NETWORK ANALYSIS AND SYNTHESIS

Faculty
ECE (RRC) Dr. Anuradha M (AM), Dr. Vamsi Krishna (VK)
Prof. Sahana S (SHK) Prof. Shashidhara M (SM)
Dr. Ashwini N (AN), Dr. Rashmi NU (RNU)

ECE (ECC) Prof. Prahalad D (PD)

EEE (RRC) Prof. K.R. Pushpa (KRP)
NETWORK ANALYSIS AND SYNTHESIS

Transient Analysis
NETWORK ANALYSIS AND SYNTHESIS
Transients

Transients Analysis:

 Electric circuits will be subjected to sudden changes which may be in the

form of opening and closing of switches or sudden changes in sources
etc.

 Whenever such a change occurs, the circuit which was in a particular

steady state condition will go to another steady state condition.

 Transient analysis is the analysis of the circuits during the time it changes
from one steady state condition to another steady state condition.
NETWORK ANALYSIS AND SYNTHESIS
Transients

Transients Analysis:

 Transient analysis will reveal how the currents and voltages

are changing during the transient period.

 To get such time responses, the mathematical models should

necessarily be a set of differential equations.
NETWORK ANALYSIS AND SYNTHESIS
Transients

RC Transients – Storage Phase and Decay Phase

NETWORK ANALYSIS AND SYNTHESIS
Transients

Transients in Capacitive Networks:

Charging Phase • The instant the switch is
closed, electrons are drawn
from the top plate and
deposited on the bottom
plate by the battery
• This results in a net positive
charge on the top plate and
negative charge on the
bottom plate.
𝑄 = 𝐶𝑉𝑐 = 𝐶𝐸
Basic Charging Network
NETWORK ANALYSIS AND SYNTHESIS
Transients

Transients in Capacitive Networks:

Charging Phase
• At t = 0s, switch is
closed.
• Rapid decay in the
current level shows
that the amount of
charge deposited on
the plates per unit
time is also decaying
rapidly.

𝒊𝒄 during charging phase

NETWORK ANALYSIS AND SYNTHESIS
Transients

Transients in Capacitive Networks:

Charging Phase 𝑞
• 𝑣𝑐 =
𝐶
• Rapid rate with which charge is
initially deposited on the plates
results in a rapid increase in 𝑣𝑐
• As the rate of flow of charge
decreases, the rate of change in
voltage will also decrease.
• The charging phase is passed.

𝒗𝒄 during charging phase

NETWORK ANALYSIS AND SYNTHESIS
Transients

Transients in Capacitive Networks: Charging Phase

Initial status
• For t < 0, vc = 0 V so capacitor acts
as short circuit
• At t = 0, E is connected,
• ic = iR = E/R
• However, vc does not change
instantaneously
• Rather, vc changes with time and
settles to approximately E
Short-circuit equivalent for a capacitor • So, we can find vc(t) for t ≥ 0
(switch closed, t = 0). • Similarly, we can find ic(t) for t ≥ 0
NETWORK ANALYSIS AND SYNTHESIS
Transients

Transients in Capacitive Networks: Charging Phase

NETWORK ANALYSIS AND SYNTHESIS
Transients

Transients in Capacitive Networks: Charging Phase

Final status (Steady state condition)
• Based on the expressions in
previous slide, we can state that
when t > 5 RC, vc(t) ≃ E and ic(t)≃0
• That means charging phase is
completed
• The term RC is referred to as time
constant, denoted by τ
• Example, R = 1 kΩ and C = 1 μF
Open-circuit equivalent for a capacitor gives τ = 1 ms and at t = 5τ, we
following the charging phase. have vc(t) = 0.993E V and ic(t) =
0.0067 (E/R) A
NETWORK ANALYSIS AND SYNTHESIS
Transients

Transients in Capacitive Networks: Charging Phase

Final status (Steady state condition)
NETWORK ANALYSIS AND SYNTHESIS
Transients

Transients in Capacitive Networks: Charging Phase

• If R is kept constant and C is increased

RC will increase and hence convergence
of vc(t) to E takes longer time.
• Capacitance of a network is a measure
of how much it will oppose a change in
voltage across the network.
• Larger the capacitance, larger the time
constant, and longer it takes to charge
up to its final value.
Effect of C on the charging phase.
NETWORK ANALYSIS AND SYNTHESIS
Transients

Transients in Capacitive Networks: Charging Phase

• Voltage across the resistor
is determined by Ohm’s
law:
𝐸 −𝑡
𝑣𝑅 = 𝑖𝑅 𝑅 = 𝑅𝑖𝑐 = 𝑅 𝑒 𝜏
𝑅
𝑡
−𝜏
Or 𝑣𝑅 = 𝐸𝑒

𝒗𝑹 versus t during the charging phase.

NETWORK ANALYSIS AND SYNTHESIS
Transients

Transients in Capacitive Networks:

Charging Phase (Switch changed from position 1 to position 2)

Effect of the leakage current on the steady-state behavior of a capacitor.

NETWORK ANALYSIS AND SYNTHESIS
Transients

Transients in Capacitive Networks: Discharge Phase

• Let us assume a reference t = 0 when

switch moves to 2 after charging phase
• For t < 0, vc = E V so capacitor acts as open
circuit
• At t = 0, capacitor begins discharging phase
via resistance R
• At t = 0, vc does not changeinstantaneously
• Rather, vc changes with time and settles to
0, so find vc(t) for t ≥ 0
• Similarly, find ic(t) for t ≥ 0
NETWORK ANALYSIS AND SYNTHESIS
Transients

Transients in Capacitive Networks: Discharge Phase

Directions of charging current (ic)
and discharging current (iR) are
opposite.
We can plot both charging and
discharging currents using the
corresponding values.
NETWORK ANALYSIS AND SYNTHESIS
Transients

Transients in Capacitive Networks: Discharge Phase

Suppose the switch alternates

between 1 and 2 every 5τ, then
the charging and discharging
behavior alternate as shown in
the figure
NETWORK ANALYSIS AND SYNTHESIS
Transients

Response to adding/removal of a source in the circuit

• Initially the network enters the transient

state
• Currents/voltages are time dependent
• Then the network enters the steady
state
• Currents/voltages are time independent
• The state of the currents / voltages
before the transient state are referred
to as initial conditions (@t=0-)
NETWORK ANALYSIS AND SYNTHESIS
Transients

Initial value and final value

• Let the initial value of the voltage across the capacitor is denoted by
Vi
• Let the final value of the voltage across the capacitor is denoted by
Vf
• Vi depends on the charge stored in the capacitor prior to the
switching action
• The switching action could lead to charging of the capacitor (i.e.,
connecting to a source)
• The switching could lead to discharging of the capacitor (i.e.,
connecting a load resistance across the capacitor)
• Vf depends on the time constant τ = RC and the time before the
next switching action
NETWORK ANALYSIS AND SYNTHESIS
Transients

Summary of charging and discharging phase

• Voltage across the capacitor based on initial and final values

• Other notations:
• Vi = vc(0-) = vc(0) = vc(0+); Vf = vc(∞)
• Similarly, the initial and final values of currents are given by ic(0) = ic(0+)
and ic(∞) respectively
NETWORK ANALYSIS AND SYNTHESIS
Transients

Summary of charging and discharging phase

• Here t = 0- refers to the time t < 0 and it represents the

time interval before switching

• Here t = 0+ refers to the time t > 0 and it represents the time
interval after switching (transient and/or steady state)
NETWORK ANALYSIS AND SYNTHESIS
Transients

Summary of charging phase

• Charging phase: The switch moves to position 1 at t = 0

after being open for a long time. Suppose initial charge
stored by the capacitor be zero
NETWORK ANALYSIS AND SYNTHESIS
Transients

Summary of discharging phase

• Discharging phase: The switch moves to position 2 at t = 0
after being at position 1 for a long time.
NETWORK ANALYSIS AND SYNTHESIS
Transients

Instantaneous values
• Given a value of t and the transient phase (charging or
discharging), the voltage or current at that time t
(referred to as instantaneous value) can be found
• Alternatively, given the value of current or voltage
corresponding to the transient phase (charging or
discharging), the corresponding value of t can be found
NETWORK ANALYSIS AND SYNTHESIS
Transients

Example
• Suppose the initial voltage across the capacitor is 4 V
• Find the expressions for the voltage and current associated with the
capacitor when the switch closes
• Find the time t at which the voltage reaches 21 V.
NETWORK ANALYSIS AND SYNTHESIS
Transients

Example – Solution
• Step 1: Calculate time constant
• τ = RC = (R1+R2)C = 11.22 ms
• Step 2: Determine initial and final values
• Hint: capacitor acts as open circuit after being closed
for a long time.
• At t=0+
• So Vi = Vc(0-) = 4 V;
• So vc(0+) = 4 V
• ic(0-)=(E–Vi)/(R1+R2) = (24-4)/(2.2+1.2 ) K ohm =5.88 mA
• At t=infinity
• ic(∞) = 0 A
• Vf = vc(∞) = E = 24 (open circuit)
NETWORK ANALYSIS AND SYNTHESIS
Transients

Example – Solution
• Step 3: Capacitor voltage and current • Step 4 time when vc(t) = 21 V
NETWORK ANALYSIS AND SYNTHESIS
Transients

Example
• The capacitor was charged to Vi = 10 V. What should be the
value of the resistance such that the capacitor is 90%
discharged in 5 ms?
NETWORK ANALYSIS AND SYNTHESIS
Transients

Example – Solution
• We have the expression below where τ = RC

• Rearranging the above to find R we get

NETWORK ANALYSIS AND SYNTHESIS

Transients-Thevenin’s Equivalent
NETWORK ANALYSIS AND SYNTHESIS
RC Transients – Thevenin’s Equivalent

Application of Thevenin’s theorem to RC transient analysis

• When the given circuit does not have a simple series RC

form, then the network part connected to the capacitor is

replaced with the Thevenin’s equivalent circuit

• The time constant τ = RThC and the Thevenin’s voltage could

be acting as the source excitation.

NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 1:
• Find the mathematical expression for the transient
behavior of the voltage vC and the current iC following the
closing of the switch (position 1 at t = 0s). Initial vC = 0 V
• Find the mathematical expression for the voltage vC and
current iC as a function of time if the switch is thrown into
position 2 at t = 9 ms
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 1 – Solution
• Step 1: Thevenin’s equivalent circuit across terminal 1
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 1 – Solution
• Step 2: Switching to position 1 (charging phase)
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

• At t = 0.009– the value of vC(t) and iC(t)

are given by 5.44 V and 0.052 mA
• When switch moves to position 2 at t = 9
ms, the capacitor voltage does not
change instantaneously so initial value of
Vi for the discharging phase (i.e., C
discharges via R4) is still 5.44 V
• However, the current changes iR
instantaneously
• The final values for discharging phase are
Vf and If are zeros
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 1 – Solution (contd.)

• The new time constant for the discharging phase is given by
τ = R4C = 2 ms
iR
• Hence, the expressions for vc(t) and ic(t) for t ≥ 9 ms are
given by
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 2
• After being zero for a long time, the value of ig(t) changes to
15 mA at t = 0 (and remains at 15 mA as time increases to
infinity).
• Find an expression for vo(t) for t ≥ 0.
• Find the current, iR, in R as a function of time
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 2 – Solution
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 3
• Find v1(t) for t ≥ 0
• Hint: To find RTh find ISC and ETh across the blue terminals
• In other words, find RTh seen by the capacitor
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 3 – Solution

Circuit for t≥0

Step 1: Find initial and final values of v1(t)

Step 2: Find ETh and ISC
Step 3: Find RTh and then τ
Step 4: Plug in the above values into
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 3 – Solution
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 3 – Solution
•To find τ, we need RTh seen by C
•Due to dependent source, we find Thevenin’s voltage and
short circuit current in the place of C given in the circuit
•Finding ETh

Similar to the analysis for t = infinity

NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 3 – Solution
•Finding ISC
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 4
The switch in the circuit is closed on position 1 at t = 0 and
then moved to 2 after one time constant, at t =τ = 250 ms.
Obtain the current for t > 0.
NETWORK ANALYSIS AND SYNTHESIS
Transients – Additional Numericals

Example 5
A capacitor in an RC circuit with R = 25 Ω and C = 50 µF is being charged
with initial zero voltage. What is the time taken for the capacitor voltage
to reach 40 % of its steady state value?
NETWORK ANALYSIS AND SYNTHESIS
Transients – Additional Numericals

Example 6
In an RC circuit, having a time constant of 2.5 ms, the capacitor discharges
with initial voltage of 80 V. (a) Find the time at which the capacitor voltage
reaches 55 V, 30 V and 10 V (b) Calculate the capacitor voltage at time 1.2 ms,
3 ms and 8 ms.
NETWORK ANALYSIS AND SYNTHESIS
Transients – Additional Numericals

Example 7:
(a) Find the values of R and C. (b) Determine the time constant.
(c) At what time the voltage vC(t) will reach half of its initial value?
NETWORK ANALYSIS AND SYNTHESIS
Transients – Additional Numericals

Example 8:
Find the time constant of the RC circuit shown in below.
NETWORK ANALYSIS AND SYNTHESIS

RL Transients – Storage Phase and Decay Phase

NETWORK ANALYSIS AND SYNTHESIS
Transients – Storage phase and decay phase

R-L Transients: Storage Cycle

• The inductor stores energy in the form
of magnetic field.
• When the switch closes at t = 0, the
inductor voltage changes
instantaneously
• However, the inductance of the coil
prevents instantaneous change in the
value of the current
NETWORK ANALYSIS AND SYNTHESIS
Transients – Storage phase and decay phase

R-L Transients: Storage Cycle

• Some observations based on switching

vL
NETWORK ANALYSIS AND SYNTHESIS
Transients – Storage phase and decay phase

R-L Transients: Storage Cycle

• Applying KVL for t ≥ 0
NETWORK ANALYSIS AND SYNTHESIS
Transients – Storage phase and decay phase

R-L Transients: Storage Cycle

• Alternatively, we can also express based on initial and final
values
NETWORK ANALYSIS AND SYNTHESIS
Transients – Storage phase and decay phase

R-L Transients: Storage Cycle

• Effect of L on duration of the transient response

• As inductance L increases, time

constant τ increases
• Therefore, the current iL takes
longer to reach the steady state
value
NETWORK ANALYSIS AND SYNTHESIS
Transients – Storage phase and decay phase

R-L Transients: Decay Cycle

• The rate of change of current when the inductor is
disconnected from the source is very high.
• This causes a high voltage at open terminals
• This can be prevented and the decay can be controlled by
introducing a shunt resistance to the source
NETWORK ANALYSIS AND SYNTHESIS
Transients – Storage phase and decay phase

R-L Transients: Decay Cycle

• The inclusion of shunt resistance R2 does not change the
storage cycle characteristics.
• When the switch opens the decay phase will depend on R1
and R2
NETWORK ANALYSIS AND SYNTHESIS
Transients – Storage phase and decay phase

R-L Transients: Decay Cycle

NETWORK ANALYSIS AND SYNTHESIS
Transients – Storage phase and decay phase

R-L Transients: Decay Cycle

2
NETWORK ANALYSIS AND SYNTHESIS
Transients – Storage phase and decay phase

R-L Transients: Decay Cycle

• Alternatively, using initial values and final values

τ'
NETWORK ANALYSIS AND SYNTHESIS
Transients – Storage phase and decay phase

Energy stored in transients: Storage phase

• Energy (electric field) stored by capacitor:

• Energy (magnetic field) stored by inductor:

Capacitor characteristics Inductor characteristics

NETWORK ANALYSIS AND SYNTHESIS
Transients – Storage phase and decay phase

Example - Energy stored in transients

• Find energy stored in the circuits shown below
NETWORK ANALYSIS AND SYNTHESIS
Transients – Storage phase and decay phase

Example - Energy stored in transients

• Find energy stored in the circuits shown below
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 1
• Find the transient current and voltage associated with the
inductor
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 1 – Solution
• By source transformation, we get 6 V source in series with
R1
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 1 – Solution (contd.)

NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 1 – Solution (contd.)

NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 2
• Find the transient current i1(t) for t ≥ 0
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 2 – Solution
• At t = 0– the inductor acts as short circuit. By KCL (red dot)
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 2 – Solution (contd.)

• Hence when switch is open dependent source gives 0 A
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 2 – Solution (contd.)

• Finding τ: Note dependent source is zero. Easy to see ETh
= 12 V and ISC = 12/4m. Hence, RTh = 4 mΩ
NETWORK ANALYSIS AND SYNTHESIS

RL Transients – Thevenin’s Equivalent

NETWORK ANALYSIS AND SYNTHESIS
RL Transients – Thevenin’s Equivalent

Application of Thevenin’s theorem to RL transient analysis

• When the given circuit does not have a simple series RL form, then
the network part connected to the inductor is replaced with the
Thevenin’s equivalent circuit
• The time constant τ = L/RTh and the Thevenin’s voltage could be
acting as the source excitation
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 1:
• Find the transient behavior of the current and the voltage.
Initial value Ii = 0 mA
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 1 – Solution
• Thevenin’s resistance and voltage are found below
NETWORK ANALYSIS AND SYNTHESIS
Transients – Thevenin’s Equivalent

Example 1 – Solution (contd.)

NETWORK ANALYSIS AND SYNTHESIS

Analysis in s-domain
Philosophy of Laplace Transform and its
applications
NETWORK ANALYSIS AND SYNTHESIS
Philosophy of Laplace transform and its properties

Philosophy of Laplace transform

• Laplace transform converts differential equations from time domain into
algebraic equations in the complex frequency domain
Algebraic X(s) x(t)
Differential equation
Transform Solve Invert
equation

• Laplace transform of f(t) and its inverse are given below

• Here, s is a complex variable expressed as s = σ + jω

NETWORK ANALYSIS AND SYNTHESIS
Philosophy of Laplace transform and its properties

Philosophy of Laplace transform

• For a function f(t) to have a Laplace transform, it must obey

• Circuit equations under transients (RL, RC or RLC) result in integro-

differential equations
• Laplace transform of an integro-differential equation results in
algebraic (polynomial) equations in s-domain
• Partial fraction expansion is applied to such transfer function to
break it into k simpler Laplace functions Fk(s)
• Inverse transform is applied to each Fk(s) to find fk(t)
• Hence, it is important to remember Laplace transform pairs
NETWORK ANALYSIS AND SYNTHESIS
Philosophy of Laplace transform and its properties

Popular Laplace transform pairs

NETWORK ANALYSIS AND SYNTHESIS
Philosophy of Laplace transform and its properties

Examples of Laplace transform

Properties of Laplace transform

• Linearity

• Real differentiation
NETWORK ANALYSIS AND SYNTHESIS
Philosophy of Laplace transform and its properties

• Example of linearity

• Example of real differential

NETWORK ANALYSIS AND SYNTHESIS
Philosophy of Laplace transform and its properties

Properties of Laplace transform

• Real integration

• Differentiation by s

• Example:
NETWORK ANALYSIS AND SYNTHESIS
Philosophy of Laplace transform and its properties

Properties of Laplace transform

• Complex translation (a is complex number)

• Examples

• Real translation (time shifting theorem)

NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Partial – Fraction Expansions

• Some simple and effective methods for partial – fraction
expansions
• Procedures for simple roots
• Procedures for complex conjugate roots
• Procedures for multiple roots
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Conditions for application of PFE

• The given function F(s) can be factorized into PFE under the
condition below

• Here, N(s) and D(s) are polynomial in s

• When the above condition is not satisfied, perform the following

• Here, Q(s) is quotient and R(s) is reminder such that the order of
R(s) ≤ order of F(s)
• R(s)/D(s) can be factorized using PFE
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Example 1: Factorize the given F(s)

• The given function F(s) can be factorized into PFE under
the condition below

Solution
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Solution (contd.)
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Partial – Fraction Expansions

Real roots
• Method for simple real roots
𝑁(𝑠)
Consider the function, 𝐹 𝑠 =
(𝑠−𝑠0 )(𝑠−𝑠1 )(𝑠−𝑠2 )
Where 𝑠0 , 𝑠1 , 𝑠2 are distinct, real roots and the degree of 𝑁 𝑠 < 3
Expanding 𝐹 𝑠 ,
𝐾0 𝐾1 𝐾2
𝐹 𝑠 = + +
(𝑠−𝑠0 ) (𝑠−𝑠1 ) (𝑠−𝑠2 )
To find 𝐾0 , multiply both sides of the equation by (𝑠 − 𝑠0 )
(𝑠 − 𝑠0 )𝐾1 (𝑠 − 𝑠0 )𝐾2
(𝑠 − 𝑠0 ) 𝐹 𝑠 = 𝐾0 + +
(𝑠 − 𝑠1 ) (𝑠 − 𝑠2 )
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Partial – Fraction Expansions

Real roots
• Method for simple real roots
If 𝑠 = 𝑠0 , 𝐾0 = 𝑠 − 𝑠0 𝐹 𝑠 |𝑠=𝑠0
Similarly other constants can be evaluated
𝐾𝑖 = 𝑠 − 𝑠𝑖 𝐹 𝑠 |𝑠=𝑠𝑖
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Example 2:
• Express the given function F(s) as a PFE

Solution
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

PFE: Complex roots

• When the denominator D(s) is factorized as simple complex
roots as given below (degree of N(s) < degree of D(s))

• We can express the above as a PFE

• Let us focus only on finding the residuals of the simple

complex roots
• Note: Complex roots always occur as conjugate pairs
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

PFE: Complex roots (contd.)

NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Example 2:
• Express the given function F(s) as a PFE

Solution
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Solution (contd.) – Method 1

NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Solution (contd.) – Method 2

• Cross multiplying and equating coefficients

A = – 0.2, B = 0.2 and C = 0.8

NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Solution (contd.) – Method 2

• Finally we get
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Example 3:
• Find y(t) given

Solution
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Solution (contd.)

Home work – Example 4:

• Find y(t) for
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Solving example 3 by Method 1

NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Solving example 3 by Method 1

NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Method
MultipleAroots
Suppose the given function,
𝑁(𝑠)
𝐹 𝑠 =
𝑠 − 𝑠0 𝑛 𝐷1 (𝑠)
With multiple roots of degree n at 𝑠 = 𝑠0 .
The partial expansion of 𝐹(𝑠) is
𝐾0 𝐾1 𝐾2 𝐾𝑛−1 𝑁1 (𝑠)
𝐹 𝑠 = + + + ⋯+ +
𝑠−𝑠0 𝑛 𝑠−𝑠0 𝑛−1 𝑠−𝑠0 𝑛−2 𝑠−𝑠0 𝐷1 (𝑠)
𝑁1 (𝑠)
Where represents the remaining terms of expansion.
𝐷1 (𝑠)
The problem is to obtain 𝐾0 , 𝐾1 , … … … . , 𝐾𝑛−1
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Method A
𝐾0 = 𝑠 − 𝑠0 𝑛 𝐹 𝑠 ቚ
𝑠=𝑠0
The same formula when used to obtain the other factors, we
arrive at the indeterminate 0/0 condition.
∴ multiply both sides of F(s) by 𝑠 − 𝑠0 𝑛 and define
𝐹1 (𝑠) ≡ 𝑠 − 𝑠0 𝑛 𝐹(𝑠)
Thus
𝐹1 𝑠
= 𝐾0 + 𝐾1 𝑠 − 𝑠0 + ⋯ + 𝐾𝑛−1 𝑠 − 𝑠0 𝑛−1 + 𝑅 𝑠 𝑠 − 𝑠0 𝑛
Where 𝑅(𝑠) indicates the remaining terms.
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Method A
Differentiating by s,
𝑑
𝐹1 𝑠
𝑑𝑠
= 𝐾1 + 2𝐾2 𝑠 − 𝑠0 + ⋯ + 𝐾𝑛−1 𝑛 − 1 𝑠 − 𝑠0 𝑛−2 +⋯
This yields
𝑑
𝐾1 = 𝐹1 𝑠 ቚ
𝑑𝑠 𝑠=𝑠0
1 𝑑2
Similarly 𝐾2 = 𝐹 𝑠 |𝑠=𝑠0
2 𝑑𝑠 2 1
1 𝑑𝑗
In general, 𝐾𝑗 = 𝐹 𝑠 |𝑠=𝑠0 ; 𝑗 = 0,1,2, … , 𝑛 − 1
𝑗! 𝑑𝑠 𝑗 1
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Example 4:
• Express as PFE

Solution
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Example 5:
• Express as PFE

Solution
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

The Initial and Final Value Theorems

The two useful theorems of Laplace transforms are
• Initial value theorem
• Final value theorem
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

 The Initial Value Theorem

 Relates the initial value of 𝑓(𝑡) at 𝑡 = 0 + to the limiting value of

𝑠𝐹 𝑠 as s approaches infinity.

 i.e. lim 𝑓 𝑡 = lim 𝑠𝐹 𝑠

𝑡→0+ 𝑠→∞

 The only restriction is that 𝑓(𝑡) must be continuous or contain, at

most, a step discontinuity at 𝑡 = 0.

 In terms of the transform, 𝐹 𝑠 = ℒ 𝑓 𝑡 ; the above restriction

implies that 𝐹 𝑠 must be a proper fraction.
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

The Initial Value Theorem

Proof:
Part 1: The function 𝒇 𝒕 is continuous at t=0,
i.e. 𝑓 0 − = 𝑓(0+)
From the relationship
∞
ℒ 𝑓′ 𝑡 = න 𝑓′(𝑡) 𝑒 𝑠𝑡 𝑑𝑡 = 𝑠𝐹 𝑠 − 𝑓(0−)
0−
We obtain lim ℒ 𝑓 ′ 𝑡 = lim 𝑠𝐹 𝑠 − 𝑓 0 − = 0
𝑠→∞ 𝑠→∞
lim 𝑠𝐹 𝑠 = 0 − = 𝑓(0+)
𝑠→∞
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

The Initial Value Theorem

Proof:
Part 2: The function 𝑓(𝑡) has a step discontinuity at t=0.
Representing 𝑓 𝑡 in terms of a continuous part 𝑓1 (𝑡) and a
step discontinuity 𝐷𝑢(𝑡) as shown in the fig.
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

The Initial Value Theorem

Proof:
Part 2:
Simplifies to 𝑠𝐹 𝑠 = 𝑠𝐹1 𝑠 + 𝐷
If we take the limit as 𝑠 → ∞ and let 𝑓 0 + − 𝑓 0 − = 𝐷,
lim 𝑠𝐹 𝑠 = lim 𝑠𝐹1 𝑠 + 𝑓 0 + − 𝑓(0−)
𝑠→∞ 𝑠→∞
lim 𝑠𝐹 𝑠 = 𝑓 0 +
𝑠→∞
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

The Final Value Theorem

This theorem states that
lim 𝑓 𝑡 = lim𝑠𝐹 𝑠
𝑡→∞ 𝑠→0
Provided the poles of the denominator of 𝐹(𝑠) must not be in
the right half of the complex frequency plane.
Proof:
∞
න 𝑓 ′ 𝑡 𝑒 −𝑠𝑡 𝑑𝑡 = 𝑠𝐹 𝑠 − 𝑓(0−)
0−
Taking the limit as 𝑠 → 0
∞
න 𝑓 ′ 𝑡 𝑑𝑡 = lim 𝑠𝐹 𝑠 − 𝑓(0−)
0− 𝑠→0
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

The Final Value Theorem

Evaluating the integral,
𝑓 ∞ − 𝑓 0 − = lim 𝑠𝐹 𝑠 − 𝑓(0−)
𝑠→0
Consequently, 𝑓 ∞ = lim 𝑠𝐹 𝑠
𝑠→0
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Example 6:
• Find the initial value of f(t) given

Solution
• Initial value theorem states that

• Verification
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Example 7:
• Find the final value of f(t) given

Solution
• Final value theorem states that

• Verification
NETWORK ANALYSIS AND SYNTHESIS
Partial Fraction Expansion, Initial and Final Value

Example 8: Failures of the initial/final value theorem

• Initial value theorem fails when function f(t) is
discontinuous

• Final value theorem fails when roots of denominator (i.e.,

poles) are on the right half of the s-plane
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Analysis in s-domain
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

The Transformed Circuit

• The transformed circuit is realized by
expressing the passive elements with their
respective circuit equivalents as shown.

• Then application of mesh/nodal analysis or

network theorems is done in the s-domain
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

The Transformed Circuit

• The voltage-current relationships of network elements in
the time domain may be represented in the complex
frequency domain.
• Ideal energy sources in time domain were 𝑣 𝑡 and 𝑖(𝑡)
may be represented by their transforms 𝑉 𝑠 = ℒ[𝑣 𝑡 ]
and 𝐼 𝑠 = ℒ[𝑖 𝑡 ]
• The resistor defined by 𝑣 − 𝑖 relationship
𝑣 𝑡 = 𝑅𝑖 𝑡
Is defined in the frequency domain by the transform
𝑉 𝑠 = 𝑅 𝐼(𝑠)
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Transformed circuit for inductor

In the KVL equation, Ls is impedance and Li(0–) is a constant

voltage

In the KCL equation, 1/Ls is admittance and i(0–)/s is constant

current
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

The Transformed Circuit

• The transformed circuit representation for an inductor is

Depiction of KVL Depiction of KCL

NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Transformed circuit for capacitor

In the KVL equation, 1/Cs is impedance and v(0–)/s is constant

voltage

In the KCL equation, Cs is admittance and Cv(0–) is constant

current
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

The Transformed Circuit

• The transformed circuit representation for a capacitor is

Depiction of KVL Depiction of KCL

NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

The Transformed Circuit

• Consider the example of a transformer as in fig. below
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

The Transformed Circuit

• Transformer equations in time domain,
𝑑𝑖1 𝑑𝑖2
𝑣1 𝑡 = 𝐿1 +𝑀
𝑑𝑡 dt
𝑑𝑖1 𝑑𝑖2
𝑣2 𝑡 = 𝑀 + 𝐿2
𝑑𝑡 dt
Transforming these set of equations,
𝑉1 𝑠 = 𝑠𝐿1 𝐼1 𝑠 − 𝐿1 𝑖1 0 − + 𝑠𝑀 𝐼2 𝑠 − 𝑀𝑖2 0 −
𝑉2 𝑠 = 𝑠𝑀 𝐼1 𝑠 − 𝑀𝑖1 0 − + 𝑠𝑀 L2 𝐼2 𝑠 − 𝑀𝑖
L2 2 (0−)
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

The Transformed Circuit

• The transformed equivalent circuit is
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Example:
• Find v0(t) given v0(0 –)=5V

Solution:
• Apply KCL
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Solution (contd.)
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Example
• Find the current i(t) for t ≥ 0. Assume initial capacitor
voltage of 1 V
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Solution
• Draw transformed circuit. Write I(s) using KVL and solve it
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Example
• Find the mesh currents in the circuit. The switch is open for
a long time before closing at t = 0
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Solution
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Example:
• Find vc(t) for t ≥ 0. The current source is activated at t = 0
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Solution
• Initial values vc(0–) = –20 V and iL(0–) = 0
• Capacitor would have completed charging and behaving as
open circuit and inductor acts as open circuit
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Solution (contd.)
• When current source is activated at t = 0, we get transient
response
• Apply source transformation so we have 6u(t) V in series
with 2 Ω.
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Solution (contd.)
• Apply KCL
NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Solution (contd.)
• Apply Laplace transform

• Take inverse Laplace transform

NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Useful table for constructing transformed circuit

NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Useful table for constructing transformed circuit

NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Useful table for constructing transformed circuit

NETWORK ANALYSIS AND SYNTHESIS
The Transformed Circuit

Constructing transformed circuit

• Two methods:
• Either apply KCL/KVL for the circuit in time domain
• Apply Laplace transform to the equations
• Draw the transformed circuit based on the KCL/KVL
equations expressed in s domain

• Alternatively, directly obtain transformed circuit using

the table in the previous slide.
• From the transformed circuit, obtain the transformed
KCL/KVL equations
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Example 1: Assuming switch is thrown from position 1 to 2 at

t = 0, write the mesh equations
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Solution 1:
• Write KVL for mesh 1 in time

• Apply Laplace transform

NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Solution 1:
• Write KVL for mesh 2 in time

• Apply Laplace transform

NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Example 2: Assuming switch is thrown from position 1 to 2 at

t = 0, write the current i(t) for t ≥ 0. Here initial values are
iL(0–) = 2 A and vC(0–) = 2 V
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Solution 2:
• Write KVL
• Note that iL(0–) = 2 A and vC(0–) = 2 V
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Example 3: Assuming switch is opened at t = 0, write the

nodal voltages for t ≥ 0
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Solution 3:
• iL(0–) = 1 A and vC(0–) = 1 V
• Write KCL at node 1

• Apply Laplace transform

NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Solution 3:
• Write KCL at node 2

• Apply Laplace transform

NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Solution 3: Transformed circuit

NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Solution 3:
• Inverse Laplace transform

• Matrix form
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Solution 3:
• Cramer’s rule

• Inverse Laplace transform

NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Thevenin’s Theorem
From the standpoint of determining the current I(s) through
an element of impedance 𝑍1 𝑠

The rest of the network N can be replaced by an equivalent

impedance 𝑍𝑒 (𝑠) in series with an equivalent voltage source
𝑉𝑒 (𝑠)
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Thevenin’s Theorem
The Thevenin’s equivalent circuit is

The only requirement for Thevenin’s theorem is that the

elements in 𝑍1 must not be magnetically coupled to any
element in 𝑁
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Thevenin’s Theorem
Proof:

The network shown above contains 𝑛 voltage and 𝑚 current

sources.
To find the current 𝐼 𝑠 through an element that is not
magnetically coupled to rest of the circuit and whose
impedance is 𝑍1 𝑠 ,
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Thevenin’s Theorem
Proof:
According to Compensation theorem, replace 𝑍1 (𝑠) by a
voltage source 𝑉 𝑠

By superposition principle,
𝐼 𝑠 = 𝐼1 𝑠 + 𝐼2 (𝑠)
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Thevenin’s Theorem
Proof:
Let 𝐼1 (𝑠) be the current due to the 𝑛 voltage and 𝑚 current
sources alone, i.e. short circuit the source 𝑉(𝑠)
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Thevenin’s Theorem
Proof:
Therefore 𝐼2 𝑠 = 𝐼𝑠𝑐
Let 𝐼2 (𝑠) be the current due to the voltage source 𝑉(𝑠) alone,
with the rest of the voltage sources short circuited and current
sources open circuited.
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Thevenin’s Theorem
Proof:
With the m current and n voltage sources removed, the
network N is passive so that 𝐼2 (𝑠) is related to the source 𝑉(𝑠)
by
𝑉 𝑠
𝐼2 𝑠 = −
𝑍𝑖𝑛 𝑠
where 𝑍𝑖𝑛 (𝑠) is the input impedance of the circuit at the
terminals of the source 𝑉 𝑠 .
𝑉 𝑠
𝐼 𝑠 = 𝐼𝑠𝑐 𝑠 −
𝑍𝑖𝑛 𝑠
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Thevenin’s Theorem
Proof:
Since the equation of 𝐼 𝑠 should be satisfied in all cases,
consider the particular case when we open circuit the branch
containing 𝑍1 (𝑠). Then 𝐼 𝑠 = 0 and 𝑉 𝑠 is the open circuit
voltage 𝑉𝑜𝑐 (𝑠).
𝑉𝑜𝑐 𝑠
𝐼𝑠𝑐 𝑠 =
𝑍𝑖𝑛 𝑠
Therefore 𝑍𝑖𝑛 𝑠 𝐼 𝑠 − 𝑉𝑜𝑐 𝑠 = −𝑉(𝑠)
In order to obtain the current 𝐼(𝑠) through 𝑍1 𝑠 , the rest of
the network N can be replaced by an equivalent source
𝑉𝑒 𝑠 = 𝑉𝑜𝑐 (𝑠) in series with an equivalent impedance
𝑍𝑒 𝑠 = 𝑍𝑖𝑛 (𝑠).
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Example 1: Find current through the capacitor using

Thevenin’s theorem
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Example 1 - Solution
• Removing the capacitor, find Thevenin’s impedance Ze(s)

• Find Thevenin’s voltage across terminals 1-2

NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Example 1 – Solution (contd.)

• By Thevenin’s theorem
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Example 2
• Find v0(t) by Thevenin’s theorem. Assume all initial values
to be zero at the by the time the switch closes at t=0
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Example 2 – Solution
• Step 1: Find Thevenin’s impedance across 1-2
• Step 2: Find Thevenin’s voltage across 1-2
• Step 3: Using voltage division find V0(s)
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Example 3
• Find the current i2(t) given v(t) = 2u(t) and iL(0–)=2 A
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Example 3 – Solution
• Thevenin’s impedance is given by

• Thevenin’s voltage is given by

NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Example 3 – Solution
• By Ohm’s law we get
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Example 3 – Solution
• Alternatively, we can also do as follows
NETWORK ANALYSIS AND SYNTHESIS
Application of Thevenin’s Theorem

Example 4
• Find i(t) using Thevenin’s theorem assuming initial values
as zeros
THANK YOU

Dr Anuradha M.
Professor
Department of Electronics and
Communication Engineering
anuradha@pes.edu