# EGGN 307 Introduction to Feedback Control Systems Transfer by rma97348

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```									              EGGN 307
Introduction to Feedback Control Systems

Transfer Functions
(Lectures 13-15)

Professor Kevin L. Moore
Spring 2010

http://engineering.mines.edu/course/eggn307a
Previously
3.0 Laplace Transform Modeling

3.1 Review of Complex Numbers
3.2 Laplace Transforms
3.3 Inverse Laplace and LODE Solutions
3.4 Transfer Functions
Recall: Laplace differentiation theorem (1)

   The differentiation theorem

   Higher order derivatives

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Differentiation Theorem (revisited)

d n 1
dn
dt n
 
f t   s n F s   s n 1 f 0   s n  2 
d
dt
f 0     n 1 f 0 
dt
 

   Differentiation Theorem when initial conditions are
zero

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Introduction to Transfer Functions

   Consider a system with input r(t), output x(t), and
zero initial conditions:

   The Laplace Transform of the output is related to the
Laplace Transform of the input by a function of s

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Definition of a Transfer Function

   Definition

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•Transfer Function of a General LODE
•Impulse Response of a General LODE

3.0 Laplace Transform Modeling

3.1 Review of Complex Numbers
3.2 Laplace Transforms
3.3 Inverse Laplace and LODE Solutions
3.4 Transfer Functions
3.5 Block Diagrams

13
Block Diagram Components

   Block diagrams are graphical representations of
algebraic relationships.

System                            Summing Junction

C (s)            R1 ( s )                     Y (s)
R(s)
G(s)                                                

R2 ( s )

C ( s )  G ( s ) R( s )              Y ( s)   R1 ( s)  R2 ( s)

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Basic Block Diagram Simplifications (1)

R(s)                       X (s )                       C (s)
G1 ( s )                    G2 ( s )

X ( s)  G1 ( s) R( s) C ( s)  G2 ( s) X ( s)

C ( s)  G2 ( s)G1 ( s) R( s)

R(s)                         C (s)
G2 ( s)G1 ( s)

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Basic Block Diagram Simplifications (2)

   Systems In Parallel

G1 ( s )

R(s)

G2 ( s )         C ( s)  G1 ( s) R( s)  G2 ( s) R( s)

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Basic Block Diagram Simplifications (3)

   Feedback Connections

R(s)         E (s)                     C (s)
G(s)


H (s )

C ( s)  G( s) E ( s)
E (s)  R(s)  H (s)C (s)

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Basic Block Diagram Simplifications (4)

G( s )
C( s)                    R( s )
1  G( s )H ( s )

R(s)         G (s)        C (s)
1  G (s) H (s)

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Example 1

   Simplify the following block diagram. All variables are
functions of s.

R                                                           C
G1                          G2


H1



H2

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Example 1 (2)


R            G1                            G2                    C


H1  H 2

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Example 1 (3)

R                                            G2                         C
G1                      1  G2 ( H1  H 2 )

R                G2G1                    C
1  G2 ( H1  H 2 )

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Other Manipulations (1)

   Sometimes additional manipulations are needed to
apply basic simplifications
   Goal: move the pick-off point so that the dotted box
has a single input and a single output.

                        
R                                                   C
G1


G2

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Other Manipulations (2)

   Pick off points can be moved around blocks

1
G1

G1                                   G1

G1

G1                                        G1

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Other Manipulations (3)

                        
R                                    C
G1


G2

   Equivalent Diagram
1
G1
                                              
R                                                                C
G1


G2

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Example 2

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Example 2 (2)

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Block diagrams just represent algebra

   If you ever get stuck – just convert back to algebraic
relationships and solve by hand
   Add variables at inputs to blocks

                                 
R                  A                                 C
G1


C  G1 A  A                              B
G2
A  R  G2 B
B  G1 A

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Block diagrams just represent algebra (2)

C  G1 A  A                  A  R  G2G1 A
A  R  G2 B                        1
A          R
B  G1 A                        1  G2G1

1           1
C  G1            R          R
1  G2G1    1  G2G1

G1  1
C          R
1  G2G1

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3.6 Transfer Functions of Physical Systems

Mesh   and Nodal Equations

29
Electrical impedance

   To work with algebraic relationships, we take the Laplace
Transform of the defining equations with zero initial conditions
– Recall: Transfer Function  zero initial conditions

resistor             capacitor             inductor
dv                       di
v  iR             C i                  vL
dt                      dt
V ( s)  RI ( s)     CsV ( s )  I ( s )   V ( s )  LsI ( s )
1
V (s)     I ( s)
Cs
   Impedance: Ratio of Laplace Transform of across variable to
Laplace Transform of through variable

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Finding Transfer Functions using mesh or nodal equations

Vout s 
 Example: Find
Vin s 
R
                                   


Vin                   C       L    Vout


                                   
   Steps
1. Convert elements to impedances
2. Either
• apply Kirchoff’s Current Law (KCL) at every node, or
• apply Kirchoff’s Voltage Law (KVL) around every mesh
3. Solve resulting system of equations for output variable

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Mesh equations from KVL

     R        
                                                       
I1 ( s )              I 2 (s)

Vin (s )

1
sL   Vout ( s )  sLI 2 ( s )
sC
 
                                                       

1
Vin ( s)  RI1 ( s)         ( I1 ( s)  I 2 ( s))
sC
1
0       ( I 2 ( s)  I1 ( s))  LsI 2 ( s)
sC
   Two equations: need to solve for I 2 ( s )

Due to Katie Johnson or Tyrone Vincent or someone                                                       32
Cramer’s Rule

    Cramer’s Rule is a tool that can be used to solve
algebraic equations.

 b1   a11 a12   x1 
If
b    a   a22   x2 
 2   21        
determinant
Then           b1     a12                    a11     b1
b2 a22                         a21 b2
x1                            x2 
a11 a12                        a11 a12
a21 a22                        a21 a22

Due to Katie Johnson or Tyrone Vincent or someone                                  33
Mesh equations in Matrix-Vector form

   The mesh equations can be re-written in matrix-
vector form
1
Vin ( s)  RI1 ( s)         ( I1 ( s)  I 2 ( s))
sC
1
0      ( I 2 ( s )  I1 ( s ))  LsI 2 ( s )
sC

Vin ( s ) R  1        
1   I ( s) 
            sC          sC    1 
             1              1         
 0   sC              Ls    I 2 ( s )
                           sC 

   Use Cramer’s Rule

Due to Katie Johnson or Tyrone Vincent or someone                                       34
Mesh equations in Matrix-Vector form (2)

Vin ( s ) R  1       
1   I ( s) 
            sC         sC    1 
             1             1         
 0    sC
          
Ls    I 2 ( s )
sC 

1
R      Vin ( s )
sC
1
          0                  1
sC                         sC Vin ( s )
I 2 ( s)                     
R
1

1      R  sC Ls  sC    sC 
1           1      1 2

sC       sC
1            1
       Ls 
sC           sC

Due to Katie Johnson or Tyrone Vincent or someone                                     35
Mesh equations in Matrix-Vector form (3)

Vout ( s )  sLI 2 ( s )

sLVin ( s )
Vout ( s )  2
s RLC  Ls  R

Vout ( s )     sL
 2
Vin ( s ) s RLC  Ls  R

Due to Katie Johnson or Tyrone Vincent or someone                                      36
Mesh equations from circuit diagram

   Mesh equations in matrix/vector form:
sum of impedances on mesh 1                     shared impedance

Vin ( s ) R  1        
1   I ( s) 
            sC          sC    1 
             1              1         
 0   sC              Ls    I 2 ( s )
                           sC 
sum of impedances on mesh 2
R
                                          
I1 ( s )    I 2 (s)
               1
Vin (s )                                sL Vout (s )

sC
                                          

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Exercise: Find the mesh equations

1
Vin  I1  I1  I 2 
                                                                           1
I1 ( s )                 I 2 (s)                                s

0   I1  I 2   3I 2  2I 3  I 2 
Vin                                1
3
1

s                                      s
I1  I 2                I3  I 2
                                                                 0  2I 3  I 2   Vout

I 3  I1
2
2s Vout
I 3 ( s)


Vin ( s ) 1  1
s                        1
s     0   I1 ( s ) 
 0 1                                               I ( s )
s 23  2
1
           s                                        2 
 0   0
                                       2   2  2s   I 3 ( s )
           
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Mechanical impedance

mass           f  M
x
X (s)
F (s)       1
Ms2
F (s)
X (s)    Ms2

damper           f  Dx
           X (s)
F (s)      1
Ds
F (s)
X (s)     Ds

spring          f  Kx
X (s)
F (s)   K
1          F (s)
X (s)   K

   Admittance, or the inverse of impedance, is more commonly
used in mechanical systems.
   Both admittances and impedances are commonly used in
electrical systems.

Due to Katie Johnson or Tyrone Vincent or someone                                     39
Nodal equations for current input

V1        R   V2


1
I in (s )                       Ls
sC   Vout ( s )  V2 ( s )


 I in ( s )  Ls  R
1    1
 R  V1 ( s ) 
1

 0   1                1         
             R Cs  R  V2 ( s )

Due to Katie Johnson or Tyrone Vincent or someone                                                   40
Nodal equations for mechanical systems

K1                             K2                               u
M1                                M2

x1                                        x2
Newton’s Laws at each node
M 11   K1 x1  K 2 ( x1  x2 )
x
M 2 2   K 2 ( x2  x1 )  u
x
0  M 1s 2 X 1 ( s)  K1 X 1 ( s)  K 2 ( X 1 ( s)  X 2 ( s))
 U ( s)  M 2 s 2 X 2 ( s)  K 2 ( X 2 ( s)  X 1 ( s))

Due to Katie Johnson or Tyrone Vincent or someone                                                    41
Nodal equations from mechanical diagram

   The same pattern as electrical nodal equations!

K1                K2              u
M1               M2

   0   M 1s 2  K1  K 2                  K 2   X 1( s) 
 U ( s )                                                   
              K2                     M 2 s  K 2   X 2 ( s )
2

Due to Katie Johnson or Tyrone Vincent or someone                                       42

•Motors

43
Electromechanical systems (1)

   What happens when you run current through a wire
in a magnetic field?
f    (out of screen)
B

i

   Lorentz force equation (SI units) f  i  B
   Key result: Magnitude of force is proportional to
current and magnetic field
f  Ki
Due to Katie Johnson or Tyrone Vincent or someone                                       44
Electromechanical Systems (2)

   What happens when you move a wire through a
magnetic field?

x         (out of screen)
B                 

Vb        i

   Conservation of energy:             iVb  xf

iVb  xKi

   Key result: induced voltage is proportional to velocity
Vb  xK

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Recall - DC Motor

   DC motors are used in many control applications
(e.g., robots, disk drives)

   Torque on load                           K f i f - magneticflux

                          

Tm  K1ia  K1 K f i f ia                 Tm - torque       
Due to Katie Johnson or Tyrone Vincent or someone                                     46
Field Controlled DC motor – armature current fixed

Tm (t )  K1K f ia i f (t )     Tm

 K mi f (t )

I f ( s)        1                  Tm ( s )                  (s)      1
                                    Km                    2
V f ( s) ( L f s  R f )           I f (s)                 Tm ( s ) Js  bs

   Overall transfer function
 ( s)                  K m / JL f

V f ( s)       s( s  b / J )(s  R f / L f )

Due to Katie Johnson or Tyrone Vincent or someone                                                  47
Armature controlled DC motor – field current fixed (1)


Vb  K b  K b
   Key relationships:
Tm  K1K t i f ia  K mia
Ra              La
Tm        b


Ia
Va    
Vb      J

Armature

Due to Katie Johnson or Tyrone Vincent or someone                                     48
Armature controlled DC motor – field current fixed (2)

Va  Vb
   Note (from circuit diagram): I a 
Ra  La s

   Thus, Tm  K m I a


Km
Va  Vb 
Ra  La s

   Since net torque = Tm  Td  TL , we have
TL        1
        ,  
Js  b     s

Due to Katie Johnson or Tyrone Vincent or someone                              49
Armature controlled DC motor – field current fixed (2)

   Closed loop transfer function
 ( s)
Km            1
1          ( Ra  La s ) ( Js b )

Va ( s ) s (1             Kb K m         1
( Ra  La s ) ( Js b )   )

 (s)                    Km

Va ( s ) s (( Ra  La s )( Js  b)  K b K m )

   When inductance is negligible:

 ( s)
Km


                            
Ra

Va ( s )       s Js  b  K b         Km
Ra

Due to Katie Johnson or Tyrone Vincent or someone                                                      50
Example

Ra
b  8 N  m  s rad

Va                   


J  7 kg  m 2

m
500
Steady state                         Va  100 V

50       m

 (s)
   Find the transfer function
Va ( s )
Due to Katie Johnson or Tyrone Vincent or someone                                              51
Example (2)

   Motor constant from stall torque
Va  Vb Va
m  0  Vb  kbm  0  ia         
Ra     Ra
K mVa
 m  K mia        @ m  0
Ra

K m 100
500 
Ra
Km
5
Ra

Due to Katie Johnson or Tyrone Vincent or someone                              52
Example (3)

   Motor constant from no-load velocity

 m  0  ia  0  Va  Vb

Vb  K b  Va  K b @  m  0

100  K b 50

2  Kb

Due to Katie Johnson or Tyrone Vincent or someone                              53
Example (4)

   Transfer function
Km
Ra   5
Kb  2
J 7
b8

 (s)                     Km

Ra

Va ( s )       s Js  b  K b K m Ra 
5

s7 s  18

Due to Katie Johnson or Tyrone Vincent or someone                                            54
Hydraulic Actuator

x                           y
Q
drain 
ps


supply                                           P

ps
Q                 


drain
   Nonlinear flow rate Q  g ( x, P)
   Piston area A
   Pressure P
Due to Katie Johnson or Tyrone Vincent or someone                                     55
Two dimensional nonlinear function

   Operating point ( x0 , P0 )

   Linearized flow
 g                 g           
Q                x                P
 x ( x , P )       P ( x , P ) 
       0 0                0 0 

 k xx  k PP
where
g
kx 
x   ( x0 , P0 )

g
kP  
P ( x0 , P0 )

Due to Katie Johnson or Tyrone Vincent or someone                                         56
Free Body Diagram

   Piston pressure/force and flow/position relationships
dy Q
f  PA                   
dt A

Y (s)
f                       N ( s)
f           F ( s)
y
Q


Y s 
 Ps 
N s  A
P
Q             

Due to Katie Johnson or Tyrone Vincent or someone                                  57
Hydraulic transfer function

   Small signal relationships in Laplace domain
Q( s)  k xX ( s)  k PP( s)
1
P ( s )          Y ( s )
N (s) A
1
Y ( s)  Q( s)
sA
         Q(s)                           Y (s)
X (s)       kx                             1
                 sA

P(s)          1
kp
N ( s) A

Due to Katie Johnson or Tyrone Vincent or someone                                          58

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