Sampling Followed by Ideal Reconstruction: 22-Elec-B1 Digital Signal Processing, December 2019, Q1

Question 1: Sampling Followed by Ideal Reconstruction (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 two-tone continuous-time signal is impulse-train sampled and then passed through the ideal reconstruction filter defined in the question.

Given data
Quantity	Symbol	Value
First tone: amplitude / radian frequency / phase	$A_1,\ \Omega_1,\ \phi_1$	$2,\ 100\pi\ \text{rad/s}\ (50\ \text{Hz}),\ -\pi/4$
Second tone: amplitude / radian frequency / phase	$A_2,\ \Omega_2,\ \phi_2$	$1,\ 300\pi\ \text{rad/s}\ (150\ \text{Hz}),\ +\pi/3$
Sampling function	$s(t)$	$\sum_n \delta(t-nT)$, period $T$
Reconstruction filter	$H_r(j\Omega)$	gain $T$ on $\|\Omega\|\le\pi/T$, zero outside

Find. The spectrum of the input, the sampled spectrum and the exact reconstructed waveform at 500 Hz and at 250 Hz, and the sampling rate (with the resulting constant) that converts the 150 Hz tone into a DC term.

Figure 1.1 — impulse-train sampling of x_c(t) followed by the ideal band-limited reconstruction filter.

Approach. Transform each cosine into its pair of impulses, build the sampled spectrum as the $1/T$-scaled sum of shifted copies of that pair, and read the reconstructed signal off whichever impulses survive the filter band $|\Omega|\le\pi/T$.

Transform each cosine into a conjugate pair of impulses. The standard pair is $$A\cos(\Omega_0 t+\phi)\ \longleftrightarrow\ \pi A\left[e^{j\phi}\delta(\Omega-\Omega_0)+e^{-j\phi}\delta(\Omega+\Omega_0)\right].$$ Applying it to both tones and adding gives the answer to part (a): $$\boxed{X_c(j\Omega)=2\pi e^{-j\pi/4}\delta(\Omega-100\pi)+2\pi e^{j\pi/4}\delta(\Omega+100\pi)+\pi e^{j\pi/3}\delta(\Omega-300\pi)+\pi e^{-j\pi/3}\delta(\Omega+300\pi)}$$ The spectrum is four impulses, of area $2\pi$ at $\pm100\pi$ and area $\pi$ at $\pm300\pi$.
Plot the input spectrum. Impulse areas, not heights, carry the information, so the arrows below are drawn with lengths proportional to the areas and are annotated with the complex weights.
Figure 1.2 — X_c(jΩ): four impulses at ±100π and ±300π rad/s.
Write the sampled spectrum. Impulse-train sampling replicates the input spectrum every $\Omega_s=2\pi/T$ and scales it by $1/T$: $$X_s(j\Omega)=\frac{1}{T}\sum_{k=-\infty}^{\infty}X_c\!\left(j(\Omega-k\Omega_s)\right).$$ Because the reconstruction filter has gain exactly $T$, the two scalings cancel and any impulse that lies inside $|\Omega|\le\pi/T$ reappears in $x_r(t)$ with its original area.
Part (b): sample at 500 Hz. Here $\Omega_s=2\pi(500)=1000\pi$ and the highest input frequency is $300\pi$, so the Nyquist condition $\Omega_s \gt 2\Omega_{\max}$ reads $1000\pi \gt 600\pi$ and is satisfied. The nearest replica lands at $1000\pi-300\pi=700\pi$, well outside the filter band $|\Omega|\le\pi/T=500\pi$.
Figure 1.3 — X_s(jΩ) at f_s = 500 Hz over −2π/T ≤ Ω ≤ 2π/T; the shaded strip is the reconstruction pass band.
Nothing is lost and nothing is added, so the reconstruction is exact: $$\boxed{x_r(t)=x_c(t)=2\cos(100\pi t-\tfrac{\pi}{4})+\cos(300\pi t+\tfrac{\pi}{3})}$$
Part (c): sample at 250 Hz. Now $\Omega_s=500\pi$ and $\Omega_s/2=250\pi \lt 300\pi$, so the 150 Hz tone is under-sampled and folds. The replica of the impulse at $+300\pi$ shifted by $-\Omega_s$ lands at $300\pi-500\pi=-200\pi$, carrying weight $\pi e^{j\pi/3}$; the mirror impulse lands at $+200\pi$ with weight $\pi e^{-j\pi/3}$. Comparing with the standard pair identifies the folded term as a cosine of amplitude 1 and phase $-\pi/3$.
Figure 1.4 — X_s(jΩ) at f_s = 250 Hz; the red lines at ±200π are the aliased image of the 150 Hz tone and they lie inside the pass band.
The 50 Hz tone is unaffected because $100\pi \lt 250\pi$, so $$\boxed{x_r(t)=2\cos(100\pi t-\tfrac{\pi}{4})+\cos(200\pi t-\tfrac{\pi}{3})}$$ The same conclusion follows in the discrete-time domain: with $T=1/250$ the second tone becomes $\cos(1.2\pi n+\pi/3)$, and subtracting $2\pi n$ gives $\cos(0.8\pi n-\pi/3)$, which maps back to $200\pi$ rad/s. Note that the sign of the folded frequency flips, so the phase changes sign as well — a detail that is easy to lose.
Part (d): force the 150 Hz tone onto DC. The requested output contains a constant plus the original 50 Hz tone, so the second tone must alias exactly to $\Omega=0$, which happens when $300\pi$ is an integer multiple of the sampling frequency: $$300\pi=k\,\Omega_s\quad\Rightarrow\quad \Omega_s=\frac{300\pi}{k},\qquad k=1,2,3,\dots$$ The first tone must survive unaliased, which needs $\Omega_s/2 \gt 100\pi$. For $k=1$, $\Omega_s=300\pi$ and $\Omega_s/2=150\pi \gt 100\pi$, so the choice is admissible; $k=2$ already gives $\Omega_s/2=75\pi \lt 100\pi$ and would destroy the wanted tone. Hence $$\boxed{f_s=\frac{\Omega_s}{2\pi}=150\ \text{samples/sec}\qquad (T=1/150\ \text{s})}$$
Figure 1.5 — at f_s = 150 Hz the two 150 Hz impulses both fold onto Ω = 0 (green), while the 100π pair stays inside the pass band and the 200π images fall outside it.
Evaluate the constant. The two impulses that land on $\Omega=0$ add coherently, $$\pi e^{j\pi/3}+\pi e^{-j\pi/3}=2\pi\cos\frac{\pi}{3}=\pi ,$$ and an impulse of area $a$ at the origin inverse-transforms to the constant $a/2\pi$. The simplest route is to sample directly: with $T=1/150$, $\cos(300\pi nT+\pi/3)=\cos(2\pi n+\pi/3)=\cos(\pi/3)$ for every $n$, so the second tone contributes the constant sequence $\cos(\pi/3)$. Either way $$\boxed{A=\cos\frac{\pi}{3}=0.5}$$

Question 1 — final results
Part	Result
(a) $X_c(j\Omega)$	$2\pi e^{-j\pi/4}\delta(\Omega-100\pi)+2\pi e^{j\pi/4}\delta(\Omega+100\pi)+\pi e^{j\pi/3}\delta(\Omega-300\pi)+\pi e^{-j\pi/3}\delta(\Omega+300\pi)$
(b) $f_s=500$ Hz	no aliasing; $x_r(t)=2\cos(100\pi t-\pi/4)+\cos(300\pi t+\pi/3)=x_c(t)$
(c) $f_s=250$ Hz	150 Hz tone folds to 100 Hz; $x_r(t)=2\cos(100\pi t-\pi/4)+\cos(200\pi t-\pi/3)$
(d) sampling rate	$f_s=150$ samples/sec
(d) constant	$A=\cos(\pi/3)=0.5$

Topic. Impulse-train sampling, spectral replication and band-limited reconstruction.

Key relations. Sampling: $X_s(j\Omega)=\tfrac{1}{T}\sum_k X_c(j(\Omega-k\Omega_s))$ with $\Omega_s=2\pi/T$. Ideal reconstruction: $X_r=H_r X_s$ with $H_r=T$ on $|\Omega|\le\pi/T$. Nyquist: exact recovery iff $\Omega_s \gt 2\Omega_{\max}$. Modulation pair: $A\cos(\Omega_0t+\phi)\leftrightarrow \pi A[e^{j\phi}\delta(\Omega-\Omega_0)+e^{-j\phi}\delta(\Omega+\Omega_0)]$.

Why this works. Multiplying by an impulse train in time is convolution with an impulse train in frequency, which copies the baseband spectrum to every multiple of $\Omega_s$. The reconstruction filter is a window that keeps exactly one Nyquist band, so the output contains one impulse pair per surviving line. When a copy of a high-frequency line falls inside that window it is indistinguishable from a genuine low-frequency line, which is precisely what aliasing means. The gain $T$ is chosen to undo the $1/T$ introduced by sampling, so amplitudes are preserved. Because the folded line is the negative-frequency copy, its phase appears with the opposite sign — the sign flip is a real part of the answer, not a bookkeeping slip.

Common pitfall. Quoting the aliased frequency but keeping the original phase, and treating a tone that folds onto $\Omega=0$ as though it simply disappeared: it becomes a DC offset whose value is $A_2\cos\phi_2$, which is zero only if the tone happens to be sampled at its zero crossings.

Source. Oppenheim & Schafer, Discrete-Time Signal Processing, 3rd ed., Ch. 4 §4.1–4.3 (sampling, reconstruction, aliasing).

Question 1 of 6: Sampling Followed by Ideal Reconstruction

Question 1: Sampling Followed by Ideal Reconstruction (12 marks)