csl3 19 j15
TRANSCRIPT
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
1
Mathematical Modeling
Mechanical translational system
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
2
Review
• Transfer Function– Def.: The ratio of output by input– Gives the mathematical characteristic of the system being analyzed
• Laplace transform– Converting time domain equation to frequency domain– Time domain – signals defined as function of time– Frequency domain –
• signals defined as function of frequency• s=σ+jω• All signals are a property of frequency rather than representing the instant of
time.
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
3
Mechanical system modeling
• Translational• RotationalExample
Automobile suspension systemAlong the road1. The vertical displacements at the tires act as the motion
excitation to the automobile suspension system2. Motion consists of a translational motion of the center of
mass3. Rotational motion about the center of mass
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
4
Mechanical system modeling
• Mechanical system model components– Mass – representing the mass of the system to be
modeled– Spring – represents the repetitive action of the
system– Damper-represents the friction of the system
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
5
Translational system
Example 1
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
6
Element Law Expression Laplace
Spring Spring Force α change in length
F(t) = K x(t)K – Stiffness constant in N/m
F(s) = K X(s)
Viscous Damper or Dashpot
Force α velocity F(t) = fv v(t)F(t) =fv (dx/dt) fv - Friction or damping coefficient ,Ns/m
F(s) = fvsX(s)
Mass Newton’s second lawForce α acceleration
F(t) = M (d2x/dt2) F(s) = Ms2X(s)
Translational system
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
7
Step 1: Decide input and output
Input variable:
)(tf
Output variable:
)(txMass position
)(txMass velocity
)(txMass acceleration
Applied force
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
8
Translational systems
Newton’s second law
𝑀𝑑2𝑥 (𝑡)𝑑𝑡 2 ¿− Kx( t)− 𝑓 𝑣
𝑑𝑥 (𝑡)𝑑𝑡 + 𝑓 (𝑡 )
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
9
Step 2: The Time and frequency response representation
dt
tdxftKxtf
dt
txdM v
)()()(
)(2
2
Taking Laplace Transform
𝑀𝑠2 𝑋 (𝑠 )+ 𝑓 𝑣 𝑠𝑋 (𝑠 )+𝐾𝑋 (𝑠 )=𝐹 (𝑠)
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
10
Step3. Write the transfer function
With x(t) as the output
𝑋 (𝑠)𝐹 (𝑠)
= 1𝑀 𝑠2+ 𝑓 𝑣 𝑠+𝐾
𝑀𝑠2 𝑋 (𝑠 )+ 𝑓 𝑣 𝑠𝑋 (𝑠 )+𝐾𝑋 (𝑠 )=𝐹 (𝑠)
Write the equation with velocity as output.
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
11
Example 2
F(t)
x1(t) x2(t)
x3(t)
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
12
Modeling Steps:
1. Decide the input and the output2. The free body diagram of the mass M (optional)3. The frequency-domain representation of the forces4. The transfer function
Example 2
F(t)
x1(t) x2(t)
x3(t)
In a multi mass system, consider each mass separately to write the equation.
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
13
Equation for Mass M1
2
12
1
)(
dt
txdM
0))()((
))()(()()()(
31
21211112
1
3
1
ssXssXf
sXsXKsXKssXfsXsM
v
v
dt
tdxfv
)(11
)(11 txK ))()(( 212 txtxK
))()(
( 313 dt
tdx
dt
tdxfv
Taking Laplace Transform
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
14
Step2: Free Body diagram of Mass M1 (Optional)
ssXfv )(33
)(1 tx
1M )(12 sXk)(11 sXk
ssXfv )(13
ssXfv )(11
211 )( ssXM
)(22 sXk
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
15
2
22
2
)(
dt
txdM
Equations for Mass M2
)(tFdt
tdxfv
)(22
))()(( 122 txtxK
))()(
( 324 dt
tdx
dt
tdxfv
)())()((
))()(()()(
32
122222
2
4
2
sFssXssXf
sXsXKssXfsXsM
TransformLaplaceTaking
v
v
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
16
)(2 tx
2M
)(tF
)(22 txK
)(2
.
4 txfv
)(2
.
2 txfv
)(2
..
2 txM
)(3
.
4 txfv
)(1
.
2 txK
Free Body Diagram for Mass M2 (Optional)
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
17
Equations for Mass M3
2
32
3
)(
dt
txdM )
)()(( 23
4 dt
tdx
dt
tdxfv )
)()(( 13
3 dt
tdx
dt
tdxfv
0))()((
))()(()(
23
1332
3
4
3
ssXssXf
ssXssXfsXsM
TransformLaplaceTaking
v
v
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
18
)(3 tx
3M
)(3
.
3 txfv )(3
.
4 txfv
)(3
..
3 txM
)(2
.
4 txfv
)(1
.
3 txfv
The free body diagram-M3
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
19
Step 2– Mass M1, M2 and M3
)(1 sX
)(1 sX
)(1 sX
)(2 sX
)(2 sX
)(2 sX
)(3 sX
)(3 sX
)(3 sX
2K 211321 sMsfsfKK vv
22422 sMsfsfK vv 2K sfv4 )(sF
0
2343 sMsfsf vv 0 sfv4 sfv3
Example 2
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
20
Example 3 – Car suspension system
Automobile suspension systemAlong the road1. The vertical displacements at the tires act as the
motion excitation to the automobile suspension system
2. Motion consists of a translational motion of the center of mass
3. Rotational motion about the center of mass
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
21
Example 3 – Car suspension system
1. Assuming that the motion xi at point P is the input to the system
2. The vertical motion x0 of the body is the output3. Transfer function = ?
X0(s)/Xi(s)
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
22
Example 3 – Car suspension system
- ¿Taking Laplace Transform
(𝑀𝑠2+𝑏𝑠+𝑘 ) 𝑋 0 (𝑠)=(𝑏𝑠+𝑘) 𝑋 𝑖(𝑠)
𝑋 0 (𝑠)𝑋𝑖 (𝑠 )
= 𝑏𝑠+𝑘𝑀 𝑠2+𝑏𝑠+𝑘
Transfer Function
ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
23
Next Class
Rotational and electrical system Mathematical modeling