5/9/2009
Generalized Generator Model
b1
Induction Generator
a2
a1
b2
c2
c1
Conversion from Fixed to Rotating Frame
q
Blondel Transformation
b
d
s
s
a
c
1
5/9/2009
Stator to dq0 frames
1
1
1
2
3
cos
sin
1⁄2
cos
sin
2
3
cos
sin
1⁄2
cos
sin
120
120
cos
sin
1⁄2
Stator to dq0 frames
120
120
1⁄2
120
120
cos
sin
1⁄2
1
cos
1
120
120 120
120
1
1⁄2
sin
cos
cos
1
1
1
1
Rotating to Rotating Frame
sin
sin
120
120
cos 2
cos 2
cos 2
120 120
1
1 1
Rotor to dq0 frames
q
∆
b
d
s


∆
2
2
2
2 cos
sin
3
1⁄2
ωs
cos
sin
120
120
1⁄2
cos
sin
120
120
1⁄2
2
2
2
a
ωs cos
cos
cos
120
120
sin
sin
sin
120
120
1
1 1
c
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5/9/2009
q
Actual to dq0 frames
1 cos
1
2
2
cos
1 sin
1
sin
2 2
All Frames
b2
b1
d
For balanced system, vo=0. Hence,
1
1
2
cos
2
cos
s
s
1
sin
2
sin 

c2
a1
a2
c1
Stator plus Rotor
1
1
1
2
System Inductances
2
2
cos
sin
2 1⁄2
0
3
0
0
cos
sin
120
120
1⁄2
0
0
0
cos
sin
120
120
1⁄2
0
0
0
0
0
0
cos
sin
1⁄2
0
0
0
cos
sin
0
0
0
120
120
1⁄2
cos
sin
1
1
120
120
1⁄2
1
2
2
2
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5/9/2009
Voltage Equation
Variable inductance
ℓ
1
1
1
2
2
2
ℓ
ℓ
ℓ
ℓ
ℓ
ℓ
1
1
1
ℓ
ℓ
ℓ
ℓ
ℓ
ℓ
ℓ
ℓ
ℓ
ℓ
ℓ
ℓ
1
1
1
ℓ
ℓ
ℓ
ℓ
ℓ
ℓ
1
1
1
2
2
2
Variable inductances are hard to use
ℓ
ℓ
ℓ
ℓ
ℓ
ℓ
2
2
2
ℓ
ℓ
ℓ
ℓ
ℓ
ℓ
1
1
1
2
2
2
2
2
2
,
Problems: 1) Full matrix; 2) inductances are time varying
Blondel Transformation of Inductances
,
Inductances on dqo Frames
2
,
1
,
2
,
1
,
11
0
12
0
12
0
12
11
0
12
0
0
22
3⁄2 0
1
12
1
0
2
22
2
11
1
1
22
2
2
Inductances are now constants and independent of the rotor position
4
5/9/2009
Objective
• To write the induction machine equation in the standard state space model form
Induction Machine Model
X: State (system) variables vector
A: Model matrix
U: Control and disturbance vector
B: Control and disturbance matrix
Y: System observation vector
H: Observation matrix
Basic Steps
•
•
•
•
Voltage Equations
Find voltage‐current relationships
Add power equation
Add torque equation
Add rotor dynamics
5
5/9/2009
Voltage Equation with winding Resistances
Voltage Equations
0
1
0
0
1
2
2
0
0
0
0
1
1
1
0
1
2
2
2
1
0
0
0
0
0
0
1
0
0
2
0
0
0
0
1
2
2
1
2
0
0
0
0
0
0
0
0
1
1
2
2
2
Voltage Equation with winding Resistances
Parks Equations
Transformer voltage
1
1
2
2
1
0
0
0
0
1
0
0
0
0
2
0
0
0
0
1
1
1
1
2
2
2
2
2
0
0
0
0
0
0
0
0
0
1
1
2
2
0
0
1
0
2
Speed voltage
1
0
0
0
0
0
0
1
0
0
2
0
0
0
0
1
2
2
1
2
1
2
0
0
0
0
0
0
0
0
1
1
2
2
1
1
2
2
11
0
12
0
0
12
11
0
12
0
0
22
0
1
12
1
0
2
22
2
Speed voltage is normally much higher than the transformer voltage plus resistance drop
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5/9/2009
Voltage‐Current Relationship‐
Rearrangement
Voltage‐Current Relationship
1
1
0
0
0
1
2
2
0
1
0
0
0
0
2
0
0
0
1
1
1
1
2
2
2
2
2
0
0
1
0
0
0
0
0
1
0
0
2
0
0
0
1
2
2
0
1
2
0
11
11
12
12
11
11
12
12
12
12
22
22
12
12
22
0
0
1
0
11
2
0
2
0
12
0
22
Voltage‐Current Relationship
0
11
0
1
12
0
12
12
0
0
0
11
0
12
0
11
12
0
0
22
0
0
0
12
11
0
12
0
0
22
1
1
12
11
0
22
0
12
1
0
11
1
12
12
0
2
22
1
0
2
22
2
1
0
2
22
2
2
0
0
22
2
1
2
2
2
2
0
22
0
22
0
0
22
2
12
12
0
12
0
12
0
11
0
11
11 22 1
1
1
12
1
12
12
1
1
12
1
0
22
12
1
0
12
2
12
0
0
12
0
11
0
2
0
22
2
0
Voltage‐Current Relationship
1
11
12
1
22
22
0
11
0
0
0
1
12
12
12
12
1
0
11
11
1 22
11 22
1 22
1 12
11 12
1
22
0
12
0
11 12
1
0
22
0
12
2
12
11 22
2
12
12
0
11
0
1
0
1 12
1
12
1
0
11
2 12
12 22
2
2
12
2 11
11 22
1
12 22
1
2
12
2 12
11 22
2 11
1
1
2
2
2
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5/9/2009
Voltage‐Current Relationship
1
State Space Equations
A
1
2
1 22
1
2
12
2
12
1 12
22
0
12
0
0
22
0
12
2 12
1 22
12 22
11 12
2 11
11 12
1
Model
2
1 12
12
0
1
12
1
0
11
0
2
1
2 12
1
2
11 22
11 22
0
12 22
X=
2
11
Observation
2
2 11
1
1
1
1
2
2
2
2
B
1
1
1
1
1
1
2
2
2
2
2
2
Output Power Equation
Electric Torque Equation
While the machine is running, the speed voltage
Is much higher than the transformer voltage, hence
2
1
3
2
1
1
1
1
1
2
2 2
3
2
2
2
2
2
3
2
2
2
2
2
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5/9/2009
Electric Torque Equation
11
0
1
0
2
12
0
1
0
2
3
2
11
12
12
0
1
12
1
22
0
2
22
2
0
0
Rotor Dynamics
1
1
12
1
22
2
2
12
1
22
2
2
3
2
12
1
22
2
2
12
1
1
22
2
2
Non‐Linear Model
• Issues:
Linearized Model
– The state variable equation is a function of the speed, which is one of the states
– The torque equation is a nonlinear function of the currents
• Problems:
– The model is difficult to analyize
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5/9/2009
Linearized Model
Linearization Process
• Objective:
Original Non‐linear System
–M
Make the model linear in a narrow range of variations
k h
d l li
i
f i i
– Update the system variables if the operating conditions changes
Linearized System
∆
• Advantage:
∆
∆
– Speed up the computation
• Disadvantage:
– Accuracy is determined by how often the model is updated
∆
Linearized Model
Δ
Δ
Δ
Δ
Linearized Power and Torque
3
2
Δ
1
1
1Δ
Δ
1
1
1Δ
1
Δ
1
1
1
2
2
2
2
12
1 22
2
12
1
1 12
22
0
12
0
2 12
1 22
12 22
11 12
11 12
1
∆ 2 11
1 12
0
22
0
12
12
0
11
0
0
12
0
11
11 22
Δ
Δ
Δ
Δ
12 22
2 12
11 22
2 11
2
12 1
2
12 1
11 12
11 12
12 22
12 22
1
1
2
2
11 22
11 22
2
2
Δ
Δ
Δ
Δ
Δ
1
Δ
1
2
2
Δ 2
3
2
2
1
1
Δ
Δ
Δ
3
2
12
12
1
12 Δ
1
22
22
1
2
2
Δ
2
22 Δ
2
2
12
12
2
1
1
12 Δ
22
22
1
2
2
Δ
2
22 Δ
2
2
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5/9/2009
Linearized Rotor Dynamics
3
2
Δ
12
12
2
1
Δ
12
2
Δ
12
1
Δ
Δ
Δ
Δ
Δ
Δ
1
Complete Model (Torque)
1
1
2
Δ
Δ
Δ
Δ
Δ
2
1
1
2
2
3
2
0
12
0 0
3
2
2
0
Δ
Δ
1
Δ
Δ
Δ
12
3
2
2
Δ
Δ
Δ
Δ
Δ
3
12
1
12
2
1
1
3
2
2
22
0
1
1
1
2
0
22
2
3
12
0
12
0
0
12
0
0
0
0
11
2
2 12
2 11
1 12
12
12
0
12
0
11
0
3
2
0
Δ
0
Δ
0
Δ
0
Δ
Δ
2
11 22
11 22
12
2
12 1
2
12 1
12 22
12 22
11 12
11 12
2
2 12
1 22
1 12
1
1
2
12
1 22
2
12
Δ
11 12
2 11
3
1
2
12
11 12
1
12 22
12 22
2
2
11 22
1
11 22
1
2
2
Δ
Δ
Δ
Δ
Δ
1
1
2
2
1
1
2
2
2
Complete Model (Power)
Δ
Δ
Δ
Δ
Δ
1
1
2
2
12
2
1 22
2 12
1 22
2
1
2 11
1 12
3
22
0
12
0
0
2
1
1 12
12
2
0
22
0
3
12
2
12
0
11
12
0
0
0
2
0
12
0
11
0
3
0
0
0
0
1
2 11
1
3
12
2
2
1
2
1
1
12
2
Δ
Δ
Δ
Δ
Δ
2
2 12
1
1
1
1
2
2
2
2
Δ
Δ
Δ
Δ
Δ
1
1
1
12 11
2
2
12 22
11 22
1
2
1
2
2
11
5/9/2009
Initial Conditions
Initial Conditions
• The model is a function of the initial currents, voltages, torques and speeds
lt
t
d
d
• These initial conditions are the steady state values before a disturbance or an input is changed.
The initial conditions are computed using the
• The initial conditions are computed using the steady state model
Steady State Voltage and Current of Stator
cos
1
1
1
1
1
1
1
Blondel Relationships
1
cos
cos
120
120
cos
1
cos
cos
1
1
1
120 120
1
1
sin
cos
cos
cos
120
120
sin
sin
120
120
1
1
1
1
1
1
e electrical angular speed
Φ is the phase shift angle
12
5/9/2009
Steady State Voltage: Stator
Steady State Current: Stator
1
1
1
1
1
1
sin
cos
1
cos
cos
120
120
sin
sin
120
120
1
1
1
1
cos
cos
cos
120
120
2
3
1
cos
sin
1⁄2
cos
sin
120
120
cos
sin
1⁄2
cos
120
120
cos
cos
1⁄2
1
1
1
120
120
1
Keep in mind that when the machine is two pole,
Keep in mind that when the machine is two pole, 1
1
2
3
1
cos
sin
1⁄2
cos
sin
120
120
cos
sin
1⁄2
cos
120
120 cos
cos
1⁄2
120
120
Steady State Variables: Stator
Steady State Voltage: Rotor
For balanced system, the zero sequence component is zero
Hence, Hence
1
1
1
1
0
1
1
cos
sin
1
2
2
2
cos
cos
120
120
cos
2
2
1
cos
2
2
2
cos
cos
2
2
2
120 120
1
Keep in mind that the rotor frequency is Sf
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5/9/2009
Blondal Relationship: Rotor
2
cos
cos
2
2
cos
120
120
sin
sin
sin
1
1
1
120
120
Steady State Voltage: Rotor
2
2
2
2
2
2
2
3
2
2
2
2
cos
cos
cos
120
120
sin
sin
sin
120
120
1
1
1
1
2
cos
sin
1⁄ 2
cos
sin
120
120
1⁄ 2
cos
sin
cos
120
120 cos
cos
1⁄ 2
120
120
cos
cos
cos
120
120
Steady State Current: Rotor
Keep in mind that when the machine is two pole
Steady State variables: Rotor
2
2
2
2
2
1
0
2
2
3
2
cos
sin
1⁄2
cos
sin
120
120
1⁄2
cos
sin
120
120
1⁄2
cos
cos
cos
2
2
2
120
120
2
2
2
cos
sin
2
2
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5/9/2009
Steady State Variables: Stator+Rotor
1
1
0
2
2
0
1
cos
sin
cos
2
2 sin
1
1
2
2
RMS Quantities
2
1
1
1
1
2
2
Blondel Transformation
RMS Quantities
1
1
cos
1
1
1
cos
2
2
cos
2
2
cos
1
1
2
cos
sin
sin
2
1
1
1
1
√2
√2
sin
1
1
cos
√2
cos
1
cos
90
cos
90
Hence, sin 1
0
1
1
sin 1
1
√2
1
1
90 1
1
1
√2
1
1
√2
15
5/9/2009
System RMS Equations
Since the dynamic model is,
1
1
0
0
0
1
1
1
1
2
2
0
0
0
1
0
0
2
2
1
1
1 2
2 1
2
2
0
1
2
0
2 2
0
0
0
0
0
0
0
0
12
0
the steady state model in RMS is
1
1
2
2
1
0
0
0
0
1
0
0
0
0
2
0
0
0
1
2
2
0
0
11
0
12
12
11
22
12
1
1
2
0
12
0
0
12
0
22
12
1
22
2
0
2
2
22
0
22
0
1
12
1
0
2
22
2
2
1
1
1
1 1
11
1
1
22
2
2
22
2
2
12
2
2
1
0
0
1
RMS Relationships
1
11
0
0
0
0
1
12
12
12
12
11
2
0
11
11
2
2
2
2
2
2
2
2
12
12
1
1
1
1
16
5/9/2009
Define,
11
11
22
22 Hence,
1
1
1
1
1
11
1
1
2
11
1
1
1
1
2
12 1
1
2
2
1
11
1
2 2
1 1
1
2
22
And
2
2
2
2
2
2
2
1
2
1
2
22
2
2
1
1
1
1
2
x1
x2
Im
2 r2(1‐S)/S
Ia2
Va1
2
r2
Ia1
2
1
Steady State Model
r1
2
1
11
1
2
Va2/S
xm
Output power
3
1
1
cos
1 For motor, the power is positive, for generator it is negative. When output power is negative, the current is negative.
i.e. the armature current reverses
17
5/9/2009
Generator Steady State Model
Power Equation
√3
x1 r1 x2 r2 Im Ia1 Va1 r2(1‐S)/S
cos
1 0
1
cos
√3
Ia2 1
⁄√3
Va2/S
xm 1
90 12
2
1
Power Equations
2
1
2
90
1
3 12
3
1
3
2
2
√2
2
2
2
2 2
3 12
1
18