2012/11/13
Sinusoidal Steady-State Analysis
•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:
–The natural response (due to initial
conditions) is ignored.
–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.
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Nodal Analysis
•Variables = Node Voltages.
•Applying KCL to each node
gives each independent
equation.
Supernode
•If supernodes included,
–Applying KCL to each
supernode gives 1 equation.
–Applying KVL at each
supernode gives 1 more
equation.
Example 1
Find ix.
20 cos 4t  200
4 rad / s
Z L jL

Z  1
C

jC

1 H  j4
0.5 H  j 2
0.1 F  j 2.5
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Example 1 (Cont
’
d)
Applying KCL at node 1,
20 V1
V
V V2
 1  1
10
j 2.5
j4
 (1 j1.5)V1 j 2.5V2 20
(1) and (2) can be put in matrix form as
(1)
1 j1.5

 11

V1  
j 2.5
20

 



V2  0 
15 

Applying KCL at node 2,
 V1 18.9718.43
V V2 V2
2I x  1

j4
j2
V
I x  1  11V1 15V2 0 (2)
j 2.5
V
I x  1 7.59108.4
j 2.5
 ix 7.59 cos( 4t 108.4
)
 V2 13.91198.3
Example 2
•Applying KCL for the
supernode gives 1
equations.
•Applying KVL at the
supernode gives 1
equations.
•2 variables solved by 2
equations.
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Example 2 (Cont
’
d)
Sol :
Applying KCL at the supernode,
V
V V
3 1  2  2
j 3 j 6 12
or 36 j 4V1 (1 j 2)V2
But V1 V2 1045
 V2 31.4187.18
 V1 V2 1045
25.7870.48
Mesh Analysis
•Variables = Mesh Currents.
•Applying KVL to each
mesh gives each
independent equation.
•If supermeshes included,
Excluded
Supermesh
–Applying KVL to each
supermesh gives 1 equation.
–Applying KCL at each
supernode gives 1 more
i2 i1 I S
equation.
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Example 1
Sol :
Find Io.
Applying KVL to mesh 1,
(8 j10 j 2)I1
(j 2)I 2 j10I 3 0
(1)
For mesh 2,
(4 j 2 j 2)I 2 (j 2)I1
(j 2)I 3 20900 (2)
For mesh 3,  I 3 5
8 8 j
j2 
I1  j50 



 j2


I2 
4 j 4
j30



 
 I 2 6.1235.22
I o I 2 6.12144.78
Example 2
Find 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
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Example 2 (Cont
’
d)
Sol :
For mesh 1,
10 (8 j 2)I1 ( j 2)I 2 8I 3 0
Applying KCL at node A gives
I 4 I 3 4
(4)
 (8 j 2)I1 j 2I 2 8I 3 10
(1)
For mesh 2,  I 2 3
(2) 4 variables can be solved by
using 4 equations.
For supermesh,
(8 j 4)I 3 8I 1 (6 j 5)I 4 j 5I 2 0 (3)
 Vo j 2(I1 I 2 )
Superposition Theorem
•Since ac circuits are linear, the superposition
theorem 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
frequency.
–Total response = summation of individual
responses in the time domain.
–Total response summation of individual
responses in the phasor domain.
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Example 1
Find Io.
=
+
I 0 I '0 I "0
Example 1 (Cont
’
d)
Sol : Z (j 2) || (8 j10) 0.25 j 2.25
j 20
j 20
I '0 

4 j 2 Z 4.25 j 4.25
2.353 j 2.353
To get I "0 ,
Applying KVL to mesh 1,
(8 j8)I1 j10I 3 j 2I 2 0
(1)
For mesh 2,
(4 j 4)I 2 (j 2)I1 (j 2)I 3 0 (2)
For mesh 3, I 3 5
(3)
I1  j 50 
8 8 j
j2 



 j2


4 j 4
I2 
j10


 

I "0 I 2 2.647 j1.176
I 0 I '0 I "0 (2.353 2.647) j (2.353 1.176)
5 j 3.529 6.12144.78
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Example 2
v0 v1 v2 v3
v1 is due to the 5 - V dc voltage source


where v2 is due to the 10 cos 2t voltage source

v3 is due to the 2 sin 5t current source

Example 2 (Cont
’
d)
Since 0,
By voltage division,
jL 0  short - circuit
1
 open circuit
jC
1
v1 
(5) 1 V
1 4
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Example 2 (Cont
’
d)
j4 
V2
j5 
100V
1
V2 
(100
)
1 j 4 (j 5 || 4)
10 cos 2t  100
, 2 rad/s
10
2 H  jL j 4 

3.439 j 2.049
1
0.1 F 
j 5 
2.49830.79
jC
 v2 2.498 cos(2t 30.79
)V
Example 2 (Cont
’
d)
j10 
V3
290A
2 sin 5t  290
5 rad/s
2 H  jL j10 
1
0.1 F 
j 2 
jC
j 2 
By current division,
j10
(290
)
j10 1 (j 2 || 4)
j10
V3 I1 1 
(j 2)
1.8 j8.4
2.3380
 v3 2.33 cos(5t 80
)V
I1 
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Source Transformation
Example 1
Find Vx.
Is
2090
Is 
490j 4
5
5(3 j 4)
Vs I s 
5 || (3 j 4)
j 4
8 j 4
j 4(2.5 j1.25) 5 j10
By voltage division,
Vs
10
Vx 
(5 j10)
2.5 j1.25 4 j13 10
5.51928V
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Thevenin & Norton Equivalent Circuits
V
Z Th Z N  Th
IN
Example 1
Z Th (8 || j 6) (4 || j12)
6.48 j 2.64
8
j12 
VTh 

(12075
)
8 j 6 4 j12 


37.95220.31V
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Example 2
Example 2 (Cont
’
d)
Applying KVL at node 1 gives
15 I 0 0.5I 0  I 0 10
Applying KVL to the loop gives
I 0 (2 j 4) 0.5I 0 (4 j 3) VTh 0
 VTh 10(2 j 4) 5(4 j 3)
j 55
5590V
Set I s 3 for simplicity,
I s 3 I 0 0.5I 0  I 0 2
Applying KVL gives
Vs I 0 (4 j 3 2 j 4) 2(6 j )
V 2(6 j )
Z Th  s 
Is
3
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Example 3
By current division,
ZN
I0 
IN
Z N (20 j15)
Example 3 (Cont
’
d)
(1) Z N can be found easily, Z N 5
(2) Apply mesh analysis to get I N .
Applying KVL for mesh 1 gives
j 40 (18 j 2)I1 (8 j 2)I 2 (10 j 4)I 3 0 (1)
Applying KVL for the supermesh gives
(13 j 2)I 2 (10 j 4)I 3 (18 j 2)I1 0 (2)
Applying KCL at node a gives
I 3 I 2 3
(3)
(1), (2), and (3) give
I N I 3 3 j8
5
 I0 
I N 1.46538.48A
5 20 j15
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OP AMP AC Circuits: Example 1
•Ideal op amps with negative feedback assumed.
–Zero input current & zero differential input voltage.
Applying KCL at node 1,
3 V1
V
V 0 V1 Vo
 1  1

10
j 5
10
20
 6 (5 j 4)V1 - Vo (1)
Applying KCL at node 2,
V1 0 0 Vo

10
j10
 V1 jVo
(2)
vs 3 cos1000t V
(1) and (2) give
6
Vo 
1.02959.04
3 j5
 vo (t ) 1.029 cos(1000t 59.04
)V
Example 2
Sol :
1
R2 ||
Zf
Vo
jC2
G   
Vs
Zi R  1
1
jC1
jC1 R2

(1 jR1C1 )(1 jR2C2 )
Find the close - loop
gain and phase shift.
R1 R2 10 k
C1 2 F
C2 1 F
200 rad/s
j 4

0.434130.6
(1 j 4)(1 j 2)
Closed loop gain : 0.434


130.6
Phase shift :
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Applications: Capacitance Multiplier
V Vo
Ii  i
jC (Vi Vo )
1 jC
V
Vi
Zi  i 

Ii
jC (Vi Vo )
1
 Vo 
jC 
1 


 Vi 
But
Vo
R
 2
Vi
R1
 Zi 
1
jCeq
 R2 
where Ceq 
1 
C


 R1 
Applications: Oscillators
v1
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Oscillation Conditions
• Barkhausen criteria must be meet for oscillators.
1.
Overall phase shift (input-to-output-to-input) = AH = 0
• The oscillation frequency can be determined.
2.
Overall gain of the oscillator = |AH| 1
• Loses must be compensated by an amplifying device.
• The minimum gain requirement can be determined.
+
+
vi
A
vf
H
A
vo
H
Example
V2
R2 || 1 jC2

Vo R1 1 jC1 
R2 || 1 jC2 
If R1 R2 R and C1 C2 C ,
V2
RC

Vo 3RC j (2 R 2C 2 1)
The phase requirement gives
1
02 R 2C 2 1 0  0 
RC
V 1
 2  for 0
Vo 3
The gain requirement gives
R
1

1 f
3
 Rg

1  R f 2 Rg


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