Circuit equations in time domain andMá a frequencyamber.feld.cvut.cz/vyu/eo2/english/files/5_Equations.pdf · Circuit equations in time domain and frequency domain EO2 –Lecture

Circuit equationsin time domain and frequency

domain

EO2 – Lecture 5

Pavel Máša

XE31EO2 - Pavel Máša - Lecture 5

XE31EO2 -

Pav

el Máš

a

• In last semester we studied circuit equations in DC and AC circuits

What is the same and what is different when we will write circuit equations in time domain or in operational form, or in DC or AC circuits?

Circuit equations, regardless of used mathematical apparatus, are always mathematical formulation of Kirchhoff’s laws:

INTRODUCTION

MESH (LOOP) ANALYSIS – KVLX

k

Uk = 0X

k

Uk = 0

NODAL ANALYSIS – KCLX

k

Ik = 0X

k

Ik = 0

voltage across R, L, C is qualified by means of current

current, passing R, L, C is qualified by means of voltage


XE31EO2 -

Pav

el Máš

a

RELATIONSHIP BETWEEN VOLTAGE AND CURRENT

Circuit element Time domain DC AC / Fourier Laplace

R

L

short

C

open circuit

usage transient analysis

usually DC/ AC analysis has to be also proceeded

Steady state withDC source

AC – steady state withsine wave source

Fourierseries – steady state with periodical excitation

transform – pulse excitation

General application

contains both steady andtransient components


XE31EO2 -

Pav

el Máš

a

( ) ( )CC

q Cudqidt

du ti t C

dt

=

=

=

( )C

C

Lidu tdt

Φ =

Φ=

Energetic initial conditions – since energy is continuous, energetic circuit variables are also continuous

INITIAL CONDITIONS

uC(0¡) = uC(0+)uC(0¡) = uC(0+)must be iC(0¡) 6= iC(0+)iC(0¡) 6= iC(0+)may be

may be uL(0¡) 6= uL(0+)uL(0¡) 6= uL(0+) must be iL(0¡) = iL(0+)iL(0¡) = iL(0+)

uL(0¡)uL(0¡)

uL(0+)uL(0+)

iL(0¡)iL(0¡) iL(0+)iL(0+)

History of capacitor describes charge stored in the capacitor voltageHistory of inductor describes magnetic flux passing the inductor electric current


XE31EO2 -

Pav

el Máš

a

• Mesh analysis– Current in the loop, where we currently write the equation has always positive sign– Other currents passing circuit element, where we currently evaluate voltage, have positive

voltage, when they have same orientation, negative, when they have opposite orientation– Voltage sources have positive sign, when the current in the loop, where we currently write

the equation, flows into positive source terminal, and negative, when it flows into negative terminal

– We could not write an equation in loops, passing current sources, but we have to count current sources in closed loops

• Nodal voltages– Voltage in the node, where we write an equation, has always positive sign– Voltages in adjacent nodes, with the exception of voltage sources, have always negative

sign (we suppose, all currents leaves the node, actual orientation results from solution of system of equations; circuit element voltage, is the difference of electric potentials)

– Voltage sources in adjacent nodes have negative sign, when they are connected by positive terminal to the adjacent node (we subtract them), positive, if they are connected by negative terminal

– Current of the current source has positive sign, if it leaves the node, negative, it runs into the node

– When the voltage source is not connected by any terminal with reference node (ground), then we refer it as floating source – we write just one equation for both nodes (where the floating source is connected.

BASIC RULES – SIGNS, …


XE31EO2 -

Pav

el Máš

a

Reduction of number of equations

DC

AC

Time domain not possible

Laplace

DC

AC

Time domain

Laplace


XE31EO2 -

Pav

el Máš

a

COUPLED INDUCTORS (TRANSFORMER)

u1(t) = L1di1(t)

dt+ M

di2(t)

dtu1(t) = L1

di1(t)

dt+ M

di2(t)

dt

u2(t) = L2di2(t)

dt+ M

di1(t)

dtu2(t) = L2

di2(t)

dt+ M

di1(t)

dt

Time domain – mesh analysis

Laplace transform – mesh analysisU1(p) = pL1I1(p)¡L1i1(0+) + pMI2(p)¡Mi2(0+)U1(p) = pL1I1(p)¡L1i1(0+) + pMI2(p)¡Mi2(0+)

U2(p) = pL2I2(p)¡L2i2(0+) + pMI1(p)¡Mi1(0+)U2(p) = pL2I2(p)¡L2i2(0+) + pMI1(p)¡Mi1(0+)

I2(p) =U2(p) + L2i2(0+)¡ pMI1(p) + Mi1(0+)

pL2I2(p) =

U2(p) + L2i2(0+)¡ pMI1(p) + Mi1(0+)

pL2

– nodal voltages

U1(p) = pL1I1(p)¡L1i1(0+)+pMU2(p) + L2i2(0+)¡ pMI1(p) + Mi1(0+)

pL2¡Mi2(0+)U1(p) = pL1I1(p)¡L1i1(0+)+pM

U2(p) + L2i2(0+)¡ pMI1(p) + Mi1(0+)

pL2¡Mi2(0+)

1

2

3

I1(p) =1

p

L2

L1L2 ¡M 2U1(p)¡ 1

p

M

L1L2 ¡M2U2(p)¡

i1(0+)

p=

1

p¡1U1(p) +

1

p¡MU2(p) ¡ i1(0+)

pI1(p) =

1

p

L2

L1L2 ¡M 2U1(p)¡ 1

p

M

L1L2 ¡M2U2(p)¡

i1(0+)

p=

1

p¡1U1(p) +

1

p¡MU2(p) ¡ i1(0+)

p

I2(p) =1

p¡MU1(p) +

1

p¡2U2(p)¡

i2(0+)

pI2(p) =

1

p¡MU1(p) +

1

p¡2U2(p)¡

i2(0+)

p


XE31EO2 -

Pav

el Máš

a

R1 L

C R2

Example:

DC:

AC:

Mesh analysis – 0 equations, voltages across R1 and R2 is evaluated by Ohm’s law

j!L I1+R2I1+1

j!C(I1¡I) = 0 ) I1 =

I

(j!)2LC + j!R2C + 1j!L I1+R2I1+

1

j!C(I1¡I) = 0 ) I1 =

I

(j!)2LC + j!R2C + 1

Time domain:

Ldi1(t)

dt+ R2i1(t) +

1

C

Z t

0

[i1(¿ )¡ i(¿ )] d¿ ¡ uC(0) = 0Ldi1(t)

dt+ R2i1(t) +

1

C

Z t

0

[i1(¿ )¡ i(¿ )] d¿ ¡ uC(0) = 0

Solution is obtained in three steps:1. Initial condition uC(0) – DC or AC analysis has to be proceeded according to the nature of exciting source before the change2. Solution of integral‐differential equation – solution of the transient3. To find steady state in the circuit after transient is over we have to proceed DC or AC analysis again

iL(0)iL(0)uC(0)uC(0)

Usage – transients – describing currents or voltages in circuit after• Source was connected• Source was disconnected• Circuit layout has been changed• Some circuit element parameters (resistivity, capacitance, inductance)

DC or AC analysis always has to be proceededXE31EO2 - Pavel Máša - Lecture 5

XE31EO2 -

Pav

el Máš

a

Laplace: pL I1(p)¡LiL(0) + R2I1(p) +1

pC[I1(p)¡ I(p)]¡uC(0)

p= 0pL I1(p)¡LiL(0) + R2I1(p) +

1

pC[I1(p)¡ I(p)]¡uC(0)

p= 0

) I1(p) =I(p)+CuC(0) + pLCiL(0)

p2LC + pR2C + 1) I1(p) =

I(p)+CuC(0) + pLCiL(0)

p2LC + pR2C + 1The solution is obtained in two steps:

1. Initial condition uC(0) DC or AC analysis has to be proceeded according to the nature of exciting source before the change (same as time domain)

2. Solve inverse transform I1(p) – it contains both transient and steady state after transient dies away

• To find initial conditions, we have to proceed DC or AC analysis as well• Inverse Laplace transform always corresponds to transient when

source I(p) is connected to the circuit (if it is present in the equation), orsource is disconnected (then the steady state is „hidden“ in the initial conditions)

The transient is integral part of Laplace transform solution, the result of time interval restriction‐ each source is zero when t < 0, because ,

• The solution of operational circuit equations contains also steady state after transient dies away• Although operational solution is formally similar to the AC analysis, it is different – „more complete“ solution

u(t) = u0(t)1(t)u(t) = u0(t)1(t) i(t) = i0(t)1(t)i(t) = i0(t)1(t)


XE31EO2 -

Pav

el Máš

a

• Now assume following values: R2 = 1 kΩ, L = 1H, C = 1 μF, R1 = 2 kΩ• DC – Circuit supplied by current source I = 10 mA

• AC – circuit supplied from sinusoidal current source

• Time domain – at time t = 0 was connected sinusoidal current source. At t < 0 the circuit was without energy.

1. Initial condition is given – uc(0) = 0

2. We solve integral‐differential equation (in detail it will be described on 8th lecture)

a) Derive

b) Solve using method of variation of parameters

U2 = R2I = 1000 ¢ 0:01 = 10 VU2 = R2I = 1000 ¢ 0:01 = 10 V U1 = R1I = 2000 ¢ 0:01 = 20 VU1 = R1I = 2000 ¢ 0:01 = 20 V

i(t) = 10 sin(1000t) mAi(t) = 10 sin(1000t) mA

I1 =I

(j!)2LC + j!R2C + 1=

0:01

(j ¢ 1000)2 ¢ 1 ¢ 10¡6 + j ¢ 1000 ¢ 1000 ¢ 10¡6 + 1= 10e¡

¼2 j mAI1 =

I

(j!)2LC + j!R2C + 1=

0:01

(j ¢ 1000)2 ¢ 1 ¢ 10¡6 + j ¢ 1000 ¢ 1000 ¢ 10¡6 + 1= 10e¡

¼2 j mA

i(t) = 10 sin(1000t) mAi(t) = 10 sin(1000t) mA

Ldi1(t)

dt+ R2i1(t) +

1

C

Z t

0

[i1(¿ )¡ i(¿ )] d¿ ¡ uC(0) = 0Ldi1(t)

dt+ R2i1(t) +

1

C

Z t

0

[i1(¿ )¡ i(¿ )] d¿ ¡ uC(0) = 0

Ld2 i1(t)

dt2+ R2

d i1(t)

dt+

i1(t)¡ i(t)

C= 0L

d2 i1(t)

dt2+ R2

d i1(t)

dt+

i1(t)¡ i(t)

C= 0

¸2 +R2

L¸ +

1

LC= 0 ) ¸2 + 1000¸ + 106 = 0¸2 +

R2

L¸ +

1

LC= 0 ) ¸2 + 1000¸ + 106 = 0

¸1;2 = ¡500§p

5002 ¡ 106 = ¡500§ 866:03j¸1;2 = ¡500§p

5002 ¡ 106 = ¡500§ 866:03j

i1(t) = [K1 cos(866:03t) + K2 sin(866:03t)] e¡500t + 0:01 sin(1000t¡ ¼

2)| {z }

AC solved above

i1(t) = [K1 cos(866:03t) + K2 sin(866:03t)] e¡500t + 0:01 sin(1000t¡ ¼

2)| {z }

AC solved above


XE31EO2 -

Pav

el Máš

a

• Laplace

I1(p) =I(p)

p2LC + pR2C + 1=

0:01¢1000p2+10002

p2 ¢ 1 ¢ 10¡6 + p ¢ 1000 ¢ 10¡6 + 1=

0:01 ¢ 1000

p2 + 10002

106

p2 + 1000p + 106

=1000 + p

p2 + 1000p + 106¡ p

p2 + 106=

= 0:01

"p + 500

(p + 500)2 + (500p

3)2+

1p3

500p

3

(p + 500)2 + (500p

3)2¡ p

p2 + 106

#I1(p) =

I(p)

p2LC + pR2C + 1=

0:01¢1000p2+10002

p2 ¢ 1 ¢ 10¡6 + p ¢ 1000 ¢ 10¡6 + 1=

0:01 ¢ 1000

p2 + 10002

106

p2 + 1000p + 106

=1000 + p

p2 + 1000p + 106¡ p

p2 + 106=

= 0:01

"p + 500

(p + 500)2 + (500p

3)2+

1p3

500p

3

(p + 500)2 + (500p

3)2¡ p

p2 + 106

#

i(t) = 10 e¡500 t

·cos

³500

p3t

´+

1p3

sin³500

p3t

´¸+10 sin

³1000 t¡ ¼

2

´mAi(t) = 10 e¡500 t

·cos

³500

p3t

´+

1p3

sin³500

p3t

´¸+10 sin

³1000 t¡ ¼

2

´mA

We don’t solve any AC, steady state is part of solution


XE31EO2 -

Pav

el Máš

a

266666664

loop I1z }| {R1 +

1

pC

branch 12z }| {¡ 1

pC

¡ 1

pC| {z }branch 21

pL + R2 +1

pC| {z }loop I2

377777775¢"I1(p)

I2(p)

#=

24 U (p)¡ uC(0+)p

uC(0+)p + LiL(0+)

35266666664

loop I1z }| {R1 +

1

pC

branch 12z }| {¡ 1

pC

¡ 1

pC| {z }branch 21

pL + R2 +1

pC| {z }loop I2

377777775¢"I1(p)

I2(p)

#=

24 U (p)¡ uC(0+)p

uC(0+)p + LiL(0+)

35

R1I1(p) +1

pC[I1(p)¡ I2(p)] +

uC(0)

p¡ U (p) = 0R1I1(p) +

1

pC[I1(p)¡ I2(p)] +

uC(0)

p¡ U (p) = 0

pLI2(p)¡ LiL(0) + R2I2(p)¡ uC(0)

p+

1

pC[I2(p)¡ I1(0)] = 0pLI2(p)¡ LiL(0) + R2I2(p)¡ uC(0)

p+

1

pC[I2(p)¡ I1(0)] = 0

R1 L

C R2

uC(0)uC(0)uC(0)

puC(0)

p

iL(0)iL(0)

LiL(0)LiL(0)

U(p)U(p) I1(p)I1(p) I2(p)I2(p)

24R1 + 1j!C ¡ 1

j!C

¡ 1j!C j!L + R2 + 1

j!C

35 ¢24I1

I2

35 =

"U

0

#24R1 + 1j!C ¡ 1

j!C

¡ 1j!C j!L + R2 + 1

j!C

35 ¢24I1

I2

35 =

"U

0

#

R1i1(t) +1

C

Z t

0

[i1(¿ )¡ i2(¿ )] d¿ + uC

(0+)¡ u(t) = 0R1i1(t) +1

C

Z t

0

[i1(¿ )¡ i2(¿ )] d¿ + uC

(0+)¡ u(t) = 0

Ldi2(t)

dt+ R2i2(t)¡ u

C(0+) +

1

C

Z t

0

[i2(¿ )¡ i1(¿ )] d¿ = 0Ldi2(t)

dt+ R2i2(t)¡ u

C(0+) +

1

C

Z t

0

[i2(¿ )¡ i1(¿ )] d¿ = 0

Time domain:

Laplace:

AC:

Matrix notation

Mesh analysis:


XE31EO2 -

Pav

el Máš

a

R1 L

C R2

266666664

node U1z }| {1

R1+ pC +

1

pL

branch between nodes 1, 2z }| {¡ 1

pL

¡ 1

pL| {z }branch between nodes 2, 1

1

pL+

1

R2| {z }node U2

377777775¢"U1(p)

U2(p)

#=

24U(p)R1

+ CuC(0+)¡ iL(0+)p

iL(0+)p

35266666664

node U1z }| {1

R1+ pC +

1

pL

branch between nodes 1, 2z }| {¡ 1

pL

¡ 1

pL| {z }branch between nodes 2, 1

1

pL+

1

R2| {z }node U2

377777775¢"U1(p)

U2(p)

#=

24U(p)R1

+ CuC(0+)¡ iL(0+)p

iL(0+)p

35

U1(p)¡ U(p)

R1+

U1(p)¡ U2(p)

pL+ pCU1(p)¡ CuC(0) +

iL(0)

p= 0

U1(p)¡ U(p)

R1+

U1(p)¡ U2(p)

pL+ pCU1(p)¡ CuC(0) +

iL(0)

p= 0

U2(p)¡ U1(p)

pL+

U2(p)

R2¡ iL(0)

p= 0

U2(p)¡ U1(p)

pL+

U2(p)

R2¡ iL(0)

p= 0

U(p)U(p)

CuC(0)CuC(0)

iL(0)p

iL(0)p

U1(p)U1(p) U2(p)U2(p)

24 1R1

+ j!C + 1j!L ¡ 1

j!L

¡ 1j!L

1j!L + 1

R2

35 ¢24U1

U2

35 =

24 UR1

0

3524 1R1

+ j!C + 1j!L ¡ 1

j!L

¡ 1j!L

1j!L + 1

R2

35 ¢24U1

U2

35 =

24 UR1

0

35

u1(t)¡ u(t)

R1+

1

L

Z t

0

[u1(¿)¡ u2(¿ )] d¿ + Cdu1(t)

dt+ iL(0) = 0

u1(t)¡ u(t)

R1+

1

L

Z t

0

[u1(¿)¡ u2(¿ )] d¿ + Cdu1(t)

dt+ iL(0) = 0

1

L

Z t

0

[u2(¿)¡ u1(¿)] d¿ +u2(t)

R2¡ iL(0) = 0

1

L

Z t

0

[u2(¿)¡ u1(¿)] d¿ +u2(t)

R2¡ iL(0) = 0

Time domain:

Laplace:

AC:

Matrix notation

Nodal analysis:

Be careful – if circuit contains controlled sources,it is necessary introduce other, more complicated rulesfor direct writing of matrices (nulor model)!!!


XE31EO2 -

Pav

el Máš

a

It is suitable for DC, AC, LaplaceGaussian elimination solution, Cramer’s rule, inverse matrix – very easy with mathematical computer programs (Matlab, Maple, …)

Could not be used in time domain

CIRCUIT EQUATIONS – MATRIX NOTATION


XE31EO2 -

Pav

el Máš

a

Documents

Circuit equations in time domain andMá a frequencyamber.feld.cvut.cz/vyu/eo2/english/files/5_Equations.pdf · Circuit equations in time domain and frequency domain EO2 –Lecture