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Sinusoidal Steady-State Analysis
Instructor: Chia-Ming TsaiElectronics Engineering
National Chiao Tung UniversityHsinchu, Taiwan, R.O.C.
Contents• Introduction
• Nodal Analysis
• Mesh Analysis
• Superposition Theorem
• Source Transformation
• Thevenin and Norton Equivalent Circuits
• OP-amp AC Circuits
• Applications
Introduction• Steps to Analyze ac Circuits:
– Transform the circuit to the phasor (frequency) domain.
– Solve the problem using circuit techniques (nodal/mesh analysis, superposition, etc.).
– Transform the resulting phasor to the time domain.
Nodal Analysis• Variables = Node Voltages
• Applying KCL to each node gives each independent equation
• If supernodes included,– Applying KCL to each supern
ode gives 1 equation.– Applying KVL at each supern
ode gives 1 more equation.
Supernode
Example 1
5.2 F 1.0
2H .50
4 H 1
1
/ 4
0204cos20
j
j
j
Cj
Lj
srad
t
C
L
Z
Z
Find ix.
Example 1 (Cont’d)
(2) 015115.2
242
2, nodeat KCL Applying
(1) 205.2)5.11(
45.210
20
1, nodeat KCL Applying
211
221
21
2111
VVV
I
VVVI
VV
VVVV
j
jj
jj
jj
x
x
)4.1084cos(59.7
4.10859.75.2
3.19891.13
43.1897.18
0
20
1511
5.25.11
as formmatrix in put becan (2) and (1)
1
2
1
2
1
ti
j
jj
x
x
VI
V
V
V
V
Example 2
• Applying KCL for the supernode gives 1 equations.
• Applying KVL at the supernode gives 1 equations.
• 2 variables solved by 2 equations.
Example 2 (Cont’d)
48.7078.25
4510
18.8741.31
4510But
)21(436or
12633
supernode, at the KCL Applying
:Sol
21
2
21
21
221
VV
V
VV
VV
VVV
jj
jj
Mesh Analysis
Supermesh
SIii 12
Excluded
• Variables = Mesh Currents
• Applying KVL to each mesh gives each independent equation
• If supermeshes included,– Applying KVL to each sup
ermesh gives 1 equation.– Applying KCL at each supe
rnode gives 1 more equation.
Example 1
Find Io.
78.14412.6
22.3512.6
30
50
442
288
5 3,mesh For
(2) 09020)2(
)2()224(
2,mesh For
(1) 010)2(
)2108(
1,mesh toKVL Applying
:Sol
2
2
2
1
3
3
12
32
1
II
I
I
I
I
I
II
II
I
o
j
j
jj
jj
j
jjj
jj
jj
Example 2Find Vo.
• Applying KVL for mesh 1 & 2 gives 2 equations.
• Applying KVL for the supermesh gives 1 equations.
• Applying KCL at node A gives 1 equations.
• 4 variables solved by 4 equations
Example 2 (Cont’d)
(3) 05)56(
8)48(
(2) 32,mesh For
(1) 1082)28(
08)2()28(10
1,mesh For
:Sol
24
13
2
321
321
II
II
I
III
III
jj
j
jj
jj
)(2
equations. 4 using
by solved becan variables4
(4) 4
givesA nodeat KCL Applying
21
34
IIV
II
jo
Superposition Theorem• Since ac circuits are linear, the superposition th
eorem applies to ac circuits as it applies to dc circuits.
• The theorem becomes important if the circuit has sources operating at different frequencies.– Different frequency-domain circuit for each frequen
cy– Total response = summation of individual responses
in the time domain– Total response summation of individual response
s in the phasor domain
Example 1
Find Io.
= +
"0
'00 III
Example 1 (Cont’d)
78.14412.6529.35
)176.1353.2()647.2353.2(
176.1647.2
10
50
442
288
(3) 5 3,mesh For
(2) 0)2()2()44(
2,mesh For
)1( 0210)88(
1,mesh toKVL Applying
,get To
353.2353.2
25.425.4
20
24
20
25.225.0)108(||)2(
:Sol
"0
'00
2"0
2
1
3
312
231
"0
'0
j
j
j
j
j
jj
jj
jjj
jjj
j
j
j
j
j
jjj
III
II
I
I
I
III
III
I
ZI
Z
Example 2
sourcecurrent 5sin2 the todue is
source voltage2cos10 the todue is
source voltagedc V-5 the todue is
where
3
2
1
3210
tv
tv
v
vvvv
Example 2 (Cont’d)
circuitopen1
circuit-short 0
,0 Since
Cj
Lj
V 1)5(41
1
division, By voltage
1
v
Example 2 (Cont’d)
2V
V 010
4j
5j
51
F 1.0
4 H 2
rad/s 2 ,0102cos10
jCj
jjωω
t
V )79.302cos(498.2
79.302.498
049.2439.3
10
)010()4||5(41
1
2
2
tv
j
jjV
Example 2 (Cont’d)
3V
10j
2jA 902
21
F 1.0
10 H 2
rad/s 5 ,9025sin2
jCj
jLj
t
V )805cos(33.2
8033.2
)2(4.88.1
101
)902()4||2(110
10
division,current By
3
13
1
tv
jj
j
jj
j
IV
I
Source Transformation
Example 1
V 28519.5
)105(1013425.15.2
10
division, By voltage
105)25.15.2(4
48
)43(54)43(||5
49045
9020
jjj
jjj
j
jjj
j
x
ss
s
V
IV
I
sI
sV
Find Vx.
Thevenin & Norton Equivalent Circuits
N
ThNTh I
VZZ
Example 1
V 31.22095.37
)75120(124
12
68
8
64.248.6
)12||4()6||8(
Th
Th
j
j
j
j
jj
V
Z
Example 2
Example 2 (Cont’d)
3
)6(2
)6(2)4234(
gives KVL Applying
25.03
,simplicityfor 3Set
sTh
0s
000
j
jjj
s
s
s
I
VZ
IV
IIII
I
V 9055
55
)34(5)42(10
0)34(5.0)42(
gives loop the toKVL Applying
105.015
gives 1 nodeat KVL Applying
Th
Th00
000
j
jj
jj
V
VII
III
Example 3
NN
N0
)1520(
division,current By
IZ
ZI
j
Example 3 (Cont’d)
A 48.38465.1 15205
5
83
give )3( and ),2( ),1(
(3) 3
gives nodeat KCL Applying
(2) 0)218()410()213(
givessupermesh for the KVL Applying
)1( 0)410()28()218(40
gives 1mesh for KVL Applying
.get toanalysismesh Apply )2(
5 ,easily found becan (1)
N0
3N
23
132
321
N
NN
II
II
II
III
III
I
ZZ
j
j
a
jjj
jjjj
OP AMP AC Circuits: Example 1• Ideal op amps assumed
– Zero input current & zero differential input voltage
V )04591000cos(0291)(
04.59029.1 53
6
give )2( and )1(
(2)
10
0
10
0
2, nodeat KCL Applying
)1( -)45(6
2010
0
510
3
1, nodeat KCL Applying
1
1
1
1111
.t.tv
j
j
j
j
j
o
o
o
o
o
o
V
VV
VV
VV
VVVVV
V 1000cos3 tvs
Example 2
.shift phase andgain
loop-close theFind
rad/s 200
F 1
F 2
k 10
2
1
21
C
C
RR
130.6 :shift Phase
434.0:gain loopOpen
6.130434.0 )21)(41(
4
)1)(1(
1
1||
:Sol
2211
21
11
22
jj
j
CRjCRj
RCj
CjR
CjR
i
f
s
o
Z
Z
V
VG
Applications: Capacitance Multiplier
i
ooi
i
i
ii
oioi
i
CjCj
CjCj
V
VVV
V
I
VZ
VVVV
I
1
1
)(
)(1
CR
RC
Cj
R
R
i
i
o
1
2eq
eq
1
2
1 where
1
But
Z
V
V
Applications: Oscillators• Barkhausen criteria must be meet for oscillators
– (1) Overall gain 1– (2) Overall phase shift = 0
gfg
f
o
o
o
RRR
R
RCCR
CRjRC
RC
CCCRRR
CjRCjR
CjR
2113
1:)1(
for 3
1
101:)2(
)1(3
, and If
1||1
1||
02
0222
0
2222
2121
2211
222
V
V
V
V
V
V
