Transcript Frequency Domain

Chap. 9 Sinusoidal Steady-State
Analysis
Contents
9.1
9.2
9.3
9.4
9.5
9.6
9.7
The Sinusoidal Source
The Sinusoidal Response
The Phasor
The Passive Circuit Elements in the Frequency Domain
Kirchhoff’s Laws in the Frequency Domain
Series, Parallel, and Delta-to-Wye Simplifications
Source Transformations &
Thévenin-Norton Equivalent Circuits
9.8 The Node-Voltage Method
9.9 The Mesh-Current Method
9.10 The Transformer
9.11 The Ideal Transformer
9.12 Phasor Diagrams
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9.1 The Sinusoidal Source
sinusoidal source
peak or amplitude
angular frequency:
ω  2πf  2π/T
period
frequency)
(Hz)
<radians/second>
<cycles/second>
phase angle: 
>0
2
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Root Mean Square, rms, Value:
the square root of the mean value of the squared periodic function
Rms, or effective value，
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EX 9.1 Finding the Characteristics of a Sinusoidal Current
A sinusoidal current has a maximum amplitude of 20 A. The current
passes through one complete cycle in 1 ms. The magnitude of the
current at zero time is 10 A.
a) What is the frequency of the current in hertz (Hz)?
b) What is the angular frequency in radians per second?
c) Write the expression for i(t) using the cosine function.
Express  in degrees.
d) What is the rms value of the current?
a)
b)
c)
d)
&
rms value =
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EX 9.2 Finding the Characteristics of a Sinusoidal Voltage
A sinusoidal voltage is given by the expression
v  300cos120πt  30.
a) What is the period of the voltage in milliseconds?
b) What is the frequency in hertz?
c) What is the magnitude of v at t = 2.778 ms?
d) What is the rms value of v ?
&
a)
b)
c)
f=
& 2.778=
d)
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EX 9.3 Translating a Sine Expression to a Cosine Expression
Translate the sine function to the cosine function by subtracting 90◦ (π/2
rad) from the argument of the sine function.
a) Verify the above translation.
b) Express sin(ωt + 30◦) as a cosine function.
a)
Let
&
b)
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EX 9.4 Calculating the rms Value of a Triangular Waveform
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9.2 The Sinusoidal Response
弦波電源
當開關閉合後，即 t  0
KVL
其解為:
暫態成分
transient component
隨時間的增加而呈指數形
式降低直至消失
穩
態
解
之
特
性
穩態成分
steady-state component
呈弦波變化之形式持續存在
1. 穩態解仍為弦波函數。
2. 對線性電路而言，響應信號之頻率與電源信號之頻率相同。
（非線性電路除外）
3. 一般而言，穩態響應之最大振幅與電源之最大振幅不同。
4. 一般而言，響應信號之相位角不同於電源之相位角。
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9.3 The Phasor
相量(phasor):
是一個包含振幅（大小）及相位角的複數，但隱藏頻率。
Euler’s identity
e
 j
 cos  j sin 
實部(real part)
虛部(imaginary part)
Vm e j 含有一已知弦波函數之振幅及相位角，將此複數
定義為弦波函數之相量表示法(phasor representation)
或相量轉換(phasor transform) ：
相量轉換
用粗黑體字代表相量，而相
量轉換將弦波函數由時域轉
換至複數領域， 或稱為頻域
(frequency domain)。
& 1  1e j
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Inverse Phasor Transform
P -1
v  100cos(ωt-26 )
相量轉換可將穩態弦波響應之最大振幅與相位角問題，轉換
為複數的代數運算。
注意：
1. 暫態成份隨時間持續而消失，穩態成份必能滿足原微分方程式。
2. 在弦波電源驅動的線性電路，其穩態響應仍為弦波形式，且具有相同
的頻率。
3. 穩態解的形式為R{Aej ejt }，其中 A 是響應之最大振幅，而  是其相
位角。
4. 當以前述穩態解的形式代入原微分方程式，其指數項次ejt 將可消去,僅
留下複數頻域下求解A和的運算式。
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Example (利用相量求穩態響應)
令其穩態解為
代入
若改以sin為電源，則
Note: If
then
重要提醒：
相量讓你輕鬆做頻域與時域的轉換；電路分析時，其解請以頻域或時域表
示，切勿同時包含頻域與時域。
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EX 9.5 Adding Cosines Using Phasors
If
and
express y = y1 + y2 as a single sinusoidal function.
a) Solve by using trigonometric identities.
b) Solve by using the phasor concept.
sin(  )  sin  cos  cos sin 
cos(  )  cos cos  sin  sin 
a)
b)
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9.4 The Passive Circuit Elements
in the Frequency Domain
The V-I Relationship for a Resistor
時域
頻域
由於電阻本身為一純量，
故跨於電阻兩端之電壓相
量應與通過電阻之電流相
量同相(in phase)。
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The V-I Relationship for an Inductor
時域
V  jω L I
頻域
電感器的電壓相量領先電
流相量90°，或是電流相
量落後電壓相量90°。
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The V-I Relationship for a Capacitor
 - ω CVm sin(ωt  θv )
時域
i  - ω CVmcos(ωt  θv - 90 )
I  jω C V
頻域
電容器的電壓相量落後電
流相量90°，或是電流相
量領先電壓相量90°。
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Impedance and Reactance
Definition of Impedance：
阻抗的定義
V
Z
I
1. 電阻器的阻抗(impedance) 為R， 電感器的阻抗為jL，
電容器的阻抗為1/jC。
2. 阻抗的單位為歐姆，要注意是，雖然阻抗可能是複數，但
它卻不是相量。
3. 阻抗的虛部稱為電抗(reactance) 。
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9.5 The Kirchhoff’s Laws
in the Frequency Domain
KVL in the Frequency Domain
時域
=0
頻域
KCL in the Frequency Domain
時域
頻域
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9.6 The Series, Parallel, and
Delta-to-Wye Simplifications
Combining Impedances in Series and Parallel
等效阻抗等於各阻抗之總和
等效阻抗之倒數等於各阻抗倒數總和
1
1
1
1
 

Z ab Z1 Z 2
Zn
見下一頁之定義
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Addmitance and Susceptance
Definition of Admittance ：
導納的定義
導納
電導(conductance)
電納(susceptance)
1. 電阻器的導納(addimitance) 為電導G， 電感器的導納
為1/jL， 電容器的導納為jC。
2. 導納的單位為西門子(siemens)，導納可能是複數，但
不是相量。
3. 導納的虛部稱為電納(susceptance) 。
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EX 9.6 Combining Impedances in Series
750 cos(5000t + 30◦) V
a) Construct the frequency-domain equivalent circuit.
b) Calculate the steady-state current i by the phasor method.
a)
b)
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EX 9.7 Combining Impedances in Series and in Parallel
8 cos200,000t A
a) Construct the frequencydomain equivalent circuit.
b) Calculate the steady-state v, i1,
i2, and i3 by the phasor method.
ω  200000 rad/s
jω L  j8
1
 -j5
jω C
a)
b)
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Delta-to-Wye Transformations
Y
Y
阻抗之Δ-Y 轉換公式與電
阻之Δ-Y 轉換關係相似。
請參考3.7節及問題3.61。
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EX 9.8 Using a Delta-to-Wye Transform in the Frequency Domain
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EX 9.8 Using a Delta-to-Wye Transform (Contd.)
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9.7 The Source Transformations and
Thévenin-Norton Equivalent Circuits
在頻域的戴維寧─諾頓等
效電路之計算方法與電源
轉換之觀念與純電阻電路
相同，除了將等效電阻用
一阻抗替代。
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EX 9.9 Performing Source Transformations
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EX 9.10 Finding a Thévenin Equivalent
KVL
Also,
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EX 9.10 Finding a Thévenin Equivalent (Contd.)
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9.8 The Node-Voltage Method
EX 9.11 Using the Node-Voltage Method in the Freq. Domain
節點1：
節點2：
控制變數：
節點1：
節點2：
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9.9 The Mesh-Current Method
EX 9.12 Using the Mesh-Current Method in the Freq. Domain
網目1：
網目2：
控制變數：
網目1：
網目2：
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9.10 The Transformer
The Analysis of a Linear Transformer Circuit
一次繞組(primary winding)
 連接至電源端
二次繞組(secondary winding)
 連接至負載端
一次繞組自感值(L1)
二次繞組自感值(L2)
互感值(M)
一次繞組電阻值(R1)
二次繞組電阻值(R2)
Let
阻抗Zab 與變壓器之磁極性無關
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Reflected Impedance
反射阻抗(reflected impedance, Zr) ：
變壓器二次側繞組及負載阻抗反射到一次側之等效阻抗
Zr 
ω2 M 2
2
Z 22

22
Z
線性變壓器將二次側自阻抗的共軛值反射至
一次側，且乘上一常數倍。
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EX 9.13 Analyzing a Linear Transformer (Frequency Domain)
The parameters of a certain linear transformer are R1 = 200 , R2 = 100 ,
L1 = 9 H, L2 = 4 H, and k = 0.5. The transformer couples an impedance
consisting of an 800  resistor in series with a 1 µF capacitor to a
sinusoidal voltage source. The 300 V (rms) source has an internal
impedance of 500 + j 100  and a frequency of 400 rad/s.
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EX 9.13 Analyzing a Linear Transformer (Contd.)
c
Open circuit:
Thevenin’s Equiv.
d
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9.11 The Ideal Transformer
理想變壓器(ideal transformer) 的特性：
1. 耦合係數k = 1。
2. 每一線圈的自感值為無限大(L1 = L2 = ∞)。
3. 因寄生電阻產生之線圈損失可忽略不計。
Exploring Limiting Values
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Exploring Limiting Values (Contd.)
k  1 L1   N1 
L2  N 2 
2
Terminal Behavior of the
ideal transformer
同理
理想變壓器：
1. 各線圈之每匝伏特數之
絕對值相等，即
2
 N1 
 R2  RL  jX L 
Z ab  R1  
 N2 
Reflected Impedance
2. 各線圈之安培-匝數之
絕對值相等，即
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Determining the Voltage and Current Ratios
&
理想變壓器的電壓關係
理想變壓器的電流關係
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Determining the Polarity of the Voltage and Current Ratios
理想變壓器的黑點規則(Dot Convention)
1. 當線圈電壓V1 及V2 在黑點端同為正或負時，
採正號，否則採負號。
2. 當線圈電流I1 及I2 同為流入或流出黑點端時，
採負號，否則採正號。
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Three ways to show the Turns Ratio
匝數比(Turns Ratio):
For a = 5
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EX 9.14 Analyzing an Ideal Transformer Ckt. (Frequency Domain)
If vg = 2500 cos 400t V, find the steady-state expressions for
(a) i1 ; (b) v1 ; (c) i2 ; and (d) v2 .
  2500 rad/s
&
a)
Also,
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EX 9.14 (Contd.)
b)
c)
d)
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The Use of an Ideal Transformer for Impedance Matching
將理想變壓器二次側負載阻抗反射至一次側時，須乘上1/a2。
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9.12 Phasor Diagrams
A graphic representation of phasors.
A phasor diagram shows the magnitude and phase angle of
each phasor quantity in the complex-number plane.
Phase angles are measured counterclockwise from the
positive real axis, and magnitudes are measured from the
origin of the axes.
Two different magnitude scales are necessary, one for
currents and one for voltages.
The complex number −7 − j 3.
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EX 9.15 Using Phasor Diagrams to Analyze a Circuit
Find the value of R that will cause the current through that resistor,
iR , to lag the source current, is , by 45◦ when  = 5 krad/s.
1
1/R  3  R  Ω
3
By KCL
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EX 9.16 Analyzing Capacitive Loading Effects
Use phasor diagrams to explore
the effect of adding a capacitor
across the terminals of the load
on the amplitude of Vs if we
adjust Vs so that the amplitude of
VL remains constant.
Utility companies use this technique to control the voltage drop on their lines.
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EX 9.16 Analyzing Capacitive Loading Effects (Contd.)
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