Lecture 11 - Control Theory · Lecture 11 - Control Theory Kim Mathiassen. Control of Manipulators...
Transcript of Lecture 11 - Control Theory · Lecture 11 - Control Theory Kim Mathiassen. Control of Manipulators...
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Lecture 11 - Control Theory
Kim Mathiassen
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Control of Manipulators and Mobile Robots
UNIK4490/TEK4030
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Lecture overview
• General introduction to control theory
– Motivation
• Self regulating systems (pendulum at equilibrium)
• Unstable systems (car speed)
– Open and closed loop systems
• Open loop (washing machine)
– Feed forward (car speed – incline as disturbance model)
• Closed loop
– Feedback (cruise control)
– Stability
• Stable systems
• Unstable systems *
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Definition input/output stability
• Asymptotically stable if:
– 𝑦 → 0 when 𝑡 → ∞ and u has a finite duration and amplitude
• Marginally stable if:
– 𝑦 < ∞ for all 𝑡 ≥ 0 and u has a finite duration and amplitude
• Unstable otherwise
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Robot Control
• We want to control the joint positions (configuration) of a
robot
– Therefore we need to figure out how we can determine the
motor/actuator inputs so as to command the robot to a desired
configuration
• In general, the input (voltage/current) does not create
instantaneous motion to a desired configuration because of
the robots dynamical properties
• There are also other real world elements that affect the
robots motion like backlash (clearance between gear tooths
- causes hysteresis) and the properties of the motor/actuator
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Robot Control
• We shall focus on single input single output systems (SISO)
meaning that we only look at each joint by itself
• We therefore assume that each joint is affected only by itself
• This means that the contributions from the other joints are
treated as a disturbance
• Before we can start to control a joint, we need to model the
joint and look at its properties.
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Robot control - Modelling
• From dynamics we have the ordinary differential equation
(ODE) for the robot. We have n non-linear equations as a
result of n joint variables, all expressed in the time domain.
• These n equations often depends on each other
• We want to transform these equations into n linear,
independent equations
• We do this by treating the non-linear effects and the
dependence of other joints as disturbance
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Example – Robot modeling
The dynamic equations are given as
We shall now give a linearization of the first equation
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Robot control - Modeling
• We now have a set of independent linear equations
• We want to analyse the properties of these systems
• Laplace transform can be used to transform the equations
into the frequency domain
– This transforms the equations from a ordinary differential equations
into a linear equations, which easier to solve
– We can analyse important system properties in the frequency
domain, such as stability and step response
• Block diagrams can be used to visualize the equations for
the system and possible feedback loops
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Laplace transform
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• The Laplace transform is defined as follows:
• For example, Laplace transform of a derivative:
– Integrating by parts:
0
dttxesx st
0
dtdt
tdxe
dt
tdxtx stLL
0
00
xssx
dttxestxedt
tdx stst
L
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• Similarly, Laplace transform of a second derivative:
• Thus, if we have a generic 2nd order system described by the following ODE:
• And we want to get a transfer function representation of the system, take the
Laplace transform of both sides:
002
0
2
2
2
2
xsxsxsdtdt
txde
dt
txdtx st
LL
tFtkxtxbtxm
sFskxxssxbxsxsxsm
tFtxktxbtxm
0002
LLLL
Laplace transform
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Laplace transform
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• Continuing:
• The transient response is the solution of the above ODE if the forcing function
F(t) = 0
• The steady state response is the solution of the above equation if the initial
conditions are zero
• This yields the equation
002 xcmsxmsFsxkbsms
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Common Laplace functions
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Review of the Laplace transform
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• Properties of the Laplace transform
– Takes an ODE to a algebraic equation
– Differentiation in the time domain is multiplication by s in the
Laplace domain
– Integration in the time domain is multiplication by 1/s in the
Laplace domain
– Considers initial conditions
• i.e. transient and steady-state response
– The Laplace transform is a linear operator
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Transfer functions
• When all initial conditions of a Laplace transform are zero,
the response Y(s) of a linear system is given by its input
X(s) and its transfer function H(s)
𝑌 𝑠 = 𝐻 𝑠 𝑋(𝑠)
𝑌(𝑠)
𝑋(𝑠)= 𝐻(𝑠)
• The denominator (X(s)) of the transfer function is called the
characteristic polynomial
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Example cont. – Robot modeling
We can now derive the transfer functions for the robot in the frequency domain
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Example – Mass spring damper system
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Block diagrams
• A block diagram visalizes one or more equations
• Makes it easier to see feedback and feed forward loops
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Block diagram symbols
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Example cont. – Robot modeling
Drawing a block diagram
The transfer function as given as:
𝐽𝑠2𝜃1 − 𝐺𝜃1 + 𝐷 = 𝜏1
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Drawing a block diagram of a mass spring damper
system
• 𝑥𝑠2 =1
𝑚(𝑢 − 𝑓𝑥𝑠 − 𝑘𝑥)
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Manipulation of block diagrams
11. april 201124Ny Powerpoint mal 2011
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Manipulation of block diagrams
11. april 201125Ny Powerpoint mal 2011
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Manipulation of block diagrams
11. april 201126Ny Powerpoint mal 2011
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Manipulation of block diagrams
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Prove on blackboard
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Zeros and poles of the transfer functions
• For rational transferfunctions we denote the roots of the
nominator zeros and roots of the denominator poles
• The poles gives important characteristics about the transfer
function
ℎ 𝑠 =𝜌𝑝𝑠
𝑝 +⋯+ 𝜌1𝑠1 + 𝜌0
𝑠𝑛 + 𝛼𝑛−1 𝑠𝑛−1 + ⋯+ 𝛼1 𝑠 + 𝛼0
ℎ 𝑠 =𝜌𝑝 𝑠 − 𝑣1 ⋯(𝑠 − 𝑣𝑝)
(𝑠 − 𝜆1) ⋯ (𝑠 − 𝜆𝑛)
• We call the polynomial in the denominator the characteristic
polynomial
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Examples – Finding roots and zeros
• Example 1: Given transfer function
• Example 2: Mass-spring-damper system
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Example cases
• Three cases depending on the poles
• Case I: Poles are real and distinct
– Over-damped system
• Case II: Poles are real and equal
– Critically damped system
• Case III: Poles are complex conjugates
– Under-damped system
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Time responses
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Effect of changes in poles
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Common transfer functions and their poles and
step respones
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Root locus plots
• Instead of using constant values in the transfer function we
now assume that we can vary one (or more) parameters
• Changing this parameter will move the poles and zeros in
the complex plane
• The paths of zeros and poles in the complex plane as a
function of changed controller parameters are called root
locus plots
• We can investigate the stability of our system by looking at
how these poles moves based on the control parameters
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Example - Root Locus Plot
Drawing on blackboard
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Stability – Frequency domain
• Find the poles (𝜆𝑖) of the transfer function
• If 𝑅𝑒(𝜆𝑖) < 0 for all 𝜆𝑖 in H(s) the system is asymptotically
stable
• If one or more poles has 𝑅𝑒(𝜆𝑖) = 0, but they are not in the
same point the system is marginally stable
• If on or more poles has 𝑅𝑒(𝜆𝑖) > 0 the system is unstable
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Example - Stability
Drawing on blackboard
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Feedback systems
• G(s) – System Model
• C(s) – Controller
• D(s) – Disturbance
• H(s) – Transducer (sensor model)
• R(s) – Reference Input
• Y(s) – Output Variable
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Feedback systems
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Feedback + Feedforward
• We want to regulate the system modeling error, therefore we add the feed forward
parts F(s) and Dc(s), where Dc(s) is a model of the systems disturbances.
• Assuming we can model the system accurately a convenient choice is F(s) = 1/G(s)
• This decreases the systems time response (feedback systems can be slower)
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Feedback + Feedforward
• The systems transfer function is then given as
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Setpoint Controllers
• We will discuss three common controllers: P, PD and PID
– All controllers attempt to drive the error (between a desired trajectory
and the actual trajectory) to zero
• The system (G(s)) can have any dynamics, but we will use the
following system as an example
Compact block diagram Block diagram with basic building blocks
𝜃 =𝑈 − 𝐷
𝐽𝑠2 + 𝐵𝑠
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Setpoint Controllers – System Model
• A generic robot model is given as
• 𝐽 𝑞 ሷ𝑞 - Inertial forces
• 𝐶 𝑞, ሶ𝑞 ሶ𝑞 - Coriolis and centrifugal forces
• 𝐵 ሶ𝑞 - Viscous friction (damping)
• 𝑔 𝑞 - Gravitational forces
• 𝜏 - Torque/Force from actuators
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𝐽 𝑞 ሷ𝑞 + 𝐶 𝑞, ሶ𝑞 ሶ𝑞 + 𝐵 ሶ𝑞 + 𝑔 𝑞 = 𝜏
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Setpoint Controllers – System Model
• In our example we assume that the following are treated as
a disturbance D
– Coriolis and centrifigal forces
– Gravitational forces
– Coupling between joints (𝐽 𝑞 ሷ𝑞 → 𝐽 ሷ𝑞, Ineria is no longer dependent
on the joint variables)
• Transforming into the frequency domain (with Laplace) gives(remember that 𝜃 is our joint variable 𝑞)
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𝐽 𝑞 ሷ𝑞 + 𝐶 𝑞, ሶ𝑞 ሶ𝑞 + 𝐵 ሶ𝑞 + 𝑔 𝑞 = 𝜏
𝐽 ሷ𝑞 + 𝐵 ሶ𝑞 + 𝐷 = 𝜏
𝐽𝑠2𝜃 + 𝐵𝑠𝜃 + 𝐷 = 𝜏
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• We will now look at different controllers for this system.
• We want the error between the reference (desired) value and the actual
output value to go to zero
• The error is defined as e 𝑡 = 𝜃𝑑 𝑡 − 𝜃(𝑡)
• The controller use the error e 𝑡 to calculate its output, also called
control effort
• We denote the control effort as u (“U” in the block diagram above)
• We will look at the following controllers:
1. Proportional (P) controller
2. Proportional Derivative (PD) controller
3. Proportional Integral Derivative (PID) controller
Setpoint Controllers – Motivation
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• Control law: 𝑈 𝑡 = 𝐾𝑃𝑒 𝑡
• Where e 𝑡 = 𝜃𝑑 𝑡 − 𝜃(𝑡)
• Taking the Laplace transformation gives:
𝑈 𝑠 = 𝐾𝑃𝐸 𝑠
• Adding this controller to our system gives the following closed-loop system
Proportional (P) Controller
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48J = 1 B = 0.7 D = 0.5 Kp = 1
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49
Increasing the gain yields:
- Faster response
- Decreased steady state error
- Increased oscillations
J = 1 B = 0.7 D = 0.5 Kp = 2.5
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Proportional Derivative (PD) Controller
• Control law: 𝑈 𝑡 = 𝐾𝑃𝑒 𝑡 + 𝐾𝑑 ሶ𝑒(𝑡)
• Where e 𝑡 = 𝜃𝑑 𝑡 − 𝜃 𝑡
• Taking the Laplace transformation gives:
𝑈 𝑠 = (𝐾𝑃+𝐾𝑑𝑠)𝐸 𝑠
• Adding this controller to our system gives the following closed-loop system:
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We see that the Derivative
term has a damping effect
J = 1 B = 0.7 D = 0.5 Kp = 2.5 Kd=2
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Proportional Derivative (PD) Controller
• Recall that this system can be described by:
• Where, again, U(s) is:
• Combining these gives us:
• Solving for Q gives:
BsJs
sDsUs
2Q
sssKKsU d
dp QQ
BsJs
sDsssKKs
d
dp
2
QQQ
pd
d
dp
d
dppd
d
dpdp
KsKBJs
sDssKKs
sDssKKsKsKBJs
sDssKKssKKsBsJs
2
2
2
QQQ
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Proportional Derivative (PD) Controller
• The denominator is the characteristic polynomial
• The roots of the characteristic polynomial determine the performance of the system
• If we think of the closed-loop system as a damped second order system, this allows us to choose values of Kp and Kd
• Thus Kp and Kd are:
• A natural choice is z = 1 (critically damped)➢ z < 1 – underdamped system
➢ z > 1 – overdamped system
02
J
Ks
J
KBs
pd
02 22 zss
BJK
JK
d
p
2
2
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Proportional Derivative (PD) Controller
• Limitations of the PD controller:
– for illustration, let our desired trajectory be a step input and our
disturbance be a constant as well:
– Plugging this into our system description gives:
– For these conditions, what is the steady-state value of the
displacement?
– Thus the steady state error is –D/Kp
– Therefore to drive the error to zero in the presence of large
disturbances, we need large gains… so we turn to another controller
s
DsD
s
Csd ,Q
pd
dp
KsKBJss
DCsKKs
2Q
pp
p
pd
dp
spd
dp
sss
K
DC
K
DCK
KsKBJs
DCsKK
KsKBJss
sDCsKKs
2020limlim
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Proportional Integral Derivative (PID) controller• Control law:
• Taking the Laplace transformation gives:
• Adding this controller to our system gives the following closed-loop system:
dtteKteKteKtu idp
sEs
KsKKsU i
dp
ipd
d
ipd
KsKsKBJs
ssDsKsKsKs
23
2 QQ
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The Ki term removes the steady
state error
J = 1 B = 0.7 D = 0.5 Kp = 2.5 Kd=2 Ki = 0.133
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57J = 1 B = 0.7 D = 0.5 Kp = 2.5 Kd=2 Ki = 0.5
To high Ki can cause overshoot,
oscillations or instability
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Proportional Integral Derivative (PID) controller
• How to determine PID gains
1. Set Ki = 0 and solve for Kp and Kd
2. Determine Ki to eliminate steady state error
• However, we need to be careful of the stability conditions
– In general real world testing we always start with determining Kp
– There are general methods for finding controller gains that could
be used (ziegler nichols methods etc.)
J
KKBK
pd
i
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Proportional Integral Derivative (PID) controller
• Stability
– The closed-loop stability of these systems is determined by the
roots of the characteristic polynomial
– If all roots (potentially complex) are in
the ‘left-half’ plane, our system is stable
• for any bounded input and disturbance
– A description of how the roots of the characteristic equation
change (as a function of controller gains) is very valuable
• Called the root locus (see example slide 36)
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Setpoint Controllers – Summary
• Proportional
– A pure proportional controller will have a steady-state error
– High gain (Kp) will produce a fast system
– High gain may cause oscillations and may make the system unstable
– High gain reduces the steady-state error
• Integral
– Removes steady-state error
– Increasing Ki accelerates the controller
– High Ki may give oscillations
– Increasing Ki will increase the settling time
• Derivative
– Larger Kd decreases oscillations
– Improves stability for low values of Kd
– May be highly sensitive to noise if one takes the derivative of a noisy error
– High noise leads to instability
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Example - Motor dynamics
• DC motors are ubiquitous in robotics applications
• Here, we develop a transfer function that describes the relationship
between the input voltage and the output angular displacement
• First, a physical description of the most common motor: permanent
magnet…
am iK 1
torque on the rotor:
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Physical instantiationstator
rotor
(armature)
commutator
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Motor dynamics
• When a conductor moves in a magnetic field, a voltage is generated
– Called back EMF:
– Where m is the rotor angular velocity
mb KV 2
armature inductance
armature resistance
baa VVRi
dt
diL
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Motor dynamics
• Since this is a permanent magnet motor, the magnetic flux is constant,
we can write:
• Similarly:
• Km and Kb are numerically equivalent, thus there is one constant
needed to characterize a motor
amam iKiK 1
dt
dKKV m
bmb
2
torque constant
back EMF constant
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Motor dynamics
• This constant is determined from torque-speed curves
– Remember, torque and displacement are work conjugates
– 0 is the blocked torque
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Single link/joint dynamics
• Now, lets take our motor and connect it to a link
• Between the motor and link there is a gear such that:
• Lump the actuator and gear inertias:
• Now we can write the dynamics of this mechanical system:
Lm r
gam JJJ
riK
rdt
dB
dt
dJ L
amL
mm
mm
m
2
2
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Motor dynamics
• Now we have the ODEs describing this system in both the electrical
and mechanical domains:
• In the Laplace domain:
riK
dt
dB
dt
dJ L
amm
mm
m
2
2
dt
dKVRi
dt
diL m
baa
r
ssIKssBsJ L
ammmm
Q 2
ssKsVsIRLs mba Q
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Motor dynamics
• These two can be combined to define, for example, the input-output
relationship for the input voltage, load torque, and output displacement:
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Motor dynamics
• Remember, we want to express the system as a transfer function from
the input to the output angular displacement
– But we have two potential inputs: the load torque and the armature voltage
– First, assume L = 0 and solve for Qm(s):
sI
K
ssBsJa
m
mmm Q2
ssKsVsK
sBsJRLsmbm
m
mm QQ 2
mbmm
mm
KKBsJRLss
K
sV
s
Q
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Motor dynamics
• Now consider that V(s) = 0 and solve for Qm(s):
• Note that this is a function of the gear ratio
– The larger the gear ratio, the less effect external torques have on the
angular displacement
mbmmL
m
KKBsJRLss
rRLs
s
s
/
Q
RLs
ssKsI mb
a
Q
r
s
RLs
ssKKssBsJ Lmbm
mmm
2
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Motor dynamics
• In this system there are two ‘time constants’
– Electrical: L/R
– Mechanical: Jm/Bm
• For intuitively obvious reasons, the electrical time constant is assumed
to be small compared to the mechanical time constant
– Thus, ignoring electrical time constant will lead to a simpler version of the
previous equations:
RKKBsJs
r
s
s
mbmmL
m
/
/1
Q
RKKBsJs
RK
sV
s
mbmm
mm
/
/
Q
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Motor dynamics
• Rewriting these in the time domain gives:
• By superposition of the solutions of these two linear 2nd order ODEs:
RKKBsJs
r
s
s
mbmmL
m
/
/1
Q
RKKBsJs
RK
sV
s
mbmm
mm
/
/
Q tVRKtRKKBtJ mmmbmmm //
tRtRKKBtJ Lmmbmmm /1/
tRtVRKtRKKBtJ Lmmmbmmm /1//
J B tu td
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Motor dynamics
• Therefore, we can write the dynamics of a DC motor attached to a load
as:
– Note that u(t) is the input and d(t) is the disturbance (e.g. the dynamic
coupling from motion of other links)
• To represent this as a transfer function, take the Laplace transform:
tdtutBtJ
sDsUsBsJs Q2