Ch.02 modeling in frequency domain

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02. Modeling in Frequency

Domain

System Dynamics and Control 2.01 Modeling in Frequency Domain

HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Chapter Objectives

After completing this chapter, the student will be able to

• Find the Laplace transform of time functions and the inverse

Laplace transform

• Find the transfer function (TF) from a differential equation and

solve the differential equation using the transfer function

• Find the transfer function for linear, time-invariant electrical networks

• Find the TF for linear, time-invariant translational mechanical systems

• Find the TF for linear, time-invariant rotational mechanical systems

• Find the TF for gear systems with no loss and for gear systems

with loss

• Find the TF for linear, time-invariant electromechanical systems

• Produce analogous electrical and mechanical circuits

• Linearize a nonlinear system in order to find the TF



§1.Introduction

- Mathematical models from schematics of physical systems

• transfer functions in the frequency domain

• state equations in the time domain



§2.Laplace Transform Review

- Transforms: a mathematical conversion from one way of

thinking to another to make a problem easier to solve

- The Laplace transform the problem in time-domain to problem in

𝑠-domain, then applying the solution in 𝑠-domain, and finally using

inverse transform to converse the solution back to the time-domain

- The Laplace transform is named in honor of mathematician and

astronomer Pierre-Simon Laplace (1749-1827)

- Others: Fourier transform, z-transform, wavelet transform, …




- The Laplace transform of the function 𝑓(𝑡) for 𝑡 > 0 is defined

by the following relationship

𝑠 : complex frequency variable, 𝑠 = 𝜎 + 𝑗𝜔 with 𝑠, 𝜔 arereal numbers, 𝑠 ∈ 𝐶 for which makes 𝐹 𝑠 convergent

ℒ : Laplace transform

𝐹(𝑠) : a complex-valued function of complex numbers

- The inverse Laplace transform of the function 𝐹(𝑠) for 𝑡 > 0 is

defined by the following relationship

𝑢(𝑡) : the unit step function, 𝑢 𝑡 = 1 𝑖𝑓 𝑡 > 00 𝑖𝑓 𝑡 < 0


𝐹 𝑠 = ℒ 𝑓(𝑡) = 0−

+∞

𝑓(𝑡)𝑒−𝑠𝑡𝑑𝑡

𝑓 𝑡 = ℒ−1 𝐹 𝑠 =1

2𝜋𝑗 𝜎−𝑗∞

𝜎+𝑗∞

𝐹 𝑠 𝑒𝑠𝑡𝑑𝑠 = 𝑓 𝑡 𝑢(𝑡)

(2.1)

(2.2)



- The Laplace transform table



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- Ex.2.1 Laplace Transform of a Time Function

Find the Laplace transform of 𝑓 𝑡 = 𝐴𝑒−𝑎𝑡𝑢(𝑡)

Solution

𝐹 𝑠 = 0

∞

𝑓(𝑡)𝑒−𝑠𝑡𝑑𝑡

= 0

∞

𝐴𝑒−𝑎𝑡𝑒−𝑠𝑡𝑑𝑡

= 𝐴 0

∞

𝑒−(𝑎+𝑠)𝑡𝑑𝑡

= −𝐴

𝑠 + 𝑎𝑒−(𝑎+𝑠)𝑡

0

∞

⟹ 𝐹 𝑠 =𝐴

𝑠 + 𝑎


(2.3)





ℒ 𝑡𝑢 𝑡 =1

𝑠2 (Table 2.1 – 3)

ℒ 𝑒−𝑎𝑡𝑓(𝑡) = 𝐹(𝑠 + 𝑎) (Table 2.2 – 4)


- Ex.2.2 Inverse Laplace Transform

Find the inverse Laplace transform of 𝐹1 𝑠 = 1/(𝑠 + 3)2

Solution

𝑓 𝑡 = ℒ−11

𝑠2 = 𝑡𝑢(𝑡)

𝑓1 𝑡 = ℒ−11

(𝑠 + 3)2

= 𝑒−3𝑡𝑓 𝑡

⟹ 𝑓1 𝑡 = 𝑒−3𝑡𝑡𝑢(𝑡)



ℒ 𝛿 𝑡 = 1 (Table 2.1 – 1)

ℒ𝑑𝑓(𝑡)

𝑑𝑡= 𝑠𝐹(𝑠) → ℒ

𝑑𝛿(𝑡)

𝑑𝑡= 𝑠 (Table 2.2 – 7)


Partial-Fraction Expansion

𝐹1 𝑠 =𝑠3 + 2𝑠2 + 6𝑠 + 7

𝑠2 + 𝑠 + 5

= (𝑠 + 1) +2

𝑠2 + 𝑠 + 5

⟹ 𝑓1 𝑡 =𝑑𝛿(𝑡)

𝑑𝑡+ 𝛿 𝑡 + ℒ−1

2

𝑠2 + 𝑠 + 5

Using partial-fraction expansion to expand function like 𝐹(𝑠)into a sum of terms and then find the inverse Laplace transform

for each term


𝐹(𝑠)



Case 1. Roots of the Denominator of 𝐹(𝑠) are Real and Distinct

𝐹 𝑠 =2

(𝑠 + 1)(𝑠 + 2)=

𝐾1

𝑠 + 1+

𝐾2

𝑠 + 2

lim𝑠→−1

[(2.8)× (𝑠 + 1)]

⟹ lim𝑠→−1

2

𝑠 + 2= lim

𝑠→−1𝐾1 +

(𝑠 + 1)𝐾2

𝑠 + 2

⟹ 𝐾1 = 2

lim𝑠→−2

[(2.8)× (𝑠 + 2)]

⟹ lim𝑠→−2

2

𝑠 + 1= lim

𝑠→−2

(𝑠 + 2)𝐾1

𝑠 + 1+ 𝐾2

⟹ 𝐾2 = −2

⟹ 𝐹 𝑠 =2

𝑠 + 1−

2

𝑠 + 2⟹ 𝑓 𝑡 = 2𝑒−𝑡 − 2𝑒−2𝑡 𝑢(𝑡)


(2.8)

(2.10)



In general, given an 𝐹(𝑠) whose denominator has real and

distinct roots, a partial-fraction expansion

𝐹 𝑠 =𝑁(𝑠)

𝐷(𝑠)

=𝑁(𝑠)

𝑠 + 𝑝1 𝑠 + 𝑝2 … 𝑠 + 𝑝𝑖 …(𝑠 + 𝑝𝑛)

=𝐾1

𝑠 + 𝑝1+

𝐾2

𝑠 + 𝑝2+ ⋯+

𝐾𝑖

𝑠 + 𝑝𝑖+ ⋯+

𝐾𝑛

𝑠 + 𝑝𝑛

To find 𝐾𝑖

• multiply (2.11) by 𝑠 + 𝑝𝑖

• let 𝑠 approach −𝑝𝑖


(2.11)


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ℒ𝑑𝑓

𝑑𝑡= 𝑠𝐹 𝑠 − 𝑓(0−) (Table 2.2 – 7)

ℒ𝑑2𝑓

𝑑𝑡2 = 𝑠2𝐹 𝑠 − 𝑠𝑓(0−) − 𝑓′(0−) (Table 2.2 – 8)


- Ex.2.3 Laplace Transform Solution of a Differential Equation

Given the following differential equation, solve for 𝑦(𝑡) if all

initial conditions are zero. Use the Laplace transform

𝑑2𝑦

𝑑𝑡2 + 12𝑑𝑦

𝑑𝑡+ 32𝑦 = 32𝑢(𝑡)

Solution

𝑠2𝑌 𝑠 + 12𝑠𝑌 𝑠 + 32𝑌 𝑠 =32

𝑠

⟹ 𝑌 𝑠 =32

𝑠 𝑠2 + 12𝑠 + 32=

32

𝑠(𝑠 + 4)(𝑠 + 8)

=𝐾1

𝑠+

𝐾2

𝑠 + 4+

𝐾3

𝑠 + 8




𝑌 𝑠 =32

𝑠(𝑠 + 4)(𝑠 + 8)=

𝐾1

𝑠+

𝐾2

𝑠 + 4+

𝐾3

𝑠 + 8

Evaluate the residue 𝐾𝑖

𝐾1 = 32

(𝑠 + 4)(𝑠 + 8)𝑠→0

= 1

𝐾2 = 32

𝑠(𝑠 + 8)𝑠→−4

= −2

𝐾2 = 32

𝑠(𝑠 + 4)𝑠→−8

= 1

⟹ 𝑌 𝑠 =1

𝑠−

2

𝑠 + 4+

1

𝑠 + 8

Hence

𝑦 𝑡 = 1 − 2𝑒−4𝑡 + 𝑒−8𝑡 𝑢(𝑡)




𝑌 𝑠 =32

𝑠 𝑠2 + 12𝑠 + 32

=1

𝑠−

2

𝑠 + 4+

1

𝑠 + 8𝑦 𝑡 = 1 − 2𝑒−4𝑡 + 𝑒−8𝑡 𝑢(𝑡) (2.20)

The 𝑢(𝑡) in (2.20) shows that the response is zero until 𝑡 = 0

Unless otherwise specified, all inputs to systems in the text

will not start until 𝑡 = 0. Thus, output responses will also be

zero until 𝑡 = 0

For convenience, the 𝑢(𝑡) notation will be eliminated from now

on. Accordingly, the output response

𝑦 𝑡 = 1 − 2𝑒−4𝑡 + 𝑒−8𝑡 (2.21)




Run ch2p1 through ch2p8 in Appendix B

Learn how to use MATLAB to

• represent polynomials

• find roots of polynomials

• multiply polynomials, and

• find partial-fraction expansions





𝑌 𝑠 =32

𝑠3 + 12𝑠2 + 32𝑠=

1

𝑠−

2

𝑠 + 4+

1

𝑠 + 8

Matlab [r,p,k] = residue([32],[1,12,32,0])

Result r = [1, −2, 1], p = [−8, −4, 0], k = [ ]

𝑌 𝑠 = 0𝑘

+ 1𝑟1

1

𝑠 − ( −8𝑝1

)+ ( −2

𝑟2

)1

𝑠 − ( −4𝑝2

)+ 1

𝑟3

1

𝑠 − ( 0𝑝3

)

=1

𝑠 + 8−

2

𝑠 + 4+

1

𝑠



Case 2. Roots of the Denominator of 𝐹(𝑠) are Real and Repeated

𝐹 𝑠 =2

(𝑠 + 1)(𝑠 + 2)2

=𝐾1

𝑠 + 1+

𝐾2

(𝑠 + 2)2 +𝐾3

𝑠 + 2

lim𝑠→−1

[(2.23)× (𝑠 + 1)]

⟹ lim𝑠→−1

2

𝑠 + 2= lim

𝑠→−1𝐾1 +

(𝑠 + 1)𝐾2

𝑠 + 2

⟹ 𝐾1 = 2

lim𝑠→−2

[(2.23)× (𝑠 + 2)]

⟹ lim𝑠→−2

2

𝑠 + 1= lim

𝑠→−2

(𝑠 + 2)𝐾1

𝑠 + 1+ 𝐾2

⟹ 𝐾2 = −2


(2.22)

(2.23)


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𝐹 𝑠 =𝐾1

𝑠 + 1+

𝐾2

(𝑠 + 2)2 +𝐾3

𝑠 + 2

𝐾1 = 2, 𝐾2 = −2

(2.23)× (𝑠 + 2)2

⟹2

𝑠 + 1=

(𝑠 + 2)2𝐾1

𝑠 + 1+ 𝐾2 + (𝑠 + 2)𝐾3

Differentiate (2.24) with respect to 𝑠−2

𝑠 + 1 2 =(𝑠 + 2)𝐾1

𝑠 + 1 2 + 𝐾3

⟹ 𝐾3 = −2

⟹ 𝑌 𝑠 =2

𝑠 + 1−

2

𝑠 + 2 2 −2

𝑠 + 2

Hence

𝑦 𝑡 = 2𝑒−𝑡 − 2𝑡𝑒−2𝑡 − 2𝑒−2𝑡 (2.26)


(2.23)

(2.24)




𝐹 𝑠 =2

(𝑠 + 1)(𝑠 + 2)2

Matlab F=zpk([], [-1 -2 -2],2)

Result F =

2

------------------

(s+1) (s+2)^2

Continuous-time zero/pole/gain model

(2.22)




𝐹 𝑠 =2

𝑠3 + 5𝑠2 + 8𝑠 + 4=

2

𝑠 + 1−

2

𝑠 + 2 2 −2

𝑠 + 2

Matlab [r,p,k] = residue([2],[1,5,8,4])

Result r = [−2, −2, 2], p = [−2, −2, −1], k = [ ]

𝐹 𝑠 = 0𝑘

+ (−2)𝑟1

1

[𝑠− ( −2𝑝1

)]2+( −2

𝑟2

)1

𝑠 −( −2𝑝2

)+ 2

𝑟3

1

𝑠 −( −1𝑝3

)

= −2

𝑠 + 2 2 −2

𝑠 + 2+

2

𝑠 + 1



In general, given an 𝐹(𝑠) whose denominator has real and

distinct roots, a partial-fraction expansion


𝐷(𝑠)

=𝑁(𝑠)

𝑠 + 𝑝1𝑟 𝑠 + 𝑝2 … 𝑠 + 𝑝𝑖 …(𝑠 + 𝑝𝑛)

=𝐾1

𝑠 + 𝑝1𝑟 +

𝐾1

𝑠 + 𝑝1𝑟−1 + ⋯+

𝐾2

𝑠 + 𝑝1

+𝐾2

𝑠 + 𝑝2+ ⋯+

𝐾𝑖

𝑠 + 𝑝𝑖+ ⋯+

𝐾𝑛

𝑠 + 𝑝𝑛

• multiply (2.27) by 𝑠 + 𝑝1𝑟 to get 𝐹1 𝑠 = 𝑠 + 𝑝1

𝑟𝐹(𝑠)

• let 𝑠 approach −𝑝𝑖


(2.27)To find 𝐾𝑖

𝐾𝑖 =1

𝑖 − 1 !

𝑑𝑖−1𝐹1(𝑠)

𝑑𝑠𝑖−1

𝑠→𝑝1

𝑖 = 1,2,… , 𝑟; 0! = 1



Case 3. Roots of the Denominator of 𝐹(𝑠) are Complex or Imaginary

𝐹 𝑠 =3

𝑠(𝑠2 + 2𝑠 + 5)

=𝐾1

𝑠+

𝐾2𝑠 + 𝐾3

𝑠2 + 2𝑠 + 5lim𝑠→0

[(2.31)× 𝑠] ⟹ 𝐾1 = 3/5

First multiplying (2.31) by the lowest common denominator,

𝑠(𝑠2 + 2𝑠 + 5), and clearing the fraction

3 = 𝐾1 𝑠2 + 2𝑠 + 5 + 𝐾2𝑠 + 𝐾3 𝑠 (2.32)

⟹ 3 = 𝐾2 +3

5𝑠2 + 𝐾3 +

6

5𝑠 + 3

Balancing the coefficients: 𝐾2 = −3/5, 𝐾3 = −6/5

𝐹 𝑠 =3

𝑠(𝑠2 + 2𝑠 + 5)=

3

5

1

𝑠−

3

5

𝑠 + 2

𝑠2 + 2𝑠 + 5


(2.30)

(2.33)

(2.31)


ℒ 𝐴𝑒−𝑎𝑡𝑐𝑜𝑠𝜔𝑡 + 𝐵𝑒−𝑎𝑡𝑠𝑖𝑛𝜔𝑡 =𝐵 𝑠+𝑎 +𝐵𝜔

(𝑠+𝑎)2+𝜔2 (Table 2.1 – 6&7)


𝐹 𝑠 =3

5

1

𝑠−

3

5

𝑠 + 2

𝑠2 + 2𝑠 + 5

=3

5

1

𝑠−

3

5

𝑠 + 1 + 1/2 2

(𝑠 + 1)2+22

⟹ 𝑓 𝑡 =3

5−

3

5𝑒−𝑡 𝑐𝑜𝑠2𝑡 +

1

2𝑠𝑖𝑛2𝑡

or

𝑓 𝑡 = 0.6 − 0.671𝑒−𝑡cos(2𝑡 − 𝜙) (2.41)


(2.38)


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𝐹 𝑠 =2

(𝑠 + 1)(𝑠 + 2)2

𝑦 𝑡 = 2𝑒−𝑡 − 2𝑡𝑒−2𝑡 − 2𝑒−2𝑡 (2.26)

Matlab numf=2;

denf=poly([-1 -2 -2]);

[r,p,k]=residue(numf,denf)

Result r = [-2 -2 2], p = [-2 -2 -1], k = []

𝐹 𝑠 = 0𝑘

+ −2𝑟1

1

[𝑠 − ( −2𝑝1

)]2+ ( −2

𝑟2

)1

𝑠 − ( −2𝑝2

)+ 2

𝑟3

1

𝑠 − ( −1𝑝3

)

= −21

(𝑠 + 2)2 − 21

𝑠 + 2+ 2

1

𝑠 + 1


(2.22)



𝐹 𝑠 =3

𝑠(𝑠2 + 2𝑠 + 5)

Matlab F=tf([3],[1 2 5 0])

Result F =

3

-----------------------

s^3 + 2 s^2 + 5 s

Continuous-time transfer function


(2.30)



In general, given an 𝐹(𝑠) whose denominator has complex or

purely imaginary roots, a partial-fraction expansion


𝐷(𝑠)

=𝑁(𝑠)

𝑠 + 𝑝1 (𝑠2 + 𝑎𝑠 + 𝑏) …

=𝐾1

(𝑠 + 𝑝1)+

𝐾2𝑠 + 𝐾3

(𝑠2 + 𝑎𝑠 + 𝑏)+ ⋯

To find 𝐾𝑖

• the 𝐾𝑖 ’s in (2.42) are found through balancing the

coefficients of the equation after clearing fractions

• put (𝐾2𝑠 + 𝐾3)/(𝑠2 + 𝑎𝑠 + 𝑏) in to the form

𝐵 𝑠 + 𝑎 + 𝐵𝜔

(𝑠 + 𝑎)2+𝜔2


(2.42)



𝐹 𝑠 =3

𝑠(𝑠2 + 2𝑠 + 5)

𝑓 𝑡 =3

5−

3

5𝑒−𝑡 𝑐𝑜𝑠2𝑡 +

1

2𝑠𝑖𝑛2𝑡

Matlab syms s; f=ilaplace(3/(s*(s^2+2*s+5))); pretty(f)

Result f =

3/5 - (3*exp(-t)*(cos(2*t) + sin(2*t)/2))/5

/ sin(2 t) \

exp(-t) | cos(2 t) + ---------- | 3

3 \ 2 /

- - --------------------------------------

5 5


(2.30)

(2.38)



𝐹 𝑠 =3

𝑠(𝑠2 + 2𝑠 + 5)

𝐹 𝑠 =3/5

𝑠−

3

20

2+𝑗1

𝑠 +1+ 𝑗2+

2−𝑗1

𝑠 +1+𝑗2

Matlab numf=3; denf=[1 2 5 0]; [r,p,k]=residue(numf,denf)

Result r=[-0.3+0.15i; -0.3-0.15i; 0.6]; p=[-1+2i; -1-2i; 0]; k=[]

𝐹 𝑠 = 0𝑘

+ (−0.3 + 𝑗0.15)𝑟1

1

𝑠 − (−1 + 𝑗2𝑝1

)

+(−0.3 − 𝑗0.15)𝑟2

1

𝑠 − (−1 − 2𝑗𝑝2

)+ ( 0.6)

𝑟3

1

𝑠 − ( 0𝑝3

)

= −0.3 − 𝑗0.15

𝑠 + 1 − 2𝑗−

0.3 + 𝑗0.15

𝑠 + 1 + 2𝑗+ 0.6

1

𝑠


(2.30)

(2.47)



Run ch2sp1 and ch2sp2 in Appendix F

Learn how to use the Symbolic Math Toolbox to

• construct symbolic objects

• find the inverse Laplace transforms of frequency

functions

• find the Laplace of time functions



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Skill-Assessment Ex.2.1

Problem Find the Laplace transform of

𝑓 𝑡 = 𝑡𝑒−5𝑡

Solution

𝐹 𝑠 = ℒ 𝑡𝑒−5𝑡 =1

(𝑠 + 5)2

Matlab syms t s F; f = t*exp(-5*t); F=laplace(f, s); pretty(F)

Result 1

---------

2

(s + 5)





Problem Find the inverse Laplace transform of

𝐹 𝑠 =10

𝑠(𝑠 + 2)(𝑠 + 3)2

Solution Expanding 𝐹(𝑠) by partial fractions

𝐹 𝑠 =𝐴

𝑠+

𝐵

𝑠 + 2+

𝐶

(𝑠 + 3)2 +𝐷

𝑠 + 3

𝐴 = 10

(𝑠 + 2)(𝑠 + 3)2𝑠→0

=5

9,𝐵 =

10

𝑠(𝑠 + 3)2𝑠→−2

= −5

𝐶 = 10

𝑠(𝑠 + 2)𝑠→−3

=10

3,𝐷 = (𝑠 + 3)2

𝑑𝐹(𝑠)

𝑑𝑠𝑠→−3

=40

9

⟹ 𝐹 𝑠 =5

9

1

𝑠− 5

1

𝑠 + 2+

10

3

1

(𝑠 + 3)2 +40

9

1

𝑠 + 3


where,



𝐹 𝑠 =5

9

1

𝑠− 5

1

𝑠 + 2+

10

3

1

(𝑠 + 3)2 +40

9

1

𝑠 + 3

Taking the inverse Laplace transform

𝑓 𝑡 =5

9− 5𝑒−2𝑡 +

10

3𝑡𝑒−3𝑡 +

40

9𝑒−3𝑡

Matlab syms s; f=ilaplace(10/(s*(s+2)*(s+3)^2)); pretty(f)

Result exp(-3 t) 40 t exp(-3 t) 10 5

--------------- - exp(-2 t) 5 + ---------------- + -

9 3 9



§3.The Transfer Function

- The transfer function of a component is the quotient of the

Laplace transform of the output divided by the Laplace

transform of the input, with all initial conditions assumed to be

zero

- Transfer functions are defined only for linear time invariant systems

- The input-output relationship of a control system 𝐺 𝑠 𝐶(𝑠)

𝑎𝑛

𝑑𝑛𝑐(𝑡)

𝑑𝑡𝑛+ 𝑎𝑛−1

𝑑𝑛−1𝑐 𝑡

𝑑𝑡𝑛−1+ ⋯+ 𝑎0𝑐 𝑡

= 𝑏𝑚

𝑑𝑚𝑟(𝑡)

𝑑𝑡𝑚+ 𝑏𝑚−1

𝑑𝑚−1𝑟(𝑡)

𝑑𝑡𝑚−1+ ⋯+ 𝑏0𝑟 𝑡

𝑐(𝑡): output 𝑟(𝑡): input 𝑎𝑖’s, 𝑏𝑖’s : constant




𝑎𝑛

𝑑𝑛𝑐(𝑡)

𝑑𝑡𝑛+ 𝑎𝑛−1

𝑑𝑛−1𝑐 𝑡

𝑑𝑡𝑛−1+ ⋯+ 𝑎0𝑐 𝑡

= 𝑏𝑚

𝑑𝑚𝑟(𝑡)

𝑑𝑡𝑚+ 𝑏𝑚−1

𝑑𝑚−1𝑟(𝑡)

𝑑𝑡𝑚−1+ ⋯+ 𝑏0𝑟 𝑡

- Taking the Laplace transform of both sides with zero initial

conditions

𝑎𝑛𝑠𝑛𝐶 𝑠 + 𝑎𝑛−1𝑠𝑛−1𝐶 𝑠 + ⋯+ 𝑎0𝐶 𝑠

= 𝑏𝑚𝑠𝑚𝑅 𝑠 + 𝑏𝑚−1𝑠𝑚−1𝑅 𝑠 + ⋯+ 𝑏0𝑅 𝑠

- The transfer function

𝐺 𝑠 =𝐶(𝑠)

𝑅(𝑠)=

𝑏𝑚𝑠𝑚 + 𝑏𝑚−1𝑠𝑚−1 + ⋯+ 𝑏0

𝑎𝑛𝑠𝑛 + 𝑎𝑛−1𝑠𝑛−1 + ⋯+ 𝑎0

- The output of the system can be written in the form

𝐶 𝑠 = 𝐺 𝑠 𝑅(𝑠) (2.54)


(2.53)



- Ex.2.4 Transfer Function for a Differential Equation

Find the transfer function represented by

𝑑𝑐(𝑡)

𝑑𝑡+ 2𝑐(𝑡) = 𝑟(𝑡)

Solution

Taking the Laplace transform with zero initial conditions

𝑠𝐶 𝑠 + 2𝐶(𝑠) = 𝑅(𝑠)

The transfer function


𝑅(𝑠)=

1

𝑠 + 2



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Run ch2p9 through ch2p12 in Appendix B

Learn how to use MATLAB to

• create transfer functions with numerators and

denominators in polynomial or factored form

• convert between polynomial and factored forms

• plot time functions




Run ch2sp3 in Appendix F


• simplify the input of complicated transfer functions as

well as improve readability

• enter a symbolic transfer function and convert it to a

linear time-invariant (LTI) object as presented in

Appendix B, ch2p9




- Ex.2.5 System Response from the Transfer Function

Given 𝐺 𝑠 = 1/(𝑠 + 2), find the response, 𝑐(𝑡) to an input,

𝑟 𝑡 = 𝑢(𝑡), a unit step, assuming zero initial conditions

Solution

For a unit step

𝑟 𝑡 = 𝑢 𝑡 ⟹ 𝑅 𝑠 = 1/𝑠

The output

𝐶 𝑠 = 𝑅 𝑠 𝐺 𝑠 =1

𝑠

1

𝑠 + 2Expanding by partial fractions

𝐶 𝑡 =1

2

1

𝑠−

1

2

1

𝑠 + 2Taking the inverse Laplace transform

𝑐 𝑡 = 0.5 − 0.5𝑒−2𝑡 (2.60)




𝐺 𝑠 =1

𝑠 + 2

𝑅 𝑠 =1

𝑠

𝑐 𝑡 =1

2−

1

2𝑒−2𝑡

Matlab syms s

C=1/(s*(s+2))

C=ilaplace(C)

Result C =

1/2 - exp(-2*t)/2

𝑐 𝑡 =1

2−

1

2𝑒−2𝑡


(2.60)



𝑐 𝑡 =1

2−

1

2𝑒−2𝑡

Matlab t=0:0.01:1;

plot(t,(1/2-1/2*exp(-2*t)))

Result


(2.60)




Problem Find the transfer function, 𝐺 𝑠 = 𝐶(𝑠)/𝑅(𝑠), corresponding

to the differential equation

𝑑3𝑐

𝑑𝑡3 + 3𝑑2𝑐

𝑑𝑡2 + 7𝑑𝑐

𝑑𝑡+ 5𝑐 =

𝑑2𝑟

𝑑𝑡2 + 4𝑑𝑟

𝑑𝑡+ 3𝑟

Solution Taking the Laplace transform with zero initial conditions

𝑠3𝐶 𝑠 + 3𝑠2𝐶 𝑠 + 7𝑠𝐶 𝑠 + 5𝐶 𝑠

= 𝑠2𝑅 𝑠 + 4𝑠𝑅 𝑠 + 3𝑅 𝑠

Collecting terms

𝑠3 + 3𝑠2 + 7𝑠 + 5 𝐶 𝑠 = 𝑠2 + 4𝑠 + 3 𝑅(𝑠)



𝑅(𝑠)=

𝑠2 + 4𝑠 + 3

𝑠3 + 3𝑠2 + 7𝑠 + 5



1/11/2016

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Problem Find the differential equation corresponding to the

transfer function

𝐺 𝑠 =2𝑠 + 1

𝑠2 + 6𝑠 + 2Solution The transfer function


𝑅(𝑠)=

2𝑠 + 1

𝑠2 + 6𝑠 + 2

Cross multiplying

𝑑2𝑐

𝑑𝑡2 + 6𝑑𝑐

𝑑𝑡+ 2𝑐 = 2

𝑑𝑟

𝑑𝑡+ 𝑟





Problem Find the ramp response for a system whose transfer

function

𝐺 𝑠 =𝑠

(𝑠 + 4)(𝑠 + 8)

Solution For a ramp response

𝑟 𝑡 = 𝑡𝑢 𝑡 ⟹ 𝑅 𝑠 =1

𝑠2

The output

𝐶 𝑠 = 𝑅 𝑠 𝐺 𝑠

=1

𝑠2

𝑠

(𝑠 + 4)(𝑠 + 8)

=𝐴

𝑠+

𝐵

𝑠 + 4+

𝐶

𝑠 + 8




𝐶 𝑠 =1

𝑠(𝑠 + 4)(𝑠 + 8)=

𝐴

𝑠+

𝐵

𝑠 + 4+

𝐶

𝑠 + 8

𝐴 = 1

(𝑠 + 4)(𝑠 + 8)𝑠→0

=1

32

𝐵 = 1

𝑠(𝑠 + 8)𝑠→−4

= −1

16

𝐶 = 1

𝑠(𝑠 + 4)𝑠→−8

=1

32

⟹ 𝐶 𝑠 =1

32

1

𝑠−

1

16

1

𝑠 + 4+

1

32

1

𝑠 + 8

The ramp response

𝑐 𝑡 =1

32−

1

16𝑒−4𝑡 +

1

32𝑒−8𝑡


where,


§4.Electrical Network Transfer Functions

Summarizes the components and the relationships between

voltage and current and between voltage and charge under zero

initial conditions




Simple Circuits via Mesh Analysis

- Ex.2.6 Transfer Function - Single Loop via the Differential Equation

Find the transfer function 𝑉𝐶(𝑠)/𝑉(𝑠)

Solution

The voltage loop

𝐿𝑑𝑖

𝑑𝑡+ 𝑅𝑖 +

1

𝐶 0

1

𝑖 𝜏 𝑑𝜏 = 𝑣(𝑡)

Using the relationships 𝑖 𝑡 = 𝑑𝑞(𝑡)/𝑑𝑡 and 𝑞 = 𝐶𝑣𝐶

𝐿𝑑2𝑞

𝑑𝑡2 + 𝑅𝑑𝑞

𝑑𝑡+

1

𝐶𝑞 = 𝑣(𝑡)

⟹ 𝐿𝐶𝑑2𝑣𝐶

𝑑𝑡2 + 𝑅𝐶𝑑𝑣𝐶

𝑑𝑡+ 𝑣𝐶 = 𝑣(𝑡)




𝐿𝐶𝑑2𝑣𝐶

𝑑𝑡2 + 𝑅𝐶𝑑𝑣𝐶

𝑑𝑡+ 𝑣𝐶 = 𝑣(𝑡)

Taking Laplace transform assuming zero initial conditions

𝐿𝐶𝑠2 + 𝑅𝐶𝑠 + 1 𝑉𝐶 𝑠 = 𝑉(𝑠)

Solving for the transfer function

𝑉𝐶(𝑠)

𝑉(𝑠)=

1𝐿𝐶

𝑠2 +𝑅𝐿

𝑠 +1𝐿𝐶

Block diagram of series RLC electrical network



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Impedance

- A resistance resists or “impedes” the flow of current. The

corresponding relation is 𝑣/𝑖 = 𝑅 . Capacitance and

inductance elements also impede the flow of current

- In electrical systems an impedance is a generalization of the

resistance concept and is defined as the ratio of a voltage

transform 𝑉(𝑠) to a current transform 𝐼(𝑠) and thus implies a

current source

- Standard symbol for impedance

𝑍(𝑠) ≡𝑉(𝑠)

𝐼(𝑠)

- Kirchhoff’s voltage law to the transformed circuit

[ Sum of Impedances ] × 𝐼 𝑠 = [ Sum of Applied Voltages ] (2.72)


(2.70)



- The impedance of a resistor is its resistance

𝑍 𝑠 = 𝑅

- For a capacitor

𝑣 𝑡 =1

𝐶 0

𝑡

𝑖𝑑𝑡 ⟹ 𝑉 𝑠 =𝐼(𝑠)

𝐶 𝑠 𝑠

The impedance of a capacitor

𝑍 𝑠 =1

𝐶𝑠- For an inductor

𝑣 𝑡 = 𝐿𝑑𝑖

𝑑𝑡⟹ 𝑉 𝑠 = 𝐿𝐼 𝑠 𝑠

The impedance of a inductor

𝑍 𝑠 = 𝐿𝑠




Series and Parallel Impedances

- The concept of impedance is useful because the impedances

of individual elements can be combined with series and

parallel laws to find the impedance at any point in the system

- The laws for combining series or parallel impedances are

extensions to the dynamic case of the laws governing series

and parallel resistance elements




- Series Impedances

• Two impedances are in series if they have the same current.

If so, the total impedance is the sum of the individual

impedances 𝑖 𝑣 𝑅 𝐶

𝑍 𝑠 = 𝑍1 𝑠 + 𝑍2(𝑠)

• Example: a resistor 𝑅 and capacitor 𝐶 in series have the

equivalent impedance

𝑍 𝑠 = 𝑅 +1

𝐶𝑠=

𝑅𝐶𝑠 + 1

𝐶𝑠

⟹𝑉(𝑠)

𝐼(𝑠)≡ 𝑍 𝑠 =

𝑅𝐶𝑠 + 1

𝐶𝑠

and the differential equation model is

𝐶𝑑𝑣

𝑑𝑡= 𝑅𝐶

𝑑𝑖

𝑑𝑡+ 𝑖(𝑡)




- Parallel Impedances

• Two impedances are in parallel if they have the same

voltage difference across them. Their impedances combine

by the reciprocal rule1

𝑍(𝑠)=

1

𝑍1(𝑠)+

1

𝑍2(𝑠)

• Example: a resistor 𝑅 and capacitor 𝐶 in parallel have the

equivalent impedance

1

𝑍(𝑠)=

1

1/𝐶𝑠+

1

𝑅⟹

𝑉(𝑠)

𝐼(𝑠)≡ 𝑍 𝑠 =

𝑅

𝑅𝐶𝑠 + 1

and the differential equation model is

𝑅𝐶𝑑𝑣

𝑑𝑡+ 𝑣 = 𝑅𝑖(𝑡)




Admittance

𝑌 𝑠 ≡1

𝑍(𝑠)=

𝐼(𝑠)

𝑉(𝑠)

In general, admittance is complex

• The real part of admittance is called conductance

𝐺 =1

𝑅• The imaginary part of admittance is called susceptance

When we take the reciprocal of resistance to obtain the

admittance, a purely real quantity results. The reciprocal of

resistance is called conductance



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Instead of taking the Laplace transform of the differential

equation, we can draw the transformed circuit and obtain the

Laplace transform of the differential equation simply by

applying Kirchhoff’s voltage law to the transformed circuit

The steps are as follows

1.Redraw the original network showing all time variables, such

as 𝑣(𝑡), 𝑖(𝑡), and 𝑣𝐶(𝑡), as Laplace transforms 𝑉(𝑠), 𝐼(𝑠), and

𝑉𝐶(𝑠), respectively

2.Replace the component values with their impedance values.

This replacement is similar to the case of dc circuits, where

we represent resistors with their resistance values




- Ex.2.7 Transfer Function - Single Loop via Transform Method


Solution

The mess equation using impedances

𝐿𝑠 + 𝑅 +1

𝐶𝑠𝐼 𝑠 = 𝑉(𝑠)

⟹𝐼(𝑠)

𝑉(𝑠)=

1

𝐿𝑠 + 𝑅 +1𝐶𝑠

The voltage across the capacitor

𝑉𝐶 𝑠 = 𝐼(𝑠)1

𝐶𝑠⟹ 𝑉𝐶(𝑠)/𝑉(𝑠)




Simple Circuits via Nodal Analysis

- Ex.2.8 Transfer Function - Single Node via Transform Method


Solution

The transfer function can be obtained by

summing currents flowing out of the node

whose voltage is 𝑉𝐶(𝑠)𝑉𝐶(𝑠)

1/𝐶𝑠+

𝑉𝐶 𝑠 − 𝑉(𝑠)

𝑅 + 𝐿𝑠= 0

: the current flowing out of the node through the

capacitor

: the current flowing out of the node through the

series resistor and inductor


⟹𝑉𝐶(𝑠)

𝑉(𝑠)𝑉𝐶(𝑠)

𝐼/𝐶𝑠

𝑉𝐶 𝑠 − 𝑉(𝑠)

𝑅 + 𝐿𝑠



Complex Circuits via Mesh Analysis

To solve complex electrical networks - those with multiple loops

and nodes – using mesh analysis

1.Replace passive element values with their impedances

2.Replace all sources and time variables with their Laplace

transform

3.Assume a transform current and a current direction in each

mesh

4.Write Kirchhoff’s voltage law around each mesh

5.Solve the simultaneous equations for the output

6.Form the transfer function




- Ex.2.10 Transfer Function – Multiple Loops

Find the transfer function 𝐼2(𝑠)/𝑉(𝑠)

Solution

Convert the network into

Laplace transforms

Summing voltages around

each mesh through which

the assumed currents flow

𝑅1𝐼1 + 𝐿𝑠𝐼1 − 𝐿𝑠𝐼2 = 𝑉

𝐿𝑠𝐼2 +𝑅2𝐼2 +1

𝐶𝑠𝐼2 −𝐿𝑠𝐼1 = 0

𝑅1 + 𝐿𝑠 𝐼1 − 𝐿𝑠𝐼2 = 𝑉

−𝐿𝑠𝐼1 + 𝐿𝑠 + 𝑅2 +1

𝐶𝑠𝐼2 = 0


or

(2.80)



Cramer’s Rule

Consider a system of 𝑛 linear equations for 𝑛 unknowns,

represented in matrix multiplication form as follows

𝐴𝑥 = 𝑏

𝐴 : (𝑛 × 𝑛) matrix has a nonzero determinant

𝑥 : the column vector of the variables 𝑥 = (𝑥1, … , 𝑥𝑛)𝑇

𝑏 : the column vector of known parameters

The system has a unique solution, whose individual values for

the unknowns are given by

𝑥𝑖 =det(𝐴𝑖)

det(𝐴), 𝑖 = 1,… , 𝑛

𝐴𝑖 : the matrix formed by replacing the 𝑖th column of 𝐴 by the

column vector 𝑏



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𝑅1 + 𝐿𝑠 𝐼1 − 𝐿𝑠𝐼2 = 𝑉

−𝐿𝑠𝐼1 + 𝐿𝑠 + 𝑅2 +1

𝐶𝑠𝐼2 = 0

Using Cramer’s rule

𝐼2 =

𝑅1 + 𝐿𝑠 𝑉−𝐿𝑠 0

𝑅1 + 𝐿𝑠 −𝐿𝑠

−𝐿𝑠 𝐿𝑠 + 𝑅2 +1𝐶𝑠

=0 − 𝐿𝑠𝑉

𝑅1 + 𝐿𝑠 𝐿𝑠 + 𝑅2 +1𝐶𝑠

− 𝐿2𝑠2

=𝐿𝐶𝑠2𝑉

𝑅1 + 𝑅2 𝐿𝐶𝑠2 + 𝑅1𝑅2𝐶 + 𝐿 𝑠 + 𝑅1


(2.80)

(2.81)



𝐼2 =𝐿𝐶𝑠2𝑉

𝑅1 + 𝑅2 𝐿𝐶𝑠2 + 𝑅1𝑅2𝐶 + 𝐿 𝑠 + 𝑅1

Forming the transfer function

𝐺 𝑠 =𝐼2 𝑠

𝑉(𝑠)

=𝐿𝐶𝑠2

𝑅1 + 𝑅2 𝐿𝐶𝑠2 + 𝑅1𝑅2𝐶 + 𝐿 𝑠 + 𝑅1

The network is now modeled as the transfer function of figure


(2.82)

(2.81)



Note

𝑅1 + 𝐿𝑠 𝐼1 − 𝐿𝑠𝐼2 = 𝑉

−𝐿𝑠𝐼1 + 𝐿𝑠 + 𝑅2 +1

𝐶𝑠𝐼2 = 0




Run ch2sp4 in Appendix F


• solve simultaneous equations using Cramer’s rule

• solve for the transfer function in Eq. (2.82) using Eq.

(2.80)




Complex Circuits via Nodal Analysis

- Ex.2.11 Transfer Function – Multiple Nodes


Solution

Sum of currents flowing from

the nodes marked 𝑉𝐿(𝑠) and

𝑉𝐶(𝑠)𝑉𝐿 − 𝑉

𝑅1+

𝑉𝐿

𝐿𝑠+

𝑉𝐿 − 𝑉𝐶

𝑅2= 0

𝑉𝐶

1/𝐶𝑠+

𝑉𝐶 − 𝑉𝐿

𝑅2= 0

𝐺1 + 𝐺2 +1

𝐿𝑠𝑉𝐿 − 𝐺2𝑉𝐶 = 𝑉𝐺1

−𝐺2𝑉𝐿 + 𝐺2 + 𝐶𝑠 𝑉𝐶 = 0


or

(2.86)

(2.85)



𝐺1 + 𝐺2 +1

𝐿𝑠𝑉𝐿 − 𝐺2𝑉𝐶 = 𝑉𝐺1

−𝐺2𝑉𝐿 + 𝐺2 + 𝐶𝑠 𝑉𝐶 = 0

Solving for the transfer function

𝑉𝐶(𝑠)

𝑉(𝑠)=

𝐺1𝐺2𝐶

𝑠

𝐺1 + 𝐺2 𝑠2 +𝐺1𝐺2𝐿 + 𝐶

𝐿𝐶𝑠 +

𝐺2𝐿𝐶

Block diagram of the network


(2.86)

(2.87)


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- Another way to write node equations is to replace voltage

sources by current sources. In order to handle multiple-node

electrical networks, we can perform the following steps

1.Replace passive element values with their admittances

2.Replace all sources and time variables with their Laplace

transform

3.Replace transformed voltage sources with transformed

current sources

4.Write Kirchhoff’s current law at each node

5.Solve the simultaneous equations for the output

6.Form the transfer function



Norton's Theorem

Any collection of batteries and resistances with two terminals is electrically equivalent

to an ideal current source 𝑖 in parallel with a single resistor 𝑟. The value of 𝑟 is the

same as that in the Thevenin equivalent and the current 𝑖 can be found by dividing the

open circuit voltage by 𝑟


- Ex.2.12 Transfer Function – Multiple Nodes with Current Sources


Solution

Convert all impedances to

admittances and all voltage

sources in series with an

impedance to current

sources in parallel with an

admittance using Norton’s

theorem




Using the general relationship 𝐼 𝑠 =𝑌 𝑠 𝑉(𝑠) and summing currents at the

node 𝑉𝐿(𝑠)

𝐺1𝑉𝐿 𝑠 +1

𝐿𝑠𝑉𝐿 𝑠 + 𝐺2 𝑉𝐿 𝑠 − 𝑉𝐶 𝑠 = 𝐺1𝑉(𝑠)

Summing the currents at the node 𝑉𝐶(𝑠)

𝐶𝑉𝐶 𝑠 + 𝐺2 𝑉𝐶 𝑠 − 𝑉𝐿 𝑠 = 0 (2.89)

Solving (2.88) and (2.89), forming the transfer function

𝑉𝐶(𝑠)

𝑉(𝑠)=

𝐺1𝐺2𝐶

𝑠

𝐺1 + 𝐺2 𝑠2 +𝐺1𝐺2𝐿 + 𝐶

𝐿𝐶𝑠 +

𝐺2𝐿𝐶


(2.88)



Note

𝐺1𝑉𝐿 𝑠 +1

𝐿𝑠𝑉𝐿 𝑠 + 𝐺2 𝑉𝐿 𝑠 − 𝑉𝐶 𝑠 = 𝐺1𝑉(𝑠)

𝐶𝑉𝐶 𝑠 + 𝐺2 𝑉𝐶 𝑠 − 𝑉𝐿 𝑠 = 0 (2.89)


(2.88)



A Problem-Solving Technique

In all of the previous examples, we have seen a repeating

pattern in the equations that we can use to our advantage. If we

recognize this pattern, we need not write the equations

component by component; we can sum impedances around a

mesh in the case of mesh equations or sum admittances at a

node in the case of node equations




- Ex.2.13 Mesh Equations via Inspection

Write the mesh equations for the

network

Solution

The mesh equations for loop 1

+ 2𝑠 + 2 𝐼1 − 2𝑠 + 1 𝐼2 − 𝐼3 = 𝑉

(2.94.a)



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− 2𝑠 + 2 𝐼1 + 9𝑠 + 1 𝐼2 − 4𝑠𝐼3 = 0

(2.94.b)





−𝐼1 − 4𝑠𝐼2 + 4𝑠 + 1 +1

𝑠𝐼3 = 0

(2.94.c)




+ 2𝑠 + 2 𝐼1 − 2𝑠 + 1 𝐼2 − 𝐼3 = 𝑉 (2.94.a)

− 2𝑠 + 2 𝐼1 + 9𝑠 + 1 𝐼2 − 4𝑠𝐼3 = 0 (2.94.b)

−𝐼1 − 4𝑠𝐼2 + 4𝑠 + 1 +1

𝑠𝐼3 = 0

Matlab syms s I1 I2 I3 V;

A=[(2*s+2) -(2*s+1) -1; -(2*s+1) (9*s+1) -4*s; -1 -4*s

(4*s+1+1/s)]; B=[I1;I2;I3]; C=[V;0;0];

B=inv(A)*C;

pretty(B)


(2.94.c)



Result / 3 2 \

| V (20 s + 13 s + 10 s + 1) |

| ----------------------------------- |

| #1 |

| 3 2 |

| V (8 s + 10 s + 3 s + 1) |

| -------------------------------- |

| #1 |

| 2 |

| V s (8 s + 13 s + 1) |

| -------------------------- |

\ #1 /

where

4 3 2

#1 == 24 s + 30 s + 17 s + 16 s + 1


20𝑠3 + 13𝑠2 + 10𝑠 + 1 𝑉

24𝑠4 + 30𝑠3 + 17𝑠2 + 16𝑠 + 1

8𝑠3 + 10𝑠2 + 3𝑠 + 1 𝑉

24𝑠4 + 30𝑠3 + 17𝑠2 + 16𝑠 + 1

𝑠 8𝑠2 + 13𝑠 + 1 𝑉

24𝑠4 + 30𝑠3 + 17𝑠2 + 16𝑠 + 1



Operational Amplifiers

- An operational amplifier (op-amp) is an electronic amplifier

used as a basic building block to implement transfer functions

- Op-amp has the following characteristics

1.Differential input, 𝑣2 𝑡 − 𝑣1 𝑡

2.High input impedance, 𝑍𝑖 = ∞ (ideal)

3.Low output impedance, 𝑍𝑜 = 0 (ideal)

4.High constant gain amplification, 𝐴 = ∞ (ideal)

- The output, 𝑣𝑜(𝑡), is given by

𝑣𝑜 𝑡 = 𝐴[𝑣2 𝑡 − 𝑣1 𝑡 ] (2.95)




Inverting Operational Amplifiers

- If 𝑣2(𝑡) is grounded, the amplifier is called an inverting

operational amplifier

- The output, 𝑣𝑜(𝑡), is given by

𝑣𝑜 𝑡 = −𝐴𝑣1 𝑡 (2.96)



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Inverting Operational Amplifiers

𝑍𝑖 𝑠 = ∞ → 𝐼𝑎(𝑠) = 0

𝐼1 𝑠 = −𝐼2 𝑠

𝐴 = ∞ → 𝑣1 𝑡 ≈ 0

𝐼1 𝑠 =𝑉𝑖 𝑠

𝑍1 𝑠= −𝐼2 𝑠 = −

𝑉𝑜 𝑠

𝑍2 𝑠

The transfer function of the inverting operational amplifier

𝑉𝑜(𝑠)

𝑉𝑖(𝑠)= −

𝑍2(𝑠)

𝑍1(𝑠)


(2.97)



- Ex.2.14 Transfer Function – Inverting Op-Amp Circuit

Find the transfer function 𝑉𝑜(𝑠)/𝑉𝑖(𝑠)

Solution

The impedances

1

𝑍1=

1

1/𝐶1𝑠+

1

𝑅1→ 𝑍1 =

1

𝐶1𝑠 +1𝑅1

=360× 103

2.016𝑠 + 1

𝑍2 = 𝑅2 +1

𝐶2𝑠= 220 × 103 +

107

𝑠


𝑉𝑜(𝑠)

𝑉𝑖(𝑠)= −

𝑍2 𝑠

𝑍1 𝑠= −

360× 103

2.016𝑠 + 1

220× 103 +107

𝑠

= −1.232𝑠2 + 45.95𝑠 + 22.55

𝑠




Noninverting Operational Amplifiers

𝑉𝑜 = 𝐴 𝑉𝑖 − 𝑉1

𝑉1 =𝑍1

𝑍1 + 𝑍2𝑉𝑜

The transfer function of the noninverting operational amplifier

𝑉𝑜(𝑠)

𝑉𝑖(𝑠)=

𝑍1 𝑠 + 𝑍2(𝑠)

𝑍1(𝑠)


(2.104)



- Ex.2.15 Transfer Function – Noninverting Op-Amp Circuit

Find the transfer function 𝑉𝑜(𝑠)/𝑉𝑖(𝑠)

Solution

The impedances

𝑍1 = 𝑅1 +1

𝐶1𝑠

𝑍2 =𝑅2

1𝐶2𝑠

𝑅2 +1

𝐶2𝑠


𝑉𝑜(𝑠)

𝑉𝑖(𝑠)=

𝑍1 𝑠 + 𝑍2(𝑠)

𝑍1(𝑠)=

𝐶2𝐶1𝑅2𝑅1𝑠2 + (𝐶2𝑅2 + 𝐶1𝑅2 + 𝐶1𝑅1)𝑠 + 1

𝐶2𝐶1𝑅2𝑅1𝑠2 + 𝐶2𝑅2 + 𝐶1𝑅1 𝑠 + 1




Skill Assessment Ex.2.6

Problem Find 𝐺 𝑠 = 𝑉𝐿(𝑠)/𝑉(𝑠) using mesh and nodal analysis

Solution

Mesh analysis

Writing the mesh equations

𝑠 + 1 𝐼1 − 𝑠𝐼2 − 𝐼3 = 𝑉

−𝑠𝐼1 + 2𝑠 + 1 𝐼2 − 𝐼3 = 0

−𝐼1 − 𝐼2 + 𝑠 + 2 𝐼3 = 0




𝑠 + 1 𝐼1 − 𝑠𝐼2 − 𝐼3 = 𝑉

−𝑠𝐼1 + 2𝑠 + 1 𝐼2 − 𝐼3 = 0

−𝐼1 − 𝐼2 + 𝑠 + 2 𝐼3 = 0

Solving the mesh equation for 𝐼2

𝐼2 =

𝑠 + 1 𝑉 −1−𝑠 0 −1−1 0 𝑠 + 2

𝑠 + 1 −𝑠 −1−𝑠 2𝑠 + 1 −1−1 −1 𝑠 + 2

=𝑠2 + 2𝑠 + 1 𝑉

𝑠(𝑠2 + 5𝑠 + 2)

The voltage across 𝐿

𝑉𝐿 = 𝑠𝐼2 =𝑠2 + 2𝑠 + 1 𝑉

𝑠2 + 5𝑠 + 2

⟹ 𝐺 𝑠 =𝑉𝐿

𝑉=

𝑠2 + 2𝑠 + 1

𝑠2 + 5𝑠 + 2



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Nodal analysis

Writing the nodal equations

1

𝑠+ 2 𝑉1 − 𝑉𝐿 = 𝑉

−𝑉1 +2

𝑠+ 1 𝑉𝐿 =

1

𝑠𝑉




1

𝑠+ 2 𝑉1 − 𝑉𝐿 = 𝑉

−𝑉1 +2

𝑠+ 1 𝑉𝐿 =

1

𝑠𝑉

Solving the nodal equation for 𝑉𝐿

𝑉𝐿 =

1𝑠

+ 2 𝑉

−11𝑠

1𝑠

+ 2 −1

−12𝑠

+ 1

=𝑠2 + 2𝑠 + 1 𝑉

𝑠2 + 5𝑠 + 2

⟹ 𝐺 𝑠 =𝑉𝐿

𝑉=

𝑠2 + 2𝑠 + 1

𝑠2 + 5𝑠 + 2




Skill Assessment Ex.2.7

Problem If 𝑍1(𝑠)is the impedance of a 10𝜇𝐹 capacitor and 𝑍2(𝑠)is the impedance of a 100𝑘Ω resistor, find the transfer

function, 𝐺 𝑠 = 𝑉𝑜(𝑠)/𝑉𝑖(𝑠) if these components are

used with (a) an inverting op-amp and (b) a

noninverting op-amp

Solution

𝑍1 = 𝑍𝐶 =1

𝐶𝑠=

1

10−5𝑠=

105

𝑠𝑍2 = 𝑍𝑅 = 𝑅 = 105




(a) Inverting Op-Amp

𝐺 𝑠 = −𝑍2

𝑍1

= −105

105

𝑠= −𝑠

(b) Noninverting Op-Amp

𝐺 𝑠 =𝑍1 + 𝑍2

𝑍1

=

105

𝑠+ 105

105

𝑠= 𝑠 + 1



§5.Translational Mechanical System Transfer Functions




- Ex.2.16 Transfer Function - One Equation of Motion

Find the transfer function 𝑋(𝑠)/𝐹(𝑠)

Solution

Free body diagram

Using Newton’s law to sum all of the forces

𝑀𝑑2𝑥(𝑡)

𝑑𝑡2 + 𝑓𝑣𝑑𝑥(𝑡)

𝑑𝑡+ 𝐾𝑥 𝑡 = 𝑓(𝑡)

Taking Laplace transform

𝑀𝑠2𝑋 𝑠 + 𝑓𝑣𝑠𝑋 𝑠 + 𝐾𝑋 𝑠 = 𝐹(𝑠)


𝐺(𝑠) =𝑋(𝑠)

𝐹(𝑠)=

𝐹 𝑠

𝑀𝑠2 + 𝑓𝑣𝑠 + 𝐾



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Impedance

- Define impedance for mechanical components

𝑍𝑀 𝑠 ≡𝐹 𝑠

𝑋 𝑠

⟹ 𝐹 𝑠 = 𝑍𝑀 𝑠 𝑋 𝑠

Sum of Impedances × 𝑋(𝑠) = Sum of Applied Forces

- The impedance of a spring is its stiffness coefficient

𝐹 𝑠 = 𝐾𝑋 𝑠 ⟹ 𝑍𝑀 𝑠 = 𝐾 (2.112)

- For the viscous damper

𝐹 𝑠 = 𝑓𝑣𝑠𝑋 𝑠 ⟹ 𝑍𝑀 𝑠 = 𝑓𝑣𝑠 (2.113)

- For the mass

𝐹 𝑠 = 𝑀𝑠2𝑋 𝑠 ⟹ 𝑍𝑀 𝑠 = 𝑀𝑠2 (2.114)




- Ex.2.17 Transfer Function - Two Degrees of Freedom

Find the transfer function

𝑋2(𝑠)/𝐹(𝑠)

Solution

Free body diagram of 𝑀1

The Laplace transform of the equation of motion of 𝑀1

+ 𝑀1𝑠2 + 𝑓𝑣1

+ 𝑓𝑣3𝑠 + 𝐾1 + 𝐾2 𝑋1 − 𝑓𝑣3

𝑠 + 𝐾2 𝑋2 = 𝐹




Free body diagram of 𝑀2

The Laplace transform of the equation of motion of 𝑀2

− 𝑓𝑣3𝑠 + 𝐾2 𝑋1 + 𝑀2𝑠

2 + 𝑓𝑣2+ 𝑓𝑣3

𝑠 + 𝐾2 + 𝐾3 𝑋2 = 0




The Laplace transform of the equations of motion

+ 𝑀1𝑠2 + 𝑓𝑣1

+ 𝑓𝑣3𝑠 + 𝐾1 + 𝐾2 𝑋1 − 𝑓𝑣3

𝑠 + 𝐾2 𝑋2 = 𝐹

− 𝑓𝑣3𝑠 + 𝐾2 𝑋1 + 𝑀2𝑠

2 + 𝑓𝑣2+ 𝑓𝑣3

𝑠 + 𝐾2 + 𝐾3 𝑋2 = 0

The Laplace transform of the equations of motion

𝑋2 =

𝑀1𝑠2 + 𝑓𝑣1

+ 𝑓𝑣3𝑠 + 𝐾1 + 𝐾2 𝐹

− 𝑓𝑣3𝑠 + 𝐾2 0

∆=

𝑓𝑣3𝑠 + 𝐾2 𝐹

∆where

∆ =𝑀1𝑠

2 + 𝑓𝑣1 +𝑓𝑣3 𝑠+ 𝐾1 +𝐾2 − 𝑓𝑣3𝑠 + 𝐾2

− 𝑓𝑣3𝑠 + 𝐾2 𝑀2𝑠

2 + 𝑓𝑣2 +𝑓𝑣3 𝑠+ 𝐾2 +𝐾3


𝐺 𝑠 =𝑋2(𝑠)

𝐹(𝑠)=

𝑓𝑣3𝑠 + 𝐾2

∆




Note

The Laplace transform of the equations of motion of 𝑀1

+ 𝑀1𝑠2 + 𝑓𝑣1

+ 𝑓𝑣3𝑠 + 𝐾1 + 𝐾2 𝑋1 − 𝑓𝑣3

𝑠 + 𝐾2 𝑋2 = 𝐹


− 𝑓𝑣3𝑠 + 𝐾2 𝑋1 + 𝑀2𝑠

2 + 𝑓𝑣2+ 𝑓𝑣3

𝑠 + 𝐾2 + 𝐾3 𝑋2 = 0




- Ex.2.18 Equations of Motion by Inspection

Write the equations of

motion for the mechanical

network

Solution


+ 𝑀1𝑠2 + 𝑓𝑣1

+ 𝑓𝑣3𝑠 + 𝐾1 + 𝐾2 𝑋1 − 𝐾2𝑋2 − 𝑓𝑣3

𝑋3 = 0



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−𝐾2𝑋1 + [𝑀2𝑠2 + 𝑓𝑣2

+ 𝑓𝑣4𝑠 + 𝐾2]𝑋2 − 𝑓𝑣4

𝑠𝑋3 = 𝐹





−𝑓𝑣3𝑠𝑋1 − 𝑓𝑣4

𝑠𝑋2 + [𝑀3𝑠2 + 𝑓𝑣3

+ 𝑓𝑣4𝑠]𝑋3 = 0





Problem Find the transfer

function

𝐺 𝑠 =𝑋2(𝑠)

𝐹(𝑠)

Solution

+ 𝑀1𝑠2 + 𝑓𝑣1

+ 𝑓𝑣2+ 𝑓𝑣3

𝑠 + 𝐾 𝑋1

− 𝑓𝑣1+ 𝑓𝑣2

+ 𝑓𝑣3𝑠 + 𝐾 𝑋2 = 𝐹

⟹ +(𝑠2 + 3𝑠 + 1)𝑋1 − (3𝑠 + 1)𝑋2 = 𝐹




− 𝑓𝑣1+ 𝑓𝑣2

+ 𝑓𝑣3𝑠 + 𝐾 𝑋1

+ 𝑀2𝑠2 + 𝑓𝑣1

+ 𝑓𝑣2+ 𝑓𝑣3

+ 𝑓𝑣4𝑠 + 𝐾 𝑋2 = 0

⟹ −(3𝑠 + 1)𝑋1 + (𝑠2 + 4𝑠 + 1)𝑋2 = 0




+ 𝑠2 + 3𝑠 + 1 𝑋1 − 3𝑠 + 1 𝑋2 = 𝐹

−(3𝑠 + 1)𝑋1 + (𝑠2 + 4𝑠 + 1)𝑋2 = 0

The solution for 𝑋2

𝑋2 =

𝑠2 + 3𝑠 + 1 𝐹− 3𝑠 + 1 0

∆=

3𝑠 + 1 𝐹

∆where

∆=𝑠2 + 3𝑠 + 1 − 3𝑠 + 1

− 3𝑠 + 1 𝑠2 + 4𝑠 + 1

= 𝑠(𝑠3 + 7𝑠2 + 5𝑠 + 1)

⟹ 𝐺 𝑠 =𝑋2(𝑠)

𝐹(𝑠)=

3𝑠 + 1

𝑠(𝑠3 + 7𝑠2 + 5𝑠 + 1)



§6.Rotational Mechanical System Transfer Functions



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- Ex.2.19 Transfer Function – Two Equations of Motion

Find the transfer function, 𝜃2(𝑠)/𝑇(𝑠), for the rotational system

shown in figure. The rod is supported by bearings at either end

and is undergoing torsion. A torque is applied at the left, and

the displacement is measured at the right

Solution

First, obtain the schematic from the physical system




Next, draw a free-body diagram of 𝐽1 và 𝐽2, using superposition

𝐽1𝑠2 + 𝐷1𝑠 + 𝐾 𝜃1 𝑠 − 𝐾𝜃2 𝑠 = 𝑇(𝑠) (2.127.a)

−𝐾1𝜃1 𝑠 + (𝐽2𝑠2 + 𝐷2𝑠 + 𝐾)𝜃2 𝑠 = 0 (2.127.b)


Final free-body diagram for 𝐽2

Torques on 𝐽2 due only

to the motion of 𝐽2



Final free-body diagram for 𝐽1







𝐽1𝑠2 + 𝐷1𝑠 + 𝐾 𝜃1 𝑠 − 𝐾𝜃2 𝑠 = 𝑇(𝑠)(2.127.a)

−𝐾1𝜃1 𝑠 + (𝐽2𝑠2 + 𝐷2𝑠 + 𝐾)𝜃2 𝑠 = 0 (2.127.b)

The solution for 𝜃2

𝜃2 =

𝐽1𝑠2 + 𝐷1𝑠 + 𝐾 𝑇

−𝐾 0∆

=𝐾𝑇

∆where

∆=𝐽1𝑠

2 + 𝐷1𝑠 + 𝐾 −𝐾

−𝐾 𝐽2𝑠2 + 𝐷2𝑠 + 𝐾

⟹ 𝐺 𝑠 =𝜃2(𝑠)

𝑇(𝑠)=

𝐾

∆




Note

𝐽1𝑠2 + 𝐷1𝑠 + 𝐾 𝜃1 𝑠 − 𝐾𝜃2 𝑠 = 𝑇(𝑠) (2.127.a)

−𝐾1𝜃1 𝑠 + (𝐽2𝑠2 + 𝐷2𝑠 + 𝐾)𝜃2 𝑠 = 0 (2.127.b)




- Ex.2.20 Equations of Motion by Inspection

Write the Laplace transform of the equations of motion for the

system shown in the figure

Solution

The Laplace transform of the equations of motion of 𝐽1

+ 𝐽1𝑠2 + 𝐷1𝑠 + 𝐾 𝜃1 − 𝐾𝜃2 − 0𝜃3 = 𝑇(𝑠) (2.131.a)





−𝐾𝜃1 + 𝐽2𝑠2 + 𝐷2𝑠 + 𝐾 𝜃2 − 𝐷2𝑠𝜃3 = 0 (2.131.b)



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−0𝜃1 − 𝐷2𝑠𝜃2 + 𝐽3𝑠2 + 𝐷3𝑠 + 𝐷2𝑠 𝜃3 = 0 (2.131.c)





Problem

Find the transfer function

𝐺 𝑠 =𝜃2(𝑠)

𝑇(𝑠)

Solution

The equations of motion

+ 𝑠2 + 𝑠 + 1 𝜃1 𝑠 − (𝑠 + 1)𝜃2 𝑠 = 𝑇(𝑠)

− 𝑠 + 1 𝜃1 𝑠 + (2𝑠 + 2)𝜃2 𝑠 = 0





𝑠2 + 𝑠 + 1 𝜃1 𝑠 − (𝑠 + 1)𝜃2 𝑠 = 𝑇(𝑠)

− 𝑠 + 1 𝜃1 𝑠 + (2𝑠 + 2)𝜃2 𝑠 = 0

Solving for 𝜃2(𝑠)

𝜃2 =

𝑠2 + 𝑠 + 1 𝑇− 𝑠 + 1 0

𝑠2 + 𝑠 + 1 −(𝑠 + 1)

− 𝑠 + 1 2𝑠 + 2

=(𝑠 + 1)𝑇

2𝑠3 + 3𝑠2 + 2𝑠 + 1

⟹ 𝐺 𝑠 =𝜃2(𝑠)

𝑇(𝑠)=

𝑠 + 1

2𝑠3 + 3𝑠2 + 2𝑠 + 1



§7.Transfer Functions for Systems with Gears

Kinematic relationship𝜃2

𝜃1=

𝑟1𝑟2

=𝑁1

𝑁2

Power on gears

𝑇1𝜃1 = 𝑇2𝜃2

The ratio of torques on two gears𝑇2

𝑇1=

𝜃1

𝜃2=

𝑁2

𝑁1

𝜃1, 𝜃2 : rotation angles of gear 1 and 2, 𝑟𝑎𝑑

𝑟1, 𝑟2 : radius of gear 1 and 2, 𝑚

𝑁1, 𝑁2 : number of teeth of gear 1 and 2

𝑇1, 𝑇2 : torques on gear 1 and 2, 𝑁𝑚




What happens to mechanical impedances that are driven by gears?

(a) : gears driving a rotational inertia, spring, and viscous damper

(b) : an equivalent system at 𝜃1 without the gears

Can the mechanical impedances be reflected from the output to

the input, thereby eliminating the gears?


b.equivalent system at the output after reflection of input torque

a.rotational system driven by gears



𝑇1 can be reflected to the output by multiplying by 𝑁2/𝑁1

𝐽𝑠2 + 𝐷𝑠 + 𝐾 𝜃2 𝑠 = 𝑇1(𝑠)𝑁2

𝑁1

Convert 𝜃2(𝑠) into an equivalent 𝜃1(𝑠), so that

𝐽𝑠2 + 𝐷𝑠 + 𝐾𝑁1

𝑁2𝜃1 𝑠 = 𝑇1(𝑠)

𝑁2

𝑁1




(2.131)

(2.132)


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20


𝐽𝑠2 + 𝐷𝑠 + 𝐾 𝜃2 𝑠 = 𝑇1(𝑠)(𝑁2/𝑁1) (2.131)

𝐽𝑠2 + 𝐷𝑠 + 𝐾 (𝑁2/𝑁1)𝜃1 𝑠 = 𝑇1(𝑠)(𝑁2/𝑁1) (2.132)

⟹ 𝐽𝑁1

𝑁2

2

𝑠2 + 𝐷𝑁1

𝑁2

2

𝑠 + 𝐾𝑁1

𝑁2

2

𝜃1 𝑠 = 𝑇1(𝑠)

Thus, the load can be thought of as having been reflected from

the output to the input


(2.133)


c.equivalent system at the input after reflection of impedances




Generalizing the results

Rotational mechanical impedances can be reflected through

gear trains by multiplying the mechanical impedance by the

ratio

where the impedance to be reflected is attached to the source

shaft and is being reflected to the destination shaft




- Ex.2.21 Transfer Function - System with Lossless Gears

Find the transfer function, 𝜃2(𝑠)/𝑇1(𝑠), for the system

Solution

Reflect the impedances (𝐽1 and 𝐷1) and torque (𝑇1) on the input

shaft to the output, where the impedances are reflected by

(𝑁1/𝑁2)2 and the torque is reflected by (𝑁1/𝑁2)

The equation of motion can now be written as

𝐽𝑒𝑠2 + 𝐷𝑒𝑠 + 𝐾𝑒 𝜃2 𝑠 = 𝑇1 𝑠 (𝑁2/𝑁1) (2.139)


b.system after reflection of torquesand impedances to the output shaft

a.rotationalmechanicalsystem withgears



𝐽𝑒𝑠2 + 𝐷𝑒𝑠 + 𝐾𝑒 𝜃2 𝑠 = 𝑇1 𝑠 (𝑁2/𝑁1) (2.139)

where, 𝐽𝑒 = 𝐽1(𝑁2/𝑁1)2+𝐽2, 𝐷𝑒 = 𝐷1(𝑁2/𝑁1)

2+𝐷2, 𝐾𝑒 = 𝐾2

Solving for 𝐺(𝑠)


𝑇1(𝑠)=

𝑁2/𝑁1

𝐽𝑒𝑠2 + 𝐷𝑒𝑠 + 𝐾𝑒


b.system after reflection of torquesand impedances to the output shaft

a.rotationalmechanicalsystem withgears



- In order to eliminate gears with large radii, a gear train is used

to implement large gear ratios by cascading smaller gear ratios.

𝜃4 =𝑁1𝑁3𝑁5

𝑁2𝑁4𝑁6𝜃1

- For gear trains, the equivalent gear ratio is the product of the

individual gear ratios




- Ex.2.22 Transfer Function – Gears with Loss

Find the transfer function, 𝜃1(𝑠)/𝑇1(𝑠), for the system

Solution

Reflect all of the impedances to the input shaft, 𝜃1

The equation of motion can now be written as

𝐽𝑒𝑠2 + 𝐷𝑒𝑠 𝜃1 𝑠 = 𝑇1 𝑠


𝐺 𝑠 = 𝜃1 𝑠 /𝑇1 𝑠 = 1/(𝐽𝑒𝑠2 + 𝐷𝑒𝑠)


b.equivalent system at the inputa.system using a gear train


1/11/2016

21



Problem Find the TF


𝑇(𝑠)

Solution Transforming the network to one without gears by

reflecting the 4𝑁𝑚/𝑟𝑎𝑑 spring to the left and multiplying

by (25/50)2

4[𝑁𝑚/𝑟𝑎𝑑] ×25

50

2

= 1[𝑁𝑚/𝑟𝑎𝑑]




The equation of motion

+ 𝑠2 + 𝑠 𝜃1 𝑠 − 𝑠𝜃𝑎 𝑠 = 𝑇(𝑠)

−𝑠𝜃1 𝑠 + (𝑠 + 1)𝜃𝑎 𝑠 = 0




The equation of motion

𝑠2 + 𝑠 𝜃1 𝑠 − 𝑠𝜃𝑎 𝑠 = 𝑇(𝑠)

−𝑠𝜃1 𝑠 + (𝑠 + 1)𝜃𝑎 𝑠 = 0

Solving for 𝜃𝑎(𝑠)

𝜃𝑎 𝑠 =

𝑠2 + 𝑠 𝑇−𝑠 0

𝑠2 + 𝑠 −𝑠−𝑠 𝑠 + 1

=𝑠𝑇(𝑠)

𝑠3 + 𝑠2 + 𝑠

⟹𝜃𝑎 𝑠

𝑇(𝑠)=

1

𝑠2 + 𝑠 + 1


𝜃2 𝑠

𝑇(𝑠)=

12

𝜃𝑎 𝑠

𝑇(𝑠)=

1/2

𝑠2 + 𝑠 + 1



§8.Electromechanical System Transfer Functions


NASA flight simulator robot arm with electromechanical control system components



- A motor is an electromechanical component that yields a

displacement output for a voltage input, that is, a mechanical

output generated by an electrical input

- Derive the transfer function for the armature-controlled dc

servomotor (Mablekos, 1980)

• Fixed field: a magnetic field is developed by stationary

permanent magnets or a stationary electromagnet

• Armature: a rotating circuit, through

which current 𝑖𝑎(𝑡) flows, passes

through this magnetic field at right

angles and feels a force

𝐹 = 𝐵𝑙𝑖𝑎(𝑡)

𝐵 : the magnetic field strength

𝑙 : the length of the conductor




• A conductor moving at right angles to a magnetic field

generates a voltage at the terminals of the conductor equal to

𝑒 = 𝐵𝑙𝑣

𝑒 : the voltage

𝑣 : the velocity of the conductor normal to the magnetic field

• The current-carrying armature is rotating in a magnetic field,

its voltage is proportional to speed

𝑣𝑏 𝑡 = 𝐾𝑏 𝜃𝑚(𝑡) (2.144)

𝑣𝑏 𝑡 : the back electromotive force

(back emf)

𝐾𝑏 : a constant of proportionality

called the back emf constant

𝜃𝑚(𝑡): the angular velocity of the

motor



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22


• Taking the Laplace transform

𝑉𝑏(𝑠) = 𝐾𝑏𝑠𝜃𝑚(𝑠) (2.145)

• The relationship between the armature current, 𝑖𝑎(𝑡) , theapplied armature voltage, 𝑒𝑎(𝑡), and the back emf, 𝑣𝑏(𝑡)

𝑅𝑎𝐼𝑎 𝑠 + 𝐿𝑎𝑠𝐼𝑎 𝑠 + 𝑉𝑏 𝑠 = 𝐸𝑎 𝑠 (2.146)

• The torque developed by the motor is proportional to the

armature current

𝑇𝑚 𝑠 = 𝐾𝑡𝐼𝑎 𝑠 (2.147)

𝑇𝑚 : the torque developed by the

motor

𝐾𝑡 : the motor torque constant,

which depends on the motor

and magnetic field characteristics



𝑉𝑏(𝑠) = 𝐾𝑏𝑠𝜃𝑚(𝑠) (2.145), 𝑅𝑎𝐼𝑎 𝑠 + 𝐿𝑎𝑠𝐼𝑎 𝑠 + 𝑉𝑏 𝑠 = 𝐸𝑎 𝑠 (2.146)


• Rearranging Eq.(2.147)

𝐼𝑎(𝑠) =1

𝐾𝑡𝑇𝑚(𝑠)

• To find the TF of the motor, first substitute Eqs. (2.145) and

(2.148) into (2.146), yielding

(𝑅𝑎 + 𝐿𝑎𝑠)𝑇𝑚(𝑠)

𝐾𝑡+ 𝐾𝑏𝑠𝜃𝑚 𝑠 = 𝐸𝑎 𝑠

• Then, find 𝑇𝑚(𝑠) in terms of 𝜃𝑚(𝑠),separate the input and output

variables and obtain the transfer

function, 𝜃𝑚(𝑠)/𝐸𝑎(𝑠)


(2.148)

(2.149)


(𝑅𝑎+𝐿𝑎𝑠)𝑇𝑚(𝑠)

𝐾𝑡+ 𝐾𝑏𝑠𝜃𝑚 𝑠 = 𝐸𝑎 𝑠 (2.149)


• A typical equivalent mechanical loading on a motor

𝐽𝑚 : the equivalent inertia at the armature and includes both the

armature inertia and, the load inertia reflected to the armature

𝐷𝑚 : the equivalent viscous damping at the armature and

includes both the armature viscous damping and, the

load viscous damping reflected to the armature

𝑇𝑚 𝑠 = (𝐽𝑚𝑠2 + 𝐷𝑚𝑠)𝜃𝑚(𝑠) (2.150)

• Substituting Eq.(2.150) into Eq.(2.149)

(𝑅𝑎 + 𝐿𝑎𝑠)(𝐽𝑚𝑠2 + 𝐷𝑚𝑠)𝜃𝑚(𝑠)

𝐾𝑡+ 𝐾𝑏𝑠𝜃𝑚 𝑠 = 𝐸𝑎(𝑠)


(2.151)


(𝑅𝑎+𝐿𝑎𝑠)(𝐽𝑚𝑠2+𝐷𝑚𝑠)𝜃𝑚(𝑠)

𝐾𝑡+ 𝐾𝑏𝑠𝜃𝑚 𝑠 = 𝐸𝑎(𝑠) (2.151)


• Assume that the armature inductance, 𝐿𝑎, is small compared

to the armature resistance, 𝑅𝑎, which is usual for a dc motor,

Eq. (2.151) becomes

𝑅𝑎

𝐾𝑡𝐽𝑚𝑠 + 𝐷𝑚 + 𝐾𝑏 𝑠𝜃𝑚 𝑠 = 𝐸𝑎(𝑠)

• After simplification

𝜃𝑚(𝑠)

𝐸𝑎(𝑠)=

𝐾𝑡𝑅𝑎

1𝐽𝑚

𝑠 𝑠 +1𝐽𝑚

𝐷𝑚 +𝐾𝑡𝑅𝑎

𝐾𝑏

• The form of Eq.(2.153)

𝜃𝑚(𝑠)

𝐸𝑎(𝑠)=

𝐾

𝑠(𝑠 + 𝛼)


(2.153)

(2.154)

(2.152)



• Let us first discuss the mechanical constants, 𝐽𝑚 and 𝐷𝑚.

Consider the figure: a motor with inertia

𝐽𝑎 and damping 𝐷𝑎 at the armature

driving a load consisting of inertia 𝐽𝐿and damping 𝐷𝐿

Assuming that all inertia and damping values shown are

known, 𝐽𝐿 and 𝐷𝐿 can be reflected back to the armature as

some equivalent inertia and damping to be added to 𝐽𝑎 and

𝐷𝑎 , respectively. Thus, the equivalent inertia, 𝐽𝑚 , and

equivalent damping, 𝐷𝑚, at the armature are

𝐽𝑚 = 𝐽𝑎 + 𝐽𝐿𝑁1

𝑁2

2

𝐷𝑚 = 𝐷𝑎 + 𝐷𝐿

𝑁1

𝑁2

2


(2.155.a)

(2.155.b)


𝑉𝑏(𝑠) = 𝐾𝑏𝑠𝜃𝑚(𝑠) (2.145), 𝑅𝑎𝐼𝑎 𝑠 + 𝐿𝑎𝑠𝐼𝑎 𝑠 + 𝑉𝑏 𝑠 = 𝐸𝑎 𝑠 (2.146), 𝐼𝑎(𝑠) =1

𝐾𝑡𝑇𝑚(𝑠) (2.148)


• Substituting Eqs.(2.145), (2.148) into Eq. (2.146), with 𝐿𝑎 = 0𝑅𝑎

𝐾𝑡𝑇𝑚 𝑠 + 𝐾𝑏𝑠𝜃𝑚 𝑠 = 𝐸𝑎 𝑠

Taking the inverse Laplace transform𝑅𝑎

𝐾𝑡𝑇𝑚 𝑡 + 𝐾𝑏𝜔𝑚 𝑡 = 𝑒𝑎 𝑡

• When the motor is operating at steady state with a dc voltage

input𝑅𝑎

𝐾𝑡𝑇𝑚 + 𝐾𝑏𝜔𝑚 = 𝑒𝑎

⟹ 𝑇𝑚 = −𝐾𝑏𝐾𝑡

𝑅𝑎𝜔𝑚 +

𝐾𝑡

𝑅𝑎𝑒𝑎


(2.156)

(2.157)

(2.158)

(2.159)


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The stall torque

𝑇𝑠𝑡𝑎𝑙𝑙 =𝐾𝑡

𝑅𝑎𝑒𝑎

The no-load speed

𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 =𝑒𝑎

𝐾𝑏

The electrical constants of the motor𝐾𝑡

𝑅𝑎=

𝑇𝑠𝑡𝑎𝑙𝑙

𝑒𝑎

𝐾𝑏 =𝑒𝑎

𝜔𝑛𝑜−𝑙𝑜𝑎𝑑

The electrical constants, 𝐾𝑡/𝑅𝑎 and 𝐾𝑏, can be found from a

dynamometer test of the motor, which would yield 𝑇𝑠𝑡𝑎𝑙𝑙 and

𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 for a given 𝑒𝑎


(2.160)

(2.161)

(2.162)

(2.163)



- Ex.2.23 Transfer Function-DC Motor and Load

Given the system and torque-speed curve, find the TF,


𝜃𝐿 𝑠

𝐸𝑎 𝑠



Solution

Find the mechanical constants


𝑁2

2

= 5 + 700 ×100

1000

2

= 12


𝑁1

𝑁2

2

= 2 + 800 ×100

1000

2

= 10




Find the electrical constants from the torque-speed curve

𝑇𝑠𝑡𝑎𝑙𝑙 = 500, 𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 = 50, 𝑒𝑎 = 100

𝐾𝑡

𝑅𝑎=


𝑒𝑎=

500

100= 5

𝐾𝑏 =𝑒𝑎

𝜔𝑛𝑜−𝑙𝑜𝑎𝑑=

100

50= 2




The transfer function 𝜃𝑚(𝑠)/𝐸𝑎(𝑠)

𝜃𝑚(𝑠)

𝐸𝑎(𝑠)=

𝐾𝑡𝑅𝑎

1𝐽𝑚



𝐾𝑏

=5 ×

112

𝑠 𝑠 +112

× 10 + 5 × 2

=0.417

𝑠(𝑠 + 1.667)

The transfer function 𝜃𝐿(𝑠)/𝐸𝑎(𝑠)

𝜃𝐿(𝑠)

𝐸𝑎(𝑠)=

𝜃𝑚(𝑠)𝑁1𝑁2

𝐸𝑎(𝑠)=

0.417 ×1001000

𝑠(𝑠 + 1.667)=

0.0417

𝑠(𝑠 + 1.667)





Problem Find the TF, 𝐺 𝑠 = 𝜃𝐿(𝑠)/𝐸𝑠(𝑠), for the motor and load

system. The torque-speed curve is given by 𝑇𝑚 =− 8𝜔𝑚 + 200 when the input voltage is 100𝑣𝑜𝑙𝑡𝑠

Solution

Find the mechanical constants


𝑁2

𝑁3

𝑁4

2

= 1 + 400 ×20

100×

25

100

2

= 2


𝑁1

𝑁2

𝑁3

𝑁4

2

= 5 + 800 ×20

100×

25

100

2

= 7



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𝑇𝑚 = −8𝜔𝑚 + 200


Find the electrical constants from the torque-speed eq.

𝜔𝑚 = 0 ⟹ 𝑇𝑚 = 200

𝑇𝑚 = 0 ⟹ 𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 = 200/8 = 25

𝐾𝑡

𝑅𝑎=


𝐸𝑎=

200

100= 2

𝐾𝑏 =𝐸𝑎

𝜔𝑛𝑜−𝑙𝑜𝑎𝑑=

100

25= 4




Substituting all values into the motor transfer function

𝜃𝑚(𝑠)

𝐸𝑎(𝑠)=

𝐾𝑡𝑅𝑎

1𝐽𝑚



𝐾𝑏

=2 ×

12

𝑠 𝑠 +12

7 + 2 × 4

=1

𝑠(𝑠 + 7.5)

The transfer function 𝜃𝐿(𝑠)/𝐸𝑎(𝑠)

𝜃𝐿(𝑠)

𝐸𝑎(𝑠)=

𝜃𝑚(𝑠)𝑁1𝑁2

𝑁3𝑁4

𝐸𝑎(𝑠)=

20100

×25100

𝑠(𝑠 + 7.5)=

0.05

𝑠(𝑠 + 7.5)



§9.Electric Circuit Analogs

- Electric circuit analog: an electric circuit that is analogous to a

system from another discipline

- In the commonality of systems from the various disciplines, the

mechanical systems can be represented by equivalent electric

circuits

- Analogs can be obtained by comparing the describing

equations, such as the equations of motion of a mechanical

system, with either electrical mesh or nodal equations

• When compared with mesh equations, the resulting electrical

circuit is called a series analog

• When compared with nodal equations, the resulting electrical

circuit is called a parallel analog




Series Analog

Consider the translational mechanical system, the equation of motion

𝑀𝑠2 + 𝑓𝑣𝑠 + 𝐾 𝑋 𝑠 = 𝐹 𝑠 =𝑀𝑠2 + 𝑓𝑣𝑠 + 𝐾

𝑠𝑠𝑋 𝑠

⟹ 𝑀𝑠 + 𝑓𝑣 +𝐾

𝑠𝑉 𝑠 = 𝐹(𝑠)

Kirchhoff’s mesh equation for the simple series RLC network

𝐿𝑠 + 𝑅 +1

𝐶𝑠𝐼 𝑠 = 𝐸(𝑠)




Parameters for series analog

mass = 𝑀 ⟹ inductor 𝐿 = 𝑀 henries

viscous damper = 𝑓𝑣 ⟹ resistor 𝑅 = 𝑓𝑣ohms

spring = 𝐾 ⟹ capacitor 𝐶 = 1/𝐾 farads

applied force = 𝑓(𝑡) ⟹ voltage source 𝑒 𝑡 = 𝑓(𝑡)

velocity = 𝑣(𝑡) ⟹ mesh current 𝑖 𝑡 = 𝑣(𝑡)




- Ex.2.24 Converting a Mechanical System to a Series Analog

Draw a series analog for the mechanical system

Solution

The equations of motion with 𝑋(𝑠) → 𝑉(𝑠)

𝑀1𝑠 + 𝑓𝑣1+ 𝑓𝑣3

+𝐾1 + 𝐾2

𝑠𝑉1 𝑠 − 𝑓𝑣3

+𝐾2

𝑠𝑉2 𝑠 = 𝐹(𝑠)

− 𝑓𝑣3+

𝐾2

𝑠𝑉1 𝑠 + 𝑀2𝑠 + 𝑓𝑣2

+ 𝑓𝑣3+

𝐾2 + 𝐾3

𝑠𝑉2 𝑠 = 0



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25

+ 𝑀1𝑠 + 𝑓𝑣1+ 𝑓𝑣3

+𝐾1+𝐾2


+𝐾2


− 𝑓𝑣3+

𝐾2


+ 𝑓𝑣3+

𝐾2+𝐾3

𝑠𝑉2 𝑠 = 0


Coefficients represent sums of electrical impedance.

Mechanical impedances associated with 𝑀1 form the first mesh,

where impedances between the two masses are common to

the two loops. Impedances associated with 𝑀2 form the second

mesh

𝑣1(𝑡) and 𝑣2(𝑡)are the velocities of 𝑀1 and 𝑀2, respectively




Parallel Analog

Consider the translational mechanical system, the equation of motion

𝑀𝑠 + 𝑓𝑣 +𝐾

𝑠𝑉 𝑠 = 𝐹(𝑠)

Kirchhoff’s nodal equation for the simple parallel RLC network

𝐶𝑠 +1

𝑅𝑠+

1

𝐿𝑠𝐸 𝑠 = 𝐼(𝑠)




Parameters for parallel analog

mass = 𝑀 ⟹ capacitor 𝐶 = 𝑀 farads

viscous damper = 𝑓𝑣 ⟹ resistor 𝑅 = 1/𝑓𝑣 ohms

spring = 𝐾 ⟹ inductor 𝐿 = 1/𝐾 henries

applied force = 𝑓(𝑡) ⟹ current source 𝑖(𝑡) = 𝑓(𝑡)

velocity = 𝑣(𝑡) ⟹ node voltage 𝑒(𝑡) = 𝑣(𝑡)




- Ex.2.25 Converting a Mechanical System to a Parallel Analog

Draw a parallel analog for the mechanical system

Solution

The equations of motion with 𝑋(𝑠) → 𝑉(𝑠)

𝑀1𝑠 + 𝑓𝑣1+ 𝑓𝑣3

+𝐾1 + 𝐾2


+𝐾2


− 𝑓𝑣3+

𝐾2


+ 𝑓𝑣3+

𝐾2 + 𝐾3

𝑠𝑉2 𝑠 = 0



+ 𝑀1𝑠 + 𝑓𝑣1+ 𝑓𝑣3

+𝐾1+𝐾2


+𝐾2


− 𝑓𝑣3+

𝐾2


+ 𝑓𝑣3+

𝐾2+𝐾3

𝑠𝑉2 𝑠 = 0


Coefficients represent sums of electrical admittances.

Admittances associated with 𝑀1 form the elements connected

to the first node, where mechanical admittances between the

two masses are common to the two nodes. Mechanical

admittances associated with 𝑀2 form the elements connected

to the second node

𝑣1(𝑡) and 𝑣2(𝑡) are the velocities of 𝑀1 and 𝑀2, respectively





Problem Draw a series and parallel analog for the rotational

mechanical system

Solution


+ 𝐽1𝑠2 + 𝐷1𝑠 + 𝐾 𝜃1 𝑠 − 𝐾𝜃2 𝑠 = 𝑇(𝑠)

−𝐾𝜃1 𝑠 + 𝐽2𝑠2 + 𝐷2𝑠 + 𝐾 𝜃2 𝑠 = 0



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𝐽1𝑠2 + 𝐷1𝑠 + 𝐾 𝜃1 𝑠 − 𝐾𝜃2 𝑠 = 𝑇(𝑠)

−𝐾𝜃1 𝑠 + 𝐽2𝑠2 + 𝐷2𝑠 + 𝐾 𝜃2 𝑠 = 0

Leting 𝜃1 𝑠 = 𝜔1(𝑠)/𝑠, 𝜃2 𝑠 = 𝜔2(𝑠)/𝑠

𝐽1𝑠 + 𝐷1 +𝐾

𝑠𝜔1 𝑠 −

𝐾

𝑠𝜔2 𝑠 = 𝑇(𝑠)

−𝐾

𝑠𝜔1 𝑠 + 𝐽2𝑠 + 𝐷2 +

𝐾

𝑠𝜔2 𝑠 = 0

From these equations, draw both series and parallel

analogs by considering these to be mesh or nodal

equations, respectively



§10.Nonlinearities

- A linear system possesses two properties

• Superposition

• Homogeneity

- Some examples of physical nonlinearities



§11.Linearization

- To obtain transfer function from a nonlinear system

• Recognize the nonlinear component and write the nonlinear

differential equation

• Find the steady-state solution is called equilibrium

• Linearize the nonlinear differential equation

• Take the Laplace transform of the linearized differential

equation, assuming zero initial conditions

• Separate input and output variables and form the transfer

function



§11.Linearization

- Assume a nonlinear system operating at point 𝐴, [𝑥0, 𝑓(𝑥0)],small changes in the input can be

related to changes in the output about

the point by way of the slope of the curve

at the point 𝐴

𝑓 𝑥 − 𝑓(𝑥0) ≈ 𝑚𝑎(𝑥 − 𝑥0)

⟹ 𝛿𝑓 𝑥 ≈ 𝑚𝑎𝛿𝑥

⟹ 𝑓 𝑥 ≈ 𝑓 𝑥0 + 𝑚𝑎 𝑥 − 𝑥0

≈ 𝑓 𝑥0 + 𝑚𝑎𝛿𝑥

𝑚𝑎 : the slope of the curve at point 𝐴

δ𝑥 : small excursions of the input about point 𝐴

𝛿𝑓(𝑥) : small changes in the output related by the slope at

point 𝐴



§11.Linearization

- Ex.2.26 Linearizing a Function

Linearize 𝑓 𝑥 = 5𝑐𝑜𝑠𝑥 about 𝑥 = 𝜋/2

Solution

Using the linearized equation

𝑓 𝑥 ≈ 𝑓 𝑥0 + 𝑚𝑎𝛿𝑥

where

𝑓𝜋

2= 5𝑐𝑜𝑠

𝜋

2= 0

𝑚𝑎 = 𝑑𝑓

𝑑𝑥𝑥=

𝜋2

= (−5𝑠𝑖𝑛𝑥)𝑥=

𝜋2

= −5

The system can be presented as

𝑓 𝑥 ≈ −5𝛿𝑥

for small excursions of 𝑥 about 𝜋/2



§11.Linearization

Taylor series expansion

Taylor series expansion expresses the value of a function in

terms of the value of that function at a particular point, the

excursion away from that point, and derivatives evaluated at

that point

𝑓 𝑥 = 𝑓 𝑥0 + 𝑑𝑓

𝑑𝑥𝑥=𝑥0

(𝑥 − 𝑥𝑜)

1!+

𝑑2𝑓

𝑑𝑥2

𝑥=𝑥0

(𝑥 − 𝑥𝑜)2

2!+ ⋯

For small excursions of 𝑥 from 𝑥0, the higher-order terms can

be neglected


𝑑𝑥𝑥=𝑥0

(𝑥 − 𝑥𝑜)



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§11.Linearization

- Ex.2.27 Linearizing a Differential Equation

Linearize the following equation for small excursion about 𝑥 = 𝜋/4

𝑑2𝑥

𝑑𝑡2 + 2𝑑𝑥

𝑑𝑡+ 𝑐𝑜𝑠𝑥 = 0

Solution

The presence of the term 𝑐𝑜𝑠𝑥 makes this equation nonlinear

Since we want to linearize the equation about 𝑥 = 𝜋/4, we let

𝑥 = 𝜋/4 + 𝛿𝑥, where 𝛿𝑥 is the small excursion about 𝜋/4

𝑑2(𝛿𝑥 + 𝜋/4)

𝑑𝑡2 + 2𝑑(𝛿𝑥 + 𝜋/4)

𝑑𝑡+ 𝑐𝑜𝑠(𝛿𝑥 + 𝜋/4) = 0

𝑑2(𝛿𝑥 + 𝜋/4)

𝑑𝑡2 =𝑑2𝛿𝑥

𝑑𝑡2

𝑑(𝛿𝑥 + 𝜋/4)

𝑑𝑡=

𝑑𝛿𝑥

𝑑𝑡



§11.Linearization


𝑑𝑥𝑥=𝑥0

(𝑥 − 𝑥0)

𝑓 𝑥 = 𝑐𝑜𝑠𝑥 = 𝑐𝑜𝑠 𝛿𝑥 + 𝜋/4

𝑓 𝑥0 = 𝑓 𝜋/4 = cos(𝜋/4) = 2/2

𝑥 − 𝑥0 = 𝛿𝑥

𝑑𝑓

𝑑𝑥𝑥=𝑥0

= 𝑑𝑐𝑜𝑠𝑥

𝑑𝑥𝑥=𝜋/4

= −sin 𝜋/4 = − 2/2

⟹ 𝑐𝑜𝑠 𝛿𝑥 + 𝜋/4 =2

2+ −

2

2𝛿𝑥

The linearized differential equation

𝑑2𝛿𝑥

𝑑𝑡2 + 2𝑑𝛿𝑥

𝑑𝑡−

2

2𝛿𝑥 = −

2

2Solve this equation for 𝛿𝑥, and obtain 𝑥 = 𝛿𝑥 + 𝜋/4



§11.Linearization

- Ex.2.28 Transfer Function-Nonlinear Electrical Network

Find the transfer function, 𝑉𝐿(𝑠)/𝑉(𝑠), for the electrical network,

which contains a nonlinear resistor whose

voltage-current relationship is defined by

𝑖𝑟 = 2𝑒0.1𝑣𝑟 , where 𝑖𝑟 and 𝑣𝑟 are the

resistor current and voltage, respectively.

Also, 𝑣(𝑡) is a small-signal source

Solution

From the voltage-current relationship

𝑖𝑟 = 2𝑒0.1𝑣𝑟

⟹ 𝑣𝑟 = 10ln(0.5𝑖𝑟) = 10ln(0.5𝑖)

Applying Kirchhoff’s voltage law around the loop

𝐿𝑑𝑖

𝑑𝑡+ 10 ln 0.5𝑖 − 20 = 𝑣(𝑡)



§11.Linearization

- Evaluate the equilibrium solution

• Set the small-signal source, 𝑣(𝑡), equal to zero

• Evaluate the steady-state current

In the steady state 𝑣𝐿 𝑡 = 𝐿𝑑𝑖/𝑑𝑡 and

𝑑𝑖/𝑑𝑡 , given a constant battery source.

Hence, the resistor voltage, 𝑣𝑟, is 20𝑉

𝑖𝑟 = 2𝑒0.1𝑣𝑟 = 2𝑒0.1×20 = 14.78𝐴

⟹ 𝑖0 = 𝑖𝑟 = 14.78𝐴

𝑖0 is the equilibrium value of the network current ⟹ 𝑖 = 𝑖0 + 𝛿𝑖

𝐿𝑑𝑖

𝑑𝑡+ 10 ln 0.5𝑖 − 20 = 𝑣(𝑡)

⟹ 𝐿𝑑(𝑖0 + 𝛿𝑖)

𝑑𝑡+ 10 ln[0.5 𝑖0 + 𝛿𝑖 ] − 20 = 𝑣(𝑡)



𝐿𝑑(𝑖0+𝛿𝑖)

𝑑𝑡+ 10 ln[0.5 𝑖0 + 𝛿𝑖 ] − 20 = 𝑣(𝑡)

§11.Linearization

𝑓 𝑖 = 𝑓 𝑖0 + 𝑑𝑓

𝑑𝑖𝑖=𝑖0

(𝑖 − 𝑖0)

𝑓 𝑖 = ln(0.5𝑖) = ln[0.5 𝑖0 + 𝛿𝑖 ]

𝑓 𝑖0 = ln(0.5𝑖0)

𝑖 − 𝑖0 = 𝛿𝑖

𝑑𝑓

𝑑𝑖𝑖=𝑖0

= 𝑑 ln(0.5𝑖)

𝑑𝑖𝑖=𝑖0

= 1

𝑖𝑖=𝑖0

=1

𝑖0

⟹ ln[0.5 𝑖0 + 𝛿𝑖 ] = ln(0.5𝑖0) +1

𝑖0𝛿𝑖

The linearized equation

𝐿𝑑𝛿𝑖

𝑑𝑡+ 10 ln(0.5𝑖0) +

1

𝑖0𝛿𝑖 − 20 = 𝑣(𝑡)



§11.Linearization

The linearized equation with 𝐿 = 1𝐻, 𝑖0 = 14.78𝐴𝑑𝛿𝑖

𝑑𝑡+ 0.677𝛿𝑖 = 𝑣(𝑡) ⟹ 𝛿𝑖 𝑠 =

𝑉(𝑠)

𝑠 + 0.677

The voltage across the inductor about the equilibrium point

𝑣𝐿 𝑡 = 𝐿𝑑(𝑖0 + 𝛿𝑖)

𝑑𝑡= 𝐿

𝑑𝛿𝑖

𝑑𝑡⟹ 𝑉𝐿 𝑠 = 𝐿𝑠𝛿𝑖 𝑠 = 𝑠𝛿𝑖 𝑠

The voltage across the inductor about the equilibrium point

𝑉𝐿 𝑠 = 𝑠𝑉(𝑠)

𝑠 + 0.677

The final transfer function

𝑉𝐿 𝑠

𝑉(𝑠)=

𝑠

𝑠 + 0.677

for small excursions about 𝑖 = 14.78𝐴 or, equivalently, about

𝑣 𝑡 = 0



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§11.Linearization


Problem Find the linearized transfer function, 𝐺 𝑠 = 𝑉(𝑠)/𝐼(𝑠),for the electrical network. The network contains a

nonlinear resistor whose voltage-current relationship is

defined by 𝑖𝑟 = 𝑒𝑣𝑟. The current source, 𝑖(𝑡), is a small-

signal generator

Solution

The nodal equation

𝐶𝑑𝑣

𝑑𝑡+ 𝑖𝑟 − 2 = 𝑖 𝑡

But 𝐶 = 1, 𝑣 = 𝑣0 + 𝛿𝑣, 𝑖𝑟 = 𝑒𝑣𝑟 = 𝑒𝑣 = 𝑒𝑣0+𝛿𝑣

𝑑(𝑣0 + 𝛿𝑣)

𝑑𝑡+ 𝑒𝑣0+𝛿𝑣 − 2 = 𝑖 𝑡



𝑑(𝑣0+𝛿𝑣)

𝑑𝑡+ 𝑒𝑣0+𝛿𝑣 − 2 = 𝑖 𝑡

§11.Linearization

Linearize 𝑒𝑣

𝑓 𝑣 = 𝑓 𝑣0 + 𝑑𝑓

𝑑𝑣𝑣=𝑣0

(𝑣 − 𝑣0)

𝑓 𝑣 = 𝑒𝑣 = 𝑒𝑣0+𝛿𝑣

𝑓 𝑣0 = 𝑒𝑣0

𝑣 − 𝑣0 = 𝛿𝑣

𝑑𝑓

𝑑𝑣𝑣=𝑣0

= 𝑑𝑒𝑣

𝑑𝑣𝑣=𝑣0

= 𝑒𝑣

𝑣=𝑣0

= 𝑒𝑣0

⟹ 𝑒𝑣0+𝛿𝑣 = 𝑒𝑣0 + 𝑒𝑣0𝛿𝑣

The linearized equation

𝑑𝛿𝑣

𝑑𝑡+ 𝑒𝑣0 + 𝑒𝑣0𝛿𝑣 − 2 = 𝑖 𝑡



§11.Linearization

Setting 𝑖 𝑡 = 0 and letting the circuit reach steady

state, the capacitor acts like an open circuit. Thus, 𝑣0 =𝑣𝑟 with 𝑖𝑟 = 2. But, 𝑖𝑟 = 𝑒𝑣𝑟 or 𝑣𝑟 = ln𝑖𝑟. Hence, 𝑣0 =ln2 = 0.693

𝑑𝛿𝑣

𝑑𝑡+ 𝑒𝑣0 + 𝑒𝑣0𝛿𝑣 − 2 = 𝑖 𝑡

⟹𝑑𝛿𝑣

𝑑𝑡+ 2𝛿𝑣 = 𝑖 𝑡

Taking the Laplace transform

𝑠 + 2 𝛿𝑣 𝑠 = 𝐼(𝑠)


𝛿𝑣(𝑠)

𝐼(𝑠)=

𝑉(𝑠)

𝐼(𝑠)=

1

𝑠 + 2

about equilibrium


HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

§12.Case Studies
