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Chap. 8 Natural and Step Responses of RLC Circuits. C ontents. 8.1 Introduction to the Natural Response of a Parallel RLC Circuit 8.2 The Forms of the Natural Response of a Parallel RLC Circuit 8.3 The Step Response of a Parallel RLC Circuit - PowerPoint PPT Presentation
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Chap. 8 Natural and Step Responses of RLC Circuits
Contents8.1 Introduction to the Natural Response of a Parallel RLC Circuit 8.2 The Forms of the Natural Response of a Parallel RLC Circuit 8.3 The Step Response of a Parallel RLC Circuit 8.4 The Natural and Step Response of a Series RLC Circuit 8.5 A Circuit with Two Integrating Amplifiers
Objectives
1.能解決出並聯 RLC電路的自然響應與步階響應。2.能解決出串聯 RLC電路的自然響應與步階響應。
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8.1 The Introduction to the Natural Response of a Parallel RLC Circuit
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KCL:
二階電路second-order circuits
The General Solution of the 2nd-Order DE
Assume that the solution is of exponential form
where A and s are unknown constants.
A 0 &est 0 for any finite st
特性方程式characteristic equation
tsts eAeAv 2121
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The General Solution of the 2nd-Order DE (Contd.)
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s1 和 s2 兩根的性質有三種情況:
1. 當 02 < 2時,兩根為相異實根,響應稱為過阻尼 (overdamped)。
2. 當 02 > 2時,兩根為共軛複數根,響應稱為欠阻尼 (underdamped)。
3. 當 02 = 2時,兩根為相等實根,響應稱為臨界阻尼 (critically damped)。
阻尼類型會影響其響應達到終值(或穩態值)的方式。
複數頻率Complex frequency
奈培頻率
諧振弳頻率
特性根
02 20
2 ωαss
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EX 8.1 Finding the Roots of the Characteristic Equation (Parallel RLC)
R = 200L = 50mHC = 0.2F
02 20
2 ωαss
underdamped
(a)
(b)
(c)
overdamped
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8.2 The Forms of the Natural Response
of a Parallel RLC Circuit
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A. The Overdamped Voltage Response
Find iC(0+) by KCL
Solve A1 and A2
220 αω
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EX 8.2 Finding the Overdamped Natural Response (Parallel RLC)
Initial currents:
KCL
Initial value of dv/dt :
010102.5 842 ss
rad/s 00020 ; rad/s 5000 21 ,-s-s
B. The Underdamped Voltage Response
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Euler identity:
Solve B1 and B2
Find iC(0+) by KCL
Damped radian frequency:
t)coefficien (damping
factor) (damping :
阻尼係數或阻尼因數
220 αω
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EX 8.4 Finding the Underdamped Natural Response (Parallel RLC)
010400 62 ssrad/s 80979 200
rad/s 80979200
2
1
.j--s
.j-s
mA 2512
0
0
0
.-I
V
00 since 00 0 VviRAlso,
dω
C. The Critically Damped Voltage Response
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Solve D1 and D2
Find iC(0+) by KCL
220 αω
EX 8.5 Finding the Critically Damped Natural Response (Parallel RLC)
a) For the circuit in EX 8.4, find the value of R that results in a critically damped voltage response.b) Calculate v(t ) for t ≥ 0.
V/s 98000
0
1
2
D
D
A Summary of the Results (Natural Response)
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特性方程式 : 兩根 :
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8.3 The Step Response of a Parallel RLC Circuit
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KCL:
The Direct Approach
對一具有定值激勵的二階微分方程式,其解為激勵響應加上自然響應之同形式函數。
Final value
The Indirect Approach
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先求電壓 v再求電流 iL
24 mA 400
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EX 8.6 Finding the Overdamped Step Response (Parallel RLC)
0101610 852 ssrad/s 000,08
rad/s 000,20
2
1
-s
-s
000 -LL ii 000 -
CC vv
Also,
The initial energy stored is zero.
0
00
L
v
dt
di CL
220ω 相異兩實根 :2
02 ω
24 mA 625
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EX 8.7 Finding the Underdamped Step Response (Parallel RLC)
000 -LL ii 000 -
CC vv
Also,
The initial energy stored is zero.
0
00
L
v
dt
di CL
01016106.4 842 ss2
20ω
共軛複數根 :20
2 ωrad/s 1042 1023.
rad/s 102.40123.44
2
441
.j--s
j-s dω
24 mA 500
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EX 8.8 Finding the Critically Damped Step Response (Parallel RLC)
000 -LL ii 000 -
CC vv
Also,
The initial energy stored is zero.
0
00
L
v
dt
di CL
01016108 842 ss2
20ω
)( :20
2 重根相同兩實根ω
rad/s 014 421 -ss
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EX 8.9 Comparing the Three-Step Response Forms
過阻尼
欠阻尼
臨界阻尼
74 97 130
90%final
24 mA 500
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EX 8.10 Finding Step Response with Initial Stored Energy (Parallel RLC)
mA 290 Li
V 500 Cv
A/s 2000
1025
5000
3
L
v
dt
di CL
Also, 01016108 842 ss)( :2
02 重根相同兩實根ω
rad/s 014 421 -ss
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8.4 The Natural and Step Response of a Series RLC Circuit
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KVL:
微分
因串聯 RLC 和並聯 RLC 電路皆以微分方程式來描述,所以串聯 RLC 電路的自然響應和步階響應求解過程和並聯 RLC 電路相同。
特性方程式Characteristic Equation
或
奈培頻率Neper Frequency
諧振弳頻率Resonant Radian Frequency
自然響應
20
Step Response of a Series RLC Circuit
KVL:
Also,
步階響應
21
A/s 100010100
1000
30
-L
V
dt
di
Also,
rad/s 9600- 220 dω
EX 8.11 Finding the Underdamped Natural Response of a Series RLC Circuit
共軛複數根 :20
2 ω
22
EX 8.12 Finding the Underdamped Step Response of a Series RLC Circuit
No energy is stored for t < 0.
8.5 A Circuit with Two Integrating Amplifiers
idealideal
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EX 8.13 Analyzing Two Cascaded Integrating Amplifiers
No energy is stored when the input voltage vg jumps instantaneously from 0 to 25 mV.
Let
However,
Two Integrating Amplifiers with Feedback Resistors
idealideal
25
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EX 8.14 Analyzing Two Cascaded Integrating Amplifiers with Feedback Resistors
100 k
500 k100 k
0.1 F
25 k
1 F
VCC1 = VCC2 = 6V
rad/s 10 rad/s; 20 21 -s-s
Since
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