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Circuit equations
in time domain and frequency
domain
XE
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EO2 – Lecture 5
Pavel Máša
XE31EO2 - Pavel Máša - Lecture 5
INTRODUCTION
In last semester we studied circuit equations in DC and AC circuits
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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?
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MESH (LOOP) ANALYSIS – KVL
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Circuit equations, regardless of used mathematical apparatus, are always mathematical formulation of Kirchhoff’s laws:
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Uk = 0
k
voltage across R, L, C is qualified by means of current
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NODAL ANALYSIS – KCL
Ik = 0
k
current, passing R, L, C is qualified by means of voltage
XE31EO2 - Pavel Máša - Lecture 5
RELATIONSHIP BETWEEN VOLTAGE AND CURRENT
Circuit element
Time domain
DC
AC / Fourier
Laplace
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C
usage
transient analysis
usually DC/ AC analysis has to be also proceeded
open circuit
Steady state with
DC source
AC – steady state with
sine wave source
Fourier
series – steady state with periodical excitation
transform – pulse excitation
General application
contains both steady and
transient components
XE31EO2 - Pavel Máša - Lecture 5
INITIAL CONDITIONS
Energetic initial conditions
– since energy is continuous, energetic circuit variables are also continuous
q = Cu
dq
i=
dt
iC (0¡) 6= iC (0+)
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duC ( t )
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dt
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iC ( t ) = C
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must be uC (0¡) = uC (0+) may be
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d ΦC
u (t ) =
dt
uL(0¡)
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ΦC = Li
uL(0¡) 6= uL(0+) must be iL(0¡) = iL(0+)
2
may be
iL(0¡)
iL(0+)
uL(0+)
History of capacitor describes charge stored in the capacitor œ voltage
History of inductor describes magnetic flux passing the inductor œ electric current
XE31EO2 - Pavel Máša - Lecture 5
BASIC RULES – SIGNS, …
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Mesh analysis
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Nodal voltages
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– 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.
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– 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
XE31EO2 - Pavel Máša - Lecture 5
Reduction of number of equations
DC
AC
not possible
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Time domain
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Time domain
Laplace
XE31EO2 - Pavel Máša - Lecture 5
COUPLED INDUCTORS (TRANSFORMER)
Time domain – mesh analysis
di2 (t)
di1 (t)
+M
dt
dt
di2 (t)
di1 (t)
u2 (t) = L2
+M
dt
dt
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u1 (t) = L1
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Laplace transform – mesh analysis
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U1 (p) = pL1 I1 (p) ¡ L1 i1 (0+ ) + pM I2 (p) ¡ M i2 (0+ )
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U2 (p) = pL2I2(p) ¡ L2 i2 (0+ ) + pM I1 (p) ¡ M i1 (0+ )
2
U2 (p) + L2 i2 (0+ ) ¡ pM I1 (p) + M i1(0+ )
pL2
U2 (p) + L2 i2 (0+ ) ¡ pM I1 (p) + M i1 (0+ )
¡M i2(0+ )
pL2
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U1 (p) = pL1 I1 (p)¡L1 i1 (0+)+pM
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I2 (p) =
– nodal voltages
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1
I1 (p) =
L2
M
1
1
i1 (0+ ) 1
1
i1 (0+ )
U1(p) ¡
U2 (p) ¡
= ¡1 U1(p) + ¡M U2 (p) ¡
2
2
p L1 L2 ¡ M
p L1 L2 ¡ M
p
p
p
p
I2(p) =
1
1
i2 (0+ )
¡M U1 (p) + ¡2 U2(p) ¡
p
p
p
XE31EO2 - Pavel Máša - Lecture 5
Example:
L
R1
uC (0)
R2
1 ^ ^
(I1 ¡ I) = 0
j!L ^I1 +R2^I1 +
j!C
)
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Mesh analysis – 0 equations, voltages across R1 and R2 is evaluated by Ohm’s law
^I1 =
^I
(j!)2 LC + j!R2 C + 1
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AC:
iL (0)
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DC:
C
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Time domain:
Z
di1 (t)
1 t
+ R2i1(t) +
L
[i1 (¿ ) ¡ i(¿ )] d¿ ¡ uC (0) = 0
dt
C 0
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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 change
2. Solution of integral‐differential equation – solution of the transient
3. To find steady state in the circuit after transient is over we have to proceed DC or AC analysis again 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 proceeded
XE31EO2 - Pavel Máša - Lecture 5
Laplace:
pL I1 (p)¡LiL(0) + R2 I1 (p) +
1
uC (0)
[I1(p) ¡ I(p)] ¡
=0
pC
p
I(p)+CuC (0) + pLCiL(0)
)
I1 (p) =
p2 LC + pR2C + 1
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The 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
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• 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), or
source is disconnected (then the steady state is „hidden“ in the initial conditions)
¾ The transient is integral part of Laplace transform solution, the result of time interval restriction
‐ each source is zero when t < 0, because u(t) = u0 (t)1(t) , i(t) = i0 (t)1(t)
• 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
XE31EO2 - Pavel Máša - Lecture 5
•
•
Now assume following values: R2 = 1 kΩ, L = 1H, C = 1 μF, R1 = 2 kΩ
DC – Circuit supplied by current source I = 10 mA
U2 = R2 I = 1000 ¢ 0:01 = 10 V
AC – circuit supplied from sinusoidal current source
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Z
t
[i1(¿ ) ¡ i(¿ )] d¿ ¡ uC (0) = 0
2
di1 (t)
1
+ R2i1(t) +
L
dt
C
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Initial condition is given – uc(0) = 0
We solve integral‐differential equation (in detail it will be described on 8th lecture)
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Time domain – at time t = 0 was connected sinusoidal current source
. i(t) = 10 sin(1000t)
mA At t < 0 the circuit was without energy.
a)
Derive
b)
Solve using method of variation of parameters
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^I
0:01
¡ ¼2 j
=
=
10e
mA
(j!)2 LC + j!R2 C + 1 (j ¢ 1000)2 ¢ 1 ¢ 10¡6 + j ¢ 1000 ¢ 1000 ¢ 10¡6 + 1
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^
I1 =
i(t) = 10 sin(1000t) mA
d2 i1 (t)
d i1(t) i1 (t) ¡ i(t)
L
+
=0
+
R
2
dt2
dt
C
R2
1
¸2 + ¸ +
= 0 ) ¸2 + 1000¸ + 106 = 0
L
LC
p
¸1;2 = ¡500 § 5002 ¡ 106 = ¡500 § 866:03j
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U1 = R1 I = 2000 ¢ 0:01 = 20 V
¼
i1 (t) = [K1 cos(866:03t) + K2 sin(866:03t)] e¡500t + 0:01 sin(1000t ¡ )
2}
|
{z
AC solved above
XE31EO2 - Pavel Máša - Lecture 5
Laplace
0:01¢1000
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106
0:01 ¢ 1000
I(p)
p2 +10002
=
=
I1(p) = 2
p LC + pR2C + 1 p2 ¢ 1 ¢ 10¡6 + p ¢ 1000 ¢ 10¡6 + 1 p2 + 10002 p2 + 1000p + 106
1000 + p
p
= 2
¡
=
p + 1000p + 106 p2 + 106
"
#
p
500 3
1
p
p + 500
p +p
p ¡ 2
= 0:01
(p + 500)2 + (500 3)2
3 (p + 500)2 + (500 3)2 p + 106
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· ³
´
´¸
³
³
p
p
1
¼´
¡500 t
cos 500 3t + p sin 500 3t +10 sin 1000 t ¡
i(t) = 10 e
mA
2
3
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We don’t solve any AC, steady state is part of solution
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XE31EO2 - Pavel Máša - Lecture 5
L
R1
uC (0)
U (p)
I1 (p)
C
LiL(0)
iL(0)
uC (0)
p
I2 (p)
Time domain:
Z
1 t
R1 i1 (t) +
[i1 (¿ ) ¡ i2 (¿ )] d¿ + uC (0+) ¡ u(t) = 0
C 0
R2
Z
di2(t)
1 t
L
[i2(¿ ) ¡ i1(¿ )] d¿ = 0
+ R2 i2 (t) ¡ uC (0+ ) +
dt
C 0
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Mesh analysis:
uC (0)
1
[I2(p) ¡ I1 (0)] = 0
+
p
pC
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pLI2 (p) ¡ LiL(0) + R2 I2 (p) ¡
el
1
uC (0)
[I1(p) ¡ I2(p)] +
¡ U (p) = 0
pC
p
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R1 I1(p) +
M
Laplace:
Matrix notation
branch 21
AC:
2
4
1
R1 + j!C
1
¡ j!C
loop I2
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2
loop I1
branch 12
z }| {
z }| {
7
6
1
1
3
7 "
6R1 +
# 2
u
¡
C (0+ )
7 I1 (p)
6
U (p) ¡ p
pC
pC
7
6
4
5
=
7¢
6
uC(0+) + Li (0 )
6
1
1 7 I2 (p)
L +
7
6 ¡
pL
+
R
+
p
2
5
4
pC
pC
| {z } |
{z
}
3 2 3 " #
^I1
^
U
5¢4 5=
1
^I2
0
j!L + R2 + j!C
1
¡ j!C
XE31EO2 - Pavel Máša - Lecture 5
Nodal analysis:
U1 (p)
R1
L
U(p)
C
iL (0)
p
R2
Z
du1(t)
u1(t) ¡ u(t) 1 t
+ iL (0) = 0
+
[u1(¿ ) ¡ u2(¿ )] d¿ + C
R1
L 0
dt
Z
u2(t)
1 t
[u2(¿ ) ¡ u1 (¿ )] d¿ +
¡ iL(0) = 0
L 0
R2
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CuC (0)
Time domain:
U2 (p)
Laplace:
z }| {
1
¡
pL
1
1
+
pL R
| {z 2}
branch between nodes 2, 1
4
1
R1
+ j!C +
1
¡ j!L
1
j!L
7
3
7"
# 2 U (p)
iL (0+ )
7 U1(p)
R1 + CuC (0+ ) ¡
p
7
4
5
=
7¢
iL (0+ )
7 U2(p)
7
p
5
node U2
AC:
2
Matrix notation
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z
}|
{
6 1
1
6
+
pC
+
6 R1
pL
6
6
6
1
6
¡
4
pL
| {z }
3
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branch between nodes 1, 2
node U1
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2
2
U2(p) ¡ U1(p) U2(p) iL (0)
+
=0
¡
pL
R2
p
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U1(p) ¡ U (p) U1(p) ¡ U2 (p)
iL(0)
+ pCU1 (p) ¡ CuC (0) +
=0
+
R1
pL
p
1
¡ j!L
1
j!L
+ R12
3 2 3 2 3
^
U
^1
U
R
5 ¢ 4 5 = 4 15
^2
0
U
Be careful – if circuit contains controlled sources,
it is necessary introduce other, more complicated rules
for direct writing of matrices (nulor model)!!!
XE31EO2 - Pavel Máša - Lecture 5
CIRCUIT EQUATIONS – MATRIX NOTATION
9 It is suitable for DC, AC, Laplace
9 Gaussian elimination solution, Cramer’s rule, inverse matrix – very easy with mathematical computer programs (Matlab, Maple, …)
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Could not be used in time domain
XE31EO2 - Pavel Máša - Lecture 5