MAT 343 ASU Plotting and Computer Animation In MATLAB Project

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MAT 343 Laboratory 4
Plotting and computer animation in MATLAB
In this laboratory session we will learn:
1. How to plot in MATLAB
2. The geometric properties of special types of matrices (rotations, dilations, reflections)
3. How to perform simple computer animations in MATLAB
Plotting in MATLAB
If x and y are vectors of the same size, the command plot(x,y) will produce a plot of all the (xi , yi )
pairs, and each point will be connected to the next by a line segment. If the x-coordinates are taken
close enough together, the graph should resemble a smooth curve. The command plot(x,y,’o’)
will mark the ordered pairs with o’s, but will not connect the points.
ex/10 sin x
For example, to plot the function f (x) =
on the interval [0, 10], set
x+1
x = 0:0.2:10;
y = exp(x/10).*sin(x)./(x+1);
(note the . before the * and /, as these operations must be done component-wise).
The command plot(x,y) will generate the graph of the function. To compare the graph with that
of sin x, we could set z = sin(x); and use the command
plot(x,y,x,z)
to plot both curves at the same time.
It is also possible to do more sophisticated types of plots in MATLAB, including polar coordinates,
contour plots, and three-dimensional surfaces.
Given the points (xi , yi ), rather than creating the vectors x and y, an equivalent approach, which
often is more convenient, is to create a 2 × n matrix of data:


x1 x2 . . . xn
A=
y1 y2 . . . yn
then typing
plot(A(1,:),A(2,:))
In other words, the first row of A gives the x’s and the second row the y’s.
EXAMPLE 1
Consider the square whose vertices are (0, 0), (1, 0), (1, 1), (0, 1).
(a) Enter the x and y coordinates of these points as MATLAB vectors by setting
x = [0, 1, 1, 0, 0]
y = [0, 0, 1, 1, 0]
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The coordinates of the first points are repeated in the fourth column of x and y because we
want to connect the fourth point back to the first. The command plot(x,y) draws this square
in the graph window. The command axis([-2,2,-2,2]) will rescale the axes so that the
square does not take up the entire figure.
(b) Now let’s store the entire figure in a matrix S by storing the x-coordinate in the first row and
the y-coordinate in the second row:
S = [0, 1, 1, 0, 0; 0, 0, 1, 1, 0]
(if x and y have already been entered we can simply type S = [x; y]).
The square can then be plotted entering
plot(S(1,:),S(2,:))
Again, set axis([-2,2,-2,2]), and add a grid by entering grid on.
Rotation Matrices in R2
Any 2-dimensional matrix of the form

Q=
cos θ − sin θ
sin θ
cos θ

is a rotation matrix. If v is a vector in R2 , the product Qv is the vector v rotated θ radians in the
counterclockwise direction around the origin (0, 0).
To rotate the vector in the clockwise direction we simply replace θ with −θ, thus


cos(−θ) − sin(−θ)
−1
.
Q =
sin(−θ)
cos(−θ)
Since cos(−θ) = cos θ and sin(−θ) = − sin(θ), we have


cos θ sin θ
−1
= QT .
Q =
− sin θ cos θ
EXAMPLE 2
We can rotate the square S from Example 1 by an angle of π/4 in the counterclockwise direction by
multiplying the matrix S by


cos(π/4) − sin(π/4)
Q=
sin(π/4) cos(π/4)
The following code implements the rotation and the resulting picture is displayed in Figure 1.
clear all ;
% clear all variables
clf ;
% clear all settings for the plot
S =[0 ,1 ,1 ,0 ,0;0 ,0 ,1 ,1 ,0];
plot ( S (1 ,:) , S (2 ,:) , ‘ linewidth ‘ ,2); % plot the square
hold on ;
theta = pi /4; % define the angle theta
Q =[ cos ( theta ) , – sin ( theta ); sin ( theta ) , cos ( theta )]; % rotation matrix
QS = Q * S ; % rotate the square
plot ( QS (1 ,:) , QS (2 ,:) , ‘ -r ‘ , ‘ linewidth ‘ ,2);
% plot the rotated square
title ( ‘ Example of Rotation ‘ ); % add a title
legend ( ‘ original square ‘ , ‘ rotated square ‘) % add a legend
axis equal ; axis ([ -1 ,2 , -1 ,2]); % set the window
grid on ; % add a grid
hold off
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Figure 1: Original square and rotated square
Remarks:
• The hold on command concatenates all subsequent plot commands on the same graph. Type
hold off to start a new plot.
• The ’-r’ plots the figure in red.
• grid on adds a grid to the axis
• axis equal sets the same scale on the x and y axes.
• ’linewidth’, 2 sets the width of the line to 2 points = 2/72 inch (the default is 0.5 points).
Dilation and Contraction
Consider the matrix


αx 0
D=
.
0 αy

T

T
If we multiply this matrix by any vector v = x y we obtain the vector αx x αy y which scales
the vector by a factor αx in the x-direction and by a factor αy in the y-direction. The scalings are a
dilation or contraction, depending on whether αx and αy are greater or smaller than one.
EXAMPLE 3
Consider the case where the dilation factors are αx = 2 and αy = 3. We can plot the original square
S together with the dilated version by executing the following commands. The output is displayed
in Figure 2.
clf
S =[0 ,1 ,1 ,0 ,0;0 ,0 ,1 ,1 ,0];
plot ( S (1 ,:) , S (2 ,:) , ‘ linewidth ‘ ,2)
hold on
D = [2 , 0; 0 , 3]; % dilation matrix
DS = D * S ;
plot ( DS (1 ,:) , DS (2 ,:) , ‘ -r ‘ , ‘ linewidth ‘ ,2)
title ( ‘ Example of Dilation ‘)
legend ( ‘ original square ‘ , ‘ dilated square ‘ , ‘ location ‘ , ‘ southeast ‘)
grid on
axis equal , axis ([ -1 ,4 , -1 ,4]) % adjust the axis and the window
hold off
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Figure 2: Original square and dilated square
Shear Matrices


1 tx
A matrix of the form T =
is a horizontal shear transformation. When we apply this
0 1
transformation, the y coordinates are unaffected, but the x coordinates are translated linearly with
y.


1 0
is a vertical shear transformation: the x coordinates are
Similarly, a matrix of the form T =
ty 1
unaffected, but the y coordinates are translated linearly with x.
EXAMPLE 4
We will
 a horizontal shear transformation of 2 units. The matrix that accomplishes that is
 apply
1 2
. We can plot the original square together with the sheared version by executing the
T =
0 1
following commands. The output is displayed in Figure 3.
clf
S =[0 ,1 ,1 ,0 ,0;0 ,0 ,1 ,1 ,0];
plot ( S (1 ,:) , S (2 ,:) , ‘ linewidth ‘ ,2)
hold on
T =[1 ,2;0 ,1];
% shear transformation matrix
TS = T * S ;
plot ( TS (1 ,:) , TS (2 ,:) , ‘ -r ‘ , ‘ linewidth ‘ ,2);
title ( ‘ Example of horizontal shear ‘)
legend ( ‘ original square ‘ , ‘ sheared square ‘ , ‘ location ‘ , ‘ southeast ‘)
axis equal , axis ([ -1 ,4 , -1 ,4]); grid on
% adjust the axis and the window
hold off
Composition of Transformations
One can perform a composition of two (or more) transformations by multiplying by the corresponding
matrices in the appropriate order.
EXAMPLE 5
We can see the effect of first rotating the square 45 degrees counterclockwise and then applying the
shear from EXAMPLE 3, by plotting the product T*Q*S . Note the order of the matrices; here we
first apply the rotation (Q is the matrix that multiplies S first) and then we apply the shear. The
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Figure 3: Original square and sheared square
output of the following code is displayed in Figure 4.
clf
S =[0 ,1 ,1 ,0 ,0;0 ,0 ,1 ,1 ,0];
plot ( S (1 ,:) , S (2 ,:) , ‘ linewidth ‘ ,2)
hold on
theta = pi /4; % define the angle
Q =[ cos ( theta ) , – sin ( theta ); sin ( theta ) , cos ( theta )];
T =[1 ,2;0 ,1];
% shear transformation matrix
TQS = T * Q * S ;
plot ( TQS (1 ,:) , TQS (2 ,:) , ‘ -r ‘ , ‘ linewidth ‘ ,2);
title ( ‘ Example of rotation and shear ‘)
legend ( ‘ original square ‘ , ‘ modified square ‘ , ‘ location ‘ , ‘ southeast ‘)
axis equal , axis ([ -1 ,4 , -1 ,4]); grid on
% adjust the axis and the window
hold off
Figure 4: Original square and square first rotated and then sheared
Animation
EXAMPLE 6
The goal of this example is to draw the square in its original size, then cause it to disappear, and
finally to redraw it as it looks after it is dilated by a factor of 9/8. If this operation is repeated ten
times in succession, the square will appear to be expanding.
To do this we need to first plot the square and store its handle in p:
p = plot(S(1,:),S(2,:))
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After we transform the matrix by multiplying by the appropriate transformation matrix, we can
erase the original figure and draw the new figure with the command
set(p,’xdata’,S(1,:),’ydata’,S(2,:));
Since we are repeating the same operation ten times, we will use a for loop. After we have gone
through this loop, we create another one that contracts the matrix by a factor of 8/9, thereby
returning it to its original size.
The following are the commands that produce the effects. Enter them in MATLAB (you need not
enter the comments that follow the % signs). The pause(0.1) command is used to adjust the speed
of the plots so that we can see the graphs before they are erased. You might need to change the 0.1
to suit the speed of your machine.
clf
% clear all settings for the plot
S =[0 ,1 ,1 ,0 ,0;0 ,0 ,1 ,1 ,0];
D1 = 9/8* eye (2);
% dilation matrix
p = plot ( S (1 ,:) , S (2 ,:));
% plot the square
axis ([ -1 ,4 , -1 ,4])
% set size of the graph
axis square , grid on
% make the display square
hold on
% hold the current graph
for i = 1:10
S = D1 * S ;
% dilate the square
set (p , ‘ xdata ‘ ,S (1 ,:) , ‘ ydata ‘ ,S (2 ,:));
% erase original figure and plot
% the transformed figure
pause (0.1)
% adjust this pause rate to suit your computer .
end
D2 = 8/9* eye (2);
% contraction matrix
for i = 1:10
S = D2 * S ;
% contract the square
set (p , ‘ xdata ‘ ,S (1 ,:) , ‘ ydata ‘ ,S (2 ,:));
% erase original figure and plot
% the transformed figure
pause (0.1)
% adjust this pause rate to suit your computer .
end
hold off
EXERCISES
Instructions: For each of the following exercises, create an M-file to store the MATLAB commands.
Copy and paste the M-file into a text document. For problems 1 and 2, include in the text document
the pictures produced by MATLAB. Resize and crop the pictures so that they do not take up too
much space. If the question requires written answers, include them in the text file in the appropriate
location. Make sure you clearly label and separate all the Exercises. For problems 3 and 4 you do
not need to include the picture.
1. Determine the shearing transformation matrix that shears 1 units in the vertical direction. Plot
the original square together with the sheared square. Use axis([-1,3,-1,3]);. Add a grid,
a legend and a title (similarly to EXAMPLE 4). Include the M-file as well as the figure.
2. Consider the original square S. First apply the shear transformation from EXAMPLE 4 and
then rotate the square 45◦ counterclockwise. Plot the resulting figure (together with the original
square) and compare with the plot in Example 5. Are the results the same? Does the order of
the transformations matter?
Include the M-file as well as the figure and don’t forget to answer all questions.
Hint: Here you want to apply the shear first, then rotate, whereas in Example 5 the square is
first rotated and then the shear is applied. To do this, TQS = T*Q*S, where Q is the rotation
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matrix and T is the shearing transformation matrix. So, if you want to first shear and then
rotate, in what order will you multiply the matrices?
3. Adapt the procedure developed in Example 6 to rotate the square counterclockwise by increments of π/7 about the origin. Stop when the square is in its original location and then rotate
it in the clockwise direction until the square is in its original location again. You may want to
rescale the axis by using axis([-2,2,-2,2]).
Include the M-file. Do not include the figure.
Hint: Since you are performing a computation several times, you will want to use two for
loops: one loop for the counterclockwise rotation and another one for the clockwise rotation.
Think about how many times you will need to go through the loops, keeping in mind that you
are rotating the square counterclockwise for a full circle by increments of π/7 and then rotating
the square clockwise back again.
4. Adapt the procedure developed in Example 6 to show the square rotating in a counterclockwise
direction about the origin by increments of π/7 for a total angle of 2π and expanding at the
same time by a factor of 8/7, then stopping and rotating in the clockwise direction as it shrinks
to its original size (with a contraction factor of 7/8). At the end of the program, the figure
should have returned to its original size and original location.
You may want to rescale the axis to axis([-7,7,-7,7]);
Include the M-file. Do not include the figure.
Hint: Similarly to Exercise 3, you might want to use two for loops: one loop for the counterclockwise rotation + dilation, and another one for the clockwise rotation + contraction. Think
about how many times you will need to go through the loops.
Homogeneous coordinates
Translations are not linear transformations and consequently cannot be accomplished via matrix
multiplication using a 2 × 2 matrix. To circumvent this problem we use homogeneous coordinates.
The homogeneous coordinates for a point (x1 , x2 )T are (x1 , x2 , 1)T .
However, when the point represented by the homogeneous coordinates (x1 , x2 , 1)T is plotted, only
the x1 and x2 are taken into consideration. The 1 is ignored. This allows us to produce the
translated coordinates for a figure by matrix multiplication. Suppose we want to translate the vector
c1 units horizontally and c2 units vertically. In homogeneous coordinates the translated vector is
(x1 + c1 , x2 + c2 , 1)T . We need a matrix M such that
M (x1 , x2 , 1)T = (x1 + c1 , x2 + c2 , 1)T
It is easy to verify that the matrix


1 0 c1
M =  0 1 c2 
0 0 1
has the desired property.
EXAMPLE 7
We will apply a horizontal translation
of 2units and a vertical translation of -3 units. The matrix

1 0 2
that accomplishes that is M = 0 1 −3. To apply the transformation to the square S, we will
0 0 1
need to convert the square to homogeneous coordinates. This is done by adding a row of ones to the
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
0

matrix S so that the new matrix is S = 0
1
with the translated version by executing the
5.
1 1 0
0 1 1
1 1 1
following

0
0. We can plot the original square together
1
commands. The output is displayed in Figure
clf
S =[0 ,1 ,1 ,0 ,0;0 ,0 ,1 ,1 ,0;1 ,1 ,1 ,1 ,1];
% square in homogeneous coordinates
M =[1 ,0 ,2;0 ,1 , -3;0 ,0 ,1];
% translation matrix
MS = M * S ;
% apply the translation to the square
plot ( S (1 ,:) , S (2 ,:) , ‘k ‘ , ‘ linewidth ‘ ,2); % plot the original square in black
hold on
plot ( MS (1 ,:) , MS (2 ,:) , ‘r ‘ , ‘ linewidth ‘ ,2); % plot the translated square in red
legend ( ‘ original square ‘ , ‘ translated square ‘ , ‘ location ‘ , ‘ southwest ‘ );
axis equal , axis ([ -1 ,4 , -4 ,4]) , grid on
% adjust the axis
hold off
Figure 5: Original square and translated square
EXAMPLE 8
Consider the matrix S representing the original square in homogeneous coordinates. In the following
M-file we translate the square horizontally using c1 = 0.2 and c2 = 0 for 40 times. We then translate
the square vertically using c1 = 0 and c2 = 0.2 and 40 iterations.
clf
S =[0 ,1 ,1 ,0 ,0;0 ,0 ,1 ,1 ,0;1 ,1 ,1 ,1 ,1]; % square in homogeneous coordinates
M1 = [1 ,0 ,0.2;0 ,1 ,0;0 ,0 ,1]; % first translation matrix
M2 = [1 ,0 ,0;0 ,1 ,0.2;0 ,0 ,1]; % the second translation matrix
p = plot ( S (1 ,:) , S (2 ,:)); % plot the original square
axis square , axis ([ -1 ,10 , -1 ,10]) , grid on
for i = 1:40
S = M1 * S ; % compute the translated square
set (p , ‘ xdata ‘ ,S (1 ,:) , ‘ ydata ‘ ,S (2 ,:)); % plot the translated square
pause (0.1)
end
for i = 1:40
S = M2 * S ; % compute the translated square
set (p , ‘ xdata ‘ ,S (1 ,:) , ‘ ydata ‘ ,S (2 ,:)); % plot the translated square
pause (0.1)
end
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Reflection Matrices
The reflection matrix about the line L with direction the unit vector u is given by R = 2uuT − I,
where I is the identity matrix.
 √ √ T
For instance, to obtain the reflection matrix about the line at 45 degrees we can take u = 22 , 22 .




1/2 1/2
0 1
T
Then uu =
and R =
. (Note that this answer makes perfect sense, since the
1/2 1/2
1 0
reflection of the vector x = (x1 , x2 )T about the line at 45 degrees is the vector (x2 , x1 )T ). In
homogeneous coordinates, the reflection matrix becomes:


0 1 0
R = 1 0 0
0 0 1
The other types of basic transformations (rotations, dilations) can all be adapted to homogeneous
coordinates by augmenting the matrix by the row (0, 0, 1) and the column (0, 0, 1)T . For instance,
in homogeneous coordinates, the dilation matrix that dilates by a factor α becomes


α 0 0
D =  0 α 0 .
0 0 1
EXERCISES
5. Consider the translated square MS in EXAMPLE 7. Apply to this square the reflection matrix that reflects about the line at 45 degrees. Plot the translated square in black and the
reflected square in red. Use axis([-4,4,-4,4]); and plot the line y = x using the command plot([-4,4],[-4,4]); (this command plots the line connecting the points (−4, −4)
and (4, 4)). Add a legend and a grid. You should reproduce Figure 6. Include the M-file and
the figure.
Figure 6: Output of Exercise 5
6. Modify the M-file in EXAMPLE 8 adding translations that bring the square to its original
position using 40 iterations and a single additional for loop.
Include the M-file (for a total of three loops) You do not need to include the figure.
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Rotations about a point different than the origin


cos θ − sin θ
performs a rotation about the origin, but what if we want
sin θ cos θ
to rotate about a point P = (Px , Py ) different than the origin? To accomplish this we need to
The rotation matrix Q =
• translate the object so that P will coincide with the origin, i.e. translate −Px units horizontally
and −Py units vertically
• rotate the object
• translate the object back, i.e. translate Px units horizonally and Py units vertically
To compose the three transformations above, we multiply the



1 0 Px
cos θ − sin θ 0 1
0 1 Py   sin θ cos θ 0 0
0 0 1
0
0
1 0
appropriate matrices:

0 −Px
1 −Py 
0
1
Note that the rotation matrix is written in homogeneous coordinates.
EXAMPLE 9
In this example we first translate the square horizontally 8 units using increments of 0.2. We then
rotate the resulting square clockwise π/2 radians around the vertex (9, 0) using increments of π/8
radians.
clf
S =[0 ,1 ,1 ,0 ,0;0 ,0 ,1 ,1 ,0;1 ,1 ,1 ,1 ,1]; % square in homogeneous coordinates
M1 = [1 ,0 ,0.2;0 ,1 ,0;0 ,0 ,1]; % first translation matrix
theta = pi /8; % define the angle theta
Q =[ cos ( theta ) , – sin ( theta ) ,0; sin ( theta ) , cos ( theta ) ,0;0 ,0 ,1]; % rotation matrix about (0 ,0)
QP =[1 ,0 ,9;0 ,1 ,0;0 ,0 ,1]* Q ‘*[1 ,0 , -9;0 ,1 ,0;0 ,0 ,1];
% rotation matrix about (9 ,0)
p = plot ( S (1 ,:) , S (2 ,:)); % plot the original square
axis equal , axis ([ -0.5 ,11 , -2 ,5]) , grid on
for i = 1:40
S = M1 * S ; % compute the translated square
set (p , ‘ xdata ‘ ,S (1 ,:) , ‘ ydata ‘ ,S (2 ,:)); % plot the translated square
pause (0.1)
end
for i = 1:4
S = QP * S ; % compute the rotated square
set (p , ‘ xdata ‘ ,S (1 ,:) , ‘ ydata ‘ ,S (2 ,:)); % plot the rotated square
pause (0.1)
end
EXERCISES
7. Consider the square in EXAMPLE 9. The goal of this exercise is to bring back the square to its
original position by first translating it horizontally to the left 8 units using 40 iterations, and
then rotating it counterclockwise π/2 radians around the point (1, 0) using 4 iterations. This
can be done by modifying the code in EXAMPLE 9 by adding two for loops. The first loop
should translate the square while the second should rotate it around the point (1, 0). Note that
the rotation is counterclockwise, while in EXAMPLE 9 it was clockwise. Include the M-file.
You do not need to include the figure.
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