Introduction to MATLAB
Introductory Problems
1. Create a vector that contains the numbers 1, 2, 3, 4, 5:
vector = [ 1 2 3 4 5 ];
vector = [ 1:5 ];
2. Clear the Workspace and Command windows:
clear
clc
3. Create a matrix with the same contents as the following table:
4 5 6
10 50 3
15 56 20
Matrix = [ 4 5 6; 10 50 3; 15 56 20 ];
Alternatively:
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V1 = [4 5 6];
V2 = [10 50 3];
V3 = [15 56 20];
Matrix1 = [V1; V2; V3];
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% The semicolon between V1, V2 and V3 specify that we want to insert V2 ...
and V3 into new rows of the matrix. Moreover, Matrix = [V1, V2, V3] ...
would insert all variables into the same row.
4. Transpose the matrix from (3) using an inverted comma (’):
4
5
6
10 15
50 56
3 20
Matrix2 = [ 4 5 6; 10 50 3; 15 56 20 ]';
Or
TransposedMatrix = Matrix';
Note: You cannot use spaces in variable names, instead remove the spaces or use an underscore.
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5. Vertically concatenate matrices 1 and 2:
Matrix3 = cat(1, Matrix1, Matrix2);
6. Horizontally concatenate matrices 1 and 2:
Matrix4 = cat(2, Matrix1, Matrix2);
7. Create a cell array that contains matrix 1 in slice 1 and matrix 2 in slice 2:
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CellArray = zeros(3, 3, 2);
CellArray(: ,: ,1) = Matrix1;
CellArray(: ,: ,2) = Matrix2;
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% zeros creates an empty matrix, as given:
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zeros(Number Columns, Number Rows, Number Slices)
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%We use a colon in the second and third lines to select all elements of ...
each column and row.
8. Determine the third power of each element within vector (a):
a = [4 6 7].ˆ3
9. Multiply each element of vector (a) by the respective element in vector (b):
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a = [4 6 7];
b = [3 4 5];
c = a.*b;
10. Multiply each element of matrix 1 by the respective element in matrix 2
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Matrix5 = Matrix1. * Matrix2;
% Removing the dot (.) will perform matrix multiplication, which ...
calculates a very different result. Check for yourself. In these ...
problems, we essentially use matrices as arrays to store data.
11. Create an empty 3 x 3 x 2 array and fill the centre of slices 1 and 2 with 6 and 7
respectively.
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Array = zeros(3,3,2);
Array(2,2,1) = 6;
Array(2,2,2) = 7;
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12. Insert numbers 5:13 into a 3 by 3 matrix using a for loop:
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TempMatrix = zeros(3,3);
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for i = 1:9;
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a = 4 + i;
TempMatrix(i) = a;
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end
13. Insert numbers 5:13 into a 3 by 3 matrix using a while loop:
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TempMatrix= zeros(3,3);
i = 1;
a =5;
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while i < 10
TempMatrix(i) = a + i;
i = i + 1;
end
14. Mulitply each row of matrix 1 by the following factors using a nested loop.
4 5 6
10 50 3
15 56 20
Row 1 Multiplication Factor = 7
Row 2 Multiplication Factor = 3
Row 3 Multiplication Factor = 4
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Matrix1 = [ 4 5 6; 10 50 3; 15 56 20];
NewMatrix = zeros(3,3);
RowNumber = 1;
ColumnNumber = 1;
RowFactor = 7;
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for i = 1: 3
for j = 1:3
NewMatrix(RowNumber, ColumnNumber) = ...
Matrix1(RowNumber, ColumnNumber) * RowFactor;
end
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if RowNumber == 3
RowFactor = 3;
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elseif RowNumber == 3 && ColumnNumber == 2;
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RowFactor = 4;
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end
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end
Alternative Method:
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Matrix1 = [ 4 5 6; 10 50 3; 15 56 20];
NewMatrix = zeros(3,3);
RowNumber = 1;
ColumnNumber = 1;
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while Column Number < 4 && Row Number < 4
NewMatrix(Row Number, Column Number) = ...
Matrix1(Row Number, Column Number) * Row Factor;
Row Number = Row Number + 1;
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if Row Number == 4;
Column Number = Column Number + 1;
Row Number = 1;
Row Factor = 3;
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elseif Row Number == 1 && Column Number == 3;
Row Factor = 4;
end
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end
Problem Section 2
Problem 1
Determine the first 100 numbers in the fibonacci sequence and plot. The first two numbers
in the sequence are:
F1 = 0
F2 = 1
Where Fn = Fn−1 + Fn−2
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f0 = 0;
f1 = 1;
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fibonacci Sequence = zeros(100,1);
fibonacci Sequence(1) = 0;
fibonacci Sequence(2) = 1;
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for i = 1:98
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fibonacci Sequence(i+2,1) = fibonacci Sequence(i+1,1) + ...
fibonacci Sequence(i,1);
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end
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plot(1:100,fibonacci Sequence);
xlabel('x coordinate');
ylabel('Fibonacci Sequence');
title('Problem 1');
Problem 2
Create a matrix, which contains the following in each column using for/ while loop (where
x = 1:100). Plot the functions on the same graph and try the subplot function.
Note: You must use the command: hold on to plot several functions on the same graph.
Use help subplot in the command window
Column
Column
Column
Column
1
1:
2:
3:
4:
y
y
y
y
=x
= x2
= x3
= x4
Matrix = zeros(20,4);
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for i = 1:20
Matrix(i,1) = i;
end
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for i = 1:20
Matrix(i,2) = iˆ2;
end
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for i = 1:20
Matrix(i,3) = iˆ3;
end
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for i = 1:20
Matrix(i,4) = iˆ4;
end
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figure(1);
plot(Matrix(:,1), Matrix(:,2),
hold on
plot(Matrix(:,1), Matrix(:,3),
plot(Matrix(:,1), Matrix(:,4),
xlabel('x')
ylabel('y')
title('Problem 2')
legend(' y = xˆ2', 'y=xˆ3', 'y
'−−r.');
'−−b.');
'−−k.');
= xˆ4')
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figure(2);
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for i = 1:3;
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subplot(3,1,i)
plot(Matrix(:,1), Matrix(:,i+1), '−−k.');
xlabel('x')
ylabel('y')
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end
Problem 3
Create a matrix, which contains the following in each column using a nested loop (where
x = 1: 100)
Column
Column
Column
Column
1
1:
2:
3:
4:
y
y
y
y
=x
= x2
= x3
= x4
Matrix = zeros(20,4);
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for j = 1:4
for i = 1:20
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h = i;
if j == 2
h = iˆ2;
elseif j == 3
h = iˆ3;
elseif j == 4
h = iˆ4;
end
Matrix(i,j) = h;
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end
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end
Problem 4
We have collected absorbance data for Cytochrome C. Import sample and control absorbance
files into MATLAB. Write a function to perform a baseline correction, plot the baseline
corrected absorbance, and calculate the concentration using the Beer Lambert Law
A=e∗l∗c
Pathlength = 1cm
Molar extinction Coefficient 280nm = 11585 M-1 cm-1
Wavelength Range: 400 to 250 nm, 0.2 nm steps.
Note: Baseline Absorbance data has not been stored the same as Cytochrome C Absorbance
data, you must either rearrange both datasets into columns or rows.
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function [Output] = ProteinConc(Wavelength, Sample Data, Control Data)
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% In the Home Tab, select Import Data; import Sample.txt and ...
Control.txt files.
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% Wavelength:
Baseline Corrected Data(:,1) = Wavelength;
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% Absorbance:
Baseline Corrected Data(:,2) = Sample Data − Control Data;
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assignin('base','Baseline Corrected Absorbance', ...
Baseline Corrected Data)
% assignin saves variables in the workspace
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% Plot:
plot(Baseline Corrected Data(:,1), Baseline Corrected Data(:,2))
xlabel('wavelength')
ylabel('Absorbance')
title('Protein Concentration')
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% Determine Concentration:
Number Data Points below 280nm = (280−250)/0.2;
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Abs 280nm = Baseline Corrected Data(Number Data Points below 280nm, 2);
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Concentration = (Abs 280nm / (1 * 11585)) * 1000000;
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Output = sprintf('Absorbance at 280 nm =%d; Cytochrome C ...
Concentration = %d micromolar', Abs 280nm, Concentration);
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end
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% Type the following in the Command Window:
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ProteinConc(Wavelengthnm,CytochromeCAbs, ControlAbs)
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