회로이론-ІI 2016년1학기 이문석 1

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−

= ⋅

→ = ⋅