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-invarianttranslational 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 are
real 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 > 0
0 𝑖𝑓 𝑡 < 0
𝐹 𝑠 = ℒ 𝑓(𝑡) =
0−
+∞
𝑓(𝑡)𝑒−𝑠𝑡
𝑑𝑡
𝑓 𝑡 = ℒ−1
𝐹 𝑠 =
1
2𝜋𝑗 𝜎−𝑗∞
𝜎+𝑗∞
𝐹 𝑠 𝑒 𝑠𝑡
𝑑𝑠 = 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−𝑎𝑡
𝑢(𝑡)
Solution
𝐹 𝑠 =
0
∞
𝑓(𝑡)𝑒−𝑠𝑡
𝑑𝑡
=
0
∞
𝐴𝑒−𝑎𝑡
𝑒−𝑠𝑡
𝑑𝑡
= 𝐴
0
∞
𝑒−(𝑎+𝑠)𝑡
𝑑𝑡
= −
𝐴
𝑠 + 𝑎
𝑒−(𝑎+𝑠)𝑡
0
∞
⟹ 𝐹 𝑠 =
𝐴
𝑠 + 𝑎
(2.3)
ℒ 𝑡𝑢 𝑡 =
1
𝑠2 (Table 2.1 – 3)
ℒ 𝑒−𝑎𝑡
𝑓(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−1
1
𝑠2 = 𝑡𝑢(𝑡)
𝑓1 𝑡 = ℒ−1
1
(𝑠 + 3)2
= 𝑒−3𝑡
𝑓 𝑡
⟹ 𝑓1 𝑡 = 𝑒−3𝑡
𝑡𝑢(𝑡)
ℒ 𝛿 𝑡 = 1 (Table 2.1 – 1)
ℒ
𝑑𝑓(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
lim
𝑠→−1
[(2.8) × (𝑠 + 1)]
⟹ lim
𝑠→−1
2
𝑠 + 2
= lim
𝑠→−1
𝐾1 +
(𝑠 + 1)𝐾2
𝑠 + 2
⟹ 𝐾1 = 2
lim
𝑠→−2
[(2.8) × (𝑠 + 2)]
⟹ lim
𝑠→−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 + 𝑝 𝑛)
=
𝐾1
𝑠 + 𝑝1
+
𝐾2
𝑠 + 𝑝2
+ ⋯ +
𝐾𝑖
𝑠 + 𝑝𝑖
+ ⋯ +
𝐾𝑛
𝑠 + 𝑝 𝑛
To find 𝐾𝑖
• multiply (2.11) by 𝑠 + 𝑝𝑖
• let 𝑠 approach −𝑝𝑖
(2.11)

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ℒ
𝑑𝑓
𝑑𝑡
= 𝑠𝐹 𝑠 − 𝑓(0−) (Table 2.2 – 7)
ℒ
𝑑2 𝑓
𝑑𝑡2 = 𝑠2
𝐹 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
𝑑2
𝑦
𝑑𝑡2 + 12
𝑑𝑦
𝑑𝑡
+ 32𝑦 = 32𝑢(𝑡)
Solution
𝑠2
𝑌 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 + 8
𝑦 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 = 0
𝑘
+ 1
𝑟1
1
𝑠 − (−8
𝑝1
)
+ (−2
𝑟2
)
1
𝑠 − (−4
𝑝2
)
+ 1
𝑟3
1
𝑠 − (0
𝑝3
)
=
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
lim
𝑠→−1
[(2.23) × (𝑠 + 1)]
⟹ lim
𝑠→−1
2
𝑠 + 2
= lim
𝑠→−1
𝐾1 +
(𝑠 + 1)𝐾2
𝑠 + 2
⟹ 𝐾1 = 2
lim
𝑠→−2
[(2.23) × (𝑠 + 2)]
⟹ lim
𝑠→−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)2
𝐾1
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 = 0
𝑘
+ (−2)
𝑟1
1
[𝑠 − (−2
𝑝1
)]2 + (−2
𝑟2
)
1
𝑠 −(−2
𝑝2
)
+ 2
𝑟3
1
𝑠 −(−1
𝑝3
)
= −
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
𝐹 𝑠 =
𝑁(𝑠)
𝐷(𝑠)
=
𝑁(𝑠)
𝑠 + 𝑝1
𝑟 𝑠 + 𝑝2 … 𝑠 + 𝑝𝑖 … (𝑠 + 𝑝 𝑛)
=
𝐾1
𝑠 + 𝑝1
𝑟 +
𝐾1
𝑠 + 𝑝1
𝑟−1 + ⋯ +
𝐾2
𝑠 + 𝑝1
+
𝐾2
𝑠 + 𝑝2
+ ⋯ +
𝐾𝑖
𝑠 + 𝑝𝑖
+ ⋯ +
𝐾𝑛
𝑠 + 𝑝 𝑛
• multiply (2.27) by 𝑠 + 𝑝1
𝑟
to get 𝐹1 𝑠 = 𝑠 + 𝑝1
𝑟
𝐹(𝑠)
• let 𝑠 approach −𝑝𝑖
(2.27)
To find 𝐾𝑖
𝐾𝑖 =
1
𝑖 − 1 !
𝑑 𝑖−1
𝐹1(𝑠)
𝑑𝑠 𝑖−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
+
𝐾2 𝑠 + 𝐾3
𝑠2 + 2𝑠 + 5
lim
𝑠→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 + 𝐾2 𝑠 + 𝐾3 𝑠 (2.32)
⟹ 3 = 𝐾2 +
3
5
𝑠2
+ 𝐾3 +
6
5
𝑠 + 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−𝑎𝑡 𝑐𝑜𝑠𝜔𝑡 + 𝐵𝑒−𝑎𝑡 𝑠𝑖𝑛𝜔𝑡 =
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
5
𝑒−𝑡
𝑐𝑜𝑠2𝑡 +
1
2
𝑠𝑖𝑛2𝑡
or
𝑓 𝑡 = 0.6 − 0.671𝑒−𝑡
cos(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 = 0
𝑘
+ −2
𝑟1
1
[𝑠 − (−2
𝑝1
)]2 + (−2
𝑟2
)
1
𝑠 − (−2
𝑝2
)
+ 2
𝑟3
1
𝑠 − (−1
𝑝3
)
= −2
1
(𝑠 + 2)2 − 2
1
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
𝐹 𝑠 =
𝑁(𝑠)
𝐷(𝑠)
=
𝑁(𝑠)
𝑠 + 𝑝1 (𝑠2 + 𝑎𝑠 + 𝑏) …
=
𝐾1
(𝑠 + 𝑝1)
+
𝐾2 𝑠 + 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 (𝐾2 𝑠 + 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
5
𝑒−𝑡
𝑐𝑜𝑠2𝑡 +
1
2
𝑠𝑖𝑛2𝑡
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 = 0
𝑘
+ (−0.3 + 𝑗0.15)
𝑟1
1
𝑠 − (−1 + 𝑗2
𝑝1
)
+(−0.3 − 𝑗0.15)
𝑟2
1
𝑠 − (−1 − 2𝑗
𝑝2
)
+ (0.6)
𝑟3
1
𝑠 − (0
𝑝3
)
= −
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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6
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)2
𝑠→0
=
5
9
, 𝐵 =
10
𝑠(𝑠 + 3)2
𝑠→−2
= −5
𝐶 =
10
𝑠(𝑠 + 2) 𝑠→−3
=
10
3
, 𝐷 = (𝑠 + 3)2
𝑑𝐹(𝑠)
𝑑𝑠 𝑠→−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
3
𝑡𝑒−3𝑡
+
40
9
𝑒−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)
𝑎 𝑛
𝑑 𝑛
𝑐(𝑡)
𝑑𝑡 𝑛
+ 𝑎 𝑛−1
𝑑 𝑛−1
𝑐 𝑡
𝑑𝑡 𝑛−1
+ ⋯ + 𝑎0 𝑐 𝑡
= 𝑏 𝑚
𝑑 𝑚
𝑟(𝑡)
𝑑𝑡 𝑚
+ 𝑏 𝑚−1
𝑑 𝑚−1
𝑟(𝑡)
𝑑𝑡 𝑚−1
+ ⋯ + 𝑏0 𝑟 𝑡
𝑐(𝑡): output 𝑟(𝑡): input 𝑎𝑖’s, 𝑏𝑖’s: constant
𝑎 𝑛
𝑑 𝑛
𝑐(𝑡)
𝑑𝑡 𝑛
+ 𝑎 𝑛−1
𝑑 𝑛−1
𝑐 𝑡
𝑑𝑡 𝑛−1
+ ⋯ + 𝑎0 𝑐 𝑡
= 𝑏 𝑚
𝑑 𝑚
𝑟(𝑡)
𝑑𝑡 𝑚
+ 𝑏 𝑚−1
𝑑 𝑚−1
𝑟(𝑡)
𝑑𝑡 𝑚−1
+ ⋯ + 𝑏0 𝑟 𝑡
- Taking the Laplace transform of both sides with zero initial
conditions
𝑎 𝑛 𝑠 𝑛
𝐶 𝑠 + 𝑎 𝑛−1 𝑠 𝑛−1
𝐶 𝑠 + ⋯ + 𝑎0 𝐶 𝑠
= 𝑏 𝑚 𝑠 𝑚
𝑅 𝑠 + 𝑏 𝑚−1 𝑠 𝑚−1
𝑅 𝑠 + ⋯ + 𝑏0 𝑅 𝑠
- The transfer function
𝐺 𝑠 =
𝐶(𝑠)
𝑅(𝑠)
=
𝑏 𝑚 𝑠 𝑚
+ 𝑏 𝑚−1 𝑠 𝑚−1
+ ⋯ + 𝑏0
𝑎 𝑛 𝑠 𝑛 + 𝑎 𝑛−1 𝑠 𝑛−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
𝐺 𝑠 =
𝐶(𝑠)
𝑅(𝑠)
=
1
𝑠 + 2

1/11/2016
7
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 + 2
Expanding by partial fractions
𝐶 𝑡 =
1
2
1
𝑠
−
1
2
1
𝑠 + 2
𝑐 𝑡 = 0.5 − 0.5𝑒−2𝑡
(2.60)
𝐺 𝑠 =
1
𝑠 + 2
𝑅 𝑠 =
1
𝑠
𝑐 𝑡 =
1
2
−
1
2
𝑒−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
2
𝑒−2𝑡
(2.60)
𝑐 𝑡 =
1
2
−
1
2
𝑒−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
𝑑3
𝑐
𝑑𝑡3 + 3
𝑑2
𝑐
𝑑𝑡2 + 7
𝑑𝑐
𝑑𝑡
+ 5𝑐 =
𝑑2
𝑟
𝑑𝑡2 + 4
𝑑𝑟
𝑑𝑡
+ 3𝑟
Solution Taking the Laplace transform with zero initial conditions
𝑠3
𝐶 𝑠 + 3𝑠2
𝐶 𝑠 + 7𝑠𝐶 𝑠 + 5𝐶 𝑠
= 𝑠2
𝑅 𝑠 + 4𝑠𝑅 𝑠 + 3𝑅 𝑠
Collecting terms
𝑠3
+ 3𝑠2
+ 7𝑠 + 5 𝐶 𝑠 = 𝑠2
+ 4𝑠 + 3 𝑅(𝑠)
𝐺 𝑠 =
𝐶(𝑠)
𝑅(𝑠)
=
𝑠2
+ 4𝑠 + 3
𝑠3 + 3𝑠2 + 7𝑠 + 5

1/11/2016
8
Problem Find the differential equation corresponding to the
transfer function
𝐺 𝑠 =
2𝑠 + 1
𝑠2 + 6𝑠 + 2
Solution The transfer function
𝐺 𝑠 =
𝐶(𝑠)
𝑅(𝑠)
=
2𝑠 + 1
𝑠2 + 6𝑠 + 2
Cross multiplying
𝑑2
𝑐
𝑑𝑡2 + 6
𝑑𝑐
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
16
𝑒−4𝑡
+
1
32
𝑒−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
𝐿
𝑑𝑖
𝑑𝑡
+ 𝑅𝑖 +
1
𝐶 0
1
𝑖 𝜏 𝑑𝜏 = 𝑣(𝑡)
Using the relationships 𝑖 𝑡 = 𝑑𝑞(𝑡)/𝑑𝑡 and 𝑞 = 𝐶𝑣 𝐶
𝐿
𝑑2
𝑞
𝑑𝑡2 + 𝑅
𝑑𝑞
𝑑𝑡
+
1
𝐶
𝑞 = 𝑣(𝑡)
⟹ 𝐿𝐶
𝑑2
𝑣 𝐶
𝑑𝑡2 + 𝑅𝐶
𝑑𝑣 𝐶
𝑑𝑡
+ 𝑣 𝐶 = 𝑣(𝑡)
𝐿𝐶
𝑑2
𝑣 𝐶
𝑑𝑡2 + 𝑅𝐶
𝑑𝑣 𝐶
𝑑𝑡
+ 𝑣 𝐶 = 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)
=
1
𝐿𝐶
𝑠2 +
𝑅
𝐿
𝑠 +
1
𝐿𝐶
Block diagram of series RLC electrical network

1/11/2016
9
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 ofApplied 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 = 𝐿
𝑑𝑖
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
𝐶
𝑑𝑣
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 rule
1
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
𝑅𝐶
𝑑𝑣
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

1/11/2016
10
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 𝑣 𝐶(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
𝐶𝑠
𝐼 𝑠 = 𝑉(𝑠)
⟹
𝐼(𝑠)
𝑉(𝑠)
=
1
𝐿𝑠 + 𝑅 +
1
𝐶𝑠
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
𝑅1 𝐼1 + 𝐿𝑠𝐼1 − 𝐿𝑠𝐼2 = 𝑉
𝐿𝑠𝐼2 + 𝑅2 𝐼2 +
1
𝐶𝑠
𝐼2 − 𝐿𝑠𝐼1 = 0
𝑅1 + 𝐿𝑠 𝐼1 − 𝐿𝑠𝐼2 = 𝑉
−𝐿𝑠𝐼1 + 𝐿𝑠 + 𝑅2 +
1
𝐶𝑠
𝐼2 = 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, …, 𝑥 𝑛) 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

1/11/2016
11
𝑅1 + 𝐿𝑠 𝐼1 − 𝐿𝑠𝐼2 = 𝑉
−𝐿𝑠𝐼1 + 𝐿𝑠 + 𝑅2 +
1
𝐶𝑠
𝐼2 = 0
Using Cramer’s rule
𝐼2 =
𝑅1 + 𝐿𝑠 𝑉
−𝐿𝑠 0
𝑅1 + 𝐿𝑠 −𝐿𝑠
−𝐿𝑠 𝐿𝑠 + 𝑅2 +
1
𝐶𝑠
=
0 − 𝐿𝑠𝑉
𝑅1 + 𝐿𝑠 𝐿𝑠 + 𝑅2 +
1
𝐶𝑠
− 𝐿2 𝑠2
=
𝐿𝐶𝑠2
𝑉
𝑅1 + 𝑅2 𝐿𝐶𝑠2 + 𝑅1 𝑅2 𝐶 + 𝐿 𝑠 + 𝑅1
(2.80)
(2.81)
𝐼2 =
𝐿𝐶𝑠2
𝑉
𝑅1 + 𝑅2 𝐿𝐶𝑠2 + 𝑅1 𝑅2 𝐶 + 𝐿 𝑠 + 𝑅1
Forming the transfer function
𝐺 𝑠 =
𝐼2 𝑠
𝑉(𝑠)
=
𝐿𝐶𝑠2
𝑅1 + 𝑅2 𝐿𝐶𝑠2 + 𝑅1 𝑅2 𝐶 + 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
𝐶𝑠
𝐼2 = 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
𝐿𝑠
𝑉𝐿 − 𝐺2 𝑉𝐶 = 𝑉𝐺1
−𝐺2 𝑉𝐿 + 𝐺2 + 𝐶𝑠 𝑉𝐶 = 0
or
(2.86)
(2.85)
𝐺1 + 𝐺2 +
1
𝐿𝑠
𝑉𝐿 − 𝐺2 𝑉𝐶 = 𝑉𝐺1
−𝐺2 𝑉𝐿 + 𝐺2 + 𝐶𝑠 𝑉𝐶 = 0
Solving for the transfer function
𝑉𝐶(𝑠)
𝑉(𝑠)
=
𝐺1 𝐺2
𝐶
𝑠
𝐺1 + 𝐺2 𝑠2 +
𝐺1 𝐺2 𝐿 + 𝐶
𝐿𝐶
𝑠 +
𝐺2
𝐿𝐶
Block diagram of the network
(2.86)
(2.87)

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12
- 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)
𝐺1 𝑉𝐿 𝑠 +
1
𝐿𝑠
𝑉𝐿 𝑠 + 𝐺2 𝑉𝐿 𝑠 − 𝑉𝐶 𝑠 = 𝐺1 𝑉(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)
=
𝐺1 𝐺2
𝐶
𝑠
𝐺1 + 𝐺2 𝑠2 +
𝐺1 𝐺2 𝐿 + 𝐶
𝐿𝐶
𝑠 +
𝐺2
𝐿𝐶
(2.88)
Note
𝐺1 𝑉𝐿 𝑠 +
1
𝐿𝑠
𝑉𝐿 𝑠 + 𝐺2 𝑉𝐿 𝑠 − 𝑉𝐶 𝑠 = 𝐺1 𝑉(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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13
− 2𝑠 + 2 𝐼1 + 9𝑠 + 1 𝐼2 − 4𝑠𝐼3 = 0
(2.94.b)
−𝐼1 − 4𝑠𝐼2 + 4𝑠 + 1 +
1
𝑠
𝐼3 = 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
𝑠
𝐼3 = 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, 𝑍 𝑜 = 0 (ideal)
4.High constant gain amplification, 𝐴 = ∞ (ideal)
- The output, 𝑣 𝑜(𝑡), is given by
𝑣 𝑜 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, 𝑣 𝑜(𝑡), is given by
𝑣 𝑜 𝑡 = −𝐴𝑣1 𝑡 (2.96)

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14
Inverting Operational Amplifiers
𝑍𝑖 𝑠 = ∞ → 𝐼 𝑎(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/𝐶1 𝑠
+
1
𝑅1
→ 𝑍1 =
1
𝐶1 𝑠 +
1
𝑅1
=
360 × 103
2.016𝑠 + 1
𝑍2 = 𝑅2 +
1
𝐶2 𝑠
= 220 × 103
+
107
𝑠
𝑉𝑜(𝑠)
𝑉𝑖(𝑠)
= −
𝑍2 𝑠
𝑍1 𝑠
= −
360 × 103
2.016𝑠 + 1
220 × 103 +
107
𝑠
= −1.232
𝑠2
+ 45.95𝑠 + 22.55
𝑠
Noninverting Operational Amplifiers
𝑉𝑜 = 𝐴 𝑉𝑖 − 𝑉1
𝑉1 =
𝑍1
𝑍1 + 𝑍2
𝑉𝑜
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
𝐶1 𝑠
𝑍2 =
𝑅2
1
𝐶2 𝑠
𝑅2 +
1
𝐶2 𝑠
𝑉𝑜(𝑠)
𝑉𝑖(𝑠)
=
𝑍1 𝑠 + 𝑍2(𝑠)
𝑍1(𝑠)
=
𝐶2 𝐶1 𝑅2 𝑅1 𝑠2
+ (𝐶2 𝑅2 + 𝐶1 𝑅2 + 𝐶1 𝑅1)𝑠 + 1
𝐶2 𝐶1 𝑅2 𝑅1 𝑠2 + 𝐶2 𝑅2 + 𝐶1 𝑅1 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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15
Nodal analysis
Writing the nodal equations
1
𝑠
+ 2 𝑉1 − 𝑉𝐿 = 𝑉
−𝑉1 +
2
𝑠
+ 1 𝑉𝐿 =
1
𝑠
𝑉
1
𝑠
+ 2 𝑉1 − 𝑉𝐿 = 𝑉
−𝑉1 +
2
𝑠
+ 1 𝑉𝐿 =
1
𝑠
𝑉
Solving the nodal equation for 𝑉𝐿
𝑉𝐿 =
1
𝑠
+ 2 𝑉
−1
1
𝑠
1
𝑠
+ 2 −1
−1
2
𝑠
+ 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 = 𝑍 𝐶 =
1
𝐶𝑠
=
1
10−5 𝑠
=
105
𝑠
𝑍2 = 𝑍 𝑅 = 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
𝑀
𝑑2
𝑥(𝑡)
𝑑𝑡2 + 𝑓𝑣
𝑑𝑥(𝑡)
𝑑𝑡
+ 𝐾𝑥 𝑡 = 𝑓(𝑡)
Taking Laplace transform
𝑀𝑠2
𝑋 𝑠 + 𝑓𝑣 𝑠𝑋 𝑠 + 𝐾𝑋 𝑠 = 𝐹(𝑠)
𝐺(𝑠) =
𝑋(𝑠)
𝐹(𝑠)
=
𝐹 𝑠
𝑀𝑠2 + 𝑓𝑣 𝑠 + 𝐾

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16
Impedance
- Define impedance for mechanical components
𝑍 𝑀 𝑠 ≡
𝐹 𝑠
𝑋 𝑠
⟹ 𝐹 𝑠 = 𝑍 𝑀 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 ⟹ 𝑍 𝑀 𝑠 = 𝐾 (2.112)
- For the viscous damper
𝐹 𝑠 = 𝑓𝑣 𝑠𝑋 𝑠 ⟹ 𝑍 𝑀 𝑠 = 𝑓𝑣 𝑠 (2.113)
- For the mass
𝐹 𝑠 = 𝑀𝑠2
𝑋 𝑠 ⟹ 𝑍 𝑀 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
+ 𝑀1 𝑠2
+ 𝑓𝑣1
+ 𝑓𝑣3
𝑠 + 𝐾1 + 𝐾2 𝑋1 − 𝑓𝑣3
𝑠 + 𝐾2 𝑋2 = 𝐹
Free body diagram of 𝑀2
The Laplace transform of the equation of motion of 𝑀2
− 𝑓𝑣3
𝑠 + 𝐾2 𝑋1 + 𝑀2 𝑠2
+ 𝑓𝑣2
+ 𝑓𝑣3
𝑠 + 𝐾2 + 𝐾3 𝑋2 = 0
The Laplace transform of the equations of motion
+ 𝑀1 𝑠2
+ 𝑓𝑣1
+ 𝑓𝑣3
𝑠 + 𝐾1 + 𝐾2 𝑋1 − 𝑓𝑣3
𝑠 + 𝐾2 𝑋2 = 𝐹
− 𝑓𝑣3
𝑠 + 𝐾2 𝑋1 + 𝑀2 𝑠2
+ 𝑓𝑣2
+ 𝑓𝑣3
𝑠 + 𝐾2 + 𝐾3 𝑋2 = 0
The Laplace transform of the equations of motion
𝑋2 =
𝑀1 𝑠2
+ 𝑓𝑣1
+ 𝑓𝑣3
𝑠 + 𝐾1 + 𝐾2 𝐹
− 𝑓𝑣3
𝑠 + 𝐾2 0
∆
=
𝑓𝑣3
𝑠 + 𝐾2 𝐹
∆
where
∆ =
𝑀1 𝑠2
+ 𝑓𝑣1
+𝑓𝑣3
𝑠+ 𝐾1 +𝐾2 − 𝑓𝑣3
𝑠 + 𝐾2
− 𝑓𝑣3
𝑠 + 𝐾2 𝑀2 𝑠2
+ 𝑓𝑣2
+𝑓𝑣3
𝑠+ 𝐾2 +𝐾3
𝐺 𝑠 =
𝑋2(𝑠)
𝐹(𝑠)
=
𝑓𝑣3
𝑠 + 𝐾2
∆
Note
The Laplace transform of the equations of motion of 𝑀1
+ 𝑀1 𝑠2
+ 𝑓𝑣1
+ 𝑓𝑣3
𝑠 + 𝐾1 + 𝐾2 𝑋1 − 𝑓𝑣3
𝑠 + 𝐾2 𝑋2 = 𝐹
− 𝑓𝑣3
𝑠 + 𝐾2 𝑋1 + 𝑀2 𝑠2
+ 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
+ 𝑀1 𝑠2
+ 𝑓𝑣1
+ 𝑓𝑣3
𝑠 + 𝐾1 + 𝐾2 𝑋1 − 𝐾2 𝑋2 − 𝑓𝑣3
𝑋3 = 0

1/11/2016
17
−𝐾2 𝑋1 + [𝑀2 𝑠2
+ 𝑓𝑣2
+ 𝑓𝑣4
𝑠 + 𝐾2]𝑋2 − 𝑓𝑣4
𝑠𝑋3 = 𝐹
−𝑓𝑣3
𝑠𝑋1 − 𝑓𝑣4
𝑠𝑋2 + [𝑀3 𝑠2
+ 𝑓𝑣3
+ 𝑓𝑣4
𝑠]𝑋3 = 0
Problem Find the transfer
function
𝐺 𝑠 =
𝑋2(𝑠)
𝐹(𝑠)
Solution
+ 𝑀1 𝑠2
+ 𝑓𝑣1
+ 𝑓𝑣2
+ 𝑓𝑣3
𝑠 + 𝐾 𝑋1
− 𝑓𝑣1
+ 𝑓𝑣2
+ 𝑓𝑣3
𝑠 + 𝐾 𝑋2 = 𝐹
⟹ +(𝑠2
+ 3𝑠 + 1)𝑋1 − (3𝑠 + 1)𝑋2 = 𝐹
− 𝑓𝑣1
+ 𝑓𝑣2
+ 𝑓𝑣3
𝑠 + 𝐾 𝑋1
+ 𝑀2 𝑠2
+ 𝑓𝑣1
+ 𝑓𝑣2
+ 𝑓𝑣3
+ 𝑓𝑣4
𝑠 + 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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18
- 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
𝐽1 𝑠2
+ 𝐷1 𝑠 + 𝐾 𝜃1 𝑠 − 𝐾𝜃2 𝑠 = 𝑇(𝑠) (2.127.a)
−𝐾1 𝜃1 𝑠 + (𝐽2 𝑠2
+ 𝐷2 𝑠 + 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
𝐽1 𝑠2
+ 𝐷1 𝑠 + 𝐾 𝜃1 𝑠 − 𝐾𝜃2 𝑠 = 𝑇(𝑠)(2.127.a)
−𝐾1 𝜃1 𝑠 + (𝐽2 𝑠2
+ 𝐷2 𝑠 + 𝐾)𝜃2 𝑠 = 0 (2.127.b)
The solution for 𝜃2
𝜃2 =
𝐽1 𝑠2
+ 𝐷1 𝑠 + 𝐾 𝑇
−𝐾 0
∆
=
𝐾𝑇
∆
where
∆=
𝐽1 𝑠2
+ 𝐷1 𝑠 + 𝐾 −𝐾
−𝐾 𝐽2 𝑠2
+ 𝐷2 𝑠 + 𝐾
⟹ 𝐺 𝑠 =
𝜃2(𝑠)
𝑇(𝑠)
=
𝐾
∆
Note
𝐽1 𝑠2
+ 𝐷1 𝑠 + 𝐾 𝜃1 𝑠 − 𝐾𝜃2 𝑠 = 𝑇(𝑠) (2.127.a)
−𝐾1 𝜃1 𝑠 + (𝐽2 𝑠2
+ 𝐷2 𝑠 + 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
+ 𝐽1 𝑠2
+ 𝐷1 𝑠 + 𝐾 𝜃1 − 𝐾𝜃2 − 0𝜃3 = 𝑇(𝑠) (2.131.a)
−𝐾𝜃1 + 𝐽2 𝑠2
+ 𝐷2 𝑠 + 𝐾 𝜃2 − 𝐷2 𝑠𝜃3 = 0 (2.131.b)

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−0𝜃1 − 𝐷2 𝑠𝜃2 + 𝐽3 𝑠2
+ 𝐷3 𝑠 + 𝐷2 𝑠 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 relationship
𝜃2
𝜃1
=
𝑟1
𝑟2
=
𝑁1
𝑁2
Power on gears
𝑇1 𝜃1 = 𝑇2 𝜃2
The ratio of torques on two gears
𝑇2
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.equivalentsystem 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 + 𝐾
𝑁1
𝑁2
𝜃1 𝑠 = 𝑇1(𝑠)
𝑁2
𝑁1
gears
(2.131)
(2.132)

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𝐽𝑠2
+ 𝐷𝑠 + 𝐾 𝜃2 𝑠 = 𝑇1(𝑠)(𝑁2/𝑁1) (2.131)
𝐽𝑠2
+ 𝐷𝑠 + 𝐾 (𝑁2/𝑁1)𝜃1 𝑠 = 𝑇1(𝑠)(𝑁2/𝑁1) (2.132)
⟹ 𝐽
𝑁1
𝑁2
2
𝑠2
+ 𝐷
𝑁1
𝑁2
2
𝑠 + 𝐾
𝑁1
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
gears
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
𝐽𝑒 𝑠2
+ 𝐷𝑒 𝑠 + 𝐾𝑒 𝜃2 𝑠 = 𝑇1 𝑠 (𝑁2/𝑁1) (2.139)
b.system after reflection of torques
and impedances to the output shaft
a.rotationalmechanicalsystemwithgears
𝐽𝑒 𝑠2
+ 𝐷𝑒 𝑠 + 𝐾𝑒 𝜃2 𝑠 = 𝑇1 𝑠 (𝑁2/𝑁1) (2.139)
where, 𝐽 𝑒 = 𝐽1(𝑁2/𝑁1)2
+𝐽2, 𝐷𝑒 = 𝐷1(𝑁2/𝑁1)2
+𝐷2, 𝐾𝑒 = 𝐾2
Solving for 𝐺(𝑠)
𝐺 𝑠 =
𝜃2(𝑠)
𝑇1(𝑠)
=
𝑁2/𝑁1
𝐽 𝑒 𝑠2 + 𝐷𝑒 𝑠 + 𝐾𝑒
b.system after reflection of torques
and impedances to the output shaft
a.rotationalmechanicalsystemwithgears
- In order to eliminate gears with large radii, a gear train is used
to implement large gear ratios by cascading smaller gear ratios.
𝜃4 =
𝑁1 𝑁3 𝑁5
𝑁2 𝑁4 𝑁6
𝜃1
- 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
𝐽𝑒 𝑠2
+ 𝐷𝑒 𝑠 𝜃1 𝑠 = 𝑇1 𝑠
𝐺 𝑠 = 𝜃1 𝑠 /𝑇1 𝑠 = 1/(𝐽 𝑒 𝑠2
+ 𝐷𝑒 𝑠)
b.equivalent system at the inputa.system using a gear train

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21
Problem Find the TF
𝐺 𝑠 =
𝜃2(𝑠)
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 − 𝑠𝜃 𝑎 𝑠 = 𝑇(𝑠)
−𝑠𝜃1 𝑠 + (𝑠 + 1)𝜃 𝑎 𝑠 = 0
The equation of motion
𝑠2
+ 𝑠 𝜃1 𝑠 − 𝑠𝜃 𝑎 𝑠 = 𝑇(𝑠)
−𝑠𝜃1 𝑠 + (𝑠 + 1)𝜃 𝑎 𝑠 = 0
Solving for 𝜃 𝑎(𝑠)
𝜃 𝑎 𝑠 =
𝑠2
+ 𝑠 𝑇
−𝑠 0
𝑠2 + 𝑠 −𝑠
−𝑠 𝑠 + 1
=
𝑠𝑇(𝑠)
𝑠3 + 𝑠2 + 𝑠
⟹
𝜃 𝑎 𝑠
𝑇(𝑠)
=
1
𝑠2 + 𝑠 + 1
𝜃2 𝑠
𝑇(𝑠)
=
1
2
𝜃 𝑎 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 𝑖 𝑎(𝑡) flows, passes
through this magnetic field at right
angles and feels a force
𝐹 = 𝐵𝑙𝑖 𝑎(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
𝑣 𝑏 𝑡 = 𝐾𝑏 𝜃 𝑚(𝑡) (2.144)
𝑣 𝑏 𝑡 : the back electromotive force
(back emf)
𝐾𝑏 : a constant of proportionality
called the back emf constant
𝜃 𝑚(𝑡): the angular velocity of the
motor

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• Taking the Laplace transform
𝑉𝑏(𝑠) = 𝐾𝑏 𝑠𝜃 𝑚(𝑠) (2.145)
• The relationship between the armature current, 𝑖 𝑎(𝑡), the
applied armature voltage, 𝑒 𝑎(𝑡), and the back emf, 𝑣 𝑏(𝑡)
𝑅 𝑎 𝐼 𝑎 𝑠 + 𝐿 𝑎 𝑠𝐼 𝑎 𝑠 + 𝑉𝑏 𝑠 = 𝐸 𝑎 𝑠 (2.146)
• The torque developed by the motor is proportional to the
armature current
𝑇 𝑚 𝑠 = 𝐾𝑡 𝐼 𝑎 𝑠 (2.147)
𝑇 𝑚 : the torque developed by the
motor
𝐾𝑡 : the motor torque constant,
which depends on the motor
and magnetic field characteristics
𝑉𝑏(𝑠) = 𝐾𝑏 𝑠𝜃 𝑚(𝑠) (2.145), 𝑅 𝑎 𝐼 𝑎 𝑠 + 𝐿 𝑎 𝑠𝐼 𝑎 𝑠 + 𝑉𝑏 𝑠 = 𝐸 𝑎 𝑠 (2.146)
• Rearranging Eq.(2.147)
𝐼 𝑎(𝑠) =
1
𝐾𝑡
𝑇 𝑚(𝑠)
• To find the TF of the motor, first substitute Eqs. (2.145) and
(2.148) into (2.146), yielding
(𝑅 𝑎 + 𝐿 𝑎 𝑠)𝑇 𝑚(𝑠)
𝐾𝑡
+ 𝐾𝑏 𝑠𝜃 𝑚 𝑠 = 𝐸 𝑎 𝑠
• Then, find 𝑇 𝑚(𝑠) in terms of 𝜃 𝑚(𝑠),
separate the input and output
variables and obtain the transfer
function, 𝜃 𝑚(𝑠)/𝐸 𝑎(𝑠)
(2.148)
(2.149)
(𝑅 𝑎+𝐿 𝑎 𝑠)𝑇 𝑚(𝑠)
𝐾𝑡
+ 𝐾𝑏 𝑠𝜃 𝑚 𝑠 = 𝐸 𝑎 𝑠 (2.149)
• A typical equivalent mechanical loading on a motor
𝐽 𝑚 : the equivalent inertia at the armature and includes both the
armature inertia and, the load inertia reflected to the armature
𝐷 𝑚 : the equivalent viscous damping at the armature and
includes both the armature viscous damping and, the
load viscous damping reflected to the armature
𝑇 𝑚 𝑠 = (𝐽 𝑚 𝑠2
+ 𝐷 𝑚 𝑠)𝜃 𝑚(𝑠) (2.150)
• Substituting Eq.(2.150) into Eq.(2.149)
(𝑅 𝑎 + 𝐿 𝑎 𝑠)(𝐽 𝑚 𝑠2
+ 𝐷 𝑚 𝑠)𝜃 𝑚(𝑠)
𝐾𝑡
+ 𝐾𝑏 𝑠𝜃 𝑚 𝑠 = 𝐸𝑎(𝑠)
(2.151)
(𝑅 𝑎+𝐿 𝑎 𝑠)(𝐽 𝑚 𝑠2+𝐷 𝑚 𝑠)𝜃 𝑚(𝑠)
𝐾𝑡
+ 𝐾𝑏 𝑠𝜃 𝑚 𝑠 = 𝐸𝑎(𝑠) (2.151)
• Assume that the armature inductance, 𝐿 𝑎, is small compared
to the armature resistance, 𝑅 𝑎, which is usual for a dc motor,
Eq. (2.151) becomes
𝑅 𝑎
𝐾𝑡
𝐽 𝑚 𝑠 + 𝐷 𝑚 + 𝐾𝑏 𝑠𝜃 𝑚 𝑠 = 𝐸 𝑎(𝑠)
• After simplification
𝜃 𝑚(𝑠)
𝐸 𝑎(𝑠)
=
𝐾𝑡
𝑅 𝑎
1
𝐽 𝑚
𝑠 𝑠 +
1
𝐽 𝑚
𝐷 𝑚 +
𝐾𝑡
𝑅 𝑎
𝐾𝑏
• The form of Eq.(2.153)
𝜃 𝑚(𝑠)
𝐸 𝑎(𝑠)
=
𝐾
𝑠(𝑠 + 𝛼)
(2.153)
(2.154)
(2.152)
• Let us first discuss the mechanical constants, 𝐽 𝑚 and 𝐷 𝑚.
Consider the figure: a motor with inertia
𝐽 𝑎 and damping 𝐷 𝑎 at the armature
driving a load consisting of inertia 𝐽 𝐿
and damping 𝐷 𝐿
Assuming that all inertia and damping values shown are
known, 𝐽 𝐿 and 𝐷 𝐿 can be reflected back to the armature as
some equivalent inertia and damping to be added to 𝐽 𝑎 and
𝐷 𝑎 , respectively. Thus, the equivalent inertia, 𝐽 𝑚 , and
equivalent damping, 𝐷 𝑚, at the armature are
𝐽 𝑚 = 𝐽 𝑎 + 𝐽𝐿
𝑁1
𝑁2
2
𝐷 𝑚 = 𝐷 𝑎 + 𝐷𝐿
𝑁1
𝑁2
2
(2.155.a)
(2.155.b)
𝑉𝑏(𝑠) = 𝐾𝑏 𝑠𝜃 𝑚(𝑠) (2.145), 𝑅 𝑎 𝐼𝑎 𝑠 + 𝐿 𝑎 𝑠𝐼𝑎 𝑠 + 𝑉𝑏 𝑠 = 𝐸𝑎 𝑠 (2.146), 𝐼𝑎(𝑠) =
1
𝐾𝑡
𝑇 𝑚(𝑠) (2.148)
• Substituting Eqs.(2.145), (2.148) into Eq. (2.146), with 𝐿 𝑎 = 0
𝑅 𝑎
𝐾𝑡
𝑇 𝑚 𝑠 + 𝐾𝑏 𝑠𝜃 𝑚 𝑠 = 𝐸 𝑎 𝑠
𝑅 𝑎
𝐾𝑡
𝑇 𝑚 𝑡 + 𝐾𝑏 𝜔 𝑚 𝑡 = 𝑒 𝑎 𝑡
• When the motor is operating at steady state with a dc voltage
input
𝑅 𝑎
𝐾𝑡
𝑇 𝑚 + 𝐾𝑏 𝜔 𝑚 = 𝑒 𝑎
⟹ 𝑇 𝑚 = −
𝐾𝑏 𝐾𝑡
𝑅 𝑎
𝜔 𝑚 +
𝐾𝑡
𝑅 𝑎
𝑒 𝑎
(2.156)
(2.157)
(2.158)
(2.159)

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The stall torque
𝑇𝑠𝑡𝑎𝑙𝑙 =
𝐾𝑡
𝑅 𝑎
𝑒 𝑎
The no-load speed
𝜔 𝑛𝑜−𝑙𝑜𝑎𝑑 =
𝑒 𝑎
𝐾𝑏
The electrical constants of the motor
𝐾𝑡
𝑅 𝑎
=
𝑇𝑠𝑡𝑎𝑙𝑙
𝑒 𝑎
𝐾𝑏 =
𝑒 𝑎
𝜔 𝑛𝑜−𝑙𝑜𝑎𝑑
The electrical constants, 𝐾𝑡/𝑅 𝑎 and 𝐾𝑏, can be found from a
dynamometer test of the motor, which would yield 𝑇𝑠𝑡𝑎𝑙𝑙 and
𝜔 𝑛𝑜−𝑙𝑜𝑎𝑑 for a given 𝑒 𝑎
(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,
𝜃 𝐿 𝑠
𝐸 𝑎 𝑠
Solution
Find the mechanical constants
𝑁1
𝑁2
2
= 5 + 700 ×
100
1000
2
= 12
𝐷 𝑚 = 𝐷 𝑎 + 𝐷 𝐿
𝑁1
𝑁2
2
= 2 + 800 ×
100
1000
2
= 10
Find the electrical constants from the torque-speed curve
𝑇𝑠𝑡𝑎𝑙𝑙 = 500, 𝜔 𝑛𝑜−𝑙𝑜𝑎𝑑 = 50, 𝑒 𝑎 = 100
𝐾𝑡
𝑅 𝑎
=
𝑒 𝑎
=
500
100
= 5
𝐾𝑏 =
𝑒 𝑎
=
100
50
= 2
The transfer function 𝜃 𝑚(𝑠)/𝐸 𝑎(𝑠)
𝜃 𝑚(𝑠)
𝐸 𝑎(𝑠)
=
𝐾𝑡
𝑅 𝑎
1
𝐽 𝑚
𝑠 𝑠 +
1
𝐽 𝑚
𝐷 𝑚 +
𝐾𝑡
𝑅 𝑎
𝐾𝑏
=
5 ×
1
12
𝑠 𝑠 +
1
12
× 10 + 5 × 2
=
0.417
𝑠(𝑠 + 1.667)
The transfer function 𝜃 𝐿(𝑠)/𝐸 𝑎(𝑠)
𝜃 𝐿(𝑠)
𝐸 𝑎(𝑠)
=
𝜃 𝑚(𝑠)
𝑁1
𝑁2
𝐸 𝑎(𝑠)
=
0.417 ×
100
1000
𝑠(𝑠 + 1.667)
=
0.0417
𝑠(𝑠 + 1.667)
Problem Find the TF, 𝐺 𝑠 = 𝜃 𝐿(𝑠)/𝐸𝑠(𝑠), for the motor and load
system. The torque-speed curve is given by 𝑇 𝑚 =
− 8𝜔 𝑚 + 200 when the input voltage is 100𝑣𝑜𝑙𝑡𝑠
Solution
Find the mechanical constants
𝑁1
𝑁2
𝑁3
𝑁4
2
= 1 + 400 ×
20
100
×
25
100
2
= 2
𝐷 𝑚 = 𝐷 𝑎 + 𝐷𝐿
𝑁1
𝑁2
𝑁3
𝑁4
2
= 5 + 800 ×
20
100
×
25
100
2
= 7

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24
𝑇 𝑚 = −8𝜔 𝑚 + 200
Find the electrical constants from the torque-speed eq.
𝜔 𝑚 = 0 ⟹ 𝑇 𝑚 = 200
𝑇 𝑚 = 0 ⟹ 𝜔 𝑛𝑜−𝑙𝑜𝑎𝑑 = 200/8 = 25
𝐾𝑡
𝑅 𝑎
=
𝐸 𝑎
=
200
100
= 2
𝐾𝑏 =
𝐸 𝑎
=
100
25
= 4
Substituting all values into the motor transfer function
𝜃 𝑚(𝑠)
𝐸 𝑎(𝑠)
=
𝐾𝑡
𝑅 𝑎
1
𝐽 𝑚
𝑠 𝑠 +
1
𝐽 𝑚
𝐷 𝑚 +
𝐾𝑡
𝑅 𝑎
𝐾𝑏
=
2 ×
1
2
𝑠 𝑠 +
1
2
7 + 2 × 4
=
1
𝑠(𝑠 + 7.5)
The transfer function 𝜃 𝐿(𝑠)/𝐸 𝑎(𝑠)
𝜃 𝐿(𝑠)
𝐸 𝑎(𝑠)
=
𝜃 𝑚(𝑠)
𝑁1
𝑁2
𝑁3
𝑁4
𝐸 𝑎(𝑠)
=
20
100
×
25
100
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
+ 𝑓𝑣 𝑠 + 𝐾 𝑋 𝑠 = 𝐹 𝑠 =
𝑀𝑠2
+ 𝑓𝑣 𝑠 + 𝐾
𝑠
𝑠𝑋 𝑠
⟹ 𝑀𝑠 + 𝑓𝑣 +
𝐾
𝑠
𝑉 𝑠 = 𝐹(𝑠)
Kirchhoff’s mesh equation for the simple series RLC network
𝐿𝑠 + 𝑅 +
1
𝐶𝑠
𝐼 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)
𝑀1 𝑠 + 𝑓𝑣1
+ 𝑓𝑣3
+
𝐾1 + 𝐾2
𝑠
𝑉1 𝑠 − 𝑓𝑣3
+
𝐾2
𝑠
𝑉2 𝑠 = 𝐹(𝑠)
− 𝑓𝑣3
+
𝐾2
𝑠
𝑉1 𝑠 + 𝑀2 𝑠 + 𝑓𝑣2
+ 𝑓𝑣3
+
𝐾2 + 𝐾3
𝑠
𝑉2 𝑠 = 0

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25
+ 𝑀1 𝑠 + 𝑓𝑣1
+ 𝑓𝑣3
+
𝐾1+𝐾2
𝑠
+
𝐾2
𝑠
𝑉2 𝑠 = 𝐹(𝑠)
− 𝑓𝑣3
+
𝐾2
𝑠
𝑉1 𝑠 + 𝑀2 𝑠 + 𝑓𝑣2
+ 𝑓𝑣3
+
𝐾2+𝐾3
𝑠
𝑉2 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
𝑠
𝑉 𝑠 = 𝐹(𝑠)
Kirchhoff’s nodal equation for the simple parallel RLC network
𝐶𝑠 +
1
𝑅𝑠
+
1
𝐿𝑠
𝐸 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)
𝑀1 𝑠 + 𝑓𝑣1
+ 𝑓𝑣3
+
𝐾1 + 𝐾2
𝑠
+
𝐾2
𝑠
𝑉2 𝑠 = 𝐹(𝑠)
− 𝑓𝑣3
+
𝐾2
𝑠
𝑉1 𝑠 + 𝑀2 𝑠 + 𝑓𝑣2
+ 𝑓𝑣3
+
𝐾2 + 𝐾3
𝑠
𝑉2 𝑠 = 0
+ 𝑀1 𝑠 + 𝑓𝑣1
+ 𝑓𝑣3
+
𝐾1+𝐾2
𝑠
+
𝐾2
𝑠
𝑉2 𝑠 = 𝐹(𝑠)
− 𝑓𝑣3
+
𝐾2
𝑠
𝑉1 𝑠 + 𝑀2 𝑠 + 𝑓𝑣2
+ 𝑓𝑣3
+
𝐾2+𝐾3
𝑠
𝑉2 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
+ 𝐽1 𝑠2
+ 𝐷1 𝑠 + 𝐾 𝜃1 𝑠 − 𝐾𝜃2 𝑠 = 𝑇(𝑠)
−𝐾𝜃1 𝑠 + 𝐽2 𝑠2
+ 𝐷2 𝑠 + 𝐾 𝜃2 𝑠 = 0

1/11/2016
26
𝐽1 𝑠2
+ 𝐷1 𝑠 + 𝐾 𝜃1 𝑠 − 𝐾𝜃2 𝑠 = 𝑇(𝑠)
−𝐾𝜃1 𝑠 + 𝐽2 𝑠2
+ 𝐷2 𝑠 + 𝐾 𝜃2 𝑠 = 0
Leting 𝜃1 𝑠 = 𝜔1(𝑠)/𝑠, 𝜃2 𝑠 = 𝜔2(𝑠)/𝑠
𝐽1 𝑠 + 𝐷1 +
𝐾
𝑠
𝜔1 𝑠 −
𝐾
𝑠
𝜔2 𝑠 = 𝑇(𝑠)
−
𝐾
𝑠
𝜔1 𝑠 + 𝐽2 𝑠 + 𝐷2 +
𝐾
𝑠
𝜔2 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) ≈ 𝑚 𝑎(𝑥 − 𝑥0)
⟹ 𝛿𝑓 𝑥 ≈ 𝑚 𝑎 𝛿𝑥
⟹ 𝑓 𝑥 ≈ 𝑓 𝑥0 + 𝑚 𝑎 𝑥 − 𝑥0
≈ 𝑓 𝑥0 + 𝑚 𝑎 𝛿𝑥
𝑚 𝑎 : 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 + 𝑚 𝑎 𝛿𝑥
where
𝑓
𝜋
2
= 5𝑐𝑜𝑠
𝜋
2
= 0
𝑚 𝑎 =
𝑑𝑓
𝑑𝑥 𝑥=
𝜋
2
= (−5𝑠𝑖𝑛𝑥)
𝑥=
𝜋
2
= −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
𝑑𝑥 𝑥=𝑥0
(𝑥 − 𝑥 𝑜)
1!
+
𝑑2
𝑓
𝑑𝑥2
𝑥=𝑥0
(𝑥 − 𝑥 𝑜)2
2!
+ ⋯
For small excursions of 𝑥 from 𝑥0, the higher-order terms can
be neglected
𝑓 𝑥 = 𝑓 𝑥0 +
𝑑𝑓
𝑑𝑥 𝑥=𝑥0
(𝑥 − 𝑥 𝑜)

1/11/2016
27
§11.Linearization
- Ex.2.27 Linearizing a Differential Equation
Linearize the following equation for small excursion about 𝑥 = 𝜋/4
𝑑2
𝑥
𝑑𝑡2 + 2
𝑑𝑥
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 + 2
𝑑(𝛿𝑥 + 𝜋/4)
𝑑𝑡
+ 𝑐𝑜𝑠(𝛿𝑥 + 𝜋/4) = 0
𝑑2
(𝛿𝑥 + 𝜋/4)
𝑑𝑡2 =
𝑑2
𝛿𝑥
𝑑𝑡2
𝑑(𝛿𝑥 + 𝜋/4)
𝑑𝑡
=
𝑑𝛿𝑥
𝑑𝑡
§11.Linearization
𝑓 𝑥 = 𝑓 𝑥0 +
𝑑𝑓
𝑑𝑥 𝑥=𝑥0
(𝑥 − 𝑥0)
𝑓 𝑥 = 𝑐𝑜𝑠𝑥 = 𝑐𝑜𝑠 𝛿𝑥 + 𝜋/4
𝑓 𝑥0 = 𝑓 𝜋/4 = cos(𝜋/4) = 2/2
𝑥 − 𝑥0 = 𝛿𝑥
𝑑𝑓
𝑑𝑥 𝑥=𝑥0
=
𝑑𝑐𝑜𝑠𝑥
𝑑𝑥 𝑥=𝜋/4
= − sin 𝜋/4 = − 2/2
⟹ 𝑐𝑜𝑠 𝛿𝑥 + 𝜋/4 =
2
2
+ −
2
2
𝛿𝑥
The linearized differential equation
𝑑2
𝛿𝑥
𝑑𝑡2 + 2
𝑑𝛿𝑥
𝑑𝑡
−
2
2
𝛿𝑥 = −
2
2
Solve 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
𝑖 𝑟 = 2𝑒0.1𝑣 𝑟 , where 𝑖 𝑟 and 𝑣𝑟 are the
resistor current and voltage, respectively.
Also, 𝑣(𝑡) is a small-signal source
Solution
From the voltage-current relationship
𝑖 𝑟 = 2𝑒0.1𝑣 𝑟
⟹ 𝑣𝑟 = 10ln(0.5𝑖 𝑟) = 10ln(0.5𝑖)
Applying Kirchhoff’s voltage law around the loop
𝐿
𝑑𝑖
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 𝑣 𝐿 𝑡 = 𝐿𝑑𝑖/𝑑𝑡 and
𝑑𝑖/𝑑𝑡, given a constant battery source.
Hence, the resistor voltage, 𝑣𝑟, is 20𝑉
𝑖 𝑟 = 2𝑒0.1𝑣 𝑟 = 2𝑒0.1×20
= 14.78𝐴
⟹ 𝑖0 = 𝑖 𝑟 = 14.78𝐴
𝑖0 is the equilibrium value of the network current ⟹ 𝑖 = 𝑖0 + 𝛿𝑖
𝐿
𝑑𝑖
𝑑𝑡
+ 10 ln 0.5𝑖 − 20 = 𝑣(𝑡)
⟹ 𝐿
𝑑(𝑖0 + 𝛿𝑖)
𝑑𝑡
+ 10ln[0.5 𝑖0 + 𝛿𝑖 ] − 20 = 𝑣(𝑡)
𝐿
𝑑(𝑖0+𝛿𝑖)
𝑑𝑡
+ 10 ln[0.5 𝑖0 + 𝛿𝑖 ] − 20 = 𝑣(𝑡)
§11.Linearization
𝑓 𝑖 = 𝑓 𝑖0 +
𝑑𝑓
𝑑𝑖 𝑖=𝑖0
(𝑖 − 𝑖0)
𝑓 𝑖 = ln(0.5𝑖) = ln[0.5 𝑖0 + 𝛿𝑖 ]
𝑓 𝑖0 = ln(0.5𝑖0)
𝑖 − 𝑖0 = 𝛿𝑖
𝑑𝑓
𝑑𝑖 𝑖=𝑖0
=
𝑑 ln(0.5𝑖)
𝑑𝑖 𝑖=𝑖0
=
1
𝑖 𝑖=𝑖0
=
1
𝑖0
⟹ ln[0.5 𝑖0 + 𝛿𝑖 ] = ln(0.5𝑖0) +
1
𝑖0
𝛿𝑖
The linearized equation
𝐿
𝑑𝛿𝑖
𝑑𝑡
+ 10 ln(0.5𝑖0) +
1
𝑖0
𝛿𝑖 − 20 = 𝑣(𝑡)
§11.Linearization
The linearized equation with 𝐿 = 1𝐻, 𝑖0 = 14.78𝐴
𝑑𝛿𝑖
𝑑𝑡
+ 0.677𝛿𝑖 = 𝑣(𝑡) ⟹ 𝛿𝑖 𝑠 =
𝑉(𝑠)
𝑠 + 0.677
The voltage across the inductor about the equilibrium point
𝑣 𝐿 𝑡 = 𝐿
𝑑(𝑖0 + 𝛿𝑖)
𝑑𝑡
= 𝐿
𝑑𝛿𝑖
𝑑𝑡
⟹ 𝑉𝐿 𝑠 = 𝐿𝑠𝛿𝑖 𝑠 = 𝑠𝛿𝑖 𝑠
The voltage across the inductor about the equilibrium point
𝑉𝐿 𝑠 = 𝑠
𝑉(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

1/11/2016
28
§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 𝑖 𝑟 = 𝑒 𝑣 𝑟. The current source, 𝑖(𝑡), is a small-
signal generator
Solution
The nodal equation
𝐶
𝑑𝑣
𝑑𝑡
+ 𝑖 𝑟 − 2 = 𝑖 𝑡
But 𝐶 = 1, 𝑣 = 𝑣0 + 𝛿𝑣, 𝑖 𝑟 = 𝑒 𝑣 𝑟 = 𝑒 𝑣
= 𝑒 𝑣0+𝛿𝑣
𝑑(𝑣0 + 𝛿𝑣)
𝑑𝑡
+ 𝑒 𝑣0+𝛿𝑣
− 2 = 𝑖 𝑡
𝑑(𝑣0+𝛿𝑣)
𝑑𝑡
+ 𝑒 𝑣0+𝛿𝑣
− 2 = 𝑖 𝑡
§11.Linearization
Linearize 𝑒 𝑣
𝑓 𝑣 = 𝑓 𝑣0 +
𝑑𝑓
𝑑𝑣 𝑣=𝑣0
(𝑣 − 𝑣0)
𝑓 𝑣 = 𝑒 𝑣
= 𝑒 𝑣0+𝛿𝑣
𝑓 𝑣0 = 𝑒 𝑣0
𝑣 − 𝑣0 = 𝛿𝑣
𝑑𝑓
𝑑𝑣 𝑣=𝑣0
=
𝑑𝑒 𝑣
𝑑𝑣 𝑣=𝑣0
= 𝑒 𝑣
𝑣=𝑣0
= 𝑒 𝑣0
⟹ 𝑒 𝑣0+𝛿𝑣
= 𝑒 𝑣0 + 𝑒 𝑣0 𝛿𝑣
The linearized equation
𝑑𝛿𝑣
𝑑𝑡
+ 𝑒 𝑣0 + 𝑒 𝑣0 𝛿𝑣 − 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 𝑖 𝑟 = 2. But, 𝑖 𝑟 = 𝑒 𝑣 𝑟 or 𝑣𝑟 = ln𝑖 𝑟. Hence, 𝑣0 =
ln2 = 0.693
𝑑𝛿𝑣
𝑑𝑡
+ 𝑒 𝑣0 + 𝑒 𝑣0 𝛿𝑣 − 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

Ch.02 modeling in frequency domain

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