Chapter 15 Circuit Analysis in the...
Transcript of Chapter 15 Circuit Analysis in the...
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Chapter 15Circuit Analysis in the s-Domain
15.1 Z(s) and Y(s)
15.2 Nodal and Mesh analysis in the s-domain
15.3 Additional Circuit Analysis Techniques
15.4 Poles, Zeros, and Transfer Functions
15.5 Convolution
15.6 The Complex-frequency Plane
15.7 Natural Response and The s Plane
15.8 A Technique for Synthesizing the Voltage Ration H(s)=Vout/Vin
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15.1 Z(s) and Y(s)
Resistors in the Frequency Domain
Ohm’s law specifies that 𝑣𝑣 𝑡𝑡 = 𝑅𝑅𝑅𝑅(𝑡𝑡)
Taking the Laplace transform of both sides : 𝑽𝑽 𝒔𝒔 = 𝑅𝑅𝑰𝑰 𝒔𝒔
The impedance Z(s) is defined as
and the admittance 𝑌𝑌 𝑠𝑠 = ⁄𝑰𝑰(𝒔𝒔) 𝑽𝑽(𝒔𝒔) is 1/R
Z(s) ≡V(s)I(s)
= R
Inductors in the Frequency Domain
( )( )
( ) ( ) (0 )
( ) ( ) for zero initial condition( ) ( ) (0 ) ( ) (0 )( ) or ( )( )
dit Ldt
V s L sI s i
V s sLI sV s V s Li V s iZ s sL I sI s sL sL s
υ
−
− −
=
= −
=
+= = = = +
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15.1 Z(s) and Y(s)
Example 15.1 Calculate the voltage 𝑣𝑣(𝑡𝑡) given an initial current 𝑅𝑅 0− = 1 𝐴𝐴
8 0.5
3 2 9.58( )1 2 ( 8)( 0.5)
2 ( 9.5)( ) 2 ( ) 2 2( 8)( 0.5)
2 8 3.2 1.2( 8)( 0.5) 8 0.5
( ) 3.2 1.2 ( ) Vt t
ssI ss s s
s sV s sI ss s
ss s s s
t e e u tυ − −
+ ++= =+ + +
+→ = − = −
+ +−
= = −+ + + +
→ = −
Capacitors in the Frequency Domain
( )( )
( ) ( ) (0 )
( ) (0 ) ( ) (0 )( )
di t Cdt
I s C sV s
I s C I sV ssC sC s
υ
υ
υ υ
−
− −
=
= −
+= = +
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15.1 Z(s) and Y(s)
Example 15.2 Determine 𝑣𝑣𝑐𝑐(𝑡𝑡) given an initial current 𝑣𝑣𝑐𝑐 0− = −2 𝑉𝑉
23
912 3
18 6 2 6 9 113 2 ( 2 3) 2 3
( ) 9 11 ( ) Vυ −
−− = +
− − +→ = = = −
+ + +
→ = −
C C
C
t
C
V V ss
s sVs s s s s
t e u t
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15.2 Nodal and Mesh Analysis in the s-Domain
Example 15.3 Determine the mesh currents 𝑅𝑅1 𝑎𝑎𝑎𝑎𝑎𝑎 𝑅𝑅2. No initially stored energy
1 1 2
2 1 2
2 2
1 4 3 2 2
2 2
2 4 3 2 2
4 3 10 10 02
2 10 10 4 01
2 (4 19 20) 2 (4 19 20)20 66 73 57 30 ( 1)( 2)(20 6 15)
30 43 6 30 43 620 66 73 57 30 ( 1)( 2)(20 6 15)
− + + − = +− + − + = +
+ + + += = + + + + + + + +→
+ + + + = =+ + + + + + + +
I I Is s
I I sIs
s s s s s sIs s s s s s s s
s s s sIs s s s s s s s
2 0.151
0.15
2 0.152
0.15
( ) 96.39 344.8 841.2 cos 0.8529
197.7 sin 0.8529( ) 481.9 241.4 723.3 cos 0.8529
472.8 sin 0.8529
− − −
−
− − −
−
= − − +
+→ = − − +
+
t t t
t
t t t
t
i t e e e te t mA
i t e e e te t mA
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15.2 Nodal and Mesh Analysis in the s-Domain
Example 15.4 Calculate 𝑣𝑣𝑥𝑥 using nodal analysis
2 2
2
1.707 0.2929
7 412 4
10 4 5 2 4 6.864 5.864(2 4 1) 2 22 2 1 11 1
2 22 2
4 6.864 5.864 ( )υ − −
− −− = + +
+ + −→ = = = + +
+ + + + + −+ + + −
→ = + −
x xx
x
t tx
V s V sVs s
s sVs s s s
s ss s s
e e u t
(0 ) 2 Vυ − =C
+ υ −C
[ ]2
2
7 4 28 3Check, (0 ) lim 2 V(2 4 1)
υ +
→∞
+ += − = → = =
+ +C x C Cs
s sV V sVs s s s
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15.2 Nodal and Mesh Analysis in the s-Domain
Example 15.5 Use nodal analysis to find 𝑣𝑣1,𝑣𝑣2, 𝑎𝑎𝑎𝑎𝑎𝑎 𝑣𝑣3. No initially stored energy
1 2
2 32 1 2
3 2 32
3 2
1 3
2
2 3
2 3
0.13 100
0100 7 6
0.26 2
100 7 150 493( 3)(30 45 14)
3 17( 3)(30 45 14)
1.4(6 5) 1.4(6 5)( 3)(30 45 14) 30( 3)( 0.2941)(( 0.14
−=
+ −−
+ + = −
− = +
+ + +=
+ + +
+→ =
+ + +− − − −
= =+ + + + + −
V Vs
V VV V Vs s
V V VVs s
s s sVs s s
sVs s s
s sVs s s s s s 2 271) 1.251 )
+
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15.3 Additional Circuit Analysis Techniques
Example 15.6 Find 𝑣𝑣(𝑡𝑡) using source transformation
2
22
22 2 2
2
2 10(9 2)9 102 9 2 99 10 2( ) ...22 2 10(9 2) 29 9 9 299 10
9 10 2
+ + + + = = = ++ + + ++ + + + +
P
P
sss s s ss s sV s
ss s sssss s s s s
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15.3 Additional Circuit Analysis Techniques
Example 15.6 Find 𝑣𝑣(𝑡𝑡) using source transformation
2 2
1 2 12 2
2 1
2
2 22
4
2 3 2
2 20 201010 2 9 9 10 22 20 2 40 4
10 2 (10 2)9 20 9( ) 40 49 9 10 2 9
(10 2)180
( 9)(90 18 40 4)
= = → = ⋅ = ⋅+ + + +
+→ = + = + =
+ +
→ = ⋅ = ⋅ ⋅++ + + ++
=+ + + +
P s sZ V Zs s s s s
sZ Zs s s s s
s s sV s V sZ s s s ss s
ss s s s
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15.3 Additional Circuit Analysis Techniques
Example 15.6 Find 𝑣𝑣(𝑡𝑡) using source transformation
4
2 3 2
5
3.912 3.912 15
180( )( 9)(90 18 40 4)
1.047 0.0716 1.047 0.0716 0.0471 0.01913 3 0.04885 0.6573
0.0471 0.0191 5.59 100.04885 0.6573 0.1023
1.049 1.049 0.050833 3
−
° − °
=+ + + +
+ − += + +
− + + −
− ×+ +
+ + +
= + +− +
j j j
sV ss s s s
j j js j s j s j
js j s
e e es j s j
7.9
157.9 5
3.912 3 3.912 3 157.9 0.04885 0.6573
157.9 0.04885 0.6573 5
0.04885 0.65730.05083 5.59 100.04885 0.6573 0.1023
1.049 1.049 0.05083( )
0.05083 5.59 10υ
°
− ° −
° − ° − ° −
− ° − − − −
+ −
×+ +
+ + +
+ +→ =
+ + ×
j
j j t j j t j t j t
j t j t
s je
s j s
e e e e e e et
e e e e 0.1023
0.04885 5 0.1023
( )
2.908cos(3 3.912 ) 0.1017 cos(0.6573 157.9 ) 5.59 10 ( ) V− − −
= + ° + + ° + ×
t
t t
u t
t e t e u t
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15.3 Additional Circuit Analysis Techniques
Example 15.7 Find the frequency-domain Thévenin equivalent network
inZ
1( )1 1
π π ππ
ππ
π
π
π π π π
= =
+ ++ = = =
→ = =+ + +
P P
inin in
in in in E E
eq EE
Ein in
E E E
VZ V
V V V r R sR r CgVZ R rR r
sCR rZ V
r R sR r C gR r
inZ
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15.4 Poles, Zeros, and Transfer Functions
𝐻𝐻 𝑠𝑠 is the transfer function of the circuit, defined as the ratio of the output to the input. No initial conditions
( ) 1 1/( )( ) 1 1/
out
in
V s RCH sV s sRC s RC
≡ = =+ +
𝐻𝐻 𝑠𝑠 has “zero” at 𝑠𝑠 = ∞
𝐻𝐻 𝑠𝑠 has “pole” at 𝑠𝑠 = ⁄−1 𝑅𝑅𝑅𝑅
[ ]1
( ) ( ) ( )
( ) ( ) ( )out in
out in
V s H s V sV t L H s V s−
= ⋅
→ = ⋅