Electric Circuit Theoryelearning.kocw.net/KOCW/document/2015/korea_sejong/minnamki1/… · CIEN 202 Electric Circuits II Nam Ki Min 010-9419-2320 [email protected] Chapter 13 The

Electric Circuit Theory

Nam Ki Min

010-9419-2320 [email protected]

The Laplace Transform in Circuit Analysis

Chapter 13

Nam Ki Min

010-9419-2320 [email protected]

CIEN 202 Electric Circuits II Nam Ki Min 010-9419-2320 [email protected]

Chapter 13 The Laplace Transform in Circuit Analysis 3 Contents and Objectives

Chapter Contents

13.1 Circuit Elements in the s Domain

13.2 Circuit Analysis in the s Domain

13.3 Applications

13.4 The Transfer Function

13.5 The Transfer Function in Partial Fraction Expansions

13.6 The Transfer Function and the Convolution Integral

13.7 The Transfer Function and the Steady-State Sinusoidal Response

13.8 The Impulse Function in Circuit Analysis

Chapter Objectives

1. Be able to transform a circuit into the s domain using Laplace transforms; be sure you understand how to represent the initial conditions on energy-storage elements in the s domain.

2. Know how to analyze a circuit in the s-domain and be able to transform an s-domain solution back to the time domain.

3. Understand the definition and significance of the transfer function and be able to calculate the transfer function for a circuit using s-domain techniques.

4. Know how to use a circuit’s transfer function to calculate the circuit’s unit impulse response, its unit step response, and its steady-state response to a sinusoidal input.


Chapter 13 The Laplace Transform in Circuit Analysis 4 Circuit Elements in the s Domain


Chapter 13 The Laplace Transform in Circuit Analysis 5 Introduction

The Laplace transform has two characteristics that make it an attractive tool in circuit analysis

• It transforms a set of linear constant-coefficient differential equations into a set of linear polynomial equations, which are easier to manipulate.

• It automatically introduces into the polynomial equations the initial values of the current and voltage variables. Thus, initial conditions are an inherent part of the transform process.

This contrasts with the classical approach to the solution of differential equations, in which initial conditions are considered when the unknown coefficients are evaluated.



Steps in Applying the Laplace Transformation to Circuits

• Transform the circuit from the time domain to the 𝑠 domain.

Having mastered how to obtain the Laplace transform and its inverse, we are now prepared to employ the Laplace transform to analyze circuits. This usually involves three steps.

• Solve the circuit using nodal analysis, mesh analysis, source transformation, superposition, or any circuit analysis technique with which we are familiar.

• Take the inverse transform of the solution and thus obtain the solution in the time domain.


Chapter 13 The Laplace Transform in Circuit Analysis 7

A Resistor in the s Domain

Circuit Elements in the s Domain

• Voltage-current relationship in the time domain

• Voltage-current relationship in the s domain

𝑣 = 𝑅𝑖

𝑉 = 𝑅𝐼

𝑉 = ℒ{𝑣}

𝐼 = ℒ{𝑖}

𝑉 = 𝑅𝐼

Phasor

(13.1)

(13.2)




𝑣 = 𝐿𝑑𝑖

𝑑𝑡

= ℒ 𝐿𝑑𝑖

𝑑𝑡= 𝑉 = ℒ{𝑣} 𝐿 𝑠𝐼 − 𝑖 0−

= 𝑠𝐿𝐼 − 𝐿𝐼0

𝑉 = 𝑠𝐿𝐼

If 𝐼0 = 0

𝑠𝐿: 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝑉 = 𝑗𝜔𝐿𝐼

Phasor

An Inductor in the s Domain

(13.3)


(13.4)



If 𝐼0 ≠ 0

𝑉 = 𝑠𝐿𝐼 − 𝐿𝐼0

𝐼 =𝑉 + 𝐿𝐼0

𝑠𝐿=

𝑉

𝑠𝐿+

𝐼0𝑠

or

(13.5)

(13.4)

An Inductor in the s Domain




𝑖 = 𝐶𝑑𝑣

𝑑𝑡

𝐶[𝑠𝑉 − 𝑣 0 ]

= 𝑠𝐶𝑉 − 𝐶𝑉0

𝐼 = ℒ 𝑖 = ℒ 𝐶𝑑𝑣

𝑑𝑡=

𝐼 = 𝑠𝐶𝑉

If 𝑉0 = 0

𝑠𝐶: 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝑉 =𝐼

𝑠𝐶


(13.6)

𝑉 =𝐼

𝑗𝜔𝐶

Phasor

(13.7)

A Capacitor in the s Domain



If 𝑉0 ≠ 0

or

𝑉 =1

𝑠𝐶𝐼 +

𝑉0

𝑠

𝐼 = 𝑠𝐶𝑉 − 𝐶𝑉0 (13.7)

=𝑉

1𝑠𝐶

− 𝐶𝑉0

(13.8)

A Capacitor in the s Domain



Summary of the s Domain Equivalent Circuits


Chapter 13 The Laplace Transform in Circuit Analysis 13

Time-domain and s-domain representations of passive elements under zero initial conditions

Circuit Elements in the s Domain

𝐼0 = 0 𝑉0 = 0

𝑉(𝑠) = 𝑅𝐼 𝑠 𝑉 𝑠 = 𝑠𝐿𝐼 𝑠 𝑉 𝑠 =1

𝑠𝐶𝐼(𝑠)

𝑍(𝑠) = 𝑅 𝑍 𝑠 = 𝑠𝐿 𝑍 𝑠 =1

𝑠𝐶

𝑍 = 𝑅 𝑍 = 𝑗𝜔𝐿 𝑍 =1

𝑗𝜔𝐶 Phasor


Chapter 13 The Laplace Transform in Circuit Analysis 14 Circuit Analysis in the s Domain

Ohm’s law in the s-domain

Impedance and Admittance in the s-domain

Kirchhoff’s laws

𝑉(𝑠) = 𝑍(𝑠)𝐼(𝑠)

𝑍(𝑠) =𝑉(𝑠)

𝐼(𝑠)

𝑌(𝑠) =1

𝑍(𝑠)=

𝐼(𝑠)

𝑉(𝑠)

alg 𝐼(𝑠) = 0 KCL:

KVL: alg 𝑉(𝑠) = 0

Basic Laws in the s Domain

(13.9)

(13.11)

(13.10)



• In the time domain

RC Circuits

KVL :

• In the s domain

𝑉0

𝑠=

1

𝑠𝐶𝐼 + 𝑅𝐼

KVL :

𝑅𝐼 −𝑉0

𝑠+

1

𝑠𝐶𝐼 = 0

Or

𝐼(𝑠) =𝐶𝑉0

𝑅𝐶𝑠 + 1=

𝑉0

𝑅 𝑠 +1

𝑅𝐶

(13.12)

(13.13)

𝑖 𝑡 =𝑉0

𝑅𝑒−

𝑡𝑅𝐶 𝑢(𝑡)

ℒ 𝑒−𝑎𝑡 =1

𝑠 + 𝑎 𝑎 =

1

𝑅𝐶

𝑖𝑅 − 𝑉0 +1

𝐶 𝑖𝑑𝑡

𝑡

0−= 0

𝑅𝑑𝑖

𝑑𝑡+

𝑖

𝐶= 0

𝑑𝑖

𝑑𝑡+

𝑖

𝑅𝐶= 0

𝑖 𝑡 = 𝑖 0 𝑒−𝑡/𝑅𝐶

𝑖 𝑡 = 0 =𝑉𝑜𝑅

𝑖 𝑡 =𝑉0

𝑅𝑒−

𝑡𝑅𝐶 𝑡 ≥ 0

(13.14)



• In the time domain

RC Circuits

𝐶𝑑𝑣

𝑑𝑡+

𝑣

𝑅= 0

KCL :

𝑑𝑣

𝑑𝑡+

𝑣

𝑅𝐶= 0

𝑣 𝑡 = 𝑣 0 𝑒−𝑡/𝑅𝐶

𝑣 𝑡 = 0 = 𝑉𝑜

𝑣 𝑡 = 𝑉𝑜 𝑒−𝑡/𝑅𝐶

= 𝑉𝑜 𝑒−𝑡/𝜏

𝜏 = 𝑅𝐶

𝑖 𝑡 =𝑣(𝑡)

𝑅=

𝑉0

𝑅𝑒−

𝑡𝑅𝐶 𝑡 ≥ 0

• In the s domain

KCL :

(13.16)

(13.18)

𝑣 𝑡 = 𝑖 𝑡 𝑅 = 𝑉0𝑒−

𝑡𝑅𝐶 𝑢(𝑡)

ℒ 𝑒−𝑎𝑡 =1

𝑠 + 𝑎 𝑎 =

1

𝑅𝐶

𝑉

𝑅+ 𝑠𝐶𝑉 = 𝐶𝑉0

𝑉(𝑠) =𝑉0

𝑠 +1

𝑅𝐶

𝑣 𝑡 = 𝑉0𝑒−

𝑡𝑅𝐶 = 𝑉0𝑒

−𝑡𝜏 𝑢(𝑡)

(13.17)

From Eq.(13.14)

(13.15)



In the circuit in Figure, the switch moves from position 𝑎 to position 𝑏 at 𝑡 = 0. Find 𝑖(𝑡) for 𝑡 > 0.

𝐿 → 𝑠𝐿 − 𝐿𝐼0

𝑉0 → 𝑉0

𝑠

𝑅 → 𝑅

• We first transform the circuit from the time domain to the 𝑠 domain.

• Using mesh analysis

𝐼 𝑅 + 𝑠𝐿 − 𝐿𝐼0 −𝑉0

𝑠= 0

𝐼 𝑠 =𝐿𝐼0

𝑅 + 𝑠𝐿+

𝑉0

𝑠(𝑅 + 𝑠𝐿)=

𝐼0

𝑠 +𝑅𝐿

+

𝑉0𝐿

𝑠 𝑠 +𝑅𝐿

RL Circuits

=𝐼0

𝑠 +𝑅𝐿

+𝑘1

𝑠+

𝑘2

𝑠 +𝑅𝐿



𝐼 𝑠 =𝐼0

𝑠 +𝑅𝐿

+𝑘1

𝑠+

𝑘2

𝑠 +𝑅𝐿

𝑉0𝐿

𝑠 𝑠 +𝑅𝐿

=𝑘1

𝑠+

𝑘2

𝑠 +𝑅𝐿

𝑉0𝐿

𝑠 +𝑅𝐿

𝑠=0

= 𝑘1 +𝑠𝑘2

𝑠 +𝑅𝐿

𝑠=0

𝑉0𝐿

𝑠 +𝑅𝐿

= 𝑘1 +𝑠𝑘2

𝑠 +𝑅𝐿

𝑉0

𝑅= 𝑘1 +0

𝑉0𝐿𝑠

=𝑘1 𝑠 +

𝑅𝐿

𝑠+ 𝑘2

𝑉0𝐿𝑠

𝑠=−𝑅𝐿

=𝑘1 𝑠 +

𝑅𝐿

𝑠

𝑠=−𝑅𝐿

+ 𝑘2

−𝑉0

𝑅= 0 + 𝑘2

RL Circuits

𝑘1 =𝑉0

𝑅 𝑘2 = −

𝑉0

𝑅



𝐼 𝑠 =𝐿𝐼0

𝑅 + 𝑠𝐿+

𝑉0

𝑠(𝑅 + 𝑠𝐿)=

𝐼0

𝑠 +𝑅𝐿

+

𝑉0𝐿

𝑠 𝑠 +𝑅𝐿

=𝐼0

𝑠 +𝑅𝐿

+𝑘1

𝑠+

𝑘2

𝑠 +𝑅𝐿

𝑘1 =𝑉0

𝑅 𝑘2 = −

𝑉0

𝑅

𝐼 𝑠 =𝐼0

𝑠 +𝑅𝐿

+𝑉0

𝑅

1

𝑠−

𝑉0

𝑅

1

𝑠 +𝑅𝐿

𝑖 𝑡 = 𝐼0𝑒−

𝑅𝐿𝑡 +

𝑉0

𝑅−

𝑉0

𝑅𝑒−

𝑅𝐿𝑡

= 𝐼0 −𝑉0

𝑅𝑒−

𝑅𝐿𝑡 +

𝑉0

𝑅 𝑡 ≥ 0

RL Circuits



• First transform the circuit from the time domain to the s domain

1 H → 𝑠𝐿 = 𝑠

𝑢 𝑡 → 1

𝑠

1

3F →

1

𝑠𝐶=

3

𝑠

• Now apply mesh analysis

For mesh 1,

→ 1

𝑠= 1 +

3

𝑠𝐼1 −

3

𝑠𝐼2

For mesh 2,

5 + 𝑠 𝐼2 +3

𝑠𝐼2 − 𝐼1 = 0 → 0 = 5 + 𝑠 𝐼2 +

3

𝑠(𝐼2 − 𝐼1)

1𝐼1 +3

𝑠𝐼1 − 𝐼2 −

1

𝑠= 0

V

→ 𝐼1 =1

3𝑠2 + 5𝑠 + 3 𝐼2

(1)

(2)

RLC Circuits



• Substituting Eq.(2) into Eq.(1),

• Multiplying through by 3s gives

1

𝑠= 1 +

3

𝑠

1

3𝑠2 + 5𝑠 + 3 𝐼2 −

3

𝑠𝐼2 (3)

(4) 3 = 𝑠3 + 8𝑠2 + 18𝑠 𝐼2

𝐼2 =3

𝑠3 + 8𝑠2 + 18𝑠

𝑉𝑜 𝑠 = 𝑠𝐼2 =3

𝑠3 + 8𝑠2 + 18𝑠 =

3

2

2

(𝑠 + 4)2+( 2)2

Taking the inverse transform yields

𝑣(𝑡) =3

2𝑒−4𝑡 sin 2𝑡

• Output voltage

(5)

(6)

(7) V, 𝑡 ≥ 0

RLC Circuits



The Step Response of a RLC Parallel Circuit

• Now apply nodal analysis

• We first transform the circuit from the time domain to the 𝑠 domain.

ℒ 𝐼𝑑𝑐𝑢(𝑡) =𝐼𝑑𝑐

𝑠

𝐼𝑑𝑐

𝑠= 𝑠𝐶𝑉 +

𝑉

𝑅+

𝑉

𝑠𝐿 (13.20)




𝐼𝑑𝑐

𝑠= 𝑠𝐶𝑉 +

𝑉

𝑅+

𝑉

𝑠𝐿 (13.20)

𝑉 =𝐼𝑑𝑐/𝐶

𝑠2 +1

𝑅𝐶 𝑠 +1𝐿𝐶

(13.21)

𝐼𝐿 =𝑉

𝑠𝐿 =

𝐼𝑑𝑐/𝐿𝐶

𝑠 𝑠2 +1


𝐼𝐿 =384 × 105

𝑠 𝑠2 + 64,000𝑠 + 16 × 108

(13.22)

(13.23)




𝐼𝐿 =384 × 105

𝑠 𝑠2 + 64,000𝑠 + 16 × 108 (13.23)

=384 × 105

𝑠 𝑠 + 32,000 − 𝑗24,000)(𝑠 + 32,000 + 𝑗24,000

(13.24)

=𝐾1

𝑠+

𝐾2

𝑠 + 32,000 − 𝑗24,000+

𝐾2∗

𝑠 + 32,000 + 𝑗24,000 (13.26)

• The partial fraction coefficients are

𝐾1 =384 × 105

16 × 108 = 24 × 10−3 𝐾2 =384 × 105

(−32,000 + 𝑗24,000)(𝑗48,000)= 20 × 10−3∠126.87°

(13.27)




𝑖𝐿(𝑡) = [24 × 10−3 + 2 × 20 × 10−3 × 𝑒−32,000𝑡 cos(24,000𝑡 + 126.87° )]𝑢(𝑡) A

• Substituting the numerical values of and into Eq. 13.26 yields

𝐼𝐿 =24 × 10−3

𝑠+

20 × 10−3∠126.87°

𝑠 + 32,000 − 𝑗24,000+

20 × 10−3∠ − 126.87°

𝑠 + 32,000 + 𝑗24,000

𝓛−1𝐾

𝑠 + 𝛼 − 𝑗𝛽+

𝐾∗

𝑠 + 𝛼 + 𝑗𝛽

= 2 𝐾 𝑒−𝛼𝑡 cos 𝛽𝑡 + 𝜃

𝛼 = 32,000

𝐾 = 20 × 10−3

𝛽 = 24,000

= [24 + 40𝑒−32,000𝑡 cos(24,000𝑡 + 126.87° )]𝑢(𝑡) mA

𝜃 = 126.87°

• Inverse transforming the resulting expression Eq.(13.28) gives

(13.28)

(13.29)




• Final value of 𝑖𝐿

𝑖𝐿(𝑡) = [24 + 40𝑒−32,000𝑡 cos(24,000𝑡 + 126.87° )]𝑢(𝑡) mA (13.29)

𝐴𝑠 𝑡 → ∞, 𝑒−32,000𝑡 → 0

𝑖𝐿 ∞ = 24 mA (13.25)

lim𝑡→∞

𝑓 𝑡 = lim𝑠→0

𝑠𝐹(𝑠) (12.94)

Or from the final-value theorem

lim𝑠→0

𝑠𝐼𝐿 =384 × 105

16 × 108 = 24 mA 𝐼𝐿 =384 × 105

𝑠 𝑠2 + 64,000𝑠 + 16 × 108 (13.23)



The Transient Response of a RLC Parallel Circuit

(13.30) 𝑖𝑔 = 𝐼𝑚cos𝜔𝑡 A

• A sinusoidal current source

𝐼𝑚 = 24 mA

𝜔 = 40,000 rad/s

The s-domain expression for the source current is

𝐼𝑔 =𝑠𝐼𝑚

𝑠2 + 𝜔2 (13.31)

𝐼𝑔 = 𝑠𝐶𝑉 +𝑉

𝑅+

𝑉

𝑠𝐿

(13.32)

• A node equation :

𝑉 =

𝐼𝑔𝐶 𝑠

𝑠2 +1


• The voltage across the parallel elements is:




(13.32) 𝑉 =

𝐼𝑔𝐶 𝑠

𝑠2 +1


• The voltage across the parallel elements is:

=

𝐼𝑚𝐶 𝑠2

𝑠2 + 𝜔2 𝑠2 +1


(13.33)

• The current 𝐼𝐿

𝐼𝐿 =𝑉

𝑠𝐿=

𝐼𝑚𝐶 𝑠2

𝑠2 + 𝜔2 𝑠2 +1


(13.34)




• The current 𝐼𝐿

𝐼𝐿 =𝑉

𝑠𝐿=

𝐼𝑚𝐶 𝑠2

𝑠2 + 𝜔2 𝑠2 +1


(13.34)

𝐼𝐿 =384 × 105𝑠

𝑠2 + 16 × 108 (𝑠2 + 64,000𝑠 + 16 × 108)

=384 × 105𝑠

(𝑠 − 𝑗𝜔)(𝑠 + 𝑗𝜔)(𝑠 + 𝛼 − 𝑗𝛽)(𝑠 + 𝛼 + 𝑗𝛽)

(13.35)

(13.36)

𝐼𝐿 =𝐾1

𝑠 − 𝑗40,000+

𝐾2∗

𝑠 + 𝑗40,000+

𝐾2

𝑠 + 32,000 − 𝑗24,000+

𝐾2∗

𝑠 + 𝑗32,000 + 𝑗24,000 (13.37)




(13.37) 𝐼𝐿 =𝐾1

𝑠 − 𝑗40,000+

𝐾2∗

𝑠 + 𝑗40,000+

𝐾2

𝑠 + 32,000 − 𝑗24,000+

𝐾2∗

𝑠 + 𝑗32,000 + 𝑗24,000

• The partial fraction coefficients are

𝐾1 =384 × 105(𝑗40,000)

(𝑗80,000)(32,000 + 𝑗16,000)(32,000 + 𝑗64,000) = 7.5 × 10−3∠ − 90°

𝐾2 =384 × 105(−32,000 + 𝑗24,000)

−𝑗32,000 − 𝑗16,000 −32,000 + 𝑗64,000 (𝑗48,000) = 12.5 × 10−3∠90°

𝑖𝐿 = [15 cos 40,000𝑡 − 90° + 25𝑒−32,000𝑡 cos(24,000𝑡 + 90°)] mA

= (15 sin 40,000𝑡 − 25𝑒−32,000𝑡 sin 24,000𝑡)𝑢(𝑡) mA

• Inverse transforming the resulting expression gives

(13.38)

(13.39)

(13.40)




• The steady-state current:

𝑖𝐿(𝑡) = (15 sin 40,000𝑡 − 25𝑒−32,000𝑡 sin 24,000𝑡)𝑢(𝑡) mA

𝑖𝐿𝑠𝑠= 15 sin 40,000𝑡 mA (13.41)

𝐴𝑠 𝑡 → ∞, 𝑒−32,000𝑡 → 0

𝑖𝑔 = 𝐼𝑚cos𝜔𝑡 A

𝐼𝑚 = 24 mA

𝜔 = 40,000 rad/s

𝑍𝑐 =1

𝑗𝜔𝐶=

1

𝑗40,000 × 25 × 10−9

𝑍𝐿 = 𝑗𝜔𝐿 = 𝑗40,000 × 25 × 10−3 = 𝑗1000

= −𝑗1

10−3= −𝑗1000

𝐼𝑔 =𝑉

𝑍𝐶+

𝑉

𝑅+

𝑉

𝑍𝐿=

𝑉

−𝑗1000+

𝑉

625+

𝑉

𝑗1000=

𝑉

625

𝐼𝑔 = 𝐼𝑚∠0°

𝑉 = 625𝐼𝑔 = 625𝐼𝑚∠0° 𝐼𝐿 =𝑉

𝑍𝐿=

625𝐼𝑚∠0°

𝑗1000=

15000𝐼𝑚∠0°

1000∠90°= 15∠ − 90°

𝑖𝐿 𝑡 = 15 cos 40,000𝑡 − 90° = 15sin40,000𝑡 mA



The Step Response of a Multiple Mesh Circuit

• s-domain equivalent circuit

ZL1 = 8.4𝑠

ZL2 = 10𝑠

• The two mesh-current equations are

336

𝑠= 42 + 8.4𝑠 𝐼1 − 42𝐼2

0 = −42𝐼1 + (90 + 10𝑠)𝐼2

VS =336

𝑠

𝑉𝑆 = 336𝑢(𝑡)

(13.42)

(13.43)




• Using Cramer’s method to solve for 𝐼1 and 𝐼2 we obtain

336

𝑠= 42 + 8.4𝑠 𝐼1 − 42𝐼2

0 = −42𝐼1 + (90 + 10𝑠)𝐼2

(13.42)

(13.43)

∆ =42 + 8.4𝑠 −42

−42 90 + 10𝑠 = 84(𝑠2 + 14𝑠 + 24)

= 84(𝑠 + 2)(𝑠 + 12)

𝑁1 =336

𝑠−42

0 90 + 10𝑠

=3360(𝑠 + 9)

𝑠

(13.44)

(13.45)

𝑁2 = 42 + 8.4𝑠336

𝑠−42 0

=14,112

𝑠 (13.46)

𝐼1 =𝑁1

∆=

40(𝑠 + 9)

𝑠(𝑠 + 2)(𝑠 + 12)

𝐼2 =𝑁2

∆=

168

𝑠(𝑠 + 2)(𝑠 + 12)

(13.47)

(13.48)




(13.50)

𝐼1 =40(𝑠 + 9)

𝑠(𝑠 + 2)(𝑠 + 12)

𝐼2 =168

𝑠(𝑠 + 2)(𝑠 + 12)

=15

𝑠−

14

𝑠 + 2−

1

𝑠 + 12

=7

𝑠−

8.4

𝑠 + 2+

1.4

𝑠 + 12

(13.49)

𝑖1 = 15 − 14𝑒−2𝑡 − 𝑒−12𝑡 𝑢 𝑡 A

𝑖2 = 7 − 8.4𝑒−2𝑡 + 1.4𝑒−12𝑡 𝑢 𝑡 A

• Expanding 𝐼1 and 𝐼2 into a sum of partial fraction gives

• By inverse-transforming Eqs. 13.49 and 13.50, respectively, we have

(13.52)

(13.51)

𝑖1 ∞ = 15 A

𝑖2 ∞ = 7 A

𝑖1 ∞ =336

42 × 4842 + 48

= 15 A

𝑖2 ∞ =42

42 + 4815 = 7 A

(13.53)

(13.54)




• Voltage across the 42 Ω

𝑣(𝑡) = 42 𝑖1 − 𝑖2 = 336 − 8.4𝑑𝑖1𝑑𝑡

= 48𝑖2 + 10𝑑𝑖2𝑑𝑡

𝑖1 = 15 − 14𝑒−2𝑡 − 𝑒−12𝑡 𝑢 𝑡 A

𝑖2 = 7 − 8.4𝑒−2𝑡 + 1.4𝑒−12𝑡 𝑢 𝑡 A

𝑣(𝑡) = 336 − 235.2𝑒−2𝑡 − 100.80𝑒−12𝑡 𝑢 𝑡 V

(13.55)



The Use of Thevenin’s Equivalent in the s Domain

• Thevenin’s Equivalent Circuit

𝑉60 = 0

𝑉𝑇ℎ = 𝑉60 + 𝑉0.002

𝑉0.002 =0.002𝑠

20 + 0.002𝑠

480

𝑠=

480

𝑠 + 104

𝑉𝑇ℎ =480

𝑠 + 104

Thévenin voltage : VTh Thévenin impedance : ZTh

𝑍𝑇ℎ =20 × 0.002𝑠

20 + 0.002𝑠+ 60

=80(𝑠 + 7500)

𝑠 + 104

(13.56)

(13.57)




• Current IC

𝐼𝐶 =

480(𝑠 + 104)

80 𝑠 + 7500𝑠 + 104 +

2 × 105

𝑠

=6𝑠

𝑠2 + 10,000𝑠 + 25 × 106 =6𝑠

(𝑠 + 50002)

= −30,000

(𝑠 + 5000)2 +6

𝑠 + 5000

𝑖𝐶(𝑡) = −30,000𝑡𝑒−5000𝑡 + 6𝑒−5000𝑡 𝑢 𝑡 A

𝑖𝐶 0 = 6 A 𝑖𝐶 0 =480

20 + 60= 6 A

(13.60)

(13.61)

(13.59)

(13.62)




• Voltage VC

𝑖𝐶(𝑡) = −30,000𝑡𝑒−5000𝑡 + 6𝑒−5000𝑡 𝑢 𝑡 A

𝑣𝑐 𝑡 =1

𝐶 𝑖 𝑡 𝑑𝑡

𝑡

0−

= 2 × 105 6 − 30,000𝑥 𝑒−5000𝑥𝑑𝑥𝑡

0−

Time domain :

s domain :

𝑉𝐶 =1

𝑠𝐶𝐼𝐶 =

2 × 105

𝑠×

6𝑠

(𝑠 + 5000)2

=12 × 105

(𝑠 + 5000)2

𝑣𝑐 = 12 × 105𝑡𝑒−5000𝑡𝑢 𝑡 V

(13.63)

(13.64)

(13.65)



The Use of Superposition

• Circuit with Vg acting alone.

𝑉𝑔 − 𝑉1′

𝑅1=

𝑉1′

𝑠𝐿+

𝑉1′ − 𝑉2

′

1𝑠𝐶

𝑉𝑔

𝑅1=

1

𝑅1+

1

𝑠𝐿+ 𝑠𝐶 𝑉1

′ − 𝑠𝐶𝑉2′ (13.73)

Node 1:

Node 2:

𝑉1′ − 𝑉2

′

1𝑠𝐶

=𝑉2

′

𝑅2

−𝑠𝐶𝑉1′ +

1

𝑅2+ 𝑠𝐶 𝑉2

′ = 0 (13.74)




• Circuit with Vg acting alone.

𝑉𝑔

𝑅1=

1

𝑅1+

1


′ − 𝑠𝐶𝑉2′ (13.73)

−𝑠𝐶𝑉1′ +

1


′ = 0 (13.74)

𝑌11 =1

𝑅1+

1

𝑠𝐿+ 𝑠𝐶;

𝑌12 = −𝑠𝐶;

𝑌22 =1

𝑅2+ 𝑠𝐶

𝑌11𝑉1′ + 𝑌12𝑉2

′ =𝑉𝑔

𝑅1

𝑌12𝑉1′ + 𝑌22𝑉2

′ = 0

(13.78)

(13.79)

𝑉2′ =

−𝑌12𝑅1

𝑌11𝑌22 − 𝑌122 𝑉𝑔 (13.80)




• Circuit with Ig acting alone.

Node 1:

Node 2:

1

𝑅1+

1


′′ + 𝑠𝐶𝑉2′ = 0

(13.81)

𝑉′2′ − 𝑉1

′′

1𝑠𝐶

=𝑉′1

′

𝑅1+

𝑉′1′

𝑠𝐿

𝑌11𝑉1′′ + 𝑌12𝑉2

′′ = 0

𝐼𝑔 =𝑉′2

′

𝑅2+

𝑉′2′ − 𝑉1

′′

1𝑠𝐶

(13.82)

−𝑠𝐶𝑉1′′ +

1


′′ = 𝐼𝑔

𝑌12𝑉1′′ + 𝑌22𝑉2

′′ = 𝐼𝑔




• Circuit with Ig acting alone.

(13.81) 𝑌11𝑉1′′ + 𝑌12𝑉2

′′ = 0

(13.82) 𝑌12𝑉1′′ + 𝑌22𝑉2

′′ = 𝐼𝑔

𝑉2′′ =

𝑌11

𝑌11𝑌22 − 𝑌122 𝐼𝑔 (13.83)




• Circuit with 𝜌/𝑠 acting alone.

(13.84)

(13.85)

(13.86)

𝑌11𝑉1′′′ + 𝑌12𝑉2

′′′ =−𝜌

𝑠

𝑌12𝑉1′′′ + 𝑌22𝑉2

′′′ = 0

𝑉2′′′ =

𝑌12𝑠

𝑌11𝑌22 − 𝑌122 𝜌




(13.87)

(13.88)

(13.89)

• Circuit with γ𝐶 acting alone.

𝑌11𝑉1′′′′ + 𝑌12𝑉2

′′′′ = 𝛾𝐶

𝑌12𝑉1′′′′ + 𝑌22𝑉2

′′′′ = −𝛾𝐶

𝑉2′′′′ =

− 𝑌11 + 𝑌12 𝐶

𝑌11𝑌22 − 𝑌122 𝛾




• The expression for V2 :

𝑉2 = 𝑉2′ + 𝑉2

′′ + 𝑉2′′′ + 𝑉2

′′′′

=−

𝑌12𝑅1

𝑌11𝑌22 − 𝑌122 𝑉𝑔 +

𝑌11

𝑌11𝑌22 − 𝑌122 𝐼𝑔 +

𝑌12𝑠

𝑌11𝑌22 − 𝑌122 𝜌 +

−𝐶(𝑌11 + 𝑌12)

𝑌11𝑌22 − 𝑌122 𝛾

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Electric Circuit Theoryelearning.kocw.net/KOCW/document/2015/korea_sejong/minnamki1/… · CIEN 202 Electric Circuits II Nam Ki Min 010-9419-2320 [email protected] Chapter 13 The