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Transcript of EE4901 musculoskeletal
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8/8/2019 EE4901 musculoskeletal
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School of Electronic and Electrical
Engineering
Part I: Musculoskeletal Control System Design
(Simulation study)
Module : EE4901
Class : FD1
Name : Tan Zhi Hao(072438L03)
Dated: Monday, 5th October 2009
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Contents
Background ................................................................................................................................ 1
Exercise 1 GTO Force Feedback Analysis .................................................................................. 2
Description ............................................................................................................................. 2
Stimulation model .................................................................................................................. 2
Stimulation Analysis ............................................................................................................... 3
Exercise 2 Examination of the Dynamics of Neuromuscular Reflex Motion ........................... .... 6
Description ............................................................................................................................. 6
Stimulation model .................................................................................................................. 6
Stimulation analysis ...............................................................................................................10
Exercise 3 Design of Functional Electrical Stimulation ............................................. ................12Description ............................................................................................................................12
Simulation model ..................................................................................................................12
Stimulation Design & Evaluation ............................................................................................14
Reference ..................................................................................................................................16
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Background
Since the ancient times, people have been finding ways to replace their lost body parts
as they are very essential components of our lives. People do surgery, build walking aids
and wheelchairs to regain their ability to move.
Through the advancement in technology, we gained more understanding about the
anatomy of our body and hence enable us to help some of unfortunates to regain their
movement. This requires comprehensive knowledge from many different aspects such
as control theory, neuroscience, physiology, robotics and others.
The objective of this module is to gain a basic understanding of the functional anatomy
of the neuromuscular system and the implementation of control theory in this aspect.
To achieve this, it is required to do simulation studies on specific cases specified in the
course notes. This design module is focused on the following topics.
y Functional anatomy of the neuromuscular systemy Motor control of human movementy Muscle modelsy Dynamics of skeletal structurey Functional electrical stimulation (FES) and Design
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Exercise 1 GTO Force FeedbackAnalysis
Description
Exercise 1 is to analyze exclusively the Golgi Tendon Organ (GTO) force feedback in the
extrafusal muscles. Intrafusal muscles and length feedback mechanisms are notconsidered.
Stimulation model
The Force Feedback System is given as shown below.
Transfer function of F(e) block
, where C is a positive constant.
Golgi Tendon Organs (GTO) block
, where H is a positive constant.
Assuming that length of the muscle is constant X(s) = 0, the muscle model can be
expressed as shown below.
With the above expressions given, force feedback system can then be expressed as
below.
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Stimulation Analysis
In this stimulation, C is fixed at 10 and R only changes from 0 to desired values at 0.1s.
A typical input (R) with value of 5
Force Output
R=5
C=10
T=0.01
H=1
Force Output
R=7
C=10
T=0.01
H=1
Force Output
R=10
C=10
T=0.01
H=1
Force Output
R=15
C=10
T=0.01
H=1
At the left most curve, we can see that when RC, the signal from the brain is higher than the maximum limit
of the muscle, hence there is not much difference compared to the previous curve.
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From the result when R=0.01 and R=0.1,
though the input signal is ten times of
the other input signal, the output only
vary by changing its frequency. As the
force of contraction depends on the rateof stimulation, if the rate of stimulation
is too slow, it may results in a series of
twitching. These results may also
indicate that the muscles cannot be
controlled like normal movement if the
distance of motion is very small (e.g.
micrometer or nano meter range).
Force Output
R=5
C=10
T=0.01H=0.6
Force Output
R=7
C=10
T=0.01H=0.6
Force Output
R=10
C=10
T=0.01H=0.6
Force Output
R=15
C=10
T=0.01H=0.6
Comparing to the previous results, we can see that by changing H from 1 to 0.6, the
muscle can generate more force with the same input. From the overall transfer function,
we will understand that it has increased the gain of the system.
Force Output
R=0.1
C=10
T=0.01
H=1
Force OutputR=0.01
C=10
T=0.01
H=1
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Force Output
R=5
C=10T=0.0001
H=1
Force Output
R=5
C=10T=0.01
H=1
Force Output
R=5
C=10T=0.1
H=1
Force Output
R=5
C=10T=1
H=1
Note: the change in x-axis (time)
With the gradual increase in time delay, the amplitude and frequency of the force
output changed. When T=0.0001, the force output is similar to the input signal. Since
the force of contraction depends on the rate stimulation, it is unknown if this kind of
input can effectively control the muscle. As T increases, the period increases. This may
produce a tremor like effect on a person.
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Exercise 2 Examination of the Dynamics of Neuromuscular Reflex
Motion
Description
Exercise 2 is to examine the dynamics of neuromuscular reflex motion. In thisexperiment, a persons arm is placed at an angle of 135 between the forearm and
upper arm. At t=0, an additional weight, represented by Mx(t), is added to the arm. (t)
is the change in angular motion. Gravitational force is ignored. In practical scenario, (t)
cannot exceed 45.
Stimulation model
A series of mathematical calculations is needed before the final simulink model can be
obtained. These steps are stated as follows.
The motion equation is given as:
Mx(t) refers to the change in external moment acting on the limb about the elbow joint.
M(t) is the net muscular torque exerted in response to the external disturbance and J is
the moment of inertia of the forearm about the elbow joint.
To simplify the question, we assume that the distance between the location of the force
acted is 1m from the joint. Therefore, in this simulation study, the value of force is equal
to the value of moment.
From the motion equation, we can obtain an incomplete model.
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From the lecture note (equation 6), we can obtain the equation of M(t) for the
extrafusal muscle without the passive tissue.
Hence, the model can be modified as shown below.
In order to illustrate the complete model, we need to model Mo(t).
Given the muscle spindle model and an equation relating Mo(t) and (t) shown above,
the following is the derivation of Mo(t).
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From the muscle spindle model,
Taking Laplace Transform,
Given = 0,
From the muscle spindle model,
Substituting Ms(t) into the equation,
Substituting into the equation given in the muscle spindle model,
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In order to substitute the values into the equation, we need to manipulate the variables.
Given J =0.2, Tdelay=0.025, B=3, K = 60,
,
, the final model is as
shown below.
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Stimulation analysis
To understand the effect of the change in weight Mx(t), the gain is fixed at 3.
Angle output
Mx(t) = 1
= 3
Angle output
Mx(t) = 5
= 3
Angle output
Mx(t) = 10
= 3
Angle output
Mx(t) = 15
= 3
When the weight of the load increases, the arm is less able to take the load. Thus, the
change in angle (t) increases. The change in angle at steady state is proportional to the
weight of the load. When Mx(t) exceeds 15N, the angle of the arm increases beyond 45
degrees. At this angle, the joint of the arm will break. Therefore, it is not meaningful to
simulate beyond 15N as the result will not match the actual scenario on a live arm.
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To understand the effect of the change in gain , the weight Mx(t) is fixed at 10.
Angle outputMx(t) = 10
= 2
Angle outputMx(t) = 10
= 3
Angle outputMx(t) = 10
= 5
Angle outputMx(t) = 10
= 10
When the gain increases, the change in angle (t) decreases. This either means that
the muscle is more able to cope with the sudden impact of the load or the muscle is
stronger than the previous setting. The change in angle at steady state is inversely
proportional to the gain . Similarly to the previous simulation, when lower than 2,
the angle of the arm increases beyond 45 degrees. At this angle, the joint of the arm will
break. Therefore, it is not meaningful to simulate as the result will not match the actual
scenario on a live arm.
Interestingly, there are some
settings which are not able
to relate to real-life situation.
However logically, it is
understandable that when
the gain goes exceedingly
high, it will make the overall
system unstable. These are
shown on the left.
Note: The change in the x-
axis and y-axis scale
Angle output
Mx(t) = 10
= 400
Angle output
Mx(t) = 10
= 450
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Exercise 3 Design of FunctionalElectricalStimulation
Description
Exercise 3 is to design the suitable control scheme for Functional Electrical Stimulation
(FES) to enable certain group of people to regain their ability to move. The PID control isused for the Control Scheme block.
Simulation model
The output of muscle force is determined by pulse width, frequency and amplitude of
the stimulation. Assume that the pulse width and amplitude is fixed and the muscle
force only varies with pulse frequency (fs).
The PID controller produces a voltage c(t) to the stimulator. The frequency (fs) of the
stimulator output is proportional to the voltage input, c(t). The input outputrelationship is expressed below.
where K1 is a constant and is chosen to be 0.3
Note: fs is limited within the range of 5 to 50Hz to express the characteristic of the
muscle in the linear region of the force-frequency curve.
Muscle activation a(t) is approximately represented with the following expression.
where Kfis a shaping factor chosen as 0.1 in this simulation.
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Output p(t) of force sensor was proportional to force f(t).
where K2 is a constant and is chosen to be 1.
The final stimulation model is shown below.
Control Scheme Block (PID Control)
Muscle Activation Block
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Stimulation Design & Evaluation
There are a few assumptions on this case. Firstly, since the output can only varies from 0
to 1, we assume that the desired force input signal is also varies from 0 to 1. Secondly,
as it is an artificial control of muscle, the input should be similar to the output. Thirdly,
the range of fs is limited between 5 to 50 Hz so that it is in the linear region of the force-frequency curve of the muscle, we should make sure that fs is within the linear region to
make sure that result match to real life.
Initial observation is when the input is between 0 and 1, the output remains at 0.2. This
may be because the input is too small to have any effects on the output. However, if we
adjust the value of Kp, we can move the operating fs between 5 and 50 Hz.
The most common method for PID tuning is Ziegler-Nichols closed-loop tuning method.
However, it cannot be employed in this case because by increasing the value of K p
cannot make the system oscillate. Without oscillation, the critical gain (Ku) cannot be
determined and hence this method cannot be used. On the other hand, Ziegler-Nichols
open-loop tuning method also cannot be used. With no feedback, the delay (L) between
the time at the transition of input and the time at the max change in output is zero in
the simulation. This will only give us infinite Kp and zero Ti and zero Td.
Cohen Coon Tuning Method in this case seems to be a viable solution. In order to make
sure that fs is in the linear region, Kp is first adjusted to 4100. This is also to make the
input as similar as the output.
The input signal is 0.8 and increase to 1 at
t=5s. Therefore, A=0.2.
From simulation, B= 0.1765, t0=5.0, t2=5.0287,
t3=5.0413.
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Kp Kp/Ti Kp/Td
P 195.90
PI 176.07 224692
PID 260.99 446646 0.0226
With the PID scheme tuned using Cohen Coon Tuning Method, t2 and t3 vary within 0.1%
if the values before tuning. Therefore, I concluded that the setting for the optimal input
and output is by adding a gain of 4100 at the control scheme. However, there are two
flaws with this setting. One, the maximum value of the output is 0.96. Two, the error of
output value is around 4%.
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The following is the model with a random signal added to the input force.
Kp=4100
= 0Kp=4100
= 0.001Kp=4100
= 0.01Kp=4100
= 0.1
Even with the noise variance at 0.01 (third diagram from the left), SNR is 2. This is not
optimistic for the system as it is very susceptible to noise.
Reference
1) Wen C Y, Biomedical Control System Design notes2) Tomas B. Co, http://www.chem.mtu.edu/~tbco/cm416/cctune.html , Cohen
Coon Tuning Method, Sept 2009
3) Xie L H, PID Control Schemes Modeling and Control notes (lecture 10),2008/2009