DIRECTIONS: READ THE FOLLOWING STUDENT POST
AND RESPOND EVALUATE ITS CONTENT. PLEASE CITE
ALL REFERENCES
Katie Kessler
Unit 2 Discussion 1
Top of Form
The word “noir” is used to remember the scaling of
measurement in psychology (Embretson, 2004). In short, the
letters stand for nominal, ordinal, interval and ratio (Embretson,
2004). To give a brief introduction of what each scale measures,
“nominal is the simplest way to measure” because it focuses on
categorizing measurements on a scale of category, according to
Embretson (2004). An example of nominal is eye color.
“Ordinal measures in terms of ranking, interval measures scores
of tests that focus on unobservable mental functioning and ratio
focuses on measuring activities in the physical world, such as
someone’s running time” (Embretson, 2004). With different
scales of measurement, there are two methods to compare sets
of data. These include norm-referenced and criterion-referenced
testing. According to Embretson (2004) norm-referenced testing
“yields information on a testtaker’s standing or ranking relative
to some comparison group of testtakers.” In other words, it
focuses on the performance of peers. Criterion-referenced
testing is a little different because it focuses on examining
individual’s scores to a set standard (Embretson, 2004).
The ability for ordinal measurement scale to be utilized on a
standardized test as a norm-referenced test is high since an
ordinal scale is based upon ranking and norm-referenced testing
gathers information on the examinees ranking compared to a
group of testtakers. For example, a study conducted on decision
making with the use of ordinal variables states that ordinal
measurement scales has the ability to be utilized by norm-
referenced testing (Barua, Kademane, Das, Gubbiyappa, Verma,
& Al-Dubai, 2014).On the other hand, ordinal scaling would not
be a strong measurement for criterion-referenced testing
because it focuses on the ranking rather than the measurement
of the scores to be close to a set standard.
Ratio scaling directs its focus on measuring objects and
activities in the physical world which would be beneficial for
criterion-referenced testing instead of norm-referenced testing.
Imagine a marathon runner who was trying to beat the world’s
fastest time running a marathon. Criterion-referenced testing
allows the runner to be aware of the set standard the marathon
runner needs to beat to be the best and set a new standard.
Norm-referenced testing would not be as useful because the
marathon runner would not have the standard measurement he or
she needs to beat. However, the marathon runner would be
aware of the relative time he or she needs to beat to be the best.
That is not as helpful as the criterion-referenced testing because
runners need an exact number instead of a relative number in
comparison to other runners.
Norm-referenced data would be collected by “the standards
relative to a group, such as means and standard deviations”
while criterion-referenced data is collected “based on absolute
standards” (Montgomery & Connolly, 1987). Using “noir” as
the scale of measurement in psychology is significant because
nominal and ordinal are thought of as categories and interval
and ratio are ways to collect quantitative data. Therefore,
“choosing a statistical analysis procedure depends on the level
of measurement of variables in the experiment” (Virginia Tech,
1999). In other words, the flexibility and rigidity of this
particular scale of measurement in psychology allows
professionals have the ability to measure in terms of categories
or quantitatively. For my career as a BCBA, I believe the norm-
referenced testing is the most appropriate use of measurement
for standardized testing because I am looking for my clients to
demonstrate a relative understanding on speech, motor skills,
social skills and leisure skills. Other tests in school may require
my clients to reach a set standard of testing such as end of the
year testing to move on to the next grade or submitting scores
for the SAT to get into college, but my goal as a BCBA is to
help clients generalize the information relatively compared to
their peers on a normal functioning level.
References
Barua, A., Kademane, K., Das, B., Gubbiyappa, K. S., Verma,
R. K., & Al-Dubai, S. (2014). A tool for decision-making in
norm-referenced survey questionnaires with items of ordinal
variables. International Journal of Collaborative Research on
Internal Medicine & Public Health, 6(3), 52-63. Retrieved from
http://search.proquest.com.library.capella.edu/docview/1524878
969?accountid=27965
Embretson, S. E. (2004). The Second Century of Ability
Testing: Some Predictions and Speculations. Measurement,
2(1), 170-174.
Montgomery C. P. & Connolly H. B. (1987). Norm-Referenced
and Criterion-Referenced Tests: Use in Pediatrics and
Appliation to Task Analysis of Motor Skill. ResearchGate
67(12), 1873-1875.
Virginia Tech. (1999). Measurement Scales in Social Science
Research. Retrieved from
http://www.simon.cs.vt.edu/SoSci/converted/Measurement/
Bottom of Form
SAMPLE QUESTION:
Exercise 1: Consider the function
f (x,C)=
sin(C x)
Cx
(a) Create a vector x with 100 elements from -3*pi to 3*pi.
Write f as an inline or anonymous function
and generate the vectors y1 = f(x,C1), y2 = f(x,C2) and y3 =
f(x,C3), where C1 = 1, C2 = 2 and
C3 = 3. Make sure you suppress the output of x and y's
vectors. Plot the function f (for the three
C's above), name the axis, give a title to the plot and include
a legend to identify the plots. Add a
grid to the plot.
(b) Without using inline or anonymous functions write a
function+function structure m-file that does
the same job as in part (a)
SAMPLE LAB WRITEUP:
MAT 275 MATLAB LAB 1
NAME: __________________________
LAB DAY and TIME:______________
Instructor: _______________________
Exercise 1
(a)
x = linspace(-3*pi,3*pi); % generating x vector - default value
for number
% of pts linspace is 100
f= @(x,C) sin(C*x)./(C*x) % C will be just a constant, no need
for ".*"
C1 = 1, C2 = 2, C3 = 3 % Using commans to separate commands
y1 = f(x,C1); y2 = f(x,C2); y3 = f(x,C3); % supressing the y's
plot(x,y1,'b.-', x,y2,'ro-', x,y3,'ks-') % using different markers
for
% black and white plots
xlabel('x'), ylabel('y') % labeling the axis
title('f(x,C) = sin(Cx)/(Cx)') % adding a title
legend('C = 1','C = 2','C = 3') % adding a legend
grid on
Command window output:
f =
@(x,C)sin(C*x)./(C*x)
C1 =
1
C2 =
2
C3 =
3
(b)
M-file of structure function+function
function ex1
x = linspace(-3*pi,3*pi); % generating x vector - default value
for number
% of pts linspace is 100
C1 = 1, C2 = 2, C3 = 3 % Using commans to separate commands
y1 = f(x,C1); y2 = f(x,C2); y3 = f(x,C3); % function f is defined
below
plot(x,y1,'b.-', x,y2,'ro-', x,y3,'ks-') % using different markers
for
% black and white plots
xlabel('x'), ylabel('y') % labeling the axis
title('f(x,C) = sin(Cx)/(Cx)') % adding a title
legend('C = 1','C = 2','C = 3') % adding a legend
grid on
end
function y = f(x,C)
y = sin(C*x)./(C*x);
end
Command window output:
C1 =
1
C2 =
2
C3 =
3
More instructions for the lab write-up:
1) You are not obligated to use the 'diary' function. It was
presented only for you convenience. You
should be copying and pasting your code, plots, and results
into some sort of "Word" type editor that
will allow you to import graphs and such. Make sure you
always include the commands to generate
what is been asked and include the outputs (from command
window and plots), unless the problem
says to suppress it.
2) Edit this document: there should be no code or MATLAB
commands that do not pertain to the
exercises you are presenting in your final submission. For
each exercise, only the relevant code that
performs the task should be included. Do not include error
messages. So once you have determined
either the command line instructions or the appropriate script
file that will perform the task you are
given for the exercise, you should only include that and the
associated output. Copy/paste these into
your final submission document followed by the output
(including plots) that these MATLAB
instructions generate.
3) All code, output and plots for an exercise are to be grouped
together. Do not put them in appendix, at
the end of the writeup, etc. In particular, put any mfiles you
write BEFORE you first call them.
Each exercise, as well as the part of the exercises, is to be
clearly demarked. Do not blend them all
together into some sort of composition style paper,
complimentary to this: do NOT double space.
You can have spacing that makes your lab report look nice,
but do not double space sections of text
as you would in a literature paper.
4) You can suppress much of the MATLAB output. If you need
to create a vector, "x = 0:0.1:10" for
example, for use, there is no need to include this as output in
your writeup. Just make sure you
include whatever result you are asked to show. Plots also do
not have to be a full, or even half page.
They just have to be large enough that the relevant structure
can be seen.
5) Before you put down any code, plots, etc. answer whatever
questions that the exercise asks first.
You will follow this with the results of your work that
support your answer.
MATLAB sessions: Laboratory 6
MAT 275 Laboratory 6
Forced Equations and Resonance
In this laboratory we take a deeper look at second-order
nonhomogeneous equations. We will concentrate
on equations with a periodic harmonic forcing term. This will
lead to a study of the phenomenon known
as resonance. The equation we consider has the form
d2y
dt2
+ c
dy
dt
+ ω20y = cos ωt. (L6.1)
This equation models the movement of a mass-spring system
similar to the one described in Laboratory
5. The forcing term on the right-hand side of (L6.1) models a
vibration, with amplitude 1 and frequency
ω (in radians per second = 1
2π
rotation per second = 60
2π
rotations per minute, or RPM) of the plate
holding the mass-spring system. All physical constants are
assumed to be positive.
Let ω1 =
√
ω20 − c2/4. When c < 2ω0 the general solution of (L6.1) is
y(t) = e−
1
2
ct(c1 cos(ω1t) + c2 sin(ω1t)) + C cos (ωt − α) (L6.2)
with
C =
1√
(ω20 − ω2)
2
+ c2ω2
, (L6.3)
α =
(
cω
ω20−ω2
)
if ω0 > ω
π + arctan
(
cω
ω20−ω2
)
if ω0 < ω
(L6.4)
and c1 and c2 determined by the initial conditions. The first
term in (L6.2) represents the complementary
solution, that is, the general solution to the homogeneous
equation (independent of ω), while the second
term represents a particular solution of the full ODE.
Note that when c > 0 the first term vanishes for large t due to
the decreasing exponential factor.
The solution then settles into a (forced) oscillation with
amplitude C given by (L6.3). The objectives of
this laboratory are then to understand
1. the effect of the forcing term on the behavior of the solution
for different values of ω, in particular
on the amplitude of the solution.
2. the phenomena of resonance and beats in the absence of
friction.
The Amplitude of Forced Oscillations
We assume here that ω0 = 3 and c = 1 are fixed. Initial
conditions are set to 0. For each value of ω, the
amplitude C can be obtained numerically by taking half the
difference between the highs and the lows
of the solution computed with a MATLAB ODE solver after a
sufficiently large time, as follows: (note
that in the M-file below we set ω = 2.4).
1 function LAB06ex1
2 omega0 = 3; c = 1; omega = 2.4;
3 param = [omega0,c,omega];
4 t0 = 0; y0 = 0; v0 = 0; Y0 = [y0;v0]; tf = 50;
5 options = odeset(’AbsTol’,1e-10,’RelTol’,1e-10);
6 [t,Y] = ode45(@f,[t0,tf],Y0,options,param);
7 y = Y(:,1); v = Y(:,2);
8 figure(1)
9 plot(t,y,’b-’); ylabel(’y’); grid on;
c©2016 Stefania Tracogna, SoMSS, ASU 1
MATLAB sessions: Laboratory 6
10 t1 = 25; i = find(t>t1);
11 C = (max(Y(i,1))-min(Y(i,1)))/2;
12 disp([’computed amplitude of forced oscillation = ’
num2str(C)]);
13 Ctheory = 1/sqrt((omega0^2-omega^2)^2+(c*omega)^2);
14 disp([’theoretical amplitude = ’ num2str(Ctheory)]);
15 %----------------------------------------------------------------
16 function dYdt = f(t,Y,param)
17 y = Y(1); v = Y(2);
18 omega0 = param(1); c = param(2); omega = param(3);
19 dYdt = [ v ; cos(omega*t)-omega0^2*y-c*v ];
When executing this program we get
computed amplitude of forced oscillation = 0.24801
theoretical amplitude = 0.24801
Lines 10-14 deserve some explanation. Line 10 defines a time t1
after which we think the contribution
of the first term in (L6.2) has become negligible compared to
the second term. This depends of course
on the parameter values, in particular c. With c = 1 we obtain
e−
1
2
ct ' 3.7 × 10−6 for t = 25, so this
is certainly small enough compared to the amplitude seen on
Figure L6a. The index i of time values
larger than t1 is then determined. The quantity Y(i,1) refers to
the values of y associated to times
larger than t1 only. The computed amplitude is simply half the
difference between the max and the min
values. This value is compared to the theoretical value (L6.3).
0 5 10 15 20 25 30 35 40 45 50
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
y
Figure L6a: Forced oscillation.
1. (a) What is the period of the forced oscillation? What is the
numerical value (modulo 2π) of the
angle α defined by (L6.4)?
(b) In this question you are asked to modify the file
LAB06ex1.m in order to plot the complemen-
tary solution of (L6.1), that is, the first term in (L6.2). First
define in the file the angle α
(alpha) using (L6.4), then evaluate the complementary solution
yc by subtracting the quan-
tity C cos(ωt − α) from the numerical solution y. Plot the
resulting quantity. Does it look
like an exponentially decreasing oscillation? Why or why not?
Include the modified M-file
and the corresponding plot.
2. We now consider C as a function of ω. We use again ω0 = 3,
c = 1 and y(0) = y
′(0) = 0. The
previous problem determined C for a specific value of ω. Here
we consider a range of values for ω
c©2016 Stefania Tracogna, SoMSS, ASU 2
MATLAB sessions: Laboratory 6
and determine numerically the corresponding amplitude C. We
then plot the result as a function
of ω, together with the theoretical amplitude from (L6.3). You
may need the following MATLAB
program.
function LAB06ex2
omega0 = 3; c = 1;
OMEGA = 2:0.02:4;
C = zeros(size(OMEGA));
Ctheory = zeros(size(OMEGA));
t0 = 0; y0 = 0; v0 = 0; Y0 = [y0;v0]; tf = 50; t1 = 25;
for k = 1:length(OMEGA)
omega = OMEGA(k);
param = [omega0,c,omega];
[t,Y] = ode45(@f,[t0,tf],Y0,[],param);
i = find(t>t1);
C(k) = (max(Y(i,1))-min(Y(i,1)))/2;
Ctheory(k) = ??; % FILL-IN
end
figure(2)
plot(??); grid on; % FILL-IN
xlabel(’omega’); ylabel(’C’);
%---------------------------------------------------------
function dYdt = f(t,Y,param)
y = Y(1); v = Y(2);
omega0 = param(1); c = param(2); omega = param(3);
dYdt = [ v ; cos(omega*t)-omega0^2*y-c*v ];
2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4
0.1
0.15
0.2
0.25
0.3
0.35
C
Figure L6b: Amplitude as a function of ω
(a) Fill in the missing parts in the M-file LAB06ex2.m and
execute it. You should get a figure like
Figure L6b. Include the modified M-file in your lab report.
(b) Examine the graph obtained by running LAB06ex2.m and
determine for what (approximate)
value of ω the amplitude of the forced oscillation, C, is
maximal. This value of ω is called the
practical resonance frequency. Give the corresponding
maximum value of C.
c©2016 Stefania Tracogna, SoMSS, ASU 3
MATLAB sessions: Laboratory 6
(c) Determine analytically the value of ω for which the
amplitude of the forced oscillation, C, is
maximal by differentiating the expression for C in (L6.3) as a
function of ω. Compare the
value you find with the value obtained in part (b).
(d) Run LAB06ex1.m with the value of ω found in part (c)
(include the graph). What is the
amplitude of the forced oscillation? How does it compare with
the amplitude of the forced
oscillation in problem 1.? If you run LAB06ex1.m with any
other value of ω, how do you
expect the amplitude of the solution to be?
(e) Are the results affected by changes in the initial conditions?
Answer this question both nu-
merically (by modifying the initial conditions in LAB06ex2.m)
and theoretically (by analyzing
the expression for C in (L6.3)). Note that the initial conditions
for the DE are y0 and v0.
Resonance
We now investigate what happens to the solution (L6.2), and
more specifically to the maximal amplitude
C of the forced oscillation, when we let c → 0. The value of ω
corresponding to this maximal amplitude is
called pure resonance frequency. When a mechanical system is
stimulated by an external force operating
at this frequency the system is said to be resonant.
3. Set c = 0 in LAB06ex2.m.
2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4
0
1
2
3
4
5
6
7
8
9
C
computed numerically
theoretical
(a) Explain what happens. What is the maximal amplitude?
What is the value of ω yielding the
maximal amplitude in the forced solution? How does this value
compare to ω0?
(b) Run LAB06ex1.m with c = 0 and ω equal to the value found
in part (a). Comment on the
behavior of the solution. Include the graph.
Beats
When c = 0 and ω 6= ω0 , the solution (L6.2) to (L6.1) reduces
to
y(t) = c1 cos(ω0t) + c2 sin(ω0t) + C cos(ωt − α)
with C = 1∣ ∣ ω20−ω2∣ ∣ . If the initial conditions are set to
zero, the solution reduces to
y(t) = C(cos(ωt) − cos(ω0t))
which can be rewritten as
y(t) = 2C sin
(
1
2
(ω0 − ω)t
)
sin
(
1
2
(ω0 + ω)t
)
.
c©2016 Stefania Tracogna, SoMSS, ASU 4
MATLAB sessions: Laboratory 6
When ω is close to ω0 we have that ω + ω0 is large in
comparison to |ω0 − ω|. Then sin
(
1
2
(ω0 + ω)t
)
is a very rapidly varying function, whereas sin
(
1
2
(ω0 − ω)t
)
is a slowly varying function. If we define
A(t) = 2C sin
(
1
2
(ω0 − ω)t
)
, then the solution can be written as
y(t) = A(t) sin
(
1
2
(ω0 + ω)t
)
and we may interpret it as a rapidly oscillating function with
period T = 4π
ω0+ω
, but with a slowly varying
amplitude A(t). This is the phenomenon known as beats. Note
that A(t) and −A(t) are the so-called
“envelope functions”. The period of A(t) is 4π|ω0−ω|, thus the
length of the beats is
2π
|ω0−ω|
.
4. To see the beats phenomenon, set c = 0 and ω = 2.8 in
LAB06ex1. Also extend the interval of
simulation to 100.
0 10 20 30 40 50 60 70 80 90 100
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
y
(a) In LAB06ex1 define the “envelope” function A = 2C sin
(
1
2
(ω0 − ω)t
)
with C = 1|ω20−ω2|
.
Plot A in red and −A in green, together with the solution. You
should obtain Figure L6c.
Include the modified M-file.
0 10 20 30 40 50 60 70 80 90 100
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
y
Figure L6c:
Solution
and envelope functions
c©2016 Stefania Tracogna, SoMSS, ASU 5
MATLAB sessions: Laboratory 6
(b) What is the period of the fast oscillation (that is, the period
of sin
(
1
2
(ω0 + ω)t
)
)? Confirm
your answer by zooming in on the graph of the solution. Include
a graph to support your
answer.
(c) What is the length of the beats? Determine the length
analytically using the envelope func-
tions, and numerically from the graph.
(d) Change the value of ω in LAB06ex1 to 2.9 (a value closer to
ω0) and then ω = 2.6 ( a value
farther away from ω0). Include the two graphs. For each of
these two values of ω find the
period of the fast oscillation and the length of the beats. How
do the periods change compared
to parts (b) and (c)?
(e) If you let ω = 1.5, is the beats phenomenon still present?
Why or why not?
c©2016 Stefania Tracogna, SoMSS, ASU 6
Exercise 1:
function LAB06ex1
clc
omega0 = 3; c = 1; omega = 2.4;
param = [omega0,c,omega];
t0 = 0; y0 = 0; v0 = 0; Y0 = [y0;v0]; tf = 50;
options = odeset('AbsTol',1e-10,'relTol',1e-10);
[t,Y] = ode45(@f,[t0,tf],Y0,options,param);
y = Y(:,1); v = Y(:,2);
figure(1)
plot(t,y,'b-'); ylabel('y'); grid on;
t1 = 25; i = find(t>t1);
C = (max(Y(i,1))-min(Y(i,1)))/2;
disp(['computed amplitude of forced oscillation = '
num2str(C)]);
Ctheory = 1/sqrt((omega0^2-omega^2)^2+(c*omega)^2);
disp(['theoretical amplitude = ' num2str(Ctheory)]);
%----------------------------------------------------------------
function dYdt = f(t,Y,param)
y = Y(1); v = Y(2);
omega0 = param(1); c = param(2); omega = param(3);
dYdt = [ v ; cos(omega*t)-omega0^2*y-c*v ];
Exercise 2:
function LAB06ex2
omega0 = 3; c = 1;
OMEGA = 2:0.02:4;
C = zeros(size(OMEGA));
Ctheory = zeros(size(OMEGA));
t0 = 0; y0 = 0; v0 = 0; Y0 = [y0;v0]; tf = 50; t1 = 25;
for k = 1:length(OMEGA)
omega = OMEGA(k);
param = [omega0,c,omega];
[t,Y] = ode45(@f,[t0,tf],Y0,[],param);
i = find(t>t1);
C(k) = (max(Y(i,1))-min(Y(i,1)))/2;
Ctheory(k) = ??; % FILL-IN
end
figure(2)
plot(??); grid on; % FILL-IN
xlabel('omega'); ylabel('C');
%---------------------------------------------------------
function dYdt = f(t,Y,param)
y = Y(1); v = Y(2);
omega0 = param(1); c = param(2); omega = param(3);
dYdt = [ v ; cos(omega*t)-omega0^2*y-c*v ];

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DIRECTIONS READ THE FOLLOWING STUDENT POST AND RESPOND EVALUATE I.docx

  • 1. DIRECTIONS: READ THE FOLLOWING STUDENT POST AND RESPOND EVALUATE ITS CONTENT. PLEASE CITE ALL REFERENCES Katie Kessler Unit 2 Discussion 1 Top of Form The word “noir” is used to remember the scaling of measurement in psychology (Embretson, 2004). In short, the letters stand for nominal, ordinal, interval and ratio (Embretson, 2004). To give a brief introduction of what each scale measures, “nominal is the simplest way to measure” because it focuses on categorizing measurements on a scale of category, according to Embretson (2004). An example of nominal is eye color. “Ordinal measures in terms of ranking, interval measures scores of tests that focus on unobservable mental functioning and ratio focuses on measuring activities in the physical world, such as someone’s running time” (Embretson, 2004). With different scales of measurement, there are two methods to compare sets of data. These include norm-referenced and criterion-referenced testing. According to Embretson (2004) norm-referenced testing “yields information on a testtaker’s standing or ranking relative to some comparison group of testtakers.” In other words, it focuses on the performance of peers. Criterion-referenced testing is a little different because it focuses on examining individual’s scores to a set standard (Embretson, 2004). The ability for ordinal measurement scale to be utilized on a standardized test as a norm-referenced test is high since an ordinal scale is based upon ranking and norm-referenced testing gathers information on the examinees ranking compared to a group of testtakers. For example, a study conducted on decision making with the use of ordinal variables states that ordinal measurement scales has the ability to be utilized by norm-
  • 2. referenced testing (Barua, Kademane, Das, Gubbiyappa, Verma, & Al-Dubai, 2014).On the other hand, ordinal scaling would not be a strong measurement for criterion-referenced testing because it focuses on the ranking rather than the measurement of the scores to be close to a set standard. Ratio scaling directs its focus on measuring objects and activities in the physical world which would be beneficial for criterion-referenced testing instead of norm-referenced testing. Imagine a marathon runner who was trying to beat the world’s fastest time running a marathon. Criterion-referenced testing allows the runner to be aware of the set standard the marathon runner needs to beat to be the best and set a new standard. Norm-referenced testing would not be as useful because the marathon runner would not have the standard measurement he or she needs to beat. However, the marathon runner would be aware of the relative time he or she needs to beat to be the best. That is not as helpful as the criterion-referenced testing because runners need an exact number instead of a relative number in comparison to other runners. Norm-referenced data would be collected by “the standards relative to a group, such as means and standard deviations” while criterion-referenced data is collected “based on absolute standards” (Montgomery & Connolly, 1987). Using “noir” as the scale of measurement in psychology is significant because nominal and ordinal are thought of as categories and interval and ratio are ways to collect quantitative data. Therefore, “choosing a statistical analysis procedure depends on the level of measurement of variables in the experiment” (Virginia Tech, 1999). In other words, the flexibility and rigidity of this particular scale of measurement in psychology allows professionals have the ability to measure in terms of categories or quantitatively. For my career as a BCBA, I believe the norm- referenced testing is the most appropriate use of measurement for standardized testing because I am looking for my clients to
  • 3. demonstrate a relative understanding on speech, motor skills, social skills and leisure skills. Other tests in school may require my clients to reach a set standard of testing such as end of the year testing to move on to the next grade or submitting scores for the SAT to get into college, but my goal as a BCBA is to help clients generalize the information relatively compared to their peers on a normal functioning level. References Barua, A., Kademane, K., Das, B., Gubbiyappa, K. S., Verma, R. K., & Al-Dubai, S. (2014). A tool for decision-making in norm-referenced survey questionnaires with items of ordinal variables. International Journal of Collaborative Research on Internal Medicine & Public Health, 6(3), 52-63. Retrieved from http://search.proquest.com.library.capella.edu/docview/1524878 969?accountid=27965 Embretson, S. E. (2004). The Second Century of Ability Testing: Some Predictions and Speculations. Measurement, 2(1), 170-174. Montgomery C. P. & Connolly H. B. (1987). Norm-Referenced and Criterion-Referenced Tests: Use in Pediatrics and Appliation to Task Analysis of Motor Skill. ResearchGate 67(12), 1873-1875. Virginia Tech. (1999). Measurement Scales in Social Science Research. Retrieved from http://www.simon.cs.vt.edu/SoSci/converted/Measurement/ Bottom of Form SAMPLE QUESTION:
  • 4. Exercise 1: Consider the function f (x,C)= sin(C x) Cx (a) Create a vector x with 100 elements from -3*pi to 3*pi. Write f as an inline or anonymous function and generate the vectors y1 = f(x,C1), y2 = f(x,C2) and y3 = f(x,C3), where C1 = 1, C2 = 2 and C3 = 3. Make sure you suppress the output of x and y's vectors. Plot the function f (for the three C's above), name the axis, give a title to the plot and include a legend to identify the plots. Add a grid to the plot. (b) Without using inline or anonymous functions write a function+function structure m-file that does the same job as in part (a) SAMPLE LAB WRITEUP: MAT 275 MATLAB LAB 1 NAME: __________________________ LAB DAY and TIME:______________ Instructor: _______________________ Exercise 1 (a) x = linspace(-3*pi,3*pi); % generating x vector - default value for number
  • 5. % of pts linspace is 100 f= @(x,C) sin(C*x)./(C*x) % C will be just a constant, no need for ".*" C1 = 1, C2 = 2, C3 = 3 % Using commans to separate commands y1 = f(x,C1); y2 = f(x,C2); y3 = f(x,C3); % supressing the y's plot(x,y1,'b.-', x,y2,'ro-', x,y3,'ks-') % using different markers for % black and white plots xlabel('x'), ylabel('y') % labeling the axis title('f(x,C) = sin(Cx)/(Cx)') % adding a title legend('C = 1','C = 2','C = 3') % adding a legend grid on Command window output: f = @(x,C)sin(C*x)./(C*x) C1 = 1 C2 = 2 C3 = 3 (b)
  • 6. M-file of structure function+function function ex1 x = linspace(-3*pi,3*pi); % generating x vector - default value for number % of pts linspace is 100 C1 = 1, C2 = 2, C3 = 3 % Using commans to separate commands y1 = f(x,C1); y2 = f(x,C2); y3 = f(x,C3); % function f is defined below plot(x,y1,'b.-', x,y2,'ro-', x,y3,'ks-') % using different markers for % black and white plots xlabel('x'), ylabel('y') % labeling the axis title('f(x,C) = sin(Cx)/(Cx)') % adding a title legend('C = 1','C = 2','C = 3') % adding a legend grid on end function y = f(x,C) y = sin(C*x)./(C*x); end Command window output: C1 = 1 C2 = 2 C3 = 3
  • 7. More instructions for the lab write-up: 1) You are not obligated to use the 'diary' function. It was presented only for you convenience. You should be copying and pasting your code, plots, and results into some sort of "Word" type editor that will allow you to import graphs and such. Make sure you always include the commands to generate what is been asked and include the outputs (from command window and plots), unless the problem says to suppress it. 2) Edit this document: there should be no code or MATLAB commands that do not pertain to the exercises you are presenting in your final submission. For each exercise, only the relevant code that performs the task should be included. Do not include error messages. So once you have determined either the command line instructions or the appropriate script file that will perform the task you are given for the exercise, you should only include that and the associated output. Copy/paste these into your final submission document followed by the output (including plots) that these MATLAB instructions generate. 3) All code, output and plots for an exercise are to be grouped together. Do not put them in appendix, at the end of the writeup, etc. In particular, put any mfiles you write BEFORE you first call them. Each exercise, as well as the part of the exercises, is to be clearly demarked. Do not blend them all
  • 8. together into some sort of composition style paper, complimentary to this: do NOT double space. You can have spacing that makes your lab report look nice, but do not double space sections of text as you would in a literature paper. 4) You can suppress much of the MATLAB output. If you need to create a vector, "x = 0:0.1:10" for example, for use, there is no need to include this as output in your writeup. Just make sure you include whatever result you are asked to show. Plots also do not have to be a full, or even half page. They just have to be large enough that the relevant structure can be seen. 5) Before you put down any code, plots, etc. answer whatever questions that the exercise asks first. You will follow this with the results of your work that support your answer. MATLAB sessions: Laboratory 6 MAT 275 Laboratory 6 Forced Equations and Resonance In this laboratory we take a deeper look at second-order nonhomogeneous equations. We will concentrate on equations with a periodic harmonic forcing term. This will lead to a study of the phenomenon known as resonance. The equation we consider has the form d2y
  • 9. dt2 + c dy dt + ω20y = cos ωt. (L6.1) This equation models the movement of a mass-spring system similar to the one described in Laboratory 5. The forcing term on the right-hand side of (L6.1) models a vibration, with amplitude 1 and frequency ω (in radians per second = 1 2π rotation per second = 60 2π rotations per minute, or RPM) of the plate holding the mass-spring system. All physical constants are assumed to be positive. Let ω1 = √ ω20 − c2/4. When c < 2ω0 the general solution of (L6.1) is y(t) = e− 1 2 ct(c1 cos(ω1t) + c2 sin(ω1t)) + C cos (ωt − α) (L6.2) with C = 1√
  • 10. (ω20 − ω2) 2 + c2ω2 , (L6.3) α = ( cω ω20−ω2 ) if ω0 > ω π + arctan ( cω ω20−ω2 ) if ω0 < ω (L6.4) and c1 and c2 determined by the initial conditions. The first term in (L6.2) represents the complementary solution, that is, the general solution to the homogeneous equation (independent of ω), while the second term represents a particular solution of the full ODE.
  • 11. Note that when c > 0 the first term vanishes for large t due to the decreasing exponential factor. The solution then settles into a (forced) oscillation with amplitude C given by (L6.3). The objectives of this laboratory are then to understand 1. the effect of the forcing term on the behavior of the solution for different values of ω, in particular on the amplitude of the solution. 2. the phenomena of resonance and beats in the absence of friction. The Amplitude of Forced Oscillations We assume here that ω0 = 3 and c = 1 are fixed. Initial conditions are set to 0. For each value of ω, the amplitude C can be obtained numerically by taking half the difference between the highs and the lows of the solution computed with a MATLAB ODE solver after a sufficiently large time, as follows: (note that in the M-file below we set ω = 2.4). 1 function LAB06ex1 2 omega0 = 3; c = 1; omega = 2.4; 3 param = [omega0,c,omega]; 4 t0 = 0; y0 = 0; v0 = 0; Y0 = [y0;v0]; tf = 50; 5 options = odeset(’AbsTol’,1e-10,’RelTol’,1e-10); 6 [t,Y] = ode45(@f,[t0,tf],Y0,options,param);
  • 12. 7 y = Y(:,1); v = Y(:,2); 8 figure(1) 9 plot(t,y,’b-’); ylabel(’y’); grid on; c©2016 Stefania Tracogna, SoMSS, ASU 1 MATLAB sessions: Laboratory 6 10 t1 = 25; i = find(t>t1); 11 C = (max(Y(i,1))-min(Y(i,1)))/2; 12 disp([’computed amplitude of forced oscillation = ’ num2str(C)]); 13 Ctheory = 1/sqrt((omega0^2-omega^2)^2+(c*omega)^2); 14 disp([’theoretical amplitude = ’ num2str(Ctheory)]); 15 %---------------------------------------------------------------- 16 function dYdt = f(t,Y,param) 17 y = Y(1); v = Y(2); 18 omega0 = param(1); c = param(2); omega = param(3); 19 dYdt = [ v ; cos(omega*t)-omega0^2*y-c*v ]; When executing this program we get computed amplitude of forced oscillation = 0.24801
  • 13. theoretical amplitude = 0.24801 Lines 10-14 deserve some explanation. Line 10 defines a time t1 after which we think the contribution of the first term in (L6.2) has become negligible compared to the second term. This depends of course on the parameter values, in particular c. With c = 1 we obtain e− 1 2 ct ' 3.7 × 10−6 for t = 25, so this is certainly small enough compared to the amplitude seen on Figure L6a. The index i of time values larger than t1 is then determined. The quantity Y(i,1) refers to the values of y associated to times larger than t1 only. The computed amplitude is simply half the difference between the max and the min values. This value is compared to the theoretical value (L6.3). 0 5 10 15 20 25 30 35 40 45 50 -0.3 -0.2 -0.1 0 0.1 0.2 0.3
  • 14. y Figure L6a: Forced oscillation. 1. (a) What is the period of the forced oscillation? What is the numerical value (modulo 2π) of the angle α defined by (L6.4)? (b) In this question you are asked to modify the file LAB06ex1.m in order to plot the complemen- tary solution of (L6.1), that is, the first term in (L6.2). First define in the file the angle α (alpha) using (L6.4), then evaluate the complementary solution yc by subtracting the quan- tity C cos(ωt − α) from the numerical solution y. Plot the resulting quantity. Does it look like an exponentially decreasing oscillation? Why or why not? Include the modified M-file and the corresponding plot. 2. We now consider C as a function of ω. We use again ω0 = 3, c = 1 and y(0) = y ′(0) = 0. The previous problem determined C for a specific value of ω. Here we consider a range of values for ω c©2016 Stefania Tracogna, SoMSS, ASU 2 MATLAB sessions: Laboratory 6 and determine numerically the corresponding amplitude C. We then plot the result as a function
  • 15. of ω, together with the theoretical amplitude from (L6.3). You may need the following MATLAB program. function LAB06ex2 omega0 = 3; c = 1; OMEGA = 2:0.02:4; C = zeros(size(OMEGA)); Ctheory = zeros(size(OMEGA)); t0 = 0; y0 = 0; v0 = 0; Y0 = [y0;v0]; tf = 50; t1 = 25; for k = 1:length(OMEGA) omega = OMEGA(k); param = [omega0,c,omega]; [t,Y] = ode45(@f,[t0,tf],Y0,[],param); i = find(t>t1); C(k) = (max(Y(i,1))-min(Y(i,1)))/2; Ctheory(k) = ??; % FILL-IN end figure(2) plot(??); grid on; % FILL-IN
  • 16. xlabel(’omega’); ylabel(’C’); %--------------------------------------------------------- function dYdt = f(t,Y,param) y = Y(1); v = Y(2); omega0 = param(1); c = param(2); omega = param(3); dYdt = [ v ; cos(omega*t)-omega0^2*y-c*v ]; 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 0.1 0.15 0.2 0.25 0.3 0.35 C Figure L6b: Amplitude as a function of ω (a) Fill in the missing parts in the M-file LAB06ex2.m and execute it. You should get a figure like Figure L6b. Include the modified M-file in your lab report. (b) Examine the graph obtained by running LAB06ex2.m and determine for what (approximate)
  • 17. value of ω the amplitude of the forced oscillation, C, is maximal. This value of ω is called the practical resonance frequency. Give the corresponding maximum value of C. c©2016 Stefania Tracogna, SoMSS, ASU 3 MATLAB sessions: Laboratory 6 (c) Determine analytically the value of ω for which the amplitude of the forced oscillation, C, is maximal by differentiating the expression for C in (L6.3) as a function of ω. Compare the value you find with the value obtained in part (b). (d) Run LAB06ex1.m with the value of ω found in part (c) (include the graph). What is the amplitude of the forced oscillation? How does it compare with the amplitude of the forced oscillation in problem 1.? If you run LAB06ex1.m with any other value of ω, how do you expect the amplitude of the solution to be? (e) Are the results affected by changes in the initial conditions? Answer this question both nu- merically (by modifying the initial conditions in LAB06ex2.m) and theoretically (by analyzing the expression for C in (L6.3)). Note that the initial conditions for the DE are y0 and v0. Resonance We now investigate what happens to the solution (L6.2), and more specifically to the maximal amplitude
  • 18. C of the forced oscillation, when we let c → 0. The value of ω corresponding to this maximal amplitude is called pure resonance frequency. When a mechanical system is stimulated by an external force operating at this frequency the system is said to be resonant. 3. Set c = 0 in LAB06ex2.m. 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 0 1 2 3 4 5 6 7 8 9 C computed numerically theoretical (a) Explain what happens. What is the maximal amplitude?
  • 19. What is the value of ω yielding the maximal amplitude in the forced solution? How does this value compare to ω0? (b) Run LAB06ex1.m with c = 0 and ω equal to the value found in part (a). Comment on the behavior of the solution. Include the graph. Beats When c = 0 and ω 6= ω0 , the solution (L6.2) to (L6.1) reduces to y(t) = c1 cos(ω0t) + c2 sin(ω0t) + C cos(ωt − α) with C = 1∣ ∣ ω20−ω2∣ ∣ . If the initial conditions are set to zero, the solution reduces to y(t) = C(cos(ωt) − cos(ω0t)) which can be rewritten as y(t) = 2C sin ( 1 2 (ω0 − ω)t ) sin ( 1 2
  • 20. (ω0 + ω)t ) . c©2016 Stefania Tracogna, SoMSS, ASU 4 MATLAB sessions: Laboratory 6 When ω is close to ω0 we have that ω + ω0 is large in comparison to |ω0 − ω|. Then sin ( 1 2 (ω0 + ω)t ) is a very rapidly varying function, whereas sin ( 1 2 (ω0 − ω)t ) is a slowly varying function. If we define A(t) = 2C sin ( 1 2 (ω0 − ω)t )
  • 21. , then the solution can be written as y(t) = A(t) sin ( 1 2 (ω0 + ω)t ) and we may interpret it as a rapidly oscillating function with period T = 4π ω0+ω , but with a slowly varying amplitude A(t). This is the phenomenon known as beats. Note that A(t) and −A(t) are the so-called “envelope functions”. The period of A(t) is 4π|ω0−ω|, thus the length of the beats is 2π |ω0−ω| . 4. To see the beats phenomenon, set c = 0 and ω = 2.8 in LAB06ex1. Also extend the interval of simulation to 100. 0 10 20 30 40 50 60 70 80 90 100 -2 -1.5
  • 22. -1 -0.5 0 0.5 1 1.5 2 y (a) In LAB06ex1 define the “envelope” function A = 2C sin ( 1 2 (ω0 − ω)t ) with C = 1|ω20−ω2| . Plot A in red and −A in green, together with the solution. You should obtain Figure L6c. Include the modified M-file. 0 10 20 30 40 50 60 70 80 90 100 -2 -1.5
  • 23. -1 -0.5 0 0.5 1 1.5 2 y Figure L6c: Solution and envelope functions c©2016 Stefania Tracogna, SoMSS, ASU 5 MATLAB sessions: Laboratory 6 (b) What is the period of the fast oscillation (that is, the period
  • 24. of sin ( 1 2 (ω0 + ω)t ) )? Confirm your answer by zooming in on the graph of the solution. Include a graph to support your answer. (c) What is the length of the beats? Determine the length analytically using the envelope func- tions, and numerically from the graph. (d) Change the value of ω in LAB06ex1 to 2.9 (a value closer to ω0) and then ω = 2.6 ( a value farther away from ω0). Include the two graphs. For each of these two values of ω find the period of the fast oscillation and the length of the beats. How do the periods change compared to parts (b) and (c)?
  • 25. (e) If you let ω = 1.5, is the beats phenomenon still present? Why or why not? c©2016 Stefania Tracogna, SoMSS, ASU 6 Exercise 1: function LAB06ex1 clc omega0 = 3; c = 1; omega = 2.4; param = [omega0,c,omega]; t0 = 0; y0 = 0; v0 = 0; Y0 = [y0;v0]; tf = 50; options = odeset('AbsTol',1e-10,'relTol',1e-10); [t,Y] = ode45(@f,[t0,tf],Y0,options,param); y = Y(:,1); v = Y(:,2); figure(1) plot(t,y,'b-'); ylabel('y'); grid on; t1 = 25; i = find(t>t1); C = (max(Y(i,1))-min(Y(i,1)))/2; disp(['computed amplitude of forced oscillation = ' num2str(C)]); Ctheory = 1/sqrt((omega0^2-omega^2)^2+(c*omega)^2); disp(['theoretical amplitude = ' num2str(Ctheory)]); %----------------------------------------------------------------
  • 26. function dYdt = f(t,Y,param) y = Y(1); v = Y(2); omega0 = param(1); c = param(2); omega = param(3); dYdt = [ v ; cos(omega*t)-omega0^2*y-c*v ]; Exercise 2: function LAB06ex2 omega0 = 3; c = 1; OMEGA = 2:0.02:4; C = zeros(size(OMEGA)); Ctheory = zeros(size(OMEGA)); t0 = 0; y0 = 0; v0 = 0; Y0 = [y0;v0]; tf = 50; t1 = 25; for k = 1:length(OMEGA) omega = OMEGA(k); param = [omega0,c,omega]; [t,Y] = ode45(@f,[t0,tf],Y0,[],param); i = find(t>t1); C(k) = (max(Y(i,1))-min(Y(i,1)))/2; Ctheory(k) = ??; % FILL-IN end figure(2) plot(??); grid on; % FILL-IN
  • 27. xlabel('omega'); ylabel('C'); %--------------------------------------------------------- function dYdt = f(t,Y,param) y = Y(1); v = Y(2); omega0 = param(1); c = param(2); omega = param(3); dYdt = [ v ; cos(omega*t)-omega0^2*y-c*v ];