授課老師:
林俊宏 老師 資料來源:
蕭子健 老師
第三章 常見的基本訊號
(一) 弦波訊號
• 前言
– 弦波訊號為許多應用的基礎,本章將從弦波之 產生、特性、到應用範疇等作說明。
• 目標
– 瞭解不同函數所產生的弦波訊號 – 瞭解弦波訊號之週期特性
• 關鍵名詞
– 正弦訊號 (Sinusoidal signal)
– 指數訊號 (Exponential signal)
– 週期性 (Periodic)
What does “sinusoidal” mean?
• The curve whose ordinates (縱座標) are proportional to sines of the abscissas (橫坐標) with the equation y=a*sin(x) (Webster’s Third New
International Dictionary)
• 數學上指三角函數之一。直角三角形中,一銳角的對邊除以斜邊所得的 值,稱為此角的「正弦」。
(
中華民國教育部 重編國語辭典修訂本)
• 三角函數中的正弦函數(sin)及餘弦函數(cos)只有相位上的差別(兩者相位相差) , 其餘性質並無不同,因此在討論訊號的時候,正弦函數及餘弦函數所表示的訊號 一律都稱為「正弦訊號」。
c
a
b
=
= +
=
=
−
( / ) tan
1 ) ( cos )
( sin
/ ) cos(
/ ) sin(
1 2 2
a b where
c a
c b
θ
θ θ
θ
θ
θ) t
Acos(
x(t) = ω + φ
A:訊號之振幅大小;
ω :角頻率,單位是每秒多少弳度(radians/s);
Φ:初始相位,亦即 t=0 之相角,單位為弳度(radians);
f:f= ω /2π,為訊號之頻率,單位是赫茲(Hz)。
首先,假設正弦訊號 x(t) 是一個週期性訊號,且 x(t) 的週期為 T,
依據週期的定義可知 ,T = 1/f = 2π /ω ,因此代入右式:
符合週期性訊號的要件,故可稱「連續時間下之弦波訊號為一週 期性訊號」
𝑥𝑥(𝑡𝑡 + 𝑇𝑇) = 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝜔𝜔(𝑡𝑡 + 𝑇𝑇) + 𝜑𝜑)
= 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝜔𝜔𝑡𝑡 + 𝜔𝜔𝑇𝑇 + 𝜑𝜑)
= 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝜔𝜔𝑡𝑡 + 𝜔𝜔2𝜋𝜋 𝜔𝜔 + 𝜑𝜑)
= 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝜔𝜔𝑡𝑡 + 2𝜋𝜋 + 𝜑𝜑)
= 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝜔𝜔𝑡𝑡 + 𝜑𝜑)
= 𝑥𝑥(𝑡𝑡)
) cos(
) ( )
( t = x t = A ω t + φ
y if
時域 (Time Domain) 頻域 (Frequency Domain)
取樣 (Sampling) 取樣週期 (Sampling Period)
一個樣本為多少秒 (second / Sample, s/S)
取樣速率 (Sampling Rate)
每秒多少樣本(Sample/second, S/s)
類比 (Analog) 秒 (second, sec) 頻率 (cycle/sec);角頻率 (rad/sec)
數位 (Digital) 次 (Time, time) 數位頻率 (cycle/S);數位角頻率 (rad/S)
時域 (Time Domain) 頻域 (Frequency Domain) 取樣
T s f s = 1/T s
類比
T f;ω = 2πf
數位
n F =f/f s ;ω=2πF=2πf/f s
假設此離散時間之正弦訊號是一週期性訊號的話,就會有一個週期N,滿足下面 情況:(由於為離散時間之狀況,因此週期N與n相同必須為整數)
𝑥𝑥[𝑛𝑛 + 𝑁𝑁] = 𝑥𝑥[𝑛𝑛]
𝑥𝑥[𝑛𝑛 + 𝑁𝑁] = 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝛺𝛺(𝑛𝑛 + 𝑁𝑁) + 𝜑𝜑)
= 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝛺𝛺𝑛𝑛 + 𝛺𝛺𝑁𝑁 + 𝜑𝜑)
由式子中可以得知,只有在 ΩN 等於2π 的整數倍時,才有可能使得 成立,亦即
𝛺𝛺𝑁𝑁 = 2𝜋𝜋𝑚𝑚 (𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑛𝑛𝐴𝐴) 𝑚𝑚 ∈ 𝑍𝑍, 𝑁𝑁 ∈ 𝑍𝑍 𝛺𝛺 =2𝜋𝜋𝑚𝑚
𝑁𝑁 �
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑛𝑛𝐴𝐴
𝐴𝐴𝑐𝑐𝐴𝐴𝑐𝑐𝑐𝑐 � 𝑚𝑚 ∈ 𝑍𝑍, 𝑁𝑁 ∈ 𝑍𝑍
Ex.. 請參考 page 3-37, 問題與討論
複數 (Complex number)
iy x
z = +
實部 (Real part)虛部 (Imaginary part)
) sin(
) ˆ cos(
ˆ ,
) ( tan )),
sin(
) (cos(
) (
2 2
1 2
2 2 2
2 2
2 2
θ θ
θ θ
θ
i z
and y
x z
where z
z z
x and y
y x
r where i
r
y x
i y y
x y x
x z
iy x z
+
= +
=
=
= +
= +
=
+ + + +
= +
=
−
複數表示法
三角函數表示法
向量空間表示法
尤拉公式 (Euler formula)
iy x
z = +
實部 (Real part)虛部 (Imaginary part)
) sin(
) cos(
,
} Im{
} Re{
) sin(
) cos(
) ( tan ),
sin(
) cos(
, )) sin(
) (cos(
2 2
1 2
2
θ θ
θ θ
θ θ
θ θ
θ
θ θ
θ
θ θ
i e
and y
x z
where e
z z
z i
z i
e
x and y
i e
y x
r where
re i
r z
iy x z
i i
i
i i
+
= +
=
=
+
= +
=
= +
= +
=
= +
= +
=
−
複數表示法
三角函數表示法 尤拉表示法
向量空間表示法
−
= +
+
= +
+
− +
=
+ +
+
=
≈ +
=
+
− +
+
− +
+
− +
+
) 2 (
) 1 sin(
) 2 (
) 1 cos(
) sin(
) cos(
) sin(
) cos(
) cos(
) (
) ( )
(
) ( )
( )
( ) (
) (
ϕ ω ϕ
ω
ϕ ω ϕ
ω ϕ
ω ϕ ω
ϕ ω
ϕ ω
ϕ ω
ϕ ω ϕ
ω
ϕ ω ϕ
ω ϕ ω
t i t
i
t i t
i t
i t i
t i
e i e
t
e e
t
t i
t e
t i
t e
Ae t
A t
x
Elementary Signals
★ 1. Exponential Signals x t ( ) = Be
a t1. Decaying exponential, for which a < 0 2. Growing exponential, for which a > 0
B and a are real parameters
Ex. Lossy capacitor:
Figure
Lossy capacitor, with the loss represented by shunt resistance R.
( ) ( ) 0 RC d v t v t
dt + =
KVL Eq.:
/( )
( ) 0 t RC
v t
=V e
−RC = Time constant
Discrete-time case:[ ]
nx n = Br
r = e α
where
Figure
(a) Decaying
exponential form of discrete-time signal.
(b) Growing
exponential form of
discrete-time signal.
Signals_and_Systems_Simon Haykin & Barry Van Veen
13
( ) ( ) 0 RC d v t v t
dt + =
/( )
( ) 0 t RC
v t
=V e
−0 )
) ( ( )
( )
(
then circuit,
RC a
s it' if
) ) (
) ( ) (
) ( (
) ) (
( )
( )
(
= +
=>
=
=
=>
=
=
=
=>
=
∫
t dt v
t RC dv
dt t C dv
R t v
dt t C dv
t C i
dt t i C
t t Q
v
R t t v
i R
t i t
v
R c
c R
c c
R
R
★ 2. Sinusoidal Signals ( ) cos( ) x t = A ω φ t +
T 2 π
= ω
where
( ) co ( ( s ) )
cos( )
cos( 2 ) cos( )
( )
x t T A t T
A t T
A t
A t
x t
ω φ
ω ω φ ω π φ ω φ
+ = + +
= + +
= + +
= +
=
periodicity
◆ Continuous-time case:
◆ Discrete-time case :
[ ] cos( )
x n = A Ω + n φ
[ ] cos( )
x n + N = A Ω + Ω + n N φ
2
N π m
Ω = 2 m radians/cycle, integer ,
N m N Ω = π
Periodic condition:
or
Ex. A discrete-time sinusoidal signal: A = 1,
φ
= 0, and N = 12.Example 1.7 Discrete-Time Sinusoidal Signal
A pair of sinusoidal signals with a common angular frequency is defined by
1 [ ] sin[5 ]
x n = π n x n 2 [ ] = 3 cos[5 π n ]
1 2
[ ] [ ] [ ] y n = x n + x n
and
(a) Both x1[n] and x2[n] are periodic. Find their common fundamental period.
(b) Express the composite sinusoidal signal
In the form y[n] = Acos(Ωn +
φ
), and evaluate the amplitude A and phaseφ
.<Sol.>
(a) Angular frequency of both x1[n] and x2[n]:
5 radians/cycle π
Ω = 2 2 2
5 5
m m m
N π π
= = π =
Ω
This can be only for m = 5, 10, 15, …, which results in N = 2, 4, 6, …
sin( ) 1 and cos( ) 3
A φ = − A φ =
1 2
sin( ) amplitude of [ ] 1 tan( )
cos( ) amplitude of [ ] 3 x n
x n φ φ
φ
= = = −
sin( ) 1 A φ = −
( 1 ) 2
sin / 6
A π
= − =
− [ ] 2cos 5
y n = π n − π 6
φ = − π / 6
Accordingly, we may express y[n] as (b) Trigonometric identity:
cos( ) cos( ) cos( ) sin( )sin( ) A Ω + n φ = A Ω n φ − A Ω n φ
x
1[n] + x2[n] with the above equation to obtain that Let Ω = 5π, then compare★ 3. Relation Between Sinusoidal and Complex Exponential Signals
1. Euler’s identity:e
jθ= cos θ + j sin θ
Complex exponential signal:
B = Ae
jφcos( ) Re{
j t} A ω φ t + = Be
ω( )
cos( ) sin( )
j t
j j t
j t
Be
Ae e Ae
A t jA t
ω
φ ω φ ω
ω φ ω φ
+
=
=
= + + +
( ) cos( ) x t = A ω φ t +
◇ Continuous-time signal in terms of sine function:
( ) sin( ) x t = A ω φ t +
sin( ) Im{
j t}
A ω φ t + = Be
ωπ / 4 π / 4
−
2. Discrete-time case:cos( ) Re{
j n}
A Ω + n φ = Be
Ω and3. Two-dimensional representation of the complex exponential e j Ω n for Ω = π/4 and n = 0, 1, 2, …, 7.
:
Projection on real axis: cos(Ωn);
Projection on imaginary axis: sin(Ωn)
Figure
Complex plane, showing eight points uniformly distributed on the unit circle.
sin( ) Im{
j n}
A Ω + n φ = Be
Ω★ 4. Exponential Damped Sinusoidal Signals
( )
tsin( ), 0
x t = Ae
−αω φ t + α >
Example for A = 60,
α
= 6, andφ
= 0.Figure
Exponentially damped sinusoidal signal Ae
−atsin( ω t), with A = 60 and α
= 6.
◆ Discrete-time case:
[ ]
nsin[ ]
x n = Br Ω + n φ
習作 3-1
• 題目:正弦訊號 (Sinusoidal signal)
• 目標:何謂正弦訊號,及如何模擬與產生?
• 程式:ex 3-1 Sinusoidal signal.vi
習作3-2
• 題目:指數訊號
• 目標:說明指數訊號之基本定義
• 程式: ex 3-2 Exponential signal.vi
習作3-3
• 題目:複數(Complex number)及正交(Orthogonal)
• 目標:說明並複習複數之表示法與性質
• 程式:ex 2-3 Addition continuous.vi & ex2-3 Addition discrete.vi
習作3-4
• 題目:實指數訊號
• 目標:說明實指數訊號之定義與意涵
• 程式:ex 3-4 exponential signal (continuous).vi
& 3-4 exponential signal (discrete).vi Case 1:當a為正數時(a>0), 隨 t 的增加成指數遞增。
Case 2:當a為負數時(a<0), 隨 t 的增加成指數遞減。
Case 3:當a為零時(a=0),,亦即是任何時間 t 下訊號值皆為常數B。
有何不同?
ex 3-4 exponential signal (multi).vi