This comprehensive text on Network Analysis and Synthesis is designed for undergraduate students of Electronics and Communication Engineering, Electrical and Electronics Engineering, Electronics and Instrumentation Engineering, Electronics and Computer Engineering and Biomedical Engineering. The book will also be useful to AMIE and IETE students. Buy Now: https://bit.ly/2WmA7is
This comprehensive text on Network Analysis and Synthesis is designed for undergraduate students of Electronics and Communication Engineering, Electrical and Electronics Engineering, Electronics and Instrumentation Engineering, Electronics and Computer Engineering and Biomedical Engineering. The book will also be useful to AMIE and IETE students. Buy Now: https://bit.ly/2WmA7is
It gives how states are representing in various canonical forms and how it it is different from transfer function approach. and finally test the system controllability and observability by kalman's test
This chapter provides complete description of two port network parameters. It also provides relationship between different parameters. Also it provides condition for symmetry and reciprocity.
Z Transform And Inverse Z Transform - Signal And SystemsMr. RahüL YøGi
The z-transform is the most general concept for the transformation of discrete-time series.
The Laplace transform is the more general concept for the transformation of continuous time processes.
For example, the Laplace transform allows you to transform a differential equation, and its corresponding initial and boundary value problems, into a space in which the equation can be solved by ordinary algebra.
The switching of spaces to transform calculus problems into algebraic operations on transforms is called operational calculus. The Laplace and z transforms are the most important methods for this purpose.
Initial and final condition for circuit
Explain the transient response of a RC circuit
As the capacitor stores energy when there is:
a transition in a unit step function source, u(t-to)
or a voltage or current source is switched into the circuit.
Explain the transient response of a RL circuit
As the inductor stores energy when there is:
a transition in a unit step function source, u(t-to)
or a voltage or current source is switched into the circuit.
RC Circuit
RL Circuit
It gives how states are representing in various canonical forms and how it it is different from transfer function approach. and finally test the system controllability and observability by kalman's test
This chapter provides complete description of two port network parameters. It also provides relationship between different parameters. Also it provides condition for symmetry and reciprocity.
Z Transform And Inverse Z Transform - Signal And SystemsMr. RahüL YøGi
The z-transform is the most general concept for the transformation of discrete-time series.
The Laplace transform is the more general concept for the transformation of continuous time processes.
For example, the Laplace transform allows you to transform a differential equation, and its corresponding initial and boundary value problems, into a space in which the equation can be solved by ordinary algebra.
The switching of spaces to transform calculus problems into algebraic operations on transforms is called operational calculus. The Laplace and z transforms are the most important methods for this purpose.
Initial and final condition for circuit
Explain the transient response of a RC circuit
As the capacitor stores energy when there is:
a transition in a unit step function source, u(t-to)
or a voltage or current source is switched into the circuit.
Explain the transient response of a RL circuit
As the inductor stores energy when there is:
a transition in a unit step function source, u(t-to)
or a voltage or current source is switched into the circuit.
RC Circuit
RL Circuit
Stochastic Processes describe the system derived by noise.
Level of graduate students in mathematics and engineering.
Probability Theory is a prerequisite.
For comments please contact me at solo.hermelin@gmail.com.
For more presentations on different subjects visit my website at http://www.solohermelin.com.
Asymptotic Behavior of Solutions of Nonlinear Neutral Delay Forced Impulsive ...IOSR Journals
Sufficient conditions are obtained for every solution of first order nonlinear neutral delay forced
impulsive differential equations with positive and negative coefficients tends to a constant as t ∞.
Mathematics Subject Classification [MSC 2010]:34A37
The purpose of this work is to formulate and investigate a boundary integral method for the solution of the internal waves/Rayleigh-Taylor problem. This problem describes the evolution of the interface between two immiscible, inviscid, incompressible, irrotational fluids of different density in three dimensions. The motion of the interface and fluids is driven by the action of a gravity force, surface tension at the interface, elastic bending and/or a prescribed far-field pressure gradient. The interface is a generalized vortex sheet, and dipole density is interpreted as the (unnormalized) vortex sheet strength. Presence of the surface tension or elastic bending effects introduces high order derivatives into the evolution equations. This makes the considered problem stiff and the application of the standard explicit time-integration methods suffers strong time-step stability constraints.
The proposed numerical method employs a special interface parameterization that enables the use of an efficient implicit time-integration method via a small-scale decomposition. This approach allows one to capture the nonlinear growth of normal modes for the case of Rayleigh-Taylor instability with the heavier fluid on top.
Validation of the results is done by comparison of numeric solution to the analytic solution of the linearized problem for a short time. We check the energy and the interface mean height preservation. The developed model and numerical method can be efficiently applied to study the motion of internal waves for doubly periodic interfacial flows with surface tension and elastic bending stress at the interface.
Courier management system project report.pdfKamal Acharya
It is now-a-days very important for the people to send or receive articles like imported furniture, electronic items, gifts, business goods and the like. People depend vastly on different transport systems which mostly use the manual way of receiving and delivering the articles. There is no way to track the articles till they are received and there is no way to let the customer know what happened in transit, once he booked some articles. In such a situation, we need a system which completely computerizes the cargo activities including time to time tracking of the articles sent. This need is fulfilled by Courier Management System software which is online software for the cargo management people that enables them to receive the goods from a source and send them to a required destination and track their status from time to time.
Overview of the fundamental roles in Hydropower generation and the components involved in wider Electrical Engineering.
This paper presents the design and construction of hydroelectric dams from the hydrologist’s survey of the valley before construction, all aspects and involved disciplines, fluid dynamics, structural engineering, generation and mains frequency regulation to the very transmission of power through the network in the United Kingdom.
Author: Robbie Edward Sayers
Collaborators and co editors: Charlie Sims and Connor Healey.
(C) 2024 Robbie E. Sayers
TECHNICAL TRAINING MANUAL GENERAL FAMILIARIZATION COURSEDuvanRamosGarzon1
AIRCRAFT GENERAL
The Single Aisle is the most advanced family aircraft in service today, with fly-by-wire flight controls.
The A318, A319, A320 and A321 are twin-engine subsonic medium range aircraft.
The family offers a choice of engines
Immunizing Image Classifiers Against Localized Adversary Attacksgerogepatton
This paper addresses the vulnerability of deep learning models, particularly convolutional neural networks
(CNN)s, to adversarial attacks and presents a proactive training technique designed to counter them. We
introduce a novel volumization algorithm, which transforms 2D images into 3D volumetric representations.
When combined with 3D convolution and deep curriculum learning optimization (CLO), itsignificantly improves
the immunity of models against localized universal attacks by up to 40%. We evaluate our proposed approach
using contemporary CNN architectures and the modified Canadian Institute for Advanced Research (CIFAR-10
and CIFAR-100) and ImageNet Large Scale Visual Recognition Challenge (ILSVRC12) datasets, showcasing
accuracy improvements over previous techniques. The results indicate that the combination of the volumetric
input and curriculum learning holds significant promise for mitigating adversarial attacks without necessitating
adversary training.
Saudi Arabia stands as a titan in the global energy landscape, renowned for its abundant oil and gas resources. It's the largest exporter of petroleum and holds some of the world's most significant reserves. Let's delve into the top 10 oil and gas projects shaping Saudi Arabia's energy future in 2024.
Student information management system project report ii.pdfKamal Acharya
Our project explains about the student management. This project mainly explains the various actions related to student details. This project shows some ease in adding, editing and deleting the student details. It also provides a less time consuming process for viewing, adding, editing and deleting the marks of the students.
Vaccine management system project report documentation..pdfKamal Acharya
The Division of Vaccine and Immunization is facing increasing difficulty monitoring vaccines and other commodities distribution once they have been distributed from the national stores. With the introduction of new vaccines, more challenges have been anticipated with this additions posing serious threat to the already over strained vaccine supply chain system in Kenya.
Industrial Training at Shahjalal Fertilizer Company Limited (SFCL)MdTanvirMahtab2
This presentation is about the working procedure of Shahjalal Fertilizer Company Limited (SFCL). A Govt. owned Company of Bangladesh Chemical Industries Corporation under Ministry of Industries.
Hybrid optimization of pumped hydro system and solar- Engr. Abdul-Azeez.pdffxintegritypublishin
Advancements in technology unveil a myriad of electrical and electronic breakthroughs geared towards efficiently harnessing limited resources to meet human energy demands. The optimization of hybrid solar PV panels and pumped hydro energy supply systems plays a pivotal role in utilizing natural resources effectively. This initiative not only benefits humanity but also fosters environmental sustainability. The study investigated the design optimization of these hybrid systems, focusing on understanding solar radiation patterns, identifying geographical influences on solar radiation, formulating a mathematical model for system optimization, and determining the optimal configuration of PV panels and pumped hydro storage. Through a comparative analysis approach and eight weeks of data collection, the study addressed key research questions related to solar radiation patterns and optimal system design. The findings highlighted regions with heightened solar radiation levels, showcasing substantial potential for power generation and emphasizing the system's efficiency. Optimizing system design significantly boosted power generation, promoted renewable energy utilization, and enhanced energy storage capacity. The study underscored the benefits of optimizing hybrid solar PV panels and pumped hydro energy supply systems for sustainable energy usage. Optimizing the design of solar PV panels and pumped hydro energy supply systems as examined across diverse climatic conditions in a developing country, not only enhances power generation but also improves the integration of renewable energy sources and boosts energy storage capacities, particularly beneficial for less economically prosperous regions. Additionally, the study provides valuable insights for advancing energy research in economically viable areas. Recommendations included conducting site-specific assessments, utilizing advanced modeling tools, implementing regular maintenance protocols, and enhancing communication among system components.
1. Network Analysis
Chapter 4
Laplace Transform and Circuit Analysis
Chien-Jung Li
Department of Electronic Engineering
National Taipei University of Technology
2. Derivative
2
y x=
• Derivative
Isaac Newton (牛頓, 1643-1727)
Method of fluxions
流量: fluent
流數: fluxion
2y x x=
i i
2
y
x
x
=
i
i
y對應x的導數(變化率)即兩個流數的比
代表對時間的微分
G. Wilhelm Leibniz (萊布尼茲, 1646-1716)
dy
dx
y
x
∆
= ∆
J-Louis Lagrange (拉格朗日, 1736-1813)
( )y f x= ( )
dy
y f x
dx
′ ′= =
2
y x= 2y x′ =
( )
2
2
d dy d y
y f x
dx dx dx
′′ ′′= = =
2/61 Department of Electronic Engineering, NTUT
3. • Operators do particular mathematical manipulations, such as
Operators
+ − × ÷
( ) ( )
dy d
y f x y
dx dx
′ ′= = = 2 2 3
2
d
x x x
dx
=
( ) ( )
2 2
2 2
d dy d y d
y f x y
dx dx dx dx
′′ ′′= = = =
d
D
dx
≜ 2 2 3
2x Dx x=
• Louis Arbogast (1759-1803) conceived the calculus as operational
symbols. The formal algebraic manipulation of series investigated by
Lagrange and Laplace.
2 5 3
20D x x=
• The derivative
n
n
n
d
D
dx
≜
Differential operator
3/61 Department of Electronic Engineering, NTUT
4. Differential Equations
x
y e= x x xd
y e De e
dx
′ = = = Dy y=
Hence, we can know the solution of the equation should be of the form
x
y e= x
y Ce=
• Oliver Heaviside (1850-1925)
Heaviside was a self-taught English electrical engineer, mathematician, and
physicist who adapted complex numbers to the study of electrical circuits,
invented mathematical techniques to the solution of differential equations
(later found to be equivalent to Laplace transforms. He changed the face of
mathematics and science for years to come.
or
C is a constant
Dy y=
kx
y e= kx kx xd
y e De ke
dx
′ = = = Dy ky=
Similarly, we can know the solution of the equation should be of the
form
Dy ky=
kx
y Ce=
4/61 Department of Electronic Engineering, NTUT
5. Solutions of the Differential Equation
• Consider the differential equation: ( )7 8 0y y y′′ ′+ − =
2
2
7 8 0
d d
y y y
dx dx
+ − =
2
7 8 0D y Dy y+ − =
(將D視為代數)
( )2
7 8 0D D y+ − =
( ) ( )( )2
7 8 1 8 0D D D D+ − = − + =
Use the differential operator D:
Take this as an algebraic equation:
The solutions:
0y = or 1D = 8D = −
Dy y= 8Dy y= − x
y Ae= 6x
y Be−
=
and
roots
6x x
y Ae Be−
= +General Solution:
put y back
and
solutions
and(/or)
5/61 Department of Electronic Engineering, NTUT
6. Fourier Transform
( ) ( ) ω
∞
−
−∞
= ∫
j t
X f x t e dt
The usefulness of the Fourier transform is limited by a series drawback:
if we try to evaluate the Fourier integral, we find that the integral does not
converge for most signals x(t), i.e.,
( ) 0sinx t tω=
( ) 00
sin j t
X f t e dtω
ω
∞
−
= ⋅∫
( ) 0 0 0
0 2 20
0 0
sin cos
sin
j t j t
j t j e t e t
X t e dt
ω ω
ω ω ω ω ω
ω ω
ω ω
∞
− −
∞
− − −
= ⋅ =
−∫
( ) 0 0
2 2
0
sin cosj j
j e e
X
ω ω ω
ω
ω ω
− ∞ − ∞
− ∞ − ∞ +
=
−
But what are andsin∞ cos ?∞
• The Fourier Transform
6/61 Department of Electronic Engineering, NTUT
7. Circumvent the Problem
The evaluation of the Fourier integral would be simpler if the function x(t)
would approach zero for every large values of t. The solution of the Fourier
integral becomes possible if x(t) is multiplied by a damping function ,
where σ is a positive real number.
( ) ( )0
, t j t
X f x t e e dtσ ω
σ
∞
− −
′ = ⋅ ∫
( ) 00
, sin t j t
X f t e e dtσ ω
σ ω
∞
− −
′ = ⋅ ⋅ ∫
( )
( ) ( ) ( )
( )
0 0 0
22
0 0
sin cos
,
j t j t
j e t e t
X
j
σ ω σ ω
σ ω ω ω ω
ω σ
ω σ ω
∞
− + − +
− + −
′ =
+ +
• The Laplace transform offers a way to circumvent this problem.
t
e σ−
( )
( ) ( ) ( )
( )
( )
( )
0 0
0 0
2 22 2
0 0
sin cos sin0 cos0
,
j j
j e e j e e
X
j j
σ ω σ ω
σ ω ω σ ω ω
ω σ
ω σ ω ω σ ω
− + ∞ − + ∞
− + ∞ − ∞ − + −
′ = −
+ + + +
( )
( )
0
22
0
,X
j
ω
ω σ
ω σ ω
′ =
+ +
For ( ) 0sinx t tω=
( ) ( )
0
j t j
t
e e
σ ω σ ω− + − + ∞
=∞
= →
7/61 Department of Electronic Engineering, NTUT
8. The Fourier Transform of the Sinusoid
( )
( )
0
22
0
,X
j
ω
ω σ
ω σ ω
′ =
+ +
• The Fourier integral X(f) is now obtained by letting
( ) 0
2 2
0
X
ω
ω
ω ω
=
−
Thus the Fourier transform of any function x(t) is obtained by first introducing
a damping function evaluating the integral for and finally
letting .
( )- 0
0 1t
e eσ
σ → =≃
- t
e σ
0σ >
0σ →
• The Fourier integral is obtained with x(t) multiplied by a damping
function:
( ),X ω σ′
8/61 Department of Electronic Engineering, NTUT
9. Definition of the Laplace Transform
( ) ( )0
, t j t
X f x t e e dtσ ω
σ
∞
− −
′ = ⋅ ∫
( ) ( ) ( )
0
,
j t
X f x t e dt
σ ω
σ
∞ − +
′ = ⋅∫
• Define s jσ ω= + , where s is called the complex frequency
( ) ( ) ( ),X f X j X sσ σ ω′ = + =
( ) ( )0
st
X s x t e dt
∞
−
= ⋅∫
This is the definition of
the Laplace transform
• As we said, the evaluation of the Fourier integral would be simpler if
the function x(t) is multiplied by a damping function , where σ is a
positive real number.
t
e σ−
• The Fourier transform can be obtained by letting , i.e.
( ) ( )s j
X X s ω
ω =
=
0σ → s jω=
9/61 Department of Electronic Engineering, NTUT
12. Definition of The Laplace Transform
Department of Electronic Engineering, NTUT
( ) ( )f t F s = L
( ) ( )-1
F s f t = L
( ) ( )
0
st
F s f t e dt
∞
−
= ∫
• The mathematical definition of the Laplace transform is
One-sided or Unilateral Laplace transform
• The process of transformation is indicated symbolically as
• The process of inverse transformation is indicated symbolically as
12/61
13. Basic Theorems of Linearity
( ) ( ) ( )Kf t K f t KF s = = L L
( ) ( ) ( ) ( ) ( ) ( )1 2 1 2 1 2f t f t f t f t F s F s + = + = + L L L
( ) ( ) ( ) ( )1 2 1 2f t f t F s F s ⋅ ≠ ⋅ L
• Consider the Laplace transform of a function f(t) is
( ) ( )f t F s = L
Let K represent an arbitrary constant, we have
Let f1(t) and f2(t) represent any arbitrary functions, we have
• It is mentioned here that
Department of Electronic Engineering, NTUT13/61
14. Step Function
( ) 0 for 0u t t= <
1 for 0t= >
( ) ( )U s u t = L
• The unit step function u(t) can be used to describe the process of
“turning on” a DC level at t=0.
1
t
( )u t
( ) ( ) 0 for 0u t Ku t t′ = = <
for 0K t= >
K
t
( )u t′
( ) ( )U s u t′ ′ = L
0
1st
e dt
s
∞
−
= =∫ 0
st K
Ke dt
s
∞
−
= =∫
Department of Electronic Engineering, NTUT14/61
15. Exponential Function
( ) 0
t t st
X s e e e dtα α
∞
− − −
= = ∫L
• The Laplace transform of the exponential function .
for 0α >1
t
( ) t
x t e α−
=
for 0α <
1
t
( ) t
x t e α−
=
( )
0
1s t
e dt
s
α
α
∞ − +
= =
+∫
t
e α−
Department of Electronic Engineering, NTUT15/61
16. Sine and Cosine Functions
• The Laplace transform of the sinusoid .0cos tω
( )
0 0
0 0
0
0 0 0 0
1
sin
2 2
j t j t
j t j tst st st ste e
F s te dt e dt e e dt e e dt
j j
ω ω
ω ω
ω
∞ ∞ ∞ ∞−
−− − − −
−
= = = −
∫ ∫ ∫ ∫
( ) ( )
( )
( )
( )
( )0 0 0 0
0 00 0 0 0
1 1 1 1
2 2
s j t s j t s j t s j t
e dt e dt e e
j j s j s j
ω ω ω ω
ω ω
∞ ∞
∞ ∞
− − − + − − − +
= − = −
− − − +
∫ ∫
( ) ( )
( )0 0 0
2 2 2 2
0 0 0 0
1 1 1 1
2 2
s j s j
j s j s j j s s
ω ω ω
ω ω ω ω
+ − −
= − = =
− + + +
( )
0 0
0 0
0
0 0 0 0
1
cos
2 2
j t j t
j t j tst st st ste e
F s te dt e dt e e dt e e dt
ω ω
ω ω
ω
∞ ∞ ∞ ∞−
−− − − −
+
= = = +
∫ ∫ ∫ ∫
( ) ( )
( )
( )
( )
( )0 0 0 0
0 00 0 0 0
1 1 1 1
2 2
s j t s j t s j t s j t
e dt e dt e e
s j s j
ω ω ω ω
ω ω
∞ ∞
∞ ∞
− − − + − − − +
= + = +
− − − +
∫ ∫
( ) ( )
( )0 0
2 2 2 2
0 0 0 0
1 1 1 1
2 2
s j s j s
s j s j s s
ω ω
ω ω ω ω
+ + −
= + = =
− + + +
• The Laplace transform of the sinusoid .0sin tω
Department of Electronic Engineering, NTUT16/61
17. Damped Sinusoidal Functions
( ) ( )
0 0
s tt st
F s e e dt e dt
αα
∞ ∞
− +− −
= =∫ ∫
( )
( )
( )0
1 1s t
e
s s
α
α α
∞
− +
= =
− + +
1
( )0cos where 0t
e tα
ω α−
>
t
e α−
( ) ( )
( )
0 0 2 2
0 0 0
cos cos
s tt st s
F s te e dt te dt
s
αα α
ω ω
α ω
∞ ∞
− +− − +
= = =
+ +
∫ ∫
( ) ( )
( )
0
0 0 2 2
0 0 0
sin sin
s tt st
F s te e dt te dt
s
αα ω
ω ω
α ω
∞ ∞
− +− −
= = =
+ +
∫ ∫
• The Laplace transform of the damped function .t
e α−
• The Laplace transform of the damped sine function .0sint
e tα
ω−
• The Laplace transform of the damped cosine function .0cost
e tα
ω−
Department of Electronic Engineering, NTUT17/61
18. Transformation Pairs Encountered in Circuit Analysis
1 ( )u t
1
s
t
e α− 1
s α+
0sin tω
0cos tω
0sint
e tα
ω−
( )
2 2
0
s
s
α
α ω
+
+ +
2
1
s
t
1
!
n
n
s +
n
t
t n
e tα−
( )
1
!
n
n
s α
+
+
( )tδ 1
( )f t ( ) ( )F s f t = L
or
0
2 2
0s
ω
ω+
2 2
0
s
s ω+
( )
0
2 2
0s
ω
α ω+ +
0cost
e tα
ω−
( )f t ( ) ( )F s f t = L
Department of Electronic Engineering, NTUT18/61
19. Transform Example
( ) 4 2
10 5 12sin3 4 cos5t t
f t e t e t− −
= + + +
( )
( )
( )
22 2 2
4 210 5 12 3
4 3 2 5
s
F s
s s s s
+⋅
= + + +
+ + + +
( )
2 2
4 210 5 36
4 9 4 29
s
s s s s s
+
= + + +
+ + + +
• Find the Laplace transform of f(t):
<Sol.>
( ) 4 2
10 5 12sin3 4 cos5t t
f t e t e t− −
= + + +
constant damped sinusoid damped cosine
Department of Electronic Engineering, NTUT19/61
20. OperationOperationOperationOperation f(t)f(t)f(t)f(t) F(s)F(s)F(s)F(s)
Laplace Transform Operations
( )f t′ ( ) ( )0sF s f−
( )0
t
f t dt∫
( )F s
s
( )t
e f tα−
( )F s α+
( ) ( )f t T u t T− − ( )sT
e F s−
( )0f ( )lim
s
sF s
→∞
( )lim
t
f t
→∞
( )0
lim
s
sF s
→
poles of sF(s) must be in left-hand
half-plane. (stable)
Differentiation
Integration
Multiplication by
t
e α−
Time shifting
(Frequency shifting)
Initial value theorem
(初值定理)
Final value theorem
(終值定理)
Department of Electronic Engineering, NTUT20/61
21. Differentiate Operation
( ) ( )
0
st
F s f t e dt
∞
−
= ∫
• If the Laplace transform of f(t) is F(s), prove that Laplace transform of
f’(t) is sF(s)-f(0).
( )
( )
0 0
st stdf t
f t e dt e dt
dt
∞ ∞
− −
′ =∫ ∫
Let ,st
u e−
= and
( )df t
dt dv
dt
= ( )v f t=
( )
( ) ( ) ( ) ( ) ( )0 00
0
st st st st stdf t
e dt f t e f t de f t e s f t e dt
dt
∞
∞ ∞∞
− − − − −
= − = − − ⋅
∫ ∫ ∫
( ) ( ) ( ) ( ) ( )0
0
0 0s s st
f e f e s f t e dt sF s f
∞
− ⋅∞ − ⋅ −
= ∞ − + = − ∫
Department of Electronic Engineering, NTUT21/61
22. Use Differential Operation to Find the Transform of Sinusoid
( ) 0
2 2
0
F s
s
ω
ω
=
+
( ) 0sinf t tω=
( ) 0 0cosf t tω ω′ =
( ) ( ) ( ) 0
2 2
0
0 0
s
f t sF s f
s
ω
ω
′ = − = − +
L
[ ] [ ] 0
0 0 0 0 2 2
0
cos cos
s
t t
s
ω
ω ω ω ω
ω
= =
+
L L
[ ] 2 2
cos
s
t
s
ω
ω
=
+
L
• Let and its Laplace transform is know as .
Find the Laplace transform of by using the differential
operation.
0cos tω
( ) 0sinf t tω=
Department of Electronic Engineering, NTUT22/61
23. Use Integral Operation to Find the Transform of Ramp Function
( ) 1f t = ( )
1
F s
s
=
( ) ( )0 0
1
t t
x t f t dt dt t= = =∫ ∫
( ) ( )
( )
20
1 1 1t F s
x t f t dt
s s s s
= = = =
∫L L
( ) ( ) [ ] 2
1
X s x t t
s
= = = L L
( ) 1f t =• Let and its Laplace transform is know as .
Find the Laplace transform of the ramp function by using the
integral operation.
( )x t t=
( )
1
F s
s
=
Department of Electronic Engineering, NTUT23/61
24. Degree or Order
( )
( )
( )
N s
F s
D s
=
( ) 1
1 0
n n
n nN s a s a s a−
−= + + +⋯
( ) 1
1 0
m m
m mD s b s b s b−
−= + + +⋯
( )
( )( )( )
5 4 3 2
2 2
19 160 1086 3896 8919 10440
4 9 4 29
s s s s s
F s
s s s s s
+ + + + +
=
+ + + +
order = m (means m roots)
order = n (means n roots)
• Most transforms of interest in circuit analysis turn out to be expressed
as ratios of polynomials in the Laplace variables s. Define the
transform function F(s) as
Numerator polynomial
Denominator polynomial
Department of Electronic Engineering, NTUT24/61
25. Zeros and Poles
( ) 0zN s = ( ) 0zF s =
( ) 0pD s = ( )pF s = ∞
( )
( )
( )
N s
F s
D s
=• For
ZerosZerosZerosZeros of F(s): roots of the numerator polynomial N(s)
PolesPolesPolesPoles of F(s): roots of the denominator polynomial D(s)
( ) 1
1 0
m m
m mD s b s b s b−
−= + + +⋯Denominator polynomial
can also be completely specified by its roots except for a constant
multiplier mb in factored form as
( ) ( )( ) ( )1 2m mD s b s s s s s s= − − −⋯
Department of Electronic Engineering, NTUT25/61
26. Classification of Poles
• The poles can be classified as either real, imaginary, or complex.
• An imaginary or complex pole is always accompanied by its complex
conjugate, i.e., jy is accompanied (-jy) and (x+jy) is with (x-jy).
x jy x+jy
• The poles can also be classified according to their order, which is the
number of times a roots is repeated in the denominator polynomial
(重根).
• The first-order (or simple-order) root is which the root appears only
once. Higher-order roots are referred to as multiple-order roots.
• The roots of second-order equations may be either real or complex. For
third- and higher-degree equations, numerical methods must often be
used.
Department of Electronic Engineering, NTUT26/61
27. Example – Inverse Laplace Transform
( ) 2
10 15 20
3
F s
s s s
= + +
+
( ) 3
10 15 20 t
f t t e−
= + +
( ) 2
8 30
25
s
F s
s
+
=
+
( ) 2 2 2 2
5
8 6
5 5
s
F s
s s
= +
+ +
( ) 8cos5 6sin5f t t t= +
( )
( )
( )
( )
( ) ( )
2 2 22 2 2
2 3 20 32 26 5
2 4
6 34 3 25 3 5 3 5
s ss
F s
s s s s s
+ + ++
= = = +
+ + + + + + + +
( ) 3 3
2 cos5 4 sin5t t
f t e t e t− −
= +
• Find the inverse Laplace transform of F(s):
<Sol.>
• Find the inverse Laplace transform of F(s):
<Sol.>
• Find the inverse Laplace transform of F(s):
<Sol.>
( ) 2
2 26
6 34
s
F s
s s
+
=
+ +
Department of Electronic Engineering, NTUT27/61
28. Example – Classify the Poles
( )
( )
( )( )( )( )2 2 2 2
3 2 16 6 34 8 16
N s
F s
s s s s s s s s
=
+ + + + + + +
9 poles 1 2 9, , ,s s s⋯
1 0s = 2
3 9 8
1
2
s
− + −
= = − 3
3 9 8
2
2
s
− − −
= = −
4 4s j= + 5 4s j= −
6
6 36 136
3 5
2
s j
− + −
= = − − 7 3 5s j= − +
8
8 64 64
4
2
s
− + −
= = − 9
8 64 64
4
2
s
− − −
= = −
Real, 1st order (3 poles): s1, s2, s3
Imaginary, 1st order (2 poles): s4, s5
Complex, 1st order (2 poles): s6, s7
Real, 2nd order (2 poles): s8, s9
• Given F(s):
and where N(s) is not specified but known that no roots coincide with
those of D(s). Classify the poles.
<Sol.>
Department of Electronic Engineering, NTUT28/61
29. Inverse Laplace Transform Step 1
• For the purpose of inverse transformation, poles will be classified in 4
categories
1. First-order real poles
2. First-order complex poles
3. Multiple-order real poles
4. Multiple-order complex poles
(purely imaginary poles will be considered as a special case of complex poles with zero real part)
• Step 1: Check Poles
( )
( )( )2 2
50 75
3 2 4 20
s
F s
s s s s
+
=
+ + + +
( )( )( )2
50 75
1 2 4 20
s
s s s s
+
=
+ + + +
real real complex conjugate
Department of Electronic Engineering, NTUT29/61
30. Inverse Laplace Transform Step 2 (I)
• Step 2: Partial Fraction Expansion
( )
( )( )( )2
50 75
1 2 4 20
s
F s
s s s s
+
=
+ + + +
( ) ( ) ( )
1 2 1 2
2
1 2 4 20
A A B s B
s s s s
+
= + +
+ + + +
A1,A2,B1,B2 are constants to be determined.
• Note that the single-pole denominator terms require only a constant in the
numerator, but the quadratic term requires a constant plus a term
proportional to s.
• Various procedures exist for determining the constants, but the results can
always be checked by combining back over a common denominator of
necessary to see if the original function is obtained.
Department of Electronic Engineering, NTUT30/61
31. Inverse Laplace Transform Step 2 (II)
( )
( ) ( ) ( )
1 2 1 2
2
1 2 4 20
A A B s B
F s
s s s s
+
= + +
+ + + +
( ) ( )2 2
1 2 sin 4t t t
f t Ae A e Be t θ− − −
= + + +
B and are determined form B1 and B2.
• A first-order real pole corresponds to an exponential time response
term.
• A quadratic factor with complex poles corresponds to a damped
sinusoidal time response term. Said differently, a pair of complex
conjugate poles corresponds to a damped sinusoidal time response
term.
• The inverse transform of :
θ
Department of Electronic Engineering, NTUT31/61
32. General Algorithm – Find Coefficients
( )
( )( ) ( )( )( ) ( )
1
2 2 2
1 2 1 1 2 2r r rk R R
A
F s
s s s s a s b s a s b s a s bα α α
=
+ + + + + + + + +⋯ ⋯
1 2, , ,r r rkα α α− − −⋯ ( ) ( ) ( )1 1 2 2, , , R Rj j jα ω α ω α ω− ± − ± − ±⋯
( ) ( ) ( ) ( ) ( ) ( ) ( )1 2 1 2e e ek s s sRf t f t f t f t f t f t f t= + + + + + + +⋯ ⋯
( ) kt
ek kf t A e α−
=
( ) ( ) k
k k s
A s F s α
α =−
= +
( ) ( )sinRt
sR R R Rf t B e tα
ω θ−
= +
( ) ( )21R
R R
j
R R R R rR
R s j
B e B s a s b F sθ
α ω
θ
ω =− +
= ∠ = + +
• Give F(s) in the factored form:
• The corresponding inverse transform:
coeff. coeff.
Department of Electronic Engineering, NTUT32/61
33. Example
( ) 2
6 42
7 10
s
F s
s s
+
=
+ +
( )
( )
( )( )
6 7
2 5
s
F s
s s
+
=
+ +
( ) ( ) ( ) 2 5
1 2 1 2
t t
e ef t f t f t Ae A e− −
= + = +
( ) ( )
( )
( )
( )
( )1 2
2
6 7 6 2 7
2 10
5 2 5s
s
s
A s F s
s=−
=−
+ − +
= + = = =
+ − +
( ) ( )
( )
( )
( )
( )2 5
5
6 7 6 5 7
5 4
2 5 2s
s
s
A s F s
s=−
=−
+ − +
= + = = = −
+ − +
( ) 2 5
10 4t t
f t e e− −
= −
• Find the inverse Laplace transform of F(s):
<Sol.>
Department of Electronic Engineering, NTUT33/61
34. Example
( )
2
3 2
10 42 24
4 3
s s
F s
s s s
+ +
=
+ +
( )
( )( )
2
10 42 24
1 3
s s
F s
s s s
+ +
=
+ +
( ) 3
1 2 3
t t
f t A A e A e− −
= + +
( )( )
2
1
0
10 42 24 24
8
1 3 1 3s
s s
A
s s =
+ +
= = =
+ + ⋅
( )
2
2
1
10 42 24 10 42 24
4
3 1 2s
s s
A
s s =−
+ + − +
= = =
+ − ⋅
( ) ( )
2
3
3
10 42 24 90 126 24
2
1 3 2s
s s
A
s s =−
+ + − +
= = = −
+ − ⋅ −
( ) 3
8 4 2t t
f t e e− −
= + −
• Find the inverse Laplace transform of F(s):
<Sol.>
Department of Electronic Engineering, NTUT34/61
35. Example
( )
( )
( )( )2
20 2
1 2 5
s
F s
s s s s
+
=
+ + +
( ) ( ) ( ) ( ) ( )1 2 1 2 sin 2t t
e e sf t f t f t f t A A e Be t θ− −
= + + = + + +
( )
( )( )1 2
0
20 2 20 2
8
1 51 2 5
s
s
A
s s s
=
+ ⋅
= = =
⋅+ + +
( )
( )2 2
1
20 2 20 1
5
1 42 5
s
s
A
s s s
=−
+ ⋅
= = = −
− ⋅+ +
( ) ( )
( )
( )
( )
( )( )
2
1 2 1 2
20 2 10 1 21 1
2 5
2 2 1 1 2 2s j s j
s j
B s s F s
s s j j
θ
=− + =− +
+ +
∠ = + + = =
+ − +
( )
( )( )
10 2.2361 63.435
5 143.13
2.2361 116.565 2 90
⋅ ∠
= = ∠ −
∠ ∠
( ) ( )8 5 5 sin 2 143.13t t
sf t e e t− −
= − + −
• Find the inverse Laplace transform of F(s):
<Sol.>
Department of Electronic Engineering, NTUT35/61
36. Example
( )
( )( )2 2
100
4 2 10
s
F s
s s s
=
+ + +
( ) ( ) ( )1 2s sf t f t f t= +
( ) ( )0
1 1 1sin 2t
sf t B e t θ−
= + ( )1 1sin 2B t θ= +
( ) ( )2 2 2sin 3t
sf t B e t θ−
= +
( ) ( )
( )
( )
2
1 1 2
2 2
50 21 1 100
4
2 2 2 10 4 10 4s j s j
js
B s F s
s s j
θ
= =
∠ = + = =
+ + − + +
( ) ( )
( )
( )
2
2 2 22
1 3 1 3
33.33 1 31 1 100
2 10 14.6176 127.875
3 3 4 1 3 4s j s j
js
B s s F s
s j
θ
=− + =− +
− +
∠ = + + = = = ∠ −
+ − + +
• Find the inverse Laplace transform of F(s):
<Sol.>
100 90
13.8675 56.3099
7.2111 33.6901
∠
= = ∠
∠
( ) ( ) ( )13.8675sin 2 56.3099 14.6176 sin 3 127.875t
sf t t e t−
= + + −
Department of Electronic Engineering, NTUT36/61
37. Inverse Transform of Multiple-order Poles (I)
• Algorithm for Multiple-order Real Poles
( )
( )
( )
i
Q s
F s
s α
=
+
( ) ( ) ( )
i
Q s s F sα= +
roots@i s α= −
The time function due to the pole of order i with a value will be
the form:
α−
( )
( ) ( ) ( )
1 2
1 2
1 ! 2 ! !
i i i k
tk
m i
C t C t C t
f t C e
i i i k
α
− − −
−
= + + + + +
− − −
⋯ ⋯
A give coefficient Ck can be determined from the expression:
( )
( )
1
1
1
1 !
k
k k
s
d
C Q s
k ds α
−
−
=−
=
−
Department of Electronic Engineering, NTUT37/61
38. Inverse Transform of Multiple-order Poles (II)
• For a first-order real pole:
( )
0 1
1
1
0!
t t
m
C t
f t e C eα α
−
− −
= =
( ) ( ) ( )1
1
0! s s
C Q s s F sα α
α=− =−
= = +
• For second-order real poles:
( ) ( ) ( )
2
Q s s F sα= +
( ) ( )1 2
t
mf t C t C e α−
= +
( )1 s
C Q s α=−
=
( )
2
s
dQ s
C
ds α=−
=
Department of Electronic Engineering, NTUT
For complex poles, find the complex coefficients Ck with s jα ω= − ±
38/61
39. s-domain Circuit Analysis
Time domain circuit for which a general solution is desired
Convert circuit to s-domain
form
Solve for desired response in
s-domain
Determine inverse transform
of desired response
Desired time domain response
Department of Electronic Engineering, NTUT39/61
40. Transform Impedances (I)
Passive
RLC Circuit
( )i t
( )v t Z(s)
( )I s
( )V s
s-domain
( ) ( )I s i t = L
( ) ( )V s v t = L
( )
( )
( )
V s
Z s
I s
=
( )
( )
( )
( )
1 I s
Y s
Z s V s
= =
Transform impedance Z(s)
Transform admittance Y(s)
Department of Electronic Engineering, NTUT40/61
41. Transform Impedances (II)
( )i t
( )v t R
( )i t
( )v t
( )i t
( )v t
C
L
( )V s
( )I s
R
1
sC
( )V s
( )I s
sL( )V s
( )I s
R
C
L
( )V ω
( )I ω
R
ω
1
j C
( )V ω
( )I ω
ωj L( )V ω
( )I ω
0σ →
0σ →
0σ →
Department of Electronic Engineering, NTUT41/61
42. Example – Transform Impedance
1 kR = Ω
0.5 FC µ=
30 mHL =
1 kR =
( )
6
6
1 1 2 10
0.5 10sC ss −
⋅
= =
⋅
( )3
30 10 0.03sL s s−
= ⋅ =
Department of Electronic Engineering, NTUT42/61
43. Models for Initially Charged Capacitor
0VC
+
−
0V
s
1
sC
+
−
0CV1
sC
0 60 VV =0.2 FC µ=
+
−
60
s
6
50 10
s
⋅
+
−
6
12 10−
⋅
6
50 10
s
⋅
Thevenin’s equivalent circuit Norton’s equivalent circuitInitially charged capacitor
Example:
Department of Electronic Engineering, NTUT43/61
44. Models for Initially Fluxed Inductor
0IL
0LI
sL
+
−
0I
s
sL
0 0.4 AI =50 mHL =
0.02
0.05s
+
−
0.4
s
0.05s
Thevenin’s equivalent circuit Norton’s equivalent circuitInitially fluxed inductor
Department of Electronic Engineering, NTUT44/61
45. Complete Circuit Models
1. Transform the complete circuit from the time-domain to the s-domain
( ) ( )v t V s→ ( ) ( )i t I s→
L sL→
1
C
sC
→
0
0
V
V
s
→ 0
0
I
I
s
→
2. Solve for the desired voltages currents using the s-domain model.
3. Using inverse Laplace transform to determine the corresponding time-
domain forms for the voltages or currents of interest.
Department of Electronic Engineering, NTUT45/61
46. Example (I)
20 V+
−10cos3t
+
−
0t =
2 H
1
F
4 3 Ω 4 Ω 5 H
+
−
8 V
1
F
6
( )1i t ( )2i t
20
s
+
−2
10
9
s
s +
+
−
2s
4
s 3 4 5s
8
s
6
s
( )1I s ( )2I s+
−
Transform from time-domain to s-domain
Department of Electronic Engineering, NTUT46/61
47. Example (II)
20
s
+
−2
10
9
s
s +
+
−
2s
4
s 3 4 5s
8
s
6
s
( )1I s ( )2I s+
−
( ) ( ) ( ) ( ) ( )1 1 1 1 22
10 4 6 8
2 3 0
9
s
sI s I s I s I s I s
s s s s
−
+ + + + − + = +
( ) ( ) ( ) ( )2 1 2 2
8 6 20
4 5 10 0I s I s I s sI s
s s s
−
+ − + + + + =
Mesh 1:
Mesh 2:
Solve for I1(s) and I2(s)
Apply the circuit laws:
Department of Electronic Engineering, NTUT47/61
48. General Forms for Solutions (I)
( ) ( ) ( )n fy t y t y t= +
• Natural and Forced Responses
Let represent some arbitrary general circuit response (either a voltage
or current). When the circuit is excited by one or more sources, a general
response may be represented by the sum of two responses as follows:
( )y t
( )y t
Natural response Forced response
• Natural Response
The form of the natural response is determined by the circuit parameter,
i.e., if the circuit has a time constant of 2 seconds, corresponding to an
exponential , such a term will appear in the response when the circuit
is excited by any type of source.
2t
e−
• Force Response
The form of the forced response is determined by the excitation source(s),
i.e., if a circuit is excited by a sinusoid having a frequency of 5 kHz, the
general response will always contain a sinusoid with a frequency of 5 kHz.
Department of Electronic Engineering, NTUT48/61
49. General Forms for Solutions (II)
( ) 10sin1000sv t t=
( ) ( )2
2 3 4sin 1000 30t t
i t e e t− −
= − + +
( ) ( ) ( )4sin 1000 30ss fi t i t t= = +( ) ( ) 2
2 3t t
t ni t i t e e− −
= = −
( ) ( ) ( )t ssy t y t y t= +
• Transient and Steady-state Responses
Transient response Steady-state response
Frequently, transient response and natural response are considered to be
equivalent, and steady-state response and forced response are considered
to be equivalent. The terms transient and steady-state relate to the
common case where the natural response is transient in nature and
eventually vanishes, whereas the forced response continuous as a steady-
state condition indefinitely.
Example:
Department of Electronic Engineering, NTUT49/61
50. First-Order Circuits
• First-order Circuit with Arbitrary Input
( ) t t
ny t Ke Keτ α− −
= =
For first-order circuits, the natural response will always be an exponential
term of the form:
( )ny t
K is a constant
is the time constantτ
is the damping factor1α τ=
Since the exponential term approaches zero as time increases, it is proper to
designate the natural response as a transient response whenever the forced
response continuous indefinitely.
Department of Electronic Engineering, NTUT50/61
51. Example of a First-Order Circuit (I)
40sin4t
+
−
0t =
4 Ω
( )i t
1
F
12
+
−
( )cv t
2
160
16s +
+
−
4 Ω
( )I s
12
s
+
−
( )cV s
( ) ( )2
160 12
4 0
16
I s I s
s s
− + + =
+
( )
( )( )2
40
3 16
s
I s
s s
=
+ +
( ) ( )3
4.8 8sin 4 36.87t
i t e t−
= − + +
( ) ( )
( )( )2
12 480
3 16
CV s I s
s s s
= =
+ +
( ) ( )3
19.2 24sin 4 53.13t
Cv t e t−
= + −
• Use Laplace transform techniques, determine the current and
voltage for .
( )i t
( )Cv t 0t >
Transform from time-domain to s-domain
Apply KVL:
Transient response Steady-state response
Transient response Steady-state response
Time constant = 1/3 = RC
Damping factor = 3
Department of Electronic Engineering, NTUT51/61
52. Example of a First-Order Circuit (II)
( ) ( )
30
6 3 0
1
I s s I s
s
− + + ⋅ =
+
( )
( )( )
10
1 2
I s
s s
=
+ +
( ) 2
10 10t t
i t e e− −
= −
• Use Laplace transform techniques, determine the for .( )i t 0t >
Transform from time-domain to s-domain
Apply KVL:
Forced response Natural response
Time constant = 1/2 = L/R
Damping factor = 2
30 t
e− +
−
0t =
6 Ω
( )i t 3 H
30
1s +
+
−
6
( )I s 3s
Department of Electronic Engineering, NTUT52/61
53. Second-Order Circuits
( ) ( ) ( )
1
0iV
sLI s RI s I s
s sC
− + + + =
( )
2 1
iV
LI s
R
s s
L LC
=
+ +
( ) ( )
2
1
1
i
C
V
LCV s I s
RsC
s s s
L LC
= =
+ +
• Second-order circuits are of special interest because they are capable
of displaying, on a simple scales, the types of responses that appear
in circuits of arbitrary order. In fact, second-order circuits and systems
occur frequently in practical applications, so their behavior is subject
of considerable interest.
• Series RLC Circuit
iV +
−
0t =
R
( )i t
+
−
( )cv t
L
C
iV
s
+
−
R
( )I s
+
−
( )CV s
sL
1
sC
Department of Electronic Engineering, NTUT53/61
54. 3 Possible Forms of the Roots
2
1
2
2
1
2 4
s R R
s L L LC
= − ± −
• The roots of the second-order circuit
• Three possibilities for the roots s1 and s2
Overdamped Case (the roots are real and different) :
2
2
1
4
R
L LC
>
Critically Damped Case (the roots are real and equal) :
2
2
1
4
R
L LC
=
Underdamped Case (the roots are complex) :
2
2
1
4
R
L LC
<
Department of Electronic Engineering, NTUT54/61
55. Overdamped Case (Series RLC)
( )
( )( )1 2
iV
LI s
s sα α
=
+ +
( )
( )( )1 2
i
C
V
LCV s
s s sα α
=
+ +
( ) 1 2
0 0
t t
i t A e A eα α− −
= −
( ) 1 2
1 2
t t
C iv t V Ae A eα α− −
= + +
• In the overdamped case, the two poles are real and different. Assume
that the poles are and , the forms for I(s) and VC(s)
can be expressed as
1 1s α= − 2 2s α= −
• The inverse transforms are of the forms
The natural response consists of two exponential
terms, each having a different damping factor or
time constant. The forced response for the current
is zero and for the capacitor voltage is the
constant final voltage across the capacitor.
Department of Electronic Engineering, NTUT55/61
56. Critically Damped Case (Series RLC)
( )
( )
2
iV
LI s
s α
=
+
( )
( )
2
i
C
V
LCV s
s s α
=
+
( ) 2
0
t Rt LiVt
i t C te e
L
α− −
= =
( ) ( ) 2
1 2
Rt L
C iv t V C t C e−
= + +
• In the critically damped case, the two poles are real and equal.
Assume that the poles are , the forms for I(s) and VC(s)
can be expressed as
1 2s s α= = −
The damping factor:
2
R
L
α =
• The inverse transforms are of the forms
0 iC V L=
The most significant aspect of the natural
response function for the critically damped case is
the form. Although the t factor increases
with increasing t, the decreases at a faster
rate, so the product eventually approaches zero.
t
te α−
t
te α−
Department of Electronic Engineering, NTUT56/61
57. Underdamped Case (Series RLC)
1 ds jα ω= − + 1 ds jα ω= − −
2
R
L
α =
2
1
2
d
R
LC L
ω
= −
( )
( )
2 2
i
d
V
LI s
s α ω
=
+ +
( )
2 1
i
C
V
LCV s
R
s s s
L LC
=
+ +
( ) sinti
d
d
V
i t e t
L
α
ω
ω
−
=
( ) ( )sint
C i dv t V Be tα
ω θ−
= + +
• In this case, the two poles are complex and denoted as
• The forms for I(s) and VC(s) can be expressed as
and
where and
is called the damped
natural oscillation frequency
dω
and
• The inverse transforms are of the forms
The natural response is oscillatory. Depending on
the value of , this response may damp out very
quickly, or it may continue for a reasonable period
of time.
α
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58. Comparison of Response Forms (I)
Overdamped Case :
Critically Damped Case :
Underdamped Case :
( ) 1 2
0 0
t t
i t A e A eα α− −
= −
( ) 2
0
t Rt LiVt
i t C te e
L
α− −
= =
( ) sinti
d
d
V
i t e t
L
α
ω
ω
−
=
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( )i t
t
Underdamped
Critically damped
Overdamped
59. Comparison of Response Forms (II)
Overdamped Case :
Critically Damped Case :
Underdamped Case :
( ) 1 2
1 2
t t
C iv t V Ae A eα α− −
= + +
( ) ( ) 2
1 2
Rt L
C iv t V C t C e−
= + +
( ) ( )sint
C i dv t V Be tα
ω θ−
= + +
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overshoot
Underdamped
Critically damped
Overdamped
Final level= iV
( )Cv t
t
60. Example (I)
40 ViV = +
−
0t =
400 Ω
( )i t
+
−
( )Cv t
2 H
0.5 Fµ
• Use Laplace transform techniques, determine the current and
voltage for .
( )i t
0t >( )Cv t
40
s
+
−
400
( )I s
+
−
( )CV s
2s
6
2 10
s
⋅
( )
6 2 6
2 10 2 400 2 10
2 400
s s
Z s s
s s
⋅ + + ⋅
= + + =
( )
( )
( ) 2 6
40
2 400 2 10
V s sI s
s sZ s
s
= =
+ + ⋅
( )
22 6 2
20 20
200 10 100 994.987s s s
= =
+ + + +
( ) ( )100
0.0201008 sin 994.987t
i t e t−
= ⋅
( ) ( ) ( )
( )
6
2 6
40 10
200 10
V s I s Z s
s s s
⋅
= =
+ +
( )
( )
−
= +
× −
100
40 40.2015
sin 994.987 95.7392
t
Cv t e
t
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61. Example (II)
( ) ( )−
= + ⋅ −100
40 40.2015 sin 994.987 95.7392t
Cv t e t
( ) ( )100
0.0201008 sin 994.987t
i t e t−
= ⋅
Department of Electronic Engineering, NTUT61/61
( ), mAi t( ), VCv t
20
10
, mst
10
20
30
60
40
20
0
−20
Final voltage = 40 V
( )i t
( )Cv t