. The 3 kHz component has not merely been attenuated or delayed; it has been moved to 2 kHz, a frequency the message never contained. No filter, equaliser or amplifier applied after the sampler can undo this, because once the replicas overlap the information about which original frequency produced a given output line is destroyed. This is the practical reason every real sampling front end places an anti-aliasing lowpass filter before the sampler: it is cheaper to discard the offending band deliberately than to have it reappear disguised as valid in-band signal.