最近突然覺得自己常常耍廢摸魚得過且過感覺未來會暴死，經過深刻的反省之後決定把這學期上課的內容整理成筆記放到 blog 裡，希望這個系列能夠安穩落地。因為目前才剛開始教所以分類會有些隨便，以後滾動調整吧～
關於訊號#
訊號：資訊的流動
訊號處理：生成、轉換和解釋資訊的關鍵技術
數位訊號處理：使用電腦處理訊號
訊號的分類與轉換# 訊號主要分為兩大類 ：
連續時間訊號 (Continuous-time, CT) ：標記為 x ( t ) x(t) x ( t ) 離散時間訊號 (Discrete-time, DT) ：標記為 x [ n ] x[n] x [ n ] n n n ( 0 , 1 , 2 , … ) (0, 1, 2, \ldots) ( 0 , 1 , 2 , … )
連續時間訊號經過取樣(Sampling)後轉換為離散訊號，取樣率是一個至關重要的參數，它決定了你每秒鐘要抓取多少個點。取樣率越高，還原出來的訊號就越接近原始的連續狀態。
類比轉數位 (A/D Conversion)# 訊號從連續時間訊號轉為類比訊號需要經過兩個步驟，分別是取樣(Sampling)以及量化(Quantization)：
C T → S a m p l i n g D T → Q u a n t i z a t i o n D i g i t a l CT \xrightarrow{Sampling} DT \xrightarrow{Quantization} Digital CT S am pl in g D T Q u an t i z a t i o n D i g i t a l
取樣（時間離散化） ：將 x ( t ) x(t) x ( t ) x [ n ] x[n] x [ n ] 頻率範圍
取樣週期(T T T
取樣率/頻率(f s f_s f s f s = 1 T f_s = \frac{1}{T} f s = T 1
關係式：t = n ⋅ T t = n \cdot T t = n ⋅ T
量化（數值離散化） ：將連續的振幅值轉為有限精度的數位資料，決定了訊號的動態範圍與細節 基礎離散時間訊號#
單位樣本序列 (Unit Sample Sequence / Impulse)
標記為 δ [ n ] δ[n] δ [ n ]
δ [ n ] = { 1 , n = 0 0 , n ≠ 0 δ[n] =
\left\{
\begin{array}{ll}
1, & n = 0 \\
0, & n \neq 0
\end{array}
\right. δ [ n ] = { 1 , 0 , n = 0 n = 0
單位步階序列 (Unit Step Sequence)
標記為 u [ n ] u[n] u [ n ]
u [ n ] = { 1 , n ≥ 0 0 , n < 0 u[n] =
\left\{
\begin{array}{ll}
1, & n \ge 0 \\
0, & n < 0
\end{array}
\right. u [ n ] = { 1 , 0 , n ≥ 0 n < 0
指數序列 (Exponential Sequence)
形式為 x [ n ] = A ⋅ α n x[n] = A \cdot \alpha^n x [ n ] = A ⋅ α n A A A α \alpha α
重要數學性質與表示法# 任意序列的表示 (Representation of Arbitrary Sequence)# 任何離散序列 x [ n ] x[n] x [ n ]
寫成一般式（a k a_k a k x [ k ] x[k] x [ k ]
x [ n ] = ∑ k = − ∞ ∞ a k ⋅ δ [ n − k ] = ∑ k = − ∞ ∞ x [ k ] ⋅ δ [ n − k ] x[n]=\sum_{k=-\infty}^\infty {{\color{#3071c4} a_k} \cdot \delta[n-k]}=\sum_{k=-\infty}^\infty {{\color{#3071c4} x[k]} \cdot \delta[n-k]} x [ n ] = k = − ∞ ∑ ∞ a k ⋅ δ [ n − k ] = k = − ∞ ∑ ∞ x [ k ] ⋅ δ [ n − k ]
u [ n ] u[n] u [ n ] δ [ n ] \delta[n] δ [ n ]
累加關係： u [ n ] = ∑ k = 0 ∞ δ [ n − k ] u[n]=\sum_{k=0}^\infty \delta[n-k] u [ n ] = ∑ k = 0 ∞ δ [ n − k ] u [ n ] = ∑ k = − ∞ n δ [ k ] u[n]=\sum_{k=-\infty}^n \delta[k] u [ n ] = ∑ k = − ∞ n δ [ k ] 差分關係： δ [ n ] = u [ n ] − u [ n − 1 ] \delta[n]=u[n]-u[n-1] δ [ n ] = u [ n ] − u [ n − 1 ] 在處理複數指數訊號時常用到：
歐拉公式：e j ϕ = cos ϕ + j sin ϕ e^{j \phi}=\cos \phi+j\sin \phi e j ϕ = cos ϕ + j sin ϕ
極座標表示：A = ∣ A ∣ e j ϕ 1 A=|A| e^{j \phi_1} A = ∣ A ∣ e j ϕ 1 α = ∣ α ∣ e j ϕ 2 \alpha= |\alpha| e^{j \phi_2} α = ∣ α ∣ e j ϕ 2
因此 x [ n ] = A α n = ∣ A ∣ ∣ α ∣ n e j ( n ϕ 2 + ϕ 1 ) x[n]=A \alpha^n= |A| |\alpha|^n e^{j(n \phi_2 + \phi_1)} x [ n ] = A α n = ∣ A ∣∣ α ∣ n e j ( n ϕ 2 + ϕ 1 )
✍️TODO 以後想到再補圖
訊號取樣的影響# 取樣與混疊 (Sampling and Aliasing)# 基本定義# 假設一個連續時間訊號 x ( t ) = cos ( ω t ) x(t) = \cos(\omega t) x ( t ) = cos ( ω t )
ω \omega ω t t t NOTE 複習一下高中內容：在圓周運動或簡諧運動的數學推導中，如果使用「圈數」，公式裡會不斷出現 2 π 2 \pi 2 π 2 π 2 \pi 2 π ω \omega ω
取樣過程（Sampling Process）#
每隔 T T T T T T
時間 t ： 0 , T , 2 T , 3 T , … t = n T n = 0 , 1 , 2 , 3 , … \begin{aligned}
{\text{時間 } t\text{：}} & 0, T, 2T, 3T, \ldots \\
t &= nT \\
n &= 0, 1, 2, 3, \ldots
\end{aligned} 時間 t ： t n 0 , T , 2 T , 3 T , … = n T = 0 , 1 , 2 , 3 , …
離散化
x ( t ) = cos ( ω t ) ↓ t = n T x ( n T ) = cos ( ω n T ) ⇓ x [ n ] = cos ( ω n T ) \begin{aligned}
x(t) &= \cos(\omega {\color{#3071c4}t}) \\
&\downarrow {\color{#3071c4}t = nT} \\
x({\color{#3071c4}nT}) &= \cos(\omega {\color{#3071c4}nT}) \\
&\Downarrow \\
x[n] &= \cos(\omega nT)
\end{aligned} x ( t ) x ( n T ) x [ n ] = cos ( ω t ) ↓ t = n T = cos ( ω n T ) ⇓ = cos ( ωn T )
取樣頻率轉換
取樣頻率：f s = 1 T f_s = \frac{1}{T} f s = T 1 ω s = 2 π T \omega_s = \frac{2 \pi}{T} ω s = T 2 π
TIP ω = 2 π f → f = 1 T ω = 2 π T \omega = 2 \pi f \xrightarrow{f = \frac{1}{T}} \omega = \frac{2 \pi}{T} ω = 2 π f f = T 1 ω = T 2 π
離散時間的週期性 (Periodicity in Discrete Time)# 假設這裡有兩個訊號：x x x x 1 x_1 x 1
{ x = cos ( ω t ) x 1 = cos ( ( ω + ω s ) t ) \left\{
\begin{array}{ll}
x &= \cos(\omega t) \\
x_1 &= \cos((\omega + {\color{#3071c4}\omega_s})t)
\end{array}
\right. { x x 1 = cos ( ω t ) = cos (( ω + ω s ) t ) 很明顯這是兩個不一樣頻率的訊號，但如果我們對它們進行採樣，會發現一個很有趣的現象：
x 1 ( t ) = cos ( ( ω + ω s ) t ) ↓ t = n T x 1 [ n ] = cos ( ( ω + ω s ) n T ) ∵ ω s = 2 π T ∴ x 1 [ n ] = cos ( ( ω + 2 π T ) ⋅ n T ) = c o s ( ω n T + 2 π T ⋅ n T ) = c o s ( ω n T + 2 π n ⏟ 2 π 的整數倍，不影響值 ) = c o s ( ω n T ) \begin{aligned}
x_1(t) &= \cos((\omega+\omega_s)t) \\
&\downarrow t = {\color{#3071c4}nT} \\
x_1[n] &= \cos((\omega+\omega_s){\color{#3071c4}nT}) \\
&\because \omega_s = \frac{2\pi}{T} \\
\therefore x_1[n] &= \cos((\omega+\frac{2\pi}{T})\cdot nT) \\
&= cos(\omega nT+\frac{2\pi}{\cancel{T}}\cdot n\cancel{T}) \\
&= cos(\omega nT+{\underbrace{2\pi n}_{\color{#e53935}{2\pi\text{ 的整數倍，不影響值}}}}) \\
&= cos(\omega nT)
\end{aligned} x 1 ( t ) x 1 [ n ] ∴ x 1 [ n ] = cos (( ω + ω s ) t ) ↓ t = n T = cos (( ω + ω s ) n T ) ∵ ω s = T 2 π = cos (( ω + T 2 π ) ⋅ n T ) = cos ( ωn T + T 2 π ⋅ n T ) = cos ( ωn T + 2 π 的整數倍，不影響值 2 πn ) = cos ( ωn T ) 可以觀察到雖然 x x x x 1 x_1 x 1 c o s ( ω n T ) cos(\omega nT) cos ( ωn T ) ∴ x [ n ] = x 1 [ n ] \therefore x[n] = x_1[n] ∴ x [ n ] = x 1 [ n ]
NOTE 這裡再假設一個 x 2 ( t ) = cos ( ( − ω + ω s ) t ) x_2(t) = \cos((-\omega+{\color{#3071c4}\omega_s})t) x 2 ( t ) = cos (( − ω + ω s ) t ) x 2 [ n ] = cos ( − ω n T ) x_2[n] = \cos(-\omega nT) x 2 [ n ] = cos ( − ωn T ) x 2 [ n ] = cos ( − ω n T ) = cos ( ω n T ) = x [ n ] x_2[n] = \cos(-\omega nT) = \cos(\omega nT) = x[n] x 2 [ n ] = cos ( − ωn T ) = cos ( ωn T ) = x [ n ]
我們可以得到一個結論，在離散的世界裡，很多不同頻率的連續波，採樣後居然會看起來一模一樣，這個就是所謂的 「混疊現象」(Aliasing Phenomenon) 。
🔗圖片來源：https://www.ni.com/docs/zh-TW/bundle/labwindows-cvi/page/advancedanalysisconcepts/aliasing.html
奈奎斯特-香農取樣定理 (Nyquist-Shannon Sampling Theorem)# 我們現在知道如果取樣頻率不夠高，就會在這種週期性的訊號上產生混疊現象，那如果不想要產生混疊現象的話，具體來說需要多高的取樣頻率呢？奈奎斯特-香農取樣定理 給出了答案，並定義了所謂的奈奎斯特頻率 (Nyquist Frequency) ，又或者稱作奈奎斯特極限 (Nyquist Limit) 。
奈奎斯特極限： 能被系統唯一表示的最高頻率為 ω s 2 \frac{\omega_s}{2} 2 ω s 發生混疊的條件： 如果輸入訊號的最高頻率 ω N > ω s 2 \omega_N > \frac{\omega_s}{2} ω N > 2 ω s 取樣定理： 假設一個連續時間訊號 x ( t ) x(t) x ( t ) 頻帶受限訊號 (band-limited signal) ，最高頻率為 ω N \omega_N ω N X ( j ω ) = 0 for ∣ ω ∣ ≥ ω N X(j\omega) = 0 \text{ for } |\omega| \ge \omega_N X ( jω ) = 0 for ∣ ω ∣ ≥ ω N ω N \omega_N ω N → X ( j ω ) = 0 \rightarrow X(j\omega) = 0 → X ( jω ) = 0 ω s ≥ 2 ω N \omega_s \ge 2\omega_N ω s ≥ 2 ω N x ( t ) x(t) x ( t ) 唯一地 (uniquely) 被它的離散取樣點 x [ n ] x[n] x [ n ] x ( t ) x(t) x ( t ) 頻帶受限訊號 (Band-Limited Signal)
指其頻譜能量集中在有限的頻率範圍內的訊號，簡單來說，它的頻率是有上限的 (ω m a x < ∞ \omega_{max} < \infty ω ma x < ∞
完美還原訊號的數學條件# if ω s ⏞ 取樣頻率 ≥ 2 ω N ⏞ 最高訊號頻率 i.e. , Nyquist Limit ω s 2 ω N ≤ ω s 2 ⇒ ω s ≥ 2 ω N ω s = 2 π T ⇒ T = 2 π ω s \begin{aligned}
&\text{if } \overbrace{\omega_s}^{\color{#3071c4}\text{取樣頻率}} \ge \overbrace{2\omega_N}^{\color{#e53935}\text{最高訊號頻率}} \\
&\text{i.e. , Nyquist Limit } \frac{\omega_s}{2} \\
&\omega_N \le \frac{\omega_s}{2} \color{#3071c4}{\Rightarrow \omega_s \ge 2\omega_N} \\
&\omega_s = \frac{2\pi}{T} \Rightarrow T = \frac{2\pi}{\omega_s}
\end{aligned} if ω s 取樣頻率 ≥ 2 ω N 最高訊號頻率 i.e. , Nyquist Limit 2 ω s ω N ≤ 2 ω s ⇒ ω s ≥ 2 ω N ω s = T 2 π ⇒ T = ω s 2 π 反混疊濾波器（Anti-aliasing Filter）# ✍️TODO 以後補
範例與練習# Ex.1# Determine which input signals to a digital filter or DSP system will be aliased by the given period Т \text{Т} Т
x ( t ) = 2 cos ( 10 t ) , T = 0.1s x(t) = 2\cos(10t)\text{ , T = 0.1s} x ( t ) = 2 cos ( 10 t ) , T = 0.1s x ( t ) = 8 cos ( 15 t ) , T = 0.2s x(t) = 8\cos(15t)\text{ , T = 0.2s} x ( t ) = 8 cos ( 15 t ) , T = 0.2s ANS
ω s = 2 π T = 2 π 0.1 = 20 π \omega_s = \frac{2\pi}{T} = \frac{2\pi}{0.1} = 20\pi ω s = T 2 π = 0.1 2 π = 20 π ω s 2 = 10 π ≈ 31.4 \frac{\omega_s}{2} = 10\pi\approx 31.4 2 ω s = 10 π ≈ 31.4 ∴ ω N = 10 < ω s 2 \therefore \omega_N = 10 < \frac{\omega_s}{2} ∴ ω N = 10 < 2 ω s 同理，ω s 2 = 2 π 0.2 2 = 5 π ≈ 15.7 \frac{\omega_s}{2} = \frac{\frac{2\pi}{0.2}}{2} = 5\pi\approx 15.7 2 ω s = 2 0.2 2 π = 5 π ≈ 15.7 ∴ ω N = 15 < ω s 2 \therefore \omega_N = 15 < \frac{\omega_s}{2} ∴ ω N = 15 < 2 ω s
Ex.2# Determine if the following signals will be aliased. If the signal is aliased into having the same sample value as a lower frequency sinusoidal signal, determine the lower sinusoidal signal.
x ( t ) = 7 cos ( 25 t ) , T = 0.1s x(t) = 7\cos(25t)\text{ , T = 0.1s} x ( t ) = 7 cos ( 25 t ) , T = 0.1s x ( t ) = 3 cos ( 160 t ) , T = 0.02s x(t) = 3\cos(160t)\text{ , T = 0.02s} x ( t ) = 3 cos ( 160 t ) , T = 0.02s ANS
ω s 2 = 2 π 0.1 2 = 10 π ≈ 31.4 \frac{\omega_s}{2} = \frac{\frac{2\pi}{0.1}}{2} = 10\pi\approx 31.4 2 ω s = 2 0.1 2 π = 10 π ≈ 31.4 ∴ ω N = 25 < ω s 2 \therefore \omega_N = 25 < \frac{\omega_s}{2} ∴ ω N = 25 < 2 ω s ω s 2 = 2 π 0.02 2 = 50 π ≈ 157 \frac{\omega_s}{2} = \frac{\frac{2\pi}{0.02}}{2} = 50\pi\approx 157 2 ω s = 2 0.02 2 π = 50 π ≈ 157 ∴ ω N = 160 > ω s 2 \therefore \omega_N = 160 > \frac{\omega_s}{2} ∴ ω N = 160 > 2 ω s 混疊高頻摺疊至低頻的現象 ，− ω 1 + ω s = 160 -\omega_1 + \omega_s = 160 − ω 1 + ω s = 160 ω s = 100 π ⇒ ω 1 = 100 π − 160 \omega_s = 100\pi \Rightarrow \omega_1 = 100\pi - 160 ω s = 100 π ⇒ ω 1 = 100 π − 160 ∴ x 1 ( t ) = 3 cos ( ( 100 π − 160 ) t ) ≈ 3 cos ( 154.2 t ) \therefore x_1(t) = 3\cos((100\pi - 160)t) \approx 3\cos(154.2t) ∴ x 1 ( t ) = 3 cos (( 100 π − 160 ) t ) ≈ 3 cos ( 154.2 t ) 數位濾波器規格# 基礎概念與公式# 在介紹數位濾波器之前，要先了解一下下面這三個東西，它們是衡量一個濾波器好不好用、有沒有效的核心概念：
增益 (Gain)
G a i n = 輸入訊號的振幅 輸出訊號的增幅 Gain = \frac{\text{輸入訊號的振幅}}{\text{輸出訊號的增幅}} G ain = 輸出訊號的增幅 輸入訊號的振幅 如果 G a i n = 1 Gain = 1 G ain = 1 通帶 (Pass Band)
如果 G a i n Gain G ain 0 0 0 阻帶 (Stop Band)
損失 (Loss)
增益的倒數，即 L o s s = ( G a i n ) − 1 Loss = {(Gain)}^{-1} L oss = ( G ain ) − 1
分貝 (dB)： 現實世界的訊號強弱差異極大，如果我們在定義濾波器規格的時候只使用倍數來表示，數值會變得極端並且難以繪圖，比如 通帶 : 阻帶 = 1 : 10 − 6 \text{通帶}:\text{阻帶} = 1 : 10^{-6} 通帶 : 阻帶 = 1 : 1 0 − 6
分貝轉換公式：G a i n d B = 20 ⋅ log 10 ( G a i n ) Gain_{\color{#3071c4}{dB}} = 20\cdot\log_{\color{#e53935}{10}}(Gain) G ai n d B = 20 ⋅ log 10 ( G ain )
無變化時（1 1 1 G a i n = 1 Gain = 1 G ain = 1 20 ⋅ log 10 ( 1 ) = 0 dB 20\cdot\log_{10}(1) = 0\text{ dB} 20 ⋅ log 10 ( 1 ) = 0 dB 0 dB 0\text{ dB} 0 dB
訊號放大時（假設 2 2 2 20 ⋅ log 10 ( 2 ) ≈ 6 dB 20\cdot\log_{10}(2) \approx 6\text{ dB} 20 ⋅ log 10 ( 2 ) ≈ 6 dB
訊號衰減時（假設 10 − 4 10^{-4} 1 0 − 4 20 ⋅ log 10 ( 10 − 4 ) = − 80 dB 20\cdot\log_{10}(10^{-4}) = -80\text{ dB} 20 ⋅ log 10 ( 1 0 − 4 ) = − 80 dB
REMARK
以分貝作為單位的時候，增益和損失的關係變得更加簡單：
∵ 20 ⋅ log 10 ( L o s s ) = 20 ⋅ log 10 ( ( g a i n ) − 1 ) ∴ L o s s = − G a i n \begin{aligned}
& \because\cancel{20}\cdot\log_{10}(Loss) = \cancel{20}\cdot\log_{10}((gain)^{-1}) \\
& \therefore Loss = -Gain
\end{aligned} ∵ 20 ⋅ log 10 ( L oss ) = 20 ⋅ log 10 (( g ain ) − 1 ) ∴ L oss = − G ain 奈奎斯特頻率 / 摺疊頻率（Nyquist / Folding frequency）# ω s 2 = 2 π ⋅ f s 2 ∵ T = 1 f s , T : Sampling Period ∴ ω s 2 = π ⋅ 1 T = π T \begin{aligned}
\frac{\omega_s}{2} &= \frac{\cancel{2}\pi \cdot f_s}{\cancel{2}} \\
\because T &= \frac{1}{f_s}, \quad T: \text{Sampling Period} \\
\therefore \frac{\omega_s}{2} &= \pi \cdot \frac{1}{T} = \frac{\pi}{T}
\end{aligned} 2 ω s ∵ T ∴ 2 ω s = 2 2 π ⋅ f s = f s 1 , T : Sampling Period = π ⋅ T 1 = T π 濾波器的實際應用#
低通濾波器
降噪，如 ECG 或 EEG 等生醫訊號及音訊處理
抗混疊濾波器
影像處理
高通濾波器
去除直流成分
影像邊緣偵測
帶通濾波器
無線通訊
生醫訊號處理
語音訊號處理
帶阻濾波器
消除電源線干擾
語音通訊
全通濾波器（增益維持1）
相位補償
多通道通訊系統中的訊號對齊
範例與練習# Ex.1# Draw the graphical specification for a digital highpass filter out to the folding frequency in rad/s that will not change the gain above 500 rad/s by more than ±3 dB, while reducing the gain below 200 rad/s by more than 40 dB.
The sampling time T = 0.001 s.
ANS
先確定 x 軸的盡頭，也就是 ω f \omega_f ω f T = 0.001 s T = 0.001\text{s} T = 0.001 s ∵ Nyquist Frequency = π T \because\text{Nyquist Frequency} = \frac{\pi}{T} ∵ Nyquist Frequency = T π 💡為什麼？ ∴ ω f = π 0.001 = 1000 π \therefore \omega_f = \frac{\pi}{0.001} = 1000\pi ∴ ω f = 0.001 π = 1000 π
畫出阻帶，題目說 200 rad/s 以下增益要衰減 40 dB，因此阻帶容許最大增益為 -40 dB，即 g s m a x = − 40 db g_{smax} = -40\text{ db} g s ma x = − 40 db
畫出通帶，題目說 500 rad/s 以上增益改變不能超過 ±3 dB，以無損增益 0 dB 為基準，可得知允許的範圍最高 g p m a x = 3 dB g_{pmax} = 3\text{ dB} g p ma x = 3 dB g p m i n = − 3 dB g_{pmin} = -3\text{ dB} g p min = − 3 dB
規格圖如下所示：
Ex.2# Draw the graphical specification for a bandstop digital filter out to its folding frequency that will reduce the gain between 1,000 rad/s and 5,000 rad/s by more than 60 dB, but will not reduce the gain above 10,000 rad/s or below 150 rad/s by more than 3 dB. The sampling rate is 10,000 Hz.
ANS
Nyquist Frequency = ω s 2 = 2 π ⋅ f s 2 = 10000 π \text{Nyquist Frequency} = \frac{\omega_s}{2} = \frac{\cancel{2}\pi\cdot f_s}{\cancel{2}} = 10000\pi Nyquist Frequency = 2 ω s = 2 2 π ⋅ f s = 10000 π
未完待續…
NOTE 部份圖片來源自：陳榮銘 教授 數位訊號處理課程投影片
