Outline
6.1 Periodic Sampling
6.2 Frequency-Domain Analysis
6.3 Nyquist Theorem & Aliasing
6.4 Reconstruction
6.5 DT Processing of CT Signals
6.6 Quantization
6.7 Summary
Exercises
DSP Study Guide · Chapter 6

Sampling of Continuous-Time Signals

How an analog signal becomes a sequence — impulse-train sampling, spectrum replication, the Nyquist limit, aliasing, sinc reconstruction, discrete-time processing of analog signals, and amplitude quantization.

6.1 Periodic Sampling

6.1.1 The Ideal Continuous-to-Discrete (C/D) Converter

Digital signal processing of real-world signals begins by converting a continuous-time (CT) signal（連續時間訊號）into a discrete-time (DT) sequence（離散時間序列）. Periodic sampling（週期取樣）records the value of $x_c(t)$ at uniformly spaced sampling instants $t = nT$:

Periodic sampling (C/D conversion) $$x[n] = x_c(nT), \qquad -\infty < n < \infty$$

Here $T$ is the sampling period（取樣週期, seconds/sample）, its reciprocal $f_s = 1/T$ is the sampling frequency（取樣頻率, samples/second or Hz）, and in radian units

Sampling frequency (rad/s) $$\Omega_s = \frac{2\pi}{T} = 2\pi f_s \quad \text{rad/s}$$
2026-06-12T22:22:44.835494 image/svg+xml Matplotlib v3.10.8, https://matplotlib.org/ 0.0 0.2 0.4 0.6 0.8 1.0 t (s) −1.0 −0.5 0.0 0.5 1.0 amplitude Periodic sampling: x[n] = x_c(nT), T = 1/12 s x t c ( )
Fig. 6-1 — Periodic sampling: the sequence $x[n]$ keeps only the values of $x_c(t)$ at $t=nT$（紅色曲線為連續訊號，藍色梗圖為每隔 $T$ 秒取出的樣本）.
Concept — why sampling is "lossy" in general
Infinitely many different CT signals pass through the same set of samples (see the aliasing demo of Fig. 6-3). The miracle of the sampling theorem (Section 6.3) is that if $x_c(t)$ is bandlimited and $T$ is small enough, the samples determine $x_c(t)$ uniquely — sampling then loses nothing.

6.1.2 Impulse-Train Model of Sampling

To analyze sampling mathematically we use a two-stage model. First multiply $x_c(t)$ by a periodic impulse train（脈衝列）$s(t)$, then convert the impulse areas to a sequence:

Impulse-train modulation $$s(t) = \sum_{n=-\infty}^{\infty} \delta(t-nT) \qquad\Longrightarrow\qquad x_s(t) = x_c(t)\,s(t) = \sum_{n=-\infty}^{\infty} x_c(nT)\,\delta(t-nT)$$

The signal $x_s(t)$ (also written $x_p(t)$ in the slides) is still a CT signal — a train of impulses whose areas equal the sample values $x_c(nT)$. The DT sequence $x[n]=x_c(nT)$ carries exactly the same information but lives on the integer axis, with the time scale $T$ removed（時間軸已正規化，不再帶有 $T$ 的資訊）.

直觀解釋（點擊展開）
為什麼要用「脈衝列相乘」這個看似多餘的中間步驟？因為 $x_s(t)$ 仍然是連續時間訊號， 我們可以對它做 CTFT，利用「時域相乘 = 頻域摺積」這個性質， 把取樣對頻譜的影響算得一清二楚（見 6.2 節）。 而 $x[n]$ 是離散序列，必須用 DTFT 分析。兩者頻譜只差一個頻率軸的縮放 $\omega = \Omega T$，所以脈衝列模型是連接「類比頻譜」與「數位頻譜」的橋樑。

6.2 Frequency-Domain Analysis of Sampling

6.2.1 Spectrum Replication at Multiples of Ωs

The impulse train $s(t)$ is periodic with period $T$, so its CTFT is itself an impulse train in frequency, with spacing $\Omega_s = 2\pi/T$ and area $2\pi/T$:

CTFT of the impulse train $$S(j\Omega) = \frac{2\pi}{T}\sum_{k=-\infty}^{\infty} \delta(\Omega - k\Omega_s)$$

Multiplication in time is convolution in frequency (divided by $2\pi$), so the spectrum of the sampled signal is a sum of shifted and scaled copies（平移且縮放的複本） of $X_c(j\Omega)$:

Spectrum of the sampled signal (replication formula) $$X_s(j\Omega) = \frac{1}{2\pi}\,X_c(j\Omega) * S(j\Omega) = \boxed{\;\frac{1}{T}\sum_{k=-\infty}^{\infty} X_c\!\big(j(\Omega - k\Omega_s)\big)\;}$$
Key result — sampling replicates the spectrum
Sampling at rate $\Omega_s$ makes the analog spectrum periodic: copies of $X_c(j\Omega)$ appear centered at every multiple of $\Omega_s$ （取樣使頻譜以 $\Omega_s$ 為週期複製）, each scaled by $1/T$. If the copies do not overlap, the original spectrum survives intact in the baseband copy; if they overlap, the overlapped sum can never be untangled — that overlap is aliasing.
2026-06-12T22:22:45.067286 image/svg+xml Matplotlib v3.10.8, https://matplotlib.org/ 0.0 0.5 1.0 1.5 X j c ( Ω ) O r i g i n a l   C T   s p e c t r u m   ( b a n d l i m i t e d   t o   Ω   =   2 ) N 0.0 0.5 1.0 1.5 T X j s ( Ω ) S a m p l e d ,   Ω = 6 > 2 Ω   :   r e p l i c a s   a t   Ω   d o   n o t   o v e r l a p s N s k −10.0 −7.5 −5.0 −2.5 0.0 2.5 5.0 7.5 10.0 Ω   ( r a d / s ) 0.0 0.5 1.0 1.5 T X j s ( Ω ) S a m p l e d ,   Ω = 3 < 2 Ω   :   r e p l i c a s   o v e r l a p   =   A L I A S I N G   ( r e d   =   s u m ) s N
Fig. 6-2 — Spectrum replication. Top: bandlimited $X_c(j\Omega)$ with highest frequency $\Omega_N$. Middle: $\Omega_s > 2\Omega_N$, the replicas at $k\Omega_s$ stay separated (green dashed lines mark $\pm\Omega_s/2$). Bottom: $\Omega_s < 2\Omega_N$, the replicas overlap and the resulting sum (red) is distorted — aliasing（頻譜複本重疊即為混疊）.

6.2.2 CT and DT Spectra — the Mapping ω = ΩT

How does the CTFT of $x_s(t)$ relate to the DTFT of the sequence $x[n]$? Take the CTFT of the impulse-train expression term by term:

From CTFT to DTFT $$X_s(j\Omega) = \sum_{n} x_c(nT)\,e^{-j\Omega T n} \qquad\text{vs}\qquad X(e^{j\omega}) = \sum_{n} x[n]\,e^{-j\omega n}$$

The two sums are identical when $\omega = \Omega T$. Hence

DT spectrum of a sampled signal $$X(e^{j\omega}) = X_s(j\Omega)\Big|_{\Omega=\omega/T} = \frac{1}{T}\sum_{k=-\infty}^{\infty} X_c\!\left(j\Big(\frac{\omega}{T} - \frac{2\pi k}{T}\Big)\right)$$
Key — normalized vs analog frequency
$$\omega = \Omega T = \frac{\Omega}{f_s}\qquad \text{(rad/sample)} = \text{(rad/s)}\times\text{(s/sample)}$$ The analog frequency $\Omega = \Omega_s$ (one full sampling rate) maps to $\omega = 2\pi$; the half sampling rate $\Omega_s/2$ maps to $\omega = \pi$, the edge of the DT baseband. This is why the DTFT is always $2\pi$-periodic — all replicas land on a $2\pi$ grid （類比頻率 $\Omega$ 乘上 $T$ 就變成數位頻率 $\omega$；$\Omega_s/2$ 對應 $\omega=\pi$）.
QuantityContinuous timeDiscrete time
Frequency variable$\Omega$ (rad/s)$\omega = \Omega T$ (rad/sample)
TransformCTFT $X_c(j\Omega)$, aperiodicDTFT $X(e^{j\omega})$, $2\pi$-periodic
Sampling rate$\Omega_s = 2\pi/T$$2\pi$
Nyquist frequency (half rate)$\Omega_s/2 = \pi/T$$\pi$
Amplitude scale$X_c$$\tfrac{1}{T}X_c$ in baseband copy

6.3 The Nyquist Sampling Theorem and Aliasing

6.3.1 The Sampling Theorem

From Fig. 6-2: if $x_c(t)$ is bandlimited（頻帶受限）, i.e. $X_c(j\Omega) = 0$ for $|\Omega| \ge \Omega_N$, the replicas remain disjoint provided $\Omega_s - \Omega_N \ge \Omega_N$. This is the Nyquist–Shannon sampling theorem:

Nyquist sampling theorem 取樣定理 $$\boxed{\;\Omega_s \;\ge\; 2\,\Omega_N\;} \qquad\Longleftrightarrow\qquad f_s \ge 2 f_N \qquad\Longleftrightarrow\qquad T \le \frac{\pi}{\Omega_N}$$
Warning — strict inequality at the band edge
A sinusoid exactly at the Nyquist frequency ($\Omega_0 = \Omega_s/2$) is a boundary case: $\sin(\Omega_s t/2)$ sampled at $t=nT$ gives $\sin(\pi n) = 0$ for all $n$ — the samples vanish entirely. For sinusoidal content you need $\Omega_s > 2\Omega_0$ strictly（正弦訊號剛好在 $\Omega_s/2$ 時相位不巧可能全取到零點，必須嚴格大於）.

6.3.2 Aliasing 混疊

If $\Omega_s < 2\Omega_N$, the shifted copies in the replication formula overlap. High frequencies fold down and masquerade as low frequencies — they take on an alias（化名）. Once the copies are summed, the overlap cannot be filtered out and $x_c(t)$ can no longer be recovered.

For a sinusoid the alias is easy to compute: frequencies $\Omega_0$ and $\Omega_0 + k\Omega_s$ produce exactly the same samples:

Sinusoid aliasing $$\cos\big((\Omega_0 + k\Omega_s)nT + \phi\big) = \cos\big(\Omega_0 nT + \phi + 2\pi k n\big) = \cos\big(\Omega_0 nT + \phi\big)$$
2026-06-12T22:22:45.244756 image/svg+xml Matplotlib v3.10.8, https://matplotlib.org/ 0.0 0.2 0.4 0.6 0.8 1.0 t (s) −1.0 −0.5 0.0 0.5 1.0 amplitude A l i a s i n g :   1   H z   a n d   9   H z   s i n u s o i d s   g i v e   I D E N T I C A L   s a m p l e s   a t     =   8   H z f s 1 Hz 9 Hz
Fig. 6-3 — Time-domain view of aliasing: a 1 Hz and a 9 Hz sinusoid sampled at $f_s = 8$ Hz pass through exactly the same samples（紅點）, since $9 = 1 + 8$. After sampling, the 9 Hz signal is indistinguishable from its 1 Hz alias（9 Hz 訊號取樣後「偽裝」成 1 Hz）.
直觀解釋：車輪倒轉效應（點擊展開）
電影裡高速旋轉的車輪有時看起來慢轉甚至倒轉，就是混疊：攝影機每秒 24 格相當於取樣率 24 Hz，輪輻轉速超過 12 Hz（奈奎斯特頻率）後，視覺上的轉速就會「摺回」低頻。 頻域裡的解釋：超過 $\Omega_s/2$ 的頻率成分被複本平移後落回 $[0,\Omega_s/2]$ 內， 與真正的低頻成分疊加且無法分離。

6.3.3 Anti-Aliasing Filter 抗混疊濾波器

Real signals (audio, sensor noise, RF interference) are never perfectly bandlimited. Before the C/D converter we therefore insert a CT lowpass filter — the anti-aliasing filter $H_{aa}(j\Omega)$ — that forcibly bandlimits the input to below the Nyquist frequency:

Ideal anti-aliasing filter $$H_{aa}(j\Omega) = \begin{cases} 1, & |\Omega| < \Omega_s/2 \\ 0, & |\Omega| \ge \Omega_s/2 \end{cases}$$
Tip — distortion trade-off
The anti-aliasing filter deliberately destroys the out-of-band content. That is the right trade: mild, known lowpass distortion of frequencies you do not care about is far better than aliasing distortion, which corrupts the in-band signal irreversibly （寧可犧牲帶外成分，也不要讓它混疊進帶內）. Practical filters are not brick-wall, so designers either leave a guard band or oversample then decimate digitally.

6.3.4 Sampling Bandpass Signals 帶通取樣

A signal need not sit around $\Omega = 0$. If its spectrum occupies a band $\Omega_L \le |\Omega| \le \Omega_H$ with bandwidth $\Delta\Omega = \Omega_H - \Omega_L$, sampling replicates the band at multiples of $\Omega_s$; as long as no replicas overlap, the signal can still be recovered, even with $\Omega_s$ far below $2\Omega_H$.

Bandpass sampling basic limit $$\frac{\Omega_s}{2} \;\ge\; \Delta\Omega = \Omega_H - \Omega_L$$

The sampling rate is set by the bandwidth, not the highest frequency — the principle behind undersampling receivers in communications（取樣率由頻寬決定， 不是由最高頻率決定，這是軟體無線電下變頻取樣的原理）. Achieving the limit requires the band edges to align favorably with $\Omega_s$, so practical designs use somewhat higher rates.

6.4 Reconstruction 訊號重建

6.4.1 Ideal Reconstruction (D/C) Filter

If sampling obeyed the Nyquist condition, the baseband replica ($k=0$ term) of $X_s(j\Omega)$ is an undistorted copy of $X_c(j\Omega)/T$. To rebuild $x_c(t)$ we pass the impulse train $x_s(t)$ through an ideal lowpass reconstruction filter with gain $T$ and cutoff at the Nyquist frequency $\pi/T$:

Ideal reconstruction filter $$H_r(j\Omega) = \begin{cases} T, & |\Omega| \le \pi/T \\[2pt] 0, & |\Omega| > \pi/T \end{cases} \qquad\Longleftrightarrow\qquad h_r(t) = \frac{\sin(\pi t/T)}{\pi t/T} = \mathrm{sinc}\!\left(\frac{t}{T}\right)$$

The gain $T$ cancels the $1/T$ amplitude scaling introduced by sampling; the cutoff $\pi/T = \Omega_s/2$ keeps only the $k=0$ replica and erases all the images at $\pm\Omega_s, \pm 2\Omega_s,\dots$（增益 $T$ 抵消取樣的 $1/T$ 縮放，截止頻率 $\pi/T$ 刪掉所有複本，只留基頻帶那一份）.

6.4.2 Sinc Interpolation 弦波內插

Convolving the impulse train with $h_r(t)$ hangs one scaled sinc on every sample:

Bandlimited (sinc) interpolation formula $$\hat{x}_a(t) = \sum_{n=-\infty}^{\infty} x[n]\; \frac{\sin\big(\pi (t-nT)/T\big)}{\pi (t-nT)/T}$$

Two properties make this work:

2026-06-12T22:22:45.350111 image/svg+xml Matplotlib v3.10.8, https://matplotlib.org/ 0 2 4 6 8 10 t / T −1.0 −0.5 0.0 0.5 1.0 1.5 amplitude Ideal bandlimited reconstruction: scaled sincs through each sample reconstruction (sum of sincs)
Fig. 6-4 — Sinc interpolation: one scaled sinc (gray) is centered on every sample (blue); their sum (red) is the unique bandlimited reconstruction. At each sample instant all other sincs cross zero（每個樣本掛一個 sinc，在取樣點上其他 sinc 都恰為零，因此重建曲線必通過所有樣本）.
Warning — ideal reconstruction is non-causal
$h_r(t)$ extends over all time and decays only like $1/t$, so the ideal D/C is unrealizable. Real DACs use a zero-order hold（零階保持）followed by a smoothing (anti-imaging) lowpass filter; the staircase hold introduces a mild $\mathrm{sinc}$-shaped droop that good designs compensate digitally. In simulations (Exercise 2) the sinc sum is also truncated to finitely many samples, which causes small edge errors.

6.5 Discrete-Time Processing of Continuous-Time Signals

6.5.1 The C/D → H(ejω) → D/C Chain

The central architecture of applied DSP: filter an analog signal by sampling it, processing the samples with a DT LTI system, and converting back:

C/D DT LTI system H(e^jω) D/C T T x_c(t) x[n] y[n] y_r(t)
Fig. 6-5 — Discrete-time processing of a continuous-time signal: sample, filter digitally, reconstruct（類比訊號經取樣、數位濾波、再重建）.

Tracking the spectra through the chain (assuming no aliasing at the C/D):

Spectra through the chain $$X(e^{j\omega}) = \frac{1}{T}X_c\!\left(j\frac{\omega}{T}\right)\;(|\omega|\le\pi) \;\;\rightarrow\;\; Y(e^{j\omega}) = H(e^{j\omega})X(e^{j\omega}) \;\;\rightarrow\;\; Y_r(j\Omega) = T\,Y(e^{j\Omega T})\;\Big(|\Omega|\le\frac{\pi}{T}\Big)$$

6.5.2 Effective Continuous-Time Frequency Response

Composing the three relations, the whole chain behaves like a single CT LTI system:

Effective analog response of a DT filter $$\boxed{\;H_{\text{eff}}(j\Omega) = \begin{cases} H(e^{j\Omega T}), & |\Omega| < \pi/T \\[2pt] 0, & |\Omega| \ge \pi/T \end{cases}\;}$$
Key — one digital filter, many analog filters
The same DT filter $H(e^{j\omega})$ implements a different analog response for every sampling rate: an ideal DT lowpass with cutoff $\omega_c$ acts as an analog lowpass with cutoff $\Omega_c = \omega_c/T$. Changing $T$ rescales the entire effective analog frequency axis（同一顆數位濾波器，改變取樣率就改變了等效的類比截止頻率 $\Omega_c=\omega_c/T$）. This is the rule you need for Homework Exercise 1: design $H(e^{j\omega})$ so that $H_{\text{eff}}(j\Omega)$ does the desired analog job, substituting $\omega = \Omega T$.
Worked Example 6-1 — DT lowpass acting on an analog signal

A DT ideal lowpass filter has cutoff $\omega_c = \pi/4$. The C/D and D/C run at $f_s = 8$ kHz. What analog filter does the chain implement?

Step 1. $T = 1/8000$ s. Frequency mapping: $\omega = \Omega T$, so $\Omega = \omega/T = \omega f_s$.
Step 2. Cutoff: $\Omega_c = \omega_c/T = (\pi/4)(8000) = 2000\pi$ rad/s, i.e. $f_c = 1$ kHz.
Step 3. So $H_{\text{eff}}(j\Omega)$ is an ideal analog lowpass with 1 kHz cutoff (and zero response beyond 4 kHz, the Nyquist frequency). Doubling $f_s$ to 16 kHz would move the cutoff to 2 kHz with the same digital filter.

6.6 Quantization 量化

6.6.1 Uniform Amplitude Quantization

Sampling discretizes time; a real A/D converter must also discretize amplitude. A uniform quantizer（均勻量化器）with $B$ bits maps each sample to the nearest of $2^B$ levels spaced $\Delta$ apart. For full-scale amplitude $m_{\max}$:

Quantizer step size $$\Delta = \frac{2\,m_{\max}}{2^{B}} = \frac{m_{\max}}{2^{B-1}}$$

Two standard conventions differ in where the levels sit:

2026-06-12T22:22:45.487690 image/svg+xml Matplotlib v3.10.8, https://matplotlib.org/ −2 −1 0 1 2 x −2 −1 0 1 2 Q[x] Midtread (round to nearest) −2 −1 0 1 2 x Midrise (round with offset) 2-bit uniform quantizers, step size 1
Fig. 6-6 — 2-bit uniform quantizer characteristics. Midtread (left) has a level at zero; midrise (right) has a threshold at zero and levels offset by $\Delta/2$（中平型在 0 有一階；中升型的階梯跨在 0 上）. Dashed line: ideal $Q[x]=x$.

6.6.2 Quantization Noise Model

The quantization error $e[n] = Q[x[n]] - x[n]$ is deterministic, but for signals that are large and busy compared to one LSB it behaves like additive white noise （加性白雜訊）, uncorrelated with itself and with the signal:

Quantization noise statistics $$e[n] \sim \mathcal{U}\!\left[-\frac{\Delta}{2}, \frac{\Delta}{2}\right] \qquad\Longrightarrow\qquad \sigma_e^2 = E[e^2] = \int_{-\Delta/2}^{\Delta/2} \frac{e^2}{\Delta}\,de = \boxed{\;\frac{\Delta^2}{12}\;}$$
2026-06-12T22:22:45.653779 image/svg+xml Matplotlib v3.10.8, https://matplotlib.org/ −1.0 −0.5 0.0 0.5 1.0 xq[n] 3-bit midrise quantization of a sinusoid (step 0.25) 0 10 20 30 40 50 60 n −0.1 0.0 0.1 e[n] Quantization error: bounded by +/- 0.125
Fig. 6-7 — A sinusoid quantized by a 3-bit midrise quantizer (top) and the resulting error sequence (bottom). The error stays within $\pm\Delta/2$ (green dashed) and looks noise-like; its measured power $\approx 6.3\times10^{-3}$ is close to the uniform-model prediction $\Delta^2/12 \approx 5.2\times10^{-3}$（誤差近似均勻分布的白雜訊）.

The slides' PSD demonstration (quantizing a sinusoid with $B = 3, 7, 11, 15$ bits) shows the noise floor dropping by about 24 dB per 4 added bits and staying flat across frequency — exactly the white, $\Delta^2/12$ model（功率頻譜近似平坦，每加 4 bit 下降約 24 dB）.

6.6.3 Quantization SNR, Bits, and Dynamic Range

With signal power $P_{\text{avg}} = \sigma_x^2$ and noise power $\sigma_e^2 = \Delta^2/12$, substituting $\Delta = 2\,m_{\max}\,2^{-B}$ gives $\sigma_e^2 = \tfrac{1}{3}\,m_{\max}^2\,2^{-2B}$, so:

Quantization SNR $$\mathrm{SNR} = \frac{\sigma_x^2}{\sigma_e^2} = \frac{3\,P_{\text{avg}}}{m_{\max}^2}\,2^{2B} \qquad\Longrightarrow\qquad \mathrm{SNR}_{\mathrm{dB}} = 6.02\,B + 4.77 + 10\log_{10}\frac{P_{\text{avg}}}{m_{\max}^2}\ \text{dB}$$
Key — 6 dB per bit
Every added bit halves $\Delta$, quarters the noise power, and adds $20\log_{10}2 \approx 6.02$ dB of SNR（每多 1 bit，SNR 提升約 6 dB）. For a full-scale sinusoid $P_{\text{avg}} = m_{\max}^2/2$, giving the famous $\mathrm{SNR} = 6.02B + 1.76$ dB. The constants 1.76 and 1.17 quoted in audio specs come from this family of formulas; they are loose upper bounds for real signals.
Worked Example 6-2 — How many bits for an audio CD?

An audio CD targets SNR around 95–98 dB. How many bits of resolution are needed?

Step 1. Use $\mathrm{SNR}_{\mathrm{dB}} \approx 6.02B + 1.76$ (full-scale sinusoid bound).
Step 2. $6.02B + 1.76 \ge 95 \Rightarrow B \ge 15.5$, so $B = 16$ bits — exactly the CD format. With $B=16$: $\mathrm{SNR} \le 98.1$ dB, matching the slide's quoted noise floor of $-98.08$ dB for the CD anti-aliasing/quantization chain.
Step 3. Real converters fall short of the bound: the TI TLV320AIC23B codec's ADC achieves 90 dB SNR ($\approx$ 14.6 effective bits) and its DAC 100 dB ($\approx$ 16 bits) — datasheet SNR divided by 6.02 estimates the effective number of bits.

Dynamic range（動態範圍）of a linear system equals this SNR: the ratio in dB between the largest representable signal and the noise floor. Powers use $10\log_{10}(\cdot)$, amplitudes $20\log_{10}(\cdot)$, since $P \propto |A|^2$. A related implementation issue is coefficient quantization（係數量化）: filter coefficients stored with finite precision move the poles/zeros of $H(z)$ slightly, perturbing the realized frequency response — a preview of finite-wordlength effects.

6.7 Chapter Summary

Sampling
$x[n] = x_c(nT)$; model as impulse-train modulation $x_s(t)=\sum_n x_c(nT)\delta(t-nT)$. Sampling rate $\Omega_s = 2\pi/T$.
Spectrum replication
$X_s(j\Omega) = \frac{1}{T}\sum_k X_c(j(\Omega-k\Omega_s))$ — copies at every multiple of $\Omega_s$, scaled by $1/T$. DT spectrum: $X(e^{j\omega}) = X_s(j\omega/T)$.
Nyquist theorem
Bandlimited to $\Omega_N$ + $\Omega_s \ge 2\Omega_N$ ⇒ perfect recovery. Below it: aliasing — high frequencies fold into the baseband irreversibly. Anti-alias filter before C/D; bandpass signals need only $\Omega_s/2 \ge \Delta\Omega$.
Reconstruction
Ideal LPF: gain $T$, cutoff $\pi/T$; equivalently sinc interpolation $\hat{x}_a(t) = \sum_n x[n]\,\mathrm{sinc}((t-nT)/T)$ — the unique bandlimited signal through the samples.
DT processing of CT signals
C/D → $H(e^{j\omega})$ → D/C $\equiv$ analog filter $H_{\text{eff}}(j\Omega) = H(e^{j\Omega T})$ for $|\Omega| < \pi/T$. Frequency mapping $\omega = \Omega T$.
Quantization
Step $\Delta = 2m_{\max}/2^B$; midtread vs midrise. Error $\approx$ uniform white noise, $\sigma_e^2 = \Delta^2/12$; SNR $= 6.02B + 4.77 + 10\log_{10}(P_{\text{avg}}/m_{\max}^2)$ dB — about 6 dB per bit.

Exercises

All problems of Homework 6 with full worked solutions. Exercise 1 is the analytical problem; Exercises 2–3 are the MATLAB simulation parts (a)–(d) and (e)–(f).

Exercise 1 — DT Processing of a CT Signal with an Anti-Aliasing Filter

Problem: The system of Fig. 6-5, preceded by a CT anti-aliasing filter, processes the input $g_c(t) = f_c(t) + e_c(t)$: $g_c(t) \to H_{aa}(j\Omega) \to \text{C/D} \to H(e^{j\omega}) \to \text{D/C} \to y_c(t)$, with both converters running at period $T$.

  • $F_c(j\Omega)$ is a triangle of height $A$, nonzero only for $|\Omega| < 400\pi$.
  • $E_c(j\Omega)$ (wideband noise) is flat with height $B$, nonzero only for $|\Omega| > 400\pi$ — so $g_c(t)$ is not bandlimited.
  • Anti-aliasing filter: $|H_{aa}(j\Omega)| = 1$ for $|\Omega| \le 400\pi$, falls linearly from 1 to 0 over $400\pi \le |\Omega| \le 800\pi$, and is 0 for $|\Omega| \ge 800\pi$; phase $\angle H_{aa}(j\Omega) = -\Omega^3$.

(a) If the sampling rate is $2\pi/T = 1600\pi$, find $H(e^{j\omega})$ such that $y_c(t) = f_c(t)$.
(b) Is $y_c(t) = f_c(t)$ still possible with $2\pi/T < 1600\pi$? If so, find the minimum $2\pi/T$ and the corresponding $H(e^{j\omega})$.

Click to reveal solution

Setup — what reaches the C/D converter. The filter output is $x_c(t)$ with spectrum $X_c(j\Omega) = H_{aa}(j\Omega)\,[F_c(j\Omega) + E_c(j\Omega)]$:

  • For $|\Omega| \le 400\pi$: only $f_c$ lives here and $|H_{aa}|=1$, so $X_c(j\Omega) = F_c(j\Omega)\,e^{-j\Omega^3}$ — the desired signal, but with a nasty nonlinear phase distortion $e^{-j\Omega^3}$.
  • For $400\pi < |\Omega| < 800\pi$: a residue of attenuated noise $E_c(j\Omega)H_{aa}(j\Omega)$ survives.
  • For $|\Omega| \ge 800\pi$: nothing — $x_c(t)$ is bandlimited to $800\pi$.

The DT system must therefore do two jobs: (i) reject everything outside the $f_c$ band, and (ii) equalize the phase $e^{-j\Omega^3}$ inside it.

(a) Sampling at $\Omega_s = 2\pi/T = 1600\pi$ (so $T = 1/800$). Since $x_c$ is bandlimited to $800\pi = \Omega_s/2$, there is no aliasing. In the DT domain, with $\omega = \Omega T$:

  • $f_c$ band: $|\Omega| \le 400\pi \Rightarrow |\omega| \le 400\pi\,T = \pi/2$.
  • noise residue: $400\pi < |\Omega| < 800\pi \Rightarrow \pi/2 < |\omega| < \pi$.

On the $f_c$ band the baseband spectrum is $X(e^{j\omega}) = \frac{1}{T} F_c(j\omega/T)\,e^{-j(\omega/T)^3}$. The D/C contributes gain $T$ and cuts at $\pi/T$, so to get $Y_c(j\Omega) = F_c(j\Omega)$ we need $H$ to cancel the phase and act as an ideal lowpass with cutoff $\pi/2$:

Answer (a) $$H(e^{j\omega}) = \begin{cases} e^{\,j(\omega/T)^3} = e^{\,j(800\,\omega)^3}, & |\omega| \le \pi/2 \\[4pt] 0, & \pi/2 < |\omega| \le \pi \end{cases}$$

(The exponent $({\omega/T})^3 = \omega^3/T^3 = 5.12\times 10^8\,\omega^3$.)

(b) Can we sample slower? Yes. The key observation: aliasing of the noise residue band is harmless as long as the folded copies do not land on top of the $f_c$ band $|\Omega| \le 400\pi$ — whatever lands in $400\pi < |\Omega|$ can still be removed by the DT lowpass.

When sampling at $\Omega_s$, the replica centered at $+\Omega_s$ occupies $[\Omega_s - 800\pi,\; \Omega_s + 800\pi]$. Its lowest frequency must not intrude below $400\pi$:

No-corruption condition $$\Omega_s - 800\pi \;\ge\; 400\pi \qquad\Longrightarrow\qquad \boxed{\;\Omega_s = \frac{2\pi}{T} \ge 1200\pi\;}$$

So the minimum rate is $2\pi/T = 1200\pi$, i.e. $T = 1/600$ — only $1.5\times$ the $800\pi$ Nyquist rate of $f_c$ itself, not $2\times$ the $800\pi$ band edge of the filtered noise. At this rate the folded noise occupies exactly $400\pi \le |\Omega| \le 600\pi$ in the baseband, touching but not entering the signal band. In DT frequency the $f_c$ band edge maps to $\omega = 400\pi\,T = 2\pi/3$:

Answer (b) $$\frac{2\pi}{T}\bigg|_{\min} = 1200\pi, \qquad H(e^{j\omega}) = \begin{cases} e^{\,j(\omega/T)^3} = e^{\,j(600\,\omega)^3}, & |\omega| \le 2\pi/3 \\[4pt] 0, & 2\pi/3 < |\omega| \le \pi \end{cases}$$

Takeaways: (1) the DT filter compensates an analog phase distortion via the substitution $\Omega = \omega/T$; (2) out-of-band noise may alias onto itself without harm — the true constraint is protecting the signal band, which permits a sub-Nyquist (relative to $800\pi$) sampling rate（雜訊混疊到雜訊帶沒關係，只要不疊進訊號帶， 取樣率就可以低於 $1600\pi$）.

Exercise 2 — MATLAB: Sampling and Sinc Reconstruction (HW6 Matlab a–d)

Problem: (a) Generate the "continuous-time" signal $x_c(t) = \sin(4\pi t)$, $0 \le t \le 1$, using the dense grid t = 0:0.001:1, and plot it. (b) Sample $x_c(t)$ with sampling periods $T = 0.01$, $0.05$, $0.1$ s and stem-plot the three sequences $x[n]$. (c) Reconstruct $y_c(t)$ from each sequence using sinc interpolation and plot the three reconstructions. (d) Compute the mean square error (MSE) between $x_c(t)$ and each $y_c(t)$.

Click to reveal solution

Frequency check first: $\sin(4\pi t)$ has $f_0 = 2$ Hz, so the Nyquist rate is 4 Hz. The three sampling rates are $f_s = 100$, $20$, and $10$ Hz — all comfortably above Nyquist, so in principle every case reconstructs perfectly; the differences we measure come from truncating the sinc sum to the finite interval $[0,1]$.

clc; clear; close all;

%% (a) "continuous-time" reference signal
t  = 0:0.001:1;
xc = sin(4*pi*t);
figure(1); plot(t, xc, 'r');
title('x_c(t) = sin(4\pi t)'); xlabel('t (s)'); ylabel('x_c(t)');

%% (b)+(c) sample with T = 0.01, 0.05, 0.1 and sinc-reconstruct
Ts  = [0.01 0.05 0.1];
mse = zeros(1,3);
figure(2);
for i = 1:3
    T  = Ts(i);
    nT = 0:T:1;                       % sampling instants
    xn = sin(4*pi*nT);                % x[n] = xc(nT)

    subplot(3,2,2*i-1);               % left column: x[n]
    stem(nT, xn);
    title(sprintf('x[n],  T = %.2f', T));
    xlabel('n'); ylabel('x[n]');

    yc = zeros(size(t));              % sinc interpolation
    for k = 1:length(nT)
        yc = yc + xn(k) * sinc((t - nT(k)) / T);
    end

    subplot(3,2,2*i);                 % right column: yc(t)
    plot(t, yc, 'r');
    title(sprintf('y_c(t),  T = %.2f', T));
    xlabel('t (s)'); ylabel('y_c(t)');

    mse(i) = mean((xc - yc).^2);      % (d) mean square error
    fprintf('MSE for T = %.2f : %e\n', Ts(i), mse(i));
end
2026-06-12T22:22:45.779233 image/svg+xml Matplotlib v3.10.8, https://matplotlib.org/ 0.0 0.2 0.4 0.6 0.8 1.0 t (s) −1.0 −0.5 0.0 0.5 1.0 x t c ( ) x t π t c ( ) = s i n ( 4 ) ,     0   < =   t   < =   1
Fig. E-1 — (a) The reference signal $x_c(t)=\sin(4\pi t)$ on the dense grid（用 0.001 s 的細網格近似連續訊號）.
2026-06-12T22:22:46.148690 image/svg+xml Matplotlib v3.10.8, https://matplotlib.org/ 0.0 0.2 0.4 0.6 0.8 1.0 t = nT (s) −1 0 1 x[n], T = 0.01 0.0 0.2 0.4 0.6 0.8 1.0 t (s) −1 0 1 y t c ( )   r e c o n s t r u c t e d ,     T   =   0 . 0 1 0.0 0.2 0.4 0.6 0.8 1.0 t = nT (s) −1 0 1 x[n], T = 0.05 0.0 0.2 0.4 0.6 0.8 1.0 t (s) −1 0 1 y t c ( )   r e c o n s t r u c t e d ,     T   =   0 . 0 5 0.0 0.2 0.4 0.6 0.8 1.0 t = nT (s) −1 0 1 x[n], T = 0.1 0.0 0.2 0.4 0.6 0.8 1.0 t (s) −1 0 1 y t c ( )   r e c o n s t r u c t e d ,     T   =   0 . 1 Sampling (left) and sinc reconstruction (right)
Fig. E-2 — (b)(c) Left: sampled sequences for $T = 0.01, 0.05, 0.1$; right: sinc reconstructions (red) vs the original (gray dashed). All three rates exceed Nyquist, so reconstruction is visually faithful even at $T=0.1$ (only 5 samples/period).

(d) Results (computed with the same algorithm in numpy):

$T$ (s)$f_s$ (Hz)samples in $[0,1]$MSE
0.01100101$2.32\times 10^{-6}$
0.052021$3.32\times 10^{-4}$
0.101011$3.24\times 10^{-3}$

Discussion. Since all three rates satisfy Nyquist, the ideal (infinite-length) sinc sum would give zero error in every case. The nonzero MSE comes from edge effects: the interpolation formula needs samples for all $n\in\mathbb{Z}$, but we only have $t\in[0,1]$. Each missing out-of-window sinc tail decays like $1/t$, and with larger $T$ there are fewer, more widely spaced samples, so the truncation error near the window edges grows — hence MSE increases by roughly $10\times$ for each step of $T$ （誤差來自 sinc 級數被截斷的邊界效應，而非混疊）.

Exercise 3 — MATLAB: Midrise Quantization and MSE vs Bits (HW6 Matlab e–f)

Problem: (e) For the $T=0.01$ sequence $x[n]$, quantize with 2 bits (4 levels), 3 bits (8 levels), and 4 bits (16 levels), rounding with offset (midrise). Stem-plot the three quantized signals $x_q[n]$. (f) Compute the MSE between $x[n]$ and each $x_q[n]$, and discuss the advantages and disadvantages of using different numbers of bits.

Click to reveal solution

Midrise rule. With full scale $m_{\max}=1$ the step is $\Delta = 2/2^B$, and rounding with offset means $x_q = \Delta\,(\lfloor x/\Delta \rfloor + \tfrac12)$, clipped to the outermost levels $\pm(1 - \Delta/2)$ — the staircase of Fig. 6-6 (right).

%% (e) midrise quantization of the T = 0.01 sequence, B = 2, 3, 4
T  = 0.01;
nT = 0:T:1;
xn = sin(4*pi*nT);

bits = [2 3 4];
figure(3);
for i = 1:3
    B  = bits(i);
    D  = 2 / 2^B;                            % step = full scale / #levels
    xq = (floor(xn / D) + 0.5) * D;          % round with offset (midrise)
    xq = min(max(xq, -1 + D/2), 1 - D/2);    % clip to outermost levels

    subplot(3,1,i);
    stem(nT, xq);
    title(sprintf('%d-bit midrise quantization (%d levels)', B, 2^B));
    xlabel('n'); ylabel('x_q[n]');

    %% (f) mean square quantization error
    mseq = mean((xn - xq).^2);
    fprintf('B = %d bits : Delta = %.4f, MSE = %e (theory %.3e)\n', ...
            B, D, mseq, D^2/12);
end
2026-06-12T22:22:46.504837 image/svg+xml Matplotlib v3.10.8, https://matplotlib.org/ −1.0 −0.5 0.0 0.5 1.0 xq[n] 2-bit midrise (4 levels), step = 0.5 −1.0 −0.5 0.0 0.5 1.0 xq[n] 3-bit midrise (8 levels), step = 0.25 0.0 0.2 0.4 0.6 0.8 1.0 t = nT (s) −1.0 −0.5 0.0 0.5 1.0 xq[n] 4-bit midrise (16 levels), step = 0.125
Fig. E-3 — (e) Midrise quantization of $x[n]$ ($T=0.01$) with 2, 3, 4 bits. The staircase approximation visibly tightens as $B$ grows（位元數越多，階梯越貼近原訊號）.

(f) Results:

$B$ (bits)levels$\Delta$MSE measured$\Delta^2/12$ (theory)
240.500$2.71\times 10^{-2}$$2.08\times 10^{-2}$
380.250$6.86\times 10^{-3}$$5.21\times 10^{-3}$
4160.125$2.02\times 10^{-3}$$1.30\times 10^{-3}$

Discussion.

  • Each added bit cuts the MSE by roughly 4× ($\approx 6$ dB), exactly the $\sigma_e^2 = \Delta^2/12 \propto 4^{-B}$ law of Section 6.6 — halving $\Delta$ quarters the error power.
  • The measured MSE sits slightly above $\Delta^2/12$ because a full-scale sinusoid spends extra time near its peaks $\pm 1$, where midrise clipping to $\pm(1-\Delta/2)$ adds overload error, and because the error of a smooth deterministic signal is not perfectly uniform.
  • Trade-off: more bits ⇒ higher fidelity (better SNR, lower distortion) but more expensive hardware, more memory/bandwidth per sample, and slower conversion; fewer bits ⇒ cheap and fast but audible/visible staircase distortion and a high noise floor（位元多：精度高但成本、儲存與傳輸量增加；位元少：便宜快速但量化雜訊明顯）. The right choice matches the application: 16 bits for CD audio, 8–12 bits for video and many sensors, 1-bit oversampled converters in sigma-delta designs.