Chapter 2 Modeling in the Frequency Domain_2nd

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Chapter 2

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• Equation of motion • To obtain in terms of • After simplification, ( ) 2 s q q₁( )s (2.136) 2 2 2 1 1 (Js Ds k) ( )s T s( ) N N q + + = 2 1 2 1 1 2 1 (Js Ds k) N ( )s T s( ) N N q N + + = 2 2 2 2 1 1 1 1 1 2 2 2 ( ) ( ) N N N J s D s k s T s N N N q é _æ _ö _æ _ö _æ _ö ù ê _ç _÷ + _ç _÷ + _ç _÷ ú = ê è ø è ø è ø ú ë û (2.137) (2.138)

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Transfer Function for System with Gears

• Generalizing the result, we can make the following statement: Rotational mechanical impedance can be reflected through gear trains by

multiplying the mechanical impedance by the ratio

2 Numver of teeth of gear on shaft Number of teeth of gear on shaft destination source æ ö ç ÷ ç ÷ ç ÷ ç ÷ è ø

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Transfer Function for System with Gears

• Ex 2.21 Transfer function – system with lossless gears

• Find the transfer function, θ2(s)/T1(s) , for the system of Figure 2.30(a).

TABLE 2.30 a. Rotational mechanical system with gears;

b. System after reflection of torques and impedances to the output; c. Block diagram

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Transfer Function for System with Gears

• The equation of motion

• Transfer function is found to be

2 2 2 1 1 (J s_e D s K_e _e) ( )s T s( ) N N q + + = 2 2 1 2 1 ; e N J J J N æ ö = _ç _÷ + è ø 2 2 1 2 1 ; e N D D D N æ ö = _ç _÷ + è ø Ke = K2

( )

2

_{( )}

( )

2 1 2 1 / e e e s N N G s T s J s D s K q = = + + (2.139) (2.140)

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Transfer Function for System with Gears

• The gear train formulation

1 3 5 4 1 2 4 6 N N N N N N q = q (2.141)

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Electromechanical System Transfer Function

• Electromechanical systems

• systems that are hybrids of electrical and mechanical variables

• DC motor system

• A motor is an electromechanical component that yields a displacement

Figure 2.31 DC motor; a. schematic; b. block

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Electromechanical System Transfer Function

• A conductor moving at right angles to a magnetic field generates

a voltage at the terminals of the conductor

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

its voltage is proportional to speed

• v_b(t) : back electromotive force (back emf)

• Kb: constant of proportionality called the back emf constant

• Tacking the Laplace transform

(2.144) (2.145) ( ) ( ) m b b d t v t K dt q = ( ) ( ) b b m V s = K sq s

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Electromechanical System Transfer Function

• Loop equation around the armature circuit

• The torque developed by the motor is proportional to the armature current

• Tm : the torque developed by the motor • Kt : motor torque constant

• Rearranging (2.146)

( )

a a a a b a

R I s

+

L sI s

+

V s

=

E s

( )

m t a

T s

=

K I s

1 ( )

( )

a m t

I s

T s

K

=

(2.147) (2.148)

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Electromechanical System Transfer Function

• To find the transfer function of the motor, we first substitute Eq.(2.145) and (2.148) into (2.146), yielding

• Typical equivalent mechanical loading on a motor

(2.149) ( ) ( ) ( ) a a m b m a t R L s T s K s s E K q + ₊ ₌

Figure 2.36 Typical equivalent mechanical loading on a

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Electromechanical System Transfer Function

• From figure 2.36

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

• If we assume that the armature induction, La , is small compared to the armature resistance, Ra, which is usual for a dc motor, Eq.(2.151)

becomes (2.150) 2 ( ) ( ) ( ) m m m m T s = J s + D s q s 2 ( )( ) ( ) ( ) ( ) a a m m m b m a t R L s J s D s s K s s E s K q _q + + ₊ ₌ (2.151) ( ) ( ) ( ) a m m b m a t R J s D K s s E s K q é ù + + = ê ú ë û (2.152)

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Electromechanical System Transfer Function

• Transfer function ( ) / ( ) ( ) 1 ( ) m t a m a t b m b m a s K R J E s K K s s D K J R q ₌ é ù + + ê ú ë û ( ) ( ) ( ) m a s K E s s s q a = + (2.153) (2.154)

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Electromechanical System Transfer Function

• DC motor driving a rotational mechanical

(2.155) 2 2 1 1 2 2 ; m a L m a L N N J J J D D D N N æ ö æ ö = + _ç _÷ = + _ç _÷ è ø è ø

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Electromechanical System Transfer Function

• Electrical constants can be obtained through a dynamometer test • To develop the relationships that dictate the use of dynamometer,

substituting Eq.(2.145) and Eq. (2.148) into Eq. (2.146)

• Tacking the inverse Laplace transform,

• When the relationship exist when the motor is operating at steady state with a dc voltage input

(2.156) ( ) ( ) ( ) a m b m a t R T s K s s E s K + q = ( ) ( ) ( ) a m b m a t R T t K t e t K + w = (2.157) a m b m a t R T K e K + w = (2.158)

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• Solving for Tm yields

• Torque speed curve

(2.159) b t t m m a a a K K K T e R w R = - +

Figure 2.38 Torque speed curves with

an armature voltage, ea, as a parameter

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• The mechanical constants of motor system

• The angular velocity: zero => the stall torque.

• The torque : zero => no-load speed

(2.160) t stall a a K T e R = a no load b e K w _- = _(2.161)

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• The electrical constant of the motor’s transfer function (2.162) t stall a a K T R = e a b no load e K w -= (2.163) ß dynamometer

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• Ex 2.23 Transfer function – dc motor and load

• Given the system and torque-speed curve of Figure 2.39(a) and (b), find the transfer function,

TABLE 2.39 a. Dc motor and load;

b. torque-speed curve; c. block diagram;

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• Begin by finding the mechanical constants, Jm and Dm in Eq.(2.153). From Eq.(2.155) , the total inertia at the armature of the motor is

• Total damping at the armature of the motor

• Now we will find the electrical , Kt / Ra and Kb. From the torque-speed curve of Figure 2.39(b),

2 ₂ 1 2 1 5 700 12 10 m a L N J J J N æ ö _æ _ö = + _ç _÷ = + _ç _÷ = è ø è ø 2 ₂ 1 2 1 2 800 10 10 m a L N D D D N æ ö _æ _ö = + _ç _÷ = + _ç _÷ = è ø è ø 500 stall T = 50 no load w _- = 100 a e = (2.164) (2.165) (2.166) (2.167) (2.168)

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Laplace Transform Review

• Electrical constants are

• Substituting Eqs.(2.164), (2.165), (2.169), and (2.170) into Eq.(2.153) yields

• We use the gear ratio, N1/N2=1/10,

500 5 100 t stall a a K T R = e = = 100 2 50 a b no load e K w -= = =

( )

_[

_]

(

)

( ) 5 /12 0.417 1 _{10 (5)(2)} 1.667 12 m a s E s _{s s} s s q ₌ ₌ + ì ₊ ₊ ü í ý î þ

( )

(

)

( ) 0.417 1.667 L a s E s s s q ₌ + (2.169) (2.170) (2.171) (2.172)

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Electrical Circuit Analogs

• Electric circuit analogy

• It is analogous to a system from another discipline

• Series Analogy

• When compared with mesh equation, the resulting electrical circuit

• Parallel analogy

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Electrical Circuit Analogs

• We can create a direct analogy by operating on Eq.(2.173)to convert displacement to velocity by dividing and multiplying the left side by s, yielding (2.175) 2 ( ) ( ) ( ) ( ) v v Ms f s K K sX s Ms f V s F s s s + + ₌ ₊ ₊ ₌

Figure 2.41 Development of series analog: a. mechanical system; b. desired electrical representation; c. series analog; d. parameters for series analog

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Electrical Circuit Analogs

• Parallel analog

• A system can also be converted to an equivalent series analog • Kirchhoff’s nodal equation for the simple parallel RLC network

• We identify the sum of admittances and draw the parallel analog circuit 1 1

(Cs ) ( )E s I s( )

R Ls

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Nonlinearities

• A linear system's properties

• Superposition : • Homogeneity: 1 1 2 2 1 2 1 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) r t c t r t c t r t r t c t c t ® ® + ® + 1 1 1 1 ( ) ( ) ( ) ( ) r t c t Ar t Ac t ® ®

Figure 2.45 a. Linear system; b. nonlinear system

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• Step 1. Recognize the nonlinear component and write the

nonlinear differential equation.

• Step 2. Linearize the nonlinear differential equation.

• Step 3. Take the Laplace transform of the linearized differential

equation, assuming zero initial conditions.

• Step 4. Separate input and output variables and form the

transfer function.

0 0 0 0

( )

(

)

or

( )

x x x x

df

f x

x x

dx

f x

m

x

d

= =

-

»

-»

(2.182) (2.183)

Linearization

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Linearization

• Ex 2.27 Linearizing a differential equation

• Linearize Eq. (2.184) for small excursions about x=π/4.

• Sol) The presence of the term cos x makes this equation nonlinear. Since we want to linearize the equation about x=π/4 , let x=δx+π/4, where δx is the small excursion about π/4, and substitute x into

Eq.(2.184). 2 2

2 cos

0 d x

dx

x

dt

+

dt

+

=

2 2

4 ₂

4 _cos

₀

4 d

x

d

x

dt

p

d

p

d

æ

₊

ö

æ

₊

ö

ç

÷

ç

÷

_æ

_ö

è

ø

₊

è

ø

₊

₌

ç

÷

è

ø

(2.184) (2.185)

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Linearization

• The term cos(δx+(π/4)) can be linearized with the truncated Taylor series. • Substituting f(x) = cos(δx+(π/4)), f(x₀)=f(π/4)=cos(π/4), and (x-x₀)= δx

into Eq.(2.182) yields 2 2 2 2 4 d x d x dt dt p d d æ ₊ ö ç ÷ è _{ø =} 4 d x d x dt dt p d d æ ₊ ö ç ÷ è _{ø =} 4 cos

cos cos sin

4 4 _x 4 d x x x x dx p p p p d d d = æ ₊ ö_- æ ö₌ _{= -} æ ö ç ÷ ç ÷ ç ÷ è ø è ø è ø (2.186) (2.187) (2.188)

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Linearization

• Solving Eq.(2.188) for cos (δx+(π/4)),

• Substituting Eqs. (2.186), (2.189) into Eq.(2.185) yields the following linearized differential equation:

2 2 cos cos sin

4 4 4 2 2 x p p p x x d d d æ ₊ ö ₌ æ ö_- æ ö ₌ -ç ÷ ç ÷ ç ÷ è ø è ø è ø 2 2 2 2 2 2 2 d x d x x dt dt d ₊ d _- _d ₌ -(2.189) (2.190)

Chapter 2 Modeling in the Frequency Domain_2nd

Chapter 2

Table of Contents

Translational Mechanical System

Transfer Function

•

The transfer function to the mathematical modeling of

translational mechanical systems.

Translational Mechanical System

Transfer Function

•

Comparing electric circuits and mechanical systems.

Translational Mechanical System

Transfer Function

•

Equation of motion

Physical system

•

Find the state equations for the translational mechanical

system.

(

) 0

(

)

( )

M s X

DsX

K X

X

K X

X

M s X

F s

+

+

-

=

-

+

=

Translational Mechanical System

Transfer Function

•

Ex 2.18 Equations of motion by inspection

Translational Mechanical System

Transfer Function

)

(

)

(

X

f

sX

M

s

X

f

s

X

X

K

X

X

K

+

+

+

-

=

-F

X

X

s

f

X

s

M

sX

f

X

_{( )}