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Dynamics and Control Problem 1 : Matrix generation i) Create the following matrixes 1  A  2

3

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7

8 

5 & 6 9

1 B  0

3



3

2

2

1

1 1

ii) Create C : a 3×3 identity matrix. Use the most convenient method.

Problem 2 :

Basic matrix operations

Use the matrixes you made in Problem 1.1. i) Compare the results of A* B and A.* B . Explain what MATLAB did. ii) Does A* B  B * A ? Explain. iii)Does A.* B  B.* A ? Explain. iv)Create matrix D  B 1 . What is B * D ? v) Compare A / B and A \ B . Can you explain what is going on? vi) Create a new 3×3 matrix E with all diagonal elements equal to 5 and other elements equal to 0.

Problem 3 :

Understanding data types

Use the matrixes you made in Problem 1.1. vii)What are the data types of the elements of matrixes A , B , and C ?

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iii)Convert A and B to 8 bit signed integer matrixes and define them as matrix F and G respectively (check out function int8( ) in the help menu). iv)Calculate F  G and F *G .

Explain results.

Problem 4 : Submatrix extraction and plotting You are given the trajectory of a ball in a MATLAB file: ball.mat. (Please, download it from MIT Server site.)

The first column contains 100 time points and the second column contains the corresponding 100 positions of the ball. The third and the fourth column contain a second trial of the same experiment. v) Extract the time data from column 1 and assign that to a new matrix “t”. vi)Extract the position data from column 2 and assign that to a new matrix “x”. vii)Plot “x” vs. “t”. viii)Extract only the first 50 time points from column 3 and make a new matrix “t2”. ix)Extract only the first 50 positions from column 4 and make a new matrix “x2”. x) Plot “x2” vs. “t2”. xi) Superimpose both trajectories. Plot them in different color.

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Solutions Problem 1 : Matrix generation i) You can make one of following methods listed below: -

A = [ [1 2 3]' [4 5 6]' [7 8 9]'];

-

A(1,:) = [1 4 7]; A(2,:) = [2 5 8]; A(3,:)=[3 6 9];

-

A(:,1) = [1 2 3]; A(:,2) = [4 5 6]; A(:,3)=[7 8 9];

- A(1,1) = 1; A(2,2) = 5; A(3,3) = 9; and so on…

A(2,1) = 2; A(3,2) = 6;

A(3,1) = 3; A(1,2) = A(1,3) = 7; A(2,3) =

4; 8;

B = [1 2 3; 0 3 1; 2 1 1]; or equivalent ways shown in creating matrix A . >> A = [1 4 7; A = 1 4 2 5 3 6 >> B = [1 2 3; B = 1 2 0 3 2 1

2 5

8; 3 6 9]

7 8 9 0 3

1; 2 1 1]

3 1 1

ii) The most convenient method is to use eye() function (Please, see eye() in help menu.) >> C = eye(3)

% eye(N) generate identity matrix with N x N

C = matlabassignmentexperts.com

1 0 0

0 1 0

0 0 1

Problem 2 : Basic matrix operations i) A*B gives matrix multiplication operation of matrices A and B (inner product between corresponding row of matrix A and corresponding column of matrix B ), whereas A.*B provides element by element multiplication only between corresponding elements of matrices A and B . >> A*B % Matrix multiplication ans = 15 21 14 18 27 19 21 33 24 >> A.*B % Element by element multiplication ans = 1 8 21 0 15 8 6 6 9 ii) In matrix multiplication, commutativity does not generally hold. >> A*B ans = 15 21 14 18 27 19 21 33 24 >> B*A % Result is not same as A*B ans = 14 32 50 9 21 33 7 19 31 iii) Following two operations are same since .* holds commutativity. >> A.*B ans =

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1 0 6 >> B.*A ans = 1 0 6

8 15 6 8 15 6

21 8 9 % Result is same as A.*B 21 8 9

iv) B*D is a 3×3 identity matrix, since B * D  B * B 1  I . >> D=B^-1 % or D=inv(B) D = -0.1667 -0.0833 0.5833 -0.1667 0.4167 0.0833 0.5000 -0.2500 -0.2500 >> B*D % Generates 3x3 identity matrix ans = 0 0 1 1 1 0 1 0 1 v) A/B is roughly same to A*inv(B) (solution to equation XB  A ) if B is a square matrix, and A\B is roughly same to inv(A)*B (solution to equation AX  B ) if A is a square matrix. (For more information see mldivide() and mrdivide() in help menu.) Since matrix A is singular, inverse of A does not exists. (determinant of A is equal to 0. Check it with det(A). ) Therefore, MATLAB gives the below warning to us for A\B. >> A/B % equal to A*inv(B) ans = 2.6667 -0.1667 -0.8333 2.8333 -0.0833 -0.4167 3.0000 0 0 >> A\B % equal to inv(A)*B Warning: Matrix is singular to working precision. matlabassignmentexperts.com

ans = NaN NaN NaN Inf -Inf Inf -Inf Inf -Inf

i, j  1, 2, 3

where aij is i-th row and j-th column element of matrix. (diagonal matrix with all 5’s) >> E=5*eye(3) % or E=diag([5 5 5]), and so on 0 0 E =5 0 5 0 0 0 5 vi) aij  5 ij

vii)

Understanding data types

i) By default, MATLAB assigns variables to ‘double’ data type variables. Therefore, A , B , and C are matrices whose elements are all ‘double’ data types. You can check it with ‘whos’ command in MATLAB. >> whos Name A

Size 3x3

Bytes Class Attributes 72 double B 3x3 72 double ii)‘double’ data type variable requires 8 bytes to store a variable value, and each element in matrix A is 8 C 3x3 72 byte long. Since matrix A has 3×3=9 elements (or variables), matrix A needs 72 bytes. Therefore, 576 bits are double required to store the information in matrix A, since 1 byte consists of 8 bits. iii)New matrices F and G are displayed same as matrices A and B respectively, but they are stored in the computer with different data format. >> F=int8(A) F = 1

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>> G=int8(B)

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G =

1 2 3 0 3 1 2 1 1 iv) Addition between only same integer data types is supported in MATLAB, but matrix multiplication between any integer data types is not. Therefore, F+G produces matrix addition and F*G generates error as below. (See ‘Arithmatic operators’ in help menu.) >> F+G % Only allowed if both are same integer types ans = 2 6 10 2 8 9 5 7 10 >> F*G % Never allowed if at least one is integer data type r ??? using ==> mtimes da a types are not fully supported for this operation. e Erro Integer t operand must be a scalar. on At least viii) Submatrix extraction and plotting i) First, saved data should be retrieved with load(). ‘:’ can be used to select either whole row or whole column in the matrix. ii) >> load ball.mat % load data from ball.mat >> t=A(:,1); % Extract 1st column data, and assign to ‘t’ iii)

>> x=A(:,2); % Extract 2nd column data, and assign to ‘x’ >> plot(t,x);

% Plot t vs. x matlabassignmentexperts.com

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iv) 1:50 means selection from first element to 50th elements v)

vi)

>> t2=A(1:50,3); %Extract first 50 elements from 3rd column of A >> x2=A(1:50,4); %Extract first 50 elements from 4th column of A >> plot(t2,x2);

% Plot t2 vs. x2 7

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vii) You can enumerate appropriate arguments for each graph (x data, y data, and linespec) sequentially in 0

the plot() function. ‘r’ means red color and ‘b’ indicates blue color. (See ‘linespec’ in help menu) Alternatively, you can also overlap several plots with ‘hold on’ and ‘hold off’ (See ‘hold’ in help menu) matlabassignmentexperts.com

>> plot(t,x,'r',t2,x2,'b'); or >>plot(t,x,'r'); hold on; plot(t2,x2,'b'); hold off;

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