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Electronic Circuits 1
Dynamic circuits:Steady-state analysis forsinusoidal excitations
Prof. C.K. Tse: Dynamic circuits:Steady-state analysis
Contents• Steady-state solution• Single-frequency excitation• Phasor diagrams• Complex Calulus
2Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Steady state
♦ Behaviour after a “long” time.
sufficiently long— depends on thenatural frequency of the system,e.g., for a circuit with componentslike µF and µH, the “long” couldbe just some fraction of a ms.
3Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
General solution♦ Consider the second-order differential equation
♦ The solution contains two parts:♦ Transient solution Steady-state solution
♦ k1 and k2 to be found from boundary conditions (initial conditions).
d x
dta
dxdt
bx f t2
2+ + = ( )
ddt
xx
a ba b
xx
f tf t
1
2
1 1
2 2
1
2
1
2
=
+
( )( )
or
x t k e k e s tt t( ) ( )= +
+1 2
1 2λ λ
4Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
General solution
♦ As t goes to ∞, only s(t) prevails.♦ If we wait sufficiently long, the solution is just s(t).♦ Obviously s(t) must be closely related to f(t).
♦ Formally it is given by taking inverse Laplace transform of
x t k e k e s tt t( ) ( )= +
+1 2
1 2λ λ
X sF s s x x
s a s b
x tF s s x x
s a s b
( ) =+ + + ′
+ +
=+ + + ′
+ +
+ +
−+ +
( ) ( ) ( ) ( )
.
( )( ) ( ) ( ) ( )
.
1 0 0
1 0 0
2
12
LHence,
5Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Special case♦ Suppose f(t) is a sine function.
♦ Then, obviously, the solution x(t) must also be a sine or cosine or acombination of sine and cosine functions of the same frequency.
♦ FACT♦ If a circuit is driven by a sinusoidal voltage or current source, then all
the voltages and currents in the circuit will be sinusoidal withdifferent phase angles and amplitudes, all at the same frequency.
d x
dta
dxdt
bx f t2
2+ + = ( )
6Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
SSSCDSFIS Steady-state solution of circuits drivenby single frequency independent sources
~+
vs(t)–
vs(t)=A sinωt
All voltages and currentsin the circuit will be sinewaves of differentamplitudes and phases:
vL(t) = A1 sin(ωt + φ1)vC(t) = A2 sin(ωt + φ2)iL(t) = A3 sin(ωt + φ3)ic(t) = A4 sin(ωt + φ4)…etc.
same freq.
7Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Solution approaches
♦ Formal solving of differential equations(as seen previously)
♦ Quick methods
♦ Phasor diagram — geometrical♦ Complex number calculus — algebraic
8Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Representing sine functions
A way to represent many sine functions of same frequency.
Consider 2 sine functions:
So, there are only two variables:amplitude and phase angle.
Alternative representation:
9Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Phasor as rotating vector
The rotating vector is called phasor
10Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
A phasor animation
11Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Phasor diagram
Consider 5 sine functions of same frequency:
12Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Example
Consider
We note that
Hence,
We can represent them as
x t t
x t t
x t t
o
o
1
2
3
2 60
45
( ) sin
( ) sin( )
( ) cos( )
=
= −
= +
ω
ω
ω
cos sin( )θ θ= + 90o
x t t o3 135( ) sin( )= +ω
13Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Basic steady-state relations
Component constitutive relations
ResistorInductorCapacitor
v i R
v Ldidt
i Cdvdt
R R
LL
cc
=
=
=
If , then v V t i V R tR R R R= =sin . sin .ω ω
If , then i I t v I L tL L L L= =sin . cos .ω ω ω
If , then v V t i V C tc C c C= =sin . cos .ω ω ω
+ v –i
14Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Phasor relation for inductor
Inductor v LdidtL
L=
If , then i I t v I L tL L L L= =sin . cos .ω ω ω
v I L tL Lo= +ω ωsin( )90or
15Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Phasor relation for capacitor
Capacitor i CdvdtC
C=
If , then v V t i V C tC C C C= =sin . cos .ω ω ω
i V C tC Co= +ω ωsin( )90or
16Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Amplitude and phase
InductorVoltage-to-current ratio = ωLVoltage leads current by 90o
CapacitorVoltage-to-current ratio = 1 / ωCVoltage lags current by 90o
like resistance,but we call it“reactance”;unit is still Ω.
17Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Using phasor diagrams for SSSCDSFIS
RULE 1 : BASIC CONSTITUTIVE RELATIONS
18Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Using phasor diagrams for SSSCDSFIS
RULE 2: BASIC MANIPULATIONS (VECTOR ALGEBRA)
19Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Example (using phasor diagram)
From Pythagoras Theorem,
Since
we have Also,
20Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Back in time domain
If then we have
21Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Example (using phasor diagram)
Find R and XL such that
Since I1 = 15 A, V = 60 V.
Hence,
V
IR I1
IXL IxIT
Applying Pythagoras theorem,
Hence,
22Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Complex number as a tool
Complex numbers can be used to represent sine functions,similar to phasors
Rectangular formz = a + jb
Polar formz = |z| φ
Here, |z| =
and
a b2 2+
φ = arctanba
Re
Im
a
jbφ
z
23Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Complex number as a tool
We can represent a sine function by a complex number.Suppose we choose v1 as the reference phase.
v1 = V1sinωt = V1 0 = V1 + j0
v2 = V2sin(ωt + 45) = V2 45 =
V V2 2
2 2+ j
24Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Multiplication in complex numbers
When one complex number is multiplied to another complexnumber, their magnitudes multiply and their phases add up.
φ1φ2
φ3 z1
z1
z3 = z1z2 |z3| = |z1|.|z2|
φ3 = φ1 + φ2
Useful tricks:
To rotate 90o anticlockwise= multiply by j
To rotate 90o clockwise= multiply by –j= divide by j
25Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Inductor relation in complex number
Clearly, in complex number domain, the voltage can be obtained bymultiplying the current by a factor of jωL.
rotation by 90 anticlockwise amplitude ratio
jωLVIL
L= like resistance,
but we call itIMPEDANCE
26Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Capacitor relation in complex number
Clearly, in complex number domain, the current can be obtained bymultiplying the voltage by a factor of jωC.
rotation by 90 clockwise amplitude ratio
1jωC
VIC
C= like resistance,
but we call itIMPEDANCE
27Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
ImpedanceIn general, we define impedance Z of an element as the ratio of its voltageand current.
Likewise, we have admittance Y, which is the dual of impedance.
+ V –I
ZResistance Reactance
Conductance Susceptance
Z and Y can alsobe expressed inpolar form.
28Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Terminologies
impedance admittance
29Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Example
Rectangular form
Polar form (convenient sometimes)
Meaning:•|V/I|=10.00000003•V leads I by –0.004584o
•V lags I by 0.004584o
Suppose excitation is 50 Hz
30Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Example
Z j= +10 10 3 Ω (frequency understood)
Z = 20 60o Ω
If i.e.,
31Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Solution approach
Replace L by jωLReplace C by 1/jωCExpress all sources in compress number
Then,solve the circuit as if it were “resistive” but in thecomplex number domain
32Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
ExampleSuppose
i.e.,
Thus,
What does this mean?
33Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Example — using mesh methodWhat you learnt before can all be used!
Suppose ω = 1 rad/sUsing mesh analysis, we get
where Answer:
34Prof. C.K. Tse: Dynamic circuits:
Steady-state analysis
Further topics
Already covered in Basic Electronics
PowerActive, reactive and apparent powerPower factor
See also Chapter 7 of my book.