electronic circuits 1 dynamic circuits: steady-state ...cktse/eie201/5.steady.pdf5 prof. c.k. tse:...

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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.



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



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,



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+ + = ( )



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.



Solution approaches

♦ Formal solving of differential equations(as seen previously)

♦ Quick methods

♦ Phasor diagram — geometrical♦ Complex number calculus — algebraic



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:



Phasor as rotating vector

The rotating vector is called phasor



A phasor animation



Phasor diagram

Consider 5 sine functions of same frequency:



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



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



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



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



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.



Using phasor diagrams for SSSCDSFIS

RULE 1 : BASIC CONSTITUTIVE RELATIONS



Using phasor diagrams for SSSCDSFIS

RULE 2: BASIC MANIPULATIONS (VECTOR ALGEBRA)



Example (using phasor diagram)

From Pythagoras Theorem,

Since

we have Also,



Back in time domain

If then we have



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,



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



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



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



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



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



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.



Terminologies

impedance admittance



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



Example

Z j= +10 10 3 Ω (frequency understood)

Z = 20 60o Ω

If i.e.,



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



ExampleSuppose

i.e.,

Thus,

What does this mean?



Example — using mesh methodWhat you learnt before can all be used!

Suppose ω = 1 rad/sUsing mesh analysis, we get

where Answer:



Further topics

Already covered in Basic Electronics

PowerActive, reactive and apparent powerPower factor

See also Chapter 7 of my book.

electronic circuits 1 dynamic circuits: steady-state ...cktse/eie201/5.steady.pdf5 prof. c.k. tse:...

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