Course 4 Modeling of dynamic systems in FD (3)

Transfer Function (2)
- Rotational Mechanical System TF √
- TF for Systems with Gears √
- Electromechanical System TF √
1
Rotational Mechanical System TF
Units:
T(t): N-m (newton-meters); 𝜃 𝑡 : rad (radians); 𝜔 𝑡 : rad/s (radians/second); K: N-m/rad
(newton-meters/radian), J: kg-𝑚2 (kilograms-meter𝑠 2 - newton-meters-second𝑠 2 /radian).
2
3
Ex.: Find the transfer function
𝜃2 𝑠
𝑇 𝑠
. The rod is undergoing torsion. A torque is
applied at the left, and the displacement is measured at the right.
Physical system.
Schematic diagram
The system is considered as a lumped parameter system. So that the torsion acts
like a spring concentrated at one particular point in the rod, with an inertia 𝐽1 to the
left and an inertia 𝐽2 to the right . The damping inside the flexible shaft is negligible.
There are two degrees of freedom, since each inertia can be rotated while the other
is held still. Hence it will take two simultaneous equations to solve the system.
4
Free-body diagram of 𝐽1 :
Torques on 𝐽1 due
only to motion of
𝐽1 . 𝐽2 is held still.
Torques on 𝐽1 due
only to motion of
𝐽2 . 𝐽1 is held still.
Final free-body
diagram for 𝐽1 .
The same process is repeated for 𝐽2 :
5
Free-body diagram of 𝐽2 :
Torques on 𝐽2 due
only to motion of
𝐽2 . 𝐽1 is held still.
Torques on 𝐽2 due
only to motion of
𝐽1 . 𝐽2 is held still.
Final free-body
diagram for 𝐽2 .
6
Summing torques:
𝐽1 𝑠 2 + 𝐷1 𝑠 + 𝐾 𝜃1 𝑠 − 𝐾𝜃2 𝑠 = 𝑇(𝑠)
−𝐾𝜃1 𝑠 + 𝐽2 𝑠 2 + 𝐷2 𝑠 + 𝐾 𝜃2 𝑠 = 0
TF:
𝜃2 (𝑠) 𝐾
=
𝑇(𝑠)
∆
𝐽1 𝑠 2 + 𝐷1 𝑠 + 𝐾
∆=
−𝐾
−𝐾
𝐽2 𝑠 2 + 𝐷2 𝑠 + 𝐾
7
TF for systems with gears
Gear 1 (input gear), radius 𝑟1 , 𝑁1 teeth, is rotated through angle 𝜃1 𝑡 due to
a torque 𝑇1 (𝑡). Gear 2 (output gear), radius 𝑟2 , 𝑁2 teeth, responds by
rotating through angle 𝜃2 (𝑡) and delivering a torque, 𝜃2 𝑡 .
Assuming that there is no backlash phenomena.
As the gear turn, the
distance traveled along each
gear’s circumference is the
same:
𝑟1 𝜃1 = 𝑟2 𝜃2
or
𝜃2 𝑟1 𝑁1
= =
𝜃1 𝑟2 𝑁2
Since the ratio of the
number of teeth along the
circumference is in the same
proportion as the ratio of the
radii.
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If we assume that the gears are lossless, that is they do not absorb or store
energy, the energy into Gear 1 equals the energy out of Gear 2 (namely inertia
and damping are negligible).
Relationship between the input torque and the delivered torque:
𝑇1 𝜃1 = 𝑇2 𝜃2
Then
𝑇2 𝜃1 𝑁2
=
=
𝑇1 𝜃2 𝑁1
Thus the torques are directly proportional to the ratio of the number of teeth.
TF for angular
displacement in
lossless gears.
TF for torque in lossless
gears.
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Mechanical impedances
driven by gears
a. Rotational system driven by gears. Gears
driving a rotational inertia, spring, and viscous
damper.
b. Equivalent system at the output after reflection of
input torque. 𝑇1 can be reflected to the output by
𝑁
multiplying by 𝑁2 . The equation of motion:
1
𝑁2
2
𝐽𝑠 + 𝐷𝑠 + 𝐾 𝜃2 𝑠 = 𝑇1 (𝑠)
𝑁1
𝑁1
𝑁2
2
𝐽𝑠 + 𝐷𝑠 + 𝐾
𝜃 𝑠 = 𝑇1 (𝑠)
𝑁2 1
𝑁1
Then
2
2
2
𝑁1
𝑁
𝑁
𝑁1
1
1
𝐽
𝑠2 + 𝐷
𝑠+𝐾
𝜃 𝑠 = 𝑇1 (𝑠)
𝑁2
𝑁2
𝑁2
𝑁2 1
c. Equivalent system at the input after reflection of
impedances without gears.
Rotational mechanical impedances can be reflected through gear
trains by multiplying the mechanical impedance by the ratio
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ 𝑜𝑓 𝑔𝑒𝑎𝑟 𝑜𝑛 𝑑𝑒𝑠𝑡𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑠ℎ𝑎𝑓𝑡
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓𝑡𝑒𝑒𝑡ℎ 𝑜𝑓 𝑔𝑒𝑎𝑎𝑟 𝑜𝑛 𝑠𝑜𝑢𝑟𝑐𝑒 𝑠ℎ𝑎𝑓𝑡
Where the impedance to be reflected is attached to the source shaft
and is being reflected to the destination shaft.
System with lossless gears
Ex.: Find the TF
𝜃2 𝑠
𝑇1 𝑠
of the rotational mechanical system with gears (a).
First, reflect the impedance 𝐽1 and 𝐷1 and torque 𝑇1 on the input shaft to the
output (b). Here the impedance are reflected by 𝑁2 /𝑁1 2 and the torque is
reflected by 𝑁2 /𝑁1 . So the equation of motion
𝑁2
𝐽𝑒 𝑠 + 𝐷𝑒 𝑠 + 𝐾𝑒 𝜃2 𝑠 = 𝑇1 (𝑠)
𝑁1
2
𝐽𝑒 =
𝑁2 2
𝐽1
𝑁1
+ 𝐽2
𝐷𝑒 =
𝑁2 2
𝐷1
𝑁1
+ 𝐷2
TF:
𝜽𝟐 (𝒔)
𝑵𝟐 /𝑵𝟏
𝑮 𝒔 =
=
𝑻𝟏 (𝒔) 𝑱𝒆 𝒔𝟐 + 𝑫𝒆 𝒔 + 𝑲𝒆
𝐾𝑒 = 𝐾2
In orderto eliminate gears with large radii, a gear train (see the figure) is used
to implement large gear ratios by cascading smaller gear ratios. Next to each
rotation, the angular displacement relative to 𝜃1 has been calculated. The
equivqlent gear ratio is the product of the individual gear ratios. So:
𝑁1 𝑁3 𝑁5
𝜃4 =
𝜃
𝑁2 𝑁4 𝑁6 1
Gears with loss
Ex.: Find the TF 𝜃1 (𝑠)/𝑇1 (𝑠).
We reflect all the impedance to the input shaft, 𝜃1 . The equation of motion:
𝐽𝑒 𝑠 2 + 𝐷𝑒 𝑠 𝜃1 𝑠 = 𝑇1 (𝑠)
𝐽𝑒 = 𝐽1 + 𝐽2 + 𝐽3
TF:
𝑮 𝒔 =
𝜽𝟏 (𝒔)
𝑻𝟏 (𝒔)
=
𝑁1 2
𝑁2
+ 𝐽4 + 𝐽5
𝑁1 𝑁3 2
;
𝑁2 𝑁4
𝐷𝑒 = 𝐷1 +
𝑁1 2
𝐷2
𝑁2
𝟏
𝑱𝒆 𝒔𝟐 +𝑫𝒆 𝒔
System using a gear train
Equivalent system at
the input
Electromechanical System TF
A motor is an electromechanical component that yields a displacement
(mechanical) output for a voltage (electrical) input.
Armature-controlled dc servomotor.
A magnetic field is developed by
stationnary permanent magnets called the
fixed field. A rotating circuit called the
armature through which current 𝑖𝑎 (𝑡) flows,
passes through this magnetic field at right
angles and feels a force,
𝐹 = 𝐵 𝑙 𝑖𝑎 (𝑡)
B is the magnetic field strength; 𝑙 is the
length of the conductor. The resulting
torque turns the rotor, the rotating member
of the motor.
A conductor moving at right angles to a
magnetic field generates a voltage at the
terminals of the conductor equal to
𝑒=𝐵𝑙𝑣
e is the voltage; v is the velocity of the
conductor normal to the magnetic field.
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Since the current-carrying armature is rotating in a magnetic field, its voltage is
proportional to speed:
𝑑𝜃𝑚 (𝑡)
𝑣𝑏 𝑡 = 𝐾𝑏
𝑑𝑡
𝑣𝑏 𝑡 : back electromotive force (back emf); 𝐾𝑏 : a constant of proportionality
(the back emf constant);
𝑑𝜃𝑚 𝑡
𝑑𝑡
= 𝜔𝑚 (𝑡): angular velocity of the motor.
Laplace transform:
𝑉𝑏 𝑠 = 𝐾𝑏 𝑠𝜃𝑚 (𝑠)
(a)
So the loop equation arround the armature circuit:
𝑅𝑎 𝐼𝑎 𝑠 + 𝐿𝑎 𝑠𝐼𝑎 𝑠 + 𝑉𝑏 𝑠 = 𝐸𝑎 (𝑠)
(b)
The torque developed by the motor is proportional to the armature current:
𝑇𝑚 𝑠 = 𝐾𝑡 𝐼𝑎 𝑠 → 𝐼𝑎 𝑠 =
1
𝑇 (𝑠)
𝐾𝑡 𝑚
(c)
𝑇𝑚 is the torque developed by the motor; 𝐾𝑡 is a constant of proportionality
called the motor torque constant which depends on the motor and magnetic
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field characteristics. The value of 𝐾𝑡 is equal to the value of 𝐾𝑏 .
Substituting (a) and (c) into (b):
𝑅𝑎 +𝐿𝑎 𝑠 𝑇𝑚 (𝑠)
+
𝐾𝑡
𝐾𝑏 𝑠𝜃𝑚 𝑠 = 𝐸𝑎 (𝑠)
(d)
We must now find 𝑇𝑚 𝑠 in term of 𝜃𝑚 (𝑠). The figure shows a typical equivalent
mechanical loading on a motor. 𝐽𝑚 is the equivalent inertia at the armature and
includes both the armature inertia and the load inertia reflected to the armature.
𝐷𝑚 is the equivalent viscous damping at the armature and includes both the
armature viscous damping and the load viscous damping reflected to the
armature. Then
𝑇𝑚 𝑠 = 𝐽𝑚 𝑠 2 + 𝐷𝑚 𝑠 𝜃𝑚 (𝑠)
(e)
(e) → (d):
𝑅𝑎 + 𝐿𝑎 𝑠 𝐽𝑚 𝑠 2 + 𝐷𝑚 𝑠 𝜃𝑚 (𝑠)
+ 𝐾𝑏 𝑠𝜃𝑚 𝑠 = 𝐸𝑎 (𝑠)
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𝐾𝑡
𝐿𝑎 , the armature inductance, is assumed to be small compared to the armature
resistance, 𝑅𝑎 , which is usual for a dc motor, so
𝑅𝑎
𝐽 𝑠 + 𝐷𝑚 + 𝐾𝑏 𝑠𝜃𝑚 𝑠 = 𝐸𝑎 (𝑠)
𝐾𝑡 𝑚
TF:
𝑲𝒕
𝑮 𝒔 =
𝜽𝒎 (𝒔)
(𝑹𝒂 𝑱𝒎 )
=
𝑬𝒂 (𝒔) 𝒔 𝒔 + 𝟏 𝑫 + 𝑲𝒕 𝑲𝒃
𝑱𝒎 𝒎
𝑹𝒂
This equation can be expressed as
𝜃𝑚 (𝑠)
𝐾
=
𝐸𝑎 (𝑠) 𝑠(𝑠 + 𝛼)
with 𝐾 =
𝐾𝑡
(𝑅𝑎 𝐽𝑚 )
𝛼=
1
𝐽𝑚
𝐷𝑚 +
How to determine the constant parameters?
𝐾𝑡 𝐾𝑏
𝑅𝑎
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Mechanical constants 𝐽𝑚 and 𝐷𝑚 .
Consider a motor with inertia 𝐽𝑎
and damping 𝐷𝑎 at the armature
driving a load consisting of inertia
𝐽𝐿 and damping 𝐷𝐿 .
𝐽𝐿 and 𝐷𝐿 can be reflected back to the armature as some equivalent inertia
and damping ratio to be added to 𝐽𝑎 and 𝐷𝑎 respectively. Thus the
equivalent inertia 𝐽𝑚 and equivalent damping 𝐷𝑚 at the armature are
𝐽𝑚 = 𝐽𝑎 +
𝑁1 2
𝐽𝐿
;
𝑁2
𝐷𝑚 = 𝐷𝑎 +
𝑁1 2
𝐷𝐿
𝑁2
Electrical constants.
Substituting (a) and (c) to (b) with 𝐿𝑎 = 0 gives
𝑅𝑎
𝑇 𝑠 + 𝐾𝑏 𝑠𝜃𝑚 𝑠 = 𝐸𝑎 (𝑠)
𝐾𝑡 𝑚
By the inverse Laplace transform, we get
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𝑅𝑎
𝑑𝜃𝑚 (𝑡)
𝑇𝑚 𝑡 + 𝐾𝑏
= 𝐸𝑎 (𝑡)
𝐾𝑡
𝑑𝑡
𝑅𝑎
𝑇 𝑡 + 𝐾𝑏 𝜔𝑚 (𝑡) = 𝑒𝑎 (𝑡)
𝐾𝑡 𝑚
If a dc voltage 𝑒𝑎 is applied, the motor will turn at a constant angular velocity
𝜔𝑚 with a constant torque 𝑇𝑚 . So when a motor is operating at a steady
state with a dc input,
𝑅𝑎
𝑇𝑚 + 𝐾𝑏 𝜔𝑚 = 𝑒𝑎
𝐾𝑡
Torque – speed curve
or
𝐾𝑏 𝐾𝑡
𝐾𝑡
𝑇𝑚 = −
𝜔 +
𝑒
𝑅𝑎 𝑚 𝑅𝑎 𝑎
𝐾𝑡
𝑲𝒕 𝑻𝒔𝒕𝒂𝒍𝒍
𝑇𝑠𝑡𝑎𝑙𝑙 =
𝑒 →
=
𝑅𝑎 𝑎 𝑹𝒂
𝑹𝒂
𝑒𝑎
𝒆𝒂
𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 =
→ 𝑲𝒃 =
𝐾𝑏
𝝎𝒏𝒐−𝒍𝒐𝒂𝒅
A dynamometer test of the motor would yield
𝑇𝑠𝑡𝑎𝑙𝑙 and 𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 for a given 𝑒𝑎 .
20
Ex.: Given the system and torque-speed curve (a) and (b), find the TF
𝜃𝐿 𝑠
𝐸𝑎 𝑠
.
Mechanical constants 𝐽𝑚 and 𝐷𝑚
𝐽𝑚 = 𝐽𝑎 + 𝐽𝐿
𝑁1
𝑁2
2
1
= 5 + 700
10
Electrical constants
From the curve:
𝐾𝑡
𝑅𝑎
2
= 12
𝐷𝑚 = 𝐷𝑎 + 𝐷𝐿
𝑁1
𝑁2
2
1
= 2 + 800
10
2
= 10
𝐾𝑡
, 𝐾𝑏
𝑅𝑎
=
𝑇𝑠𝑡𝑎𝑙𝑙
𝑒𝑎
=
500
100
= 5 and 𝐾𝑏 =
𝑒𝑎
𝜔𝑛𝑜−𝑙𝑜𝑎𝑑
=
100
50
=2
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Then
𝐾𝑡
𝜃𝑚 (𝑠)
(𝑅𝑎 𝐽𝑚 )
=
𝐸𝑎 (𝑠) 𝑠 𝑠 + 1 𝐷 + 𝐾𝑡 𝐾𝑏
𝐽𝑚 𝑚
𝑅𝑎
We know that
𝑁1
𝑁2
=
1
,
10
5
=
𝑠 𝑠+
12
1
10 + (5)(2)
12
0.417
=
𝑠(𝑠 + 1.667)
so the TF:
𝜃𝐿 (𝑠)
0.0417
=
𝐸𝑎 (𝑠) 𝑠(𝑠 + 1.667)
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HW
• Problems no. 33 and 44 Chapter 2 [Nise]
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