Question

matlab help!!!

Answer 1

@phi

Answer 2

%% 4.2
% Load the file thermo_scores.dat provided by your instructor, or enter
% the matrix below and name it thermo_scores. (Enter only the numbers.)

load('thermo'); % Loads the thermo_scores data

% (a) Extract the scores and student number for student 5 into a row
% vector named student_5.
student_5 = data(5,:);

% (b) Extract the scores for Test 1 into a column vector named test_1.
test_1 = data(5,2);

% (c) Find the standard deviation and variance for each test.
Test_1 = data(:, 2);
Test_2 = data(:, 3);
Test_3 = data(:, 4);

% standard deviation and variance for test scores 1-3
S1 = std(Test_1);
S2 = std(Test_2);
S3 = std(Test_3);

% variance for tests 1-3
var1 = var(Test_1);
var2 = var(Test_2);
var3 = var(Test_3);

% (d) Assuming that each test was worth 100 points, find each student's
% final total score and final percentage. (Be careful not to add in the
% student number.)

total_sum1 = sum(data(1, 2:4))
total_sum2 = sum(data(2, 2:4))
total_sum3 = sum(data(3, 2:4))
total_sum4 = sum(data(4, 2:4))
total_sum5 = sum(data(5, 2:4))
total_sum6 = sum(data

% (e) Create a table that includes the final percentages and the scores
% from the original table.

Answer 3

part d @phi

Answer 4

so I could answer part d just making total_sum1 though total_sum20 but that seems like a waste of time.

Answer 5

data =

1 68 45 92
2 83 54 93
3 61 67 91
4 70 66 92
5 75 68 96
6 82 67 90
7 57 65 89
8 5 69 89
9 76 62 97
10 85 52 94
11 62 34 87
12 71 45 85
13 96 56 45
14 78 65 87
15 76 43 97
16 68 76 95
17 72 65 89
18 75 67 88
19 83 68 91
20 93 90 92

Answer 6

col 1 is number of students, col 2 is test score 1, col 2 is test score 2, and col 4 is test score 3

Answer 7

col3 is test score 2

Answer 8

rather than putting the sums into different variables, I would use a vector. Example
ss=sum(data(:,2:4),2)

Answer 9

@phi thank you so much.

Answer 10

what does the 2 at the end indicate?

Answer 11

do
help sum

Answer 12

you can put all the info into one matrix
sc=[ data(:,1) ss ss/3]

Answer 13

okay, sorry I was experimenting with the 2 at the end of the last command you showed me.

Answer 14

why SS/3 ?

Answer 15

I was assuming the final score was the total divided by 3

I guess we could do total/300 then multiply by 100 to change to percent.
that works out to the same thing

Answer 16

I see what you are saying. I misinterpreted the question. I thought total score equaled 100. My bad.

Answer 17

I'm not sure what they want for score. It was just a guess

if you want to print it out, you could do something like
printf('Student total Score Percent\n');
printf(' %2d\t\t%5.1f\t%5.1f%% \n',sc');

Answer 18

let me do that

Answer 19

is it fprintf ? not printf

Answer 20

try fprintf
I'm using Octave, and there are slight differences

It's not important, but I thought you might want to know about how to format printouts

Answer 21

fprintf('data(:, 2:4), sum(data(:, 2:4), 2)), sum(data(:, 2:4), 2)) / 3\n')

Answer 22

fprintf(' %2d\t\t%5.1f\t%5.1f%% \n',sc')

what does that do ?

Answer 23

do I need to replace 2d and 5.1f with my commands?

Answer 24

fprintf(' %2d\t\t%5.1f\t%5.1f%% \n',sc')
1 205.0 68.3%
2 230.0 76.7%
3 219.0 73.0%
4 228.0 76.0%
5 239.0 79.7%
6 239.0 79.7%
7 211.0 70.3%
8 163.0 54.3%
9 235.0 78.3%
10 231.0 77.0%
11 183.0 61.0%
12 201.0 67.0%
13 197.0 65.7%
14 230.0 76.7%
15 216.0 72.0%
16 239.0 79.7%
17 226.0 75.3%
18 230.0 76.7%
19 242.0 80.7%
20 275.0 91.7%
>>

Answer 25

did you just assign variables?

Answer 26

yes, that is what it's supposed to do.

the "format" is given by for example %2d (2 digits of an integer)
%5.1f 5 digits of a "float" including the decimal point and 1 digit to the right of the decimal

the data in sc transposed (notice it's sc')
matches up with the format statement.

you can play with it when you have time.

Answer 27

okay let me put that in. I've got one more question. It's my lab.

Answer 28

% defining input variables for RLC circuit

f_volt = 60; % measured in hertz
W_volt = 120; % measured in volts rms
R = 100; % measured in ohms

L_o = 0.005:0.005:0.245; % matrix L 2D matrix incr. from left to right
C_o = 5e-6:10e-10:9.5e-5; % matrix C 2D matrix incr. from top to bottom

[L_o, C_o] = meshgrid(L_o,C_o);

% current output for RLC circuit (I) in RLC circuit (note: j = sqrt(-1))
I = (120)./(100 + 1i*2*pi*60*L_o - ((1i)./(2*pi*60*C_o)));

% power obsorbed by resistor
P_R = 100*abs(I.^2);

% stores max power inside P_R matrix
max_PR = max(P_R);

Max_L = find(max(L_o)) % specifies maximum for inductor L
Max_C = find(max(C_o))

Answer 29

so the last little bit of the assignment. max_PR, max_L, and max_C. I just wanted to make sure I am heading in the right direction?

Answer 30

depends. what are you trying to do ?

Answer 31

I need max_power for each L and C within the matrix

Answer 32

Use your results to report which specific value for L and value for C resulted in max PR. Store
these two results in the variables max L and max C, respectively. The find function should be
instrumental here. More specifically, see its usage near the top of p. 178 in your textbook.

Answer 33

btw, j should also be pre-defined to equal sqr(-1)

Answer 34

I have a note on the side. I used j instead but matlab didn't like that.

Answer 35

% current output for RLC circuit (I) in RLC circuit (note: j = sqrt(-1))

Answer 36

so my understanding for the last part is that I need to find the max power for the inductor and max power for the capacitor in the array/matrix?

Answer 37

is my assumption right? Don't write my code for me I just want to make sure my logic is correct.

Answer 38

are you sure? you may have overwritten it . but if you start matlab clean, j is defined

Answer 39

hold on let me try that

Answer 40

well it works... That is strange. For the longest time it wouldn't accept j. Matlab kept telling me to change it to 1i

Answer 41

so back to my other question do I need to find max values for L and C in the power matrix

Answer 42

Use your results to report which specific value for L and value for C resulted in max PR.

you first need to find the max power
you have to use max twice on your 2d power matrix
also keep track of the indices

Answer 43

why do I need to use max() twice on 2d power matrix?

Answer 44

are you looking for the max value in the whole matrix ?

max works on one dimension at a time

Answer 45

but how do I distinguish the max power for L and C in my power matrix? Is this where the find() command comes into play? when I use max(P_R) it shows.... max_PR =

Columns 1 through 11

134.8577 136.0632 137.1975 138.2564 139.2361 140.1328 140.9431 141.6636 142.2916 142.8244 143.2599

Columns 12 through 22

143.5963 143.8320 143.9662 144.0000 144.0000 144.0000 144.0000 144.0000 144.0000 144.0000 144.0000

Columns 23 through 33

144.0000 144.0000 144.0000 144.0000 144.0000 144.0000 144.0000 144.0000 144.0000 144.0000 144.0000

Columns 34 through 44

144.0000 144.0000 144.0000 144.0000 144.0000 144.0000 144.0000 144.0000 144.0000 144.0000 144.0000

Columns 45 through 49

144.0000 144.0000 144.0000 144.0000 144.0000

Answer 46

it max. to 144.000000

Answer 47

you should call it with
[val, ix]= max(P_R)
val (printed out up above) is the max value in each of the 49 columns
if you want the max of those 49 values you now do
[val2, ix2]= max(val)
val2 will have the 144
ix2 will have the column that it was in
ix(ix2) will be the row index

i.e. P_R( ix(ix2), ix2)
should have a 144.0 in it.

the double indexing can be confusing.

Answer 48

Okay I got to run. When I return I will play around with double index. I watched a few vids on utube early to try to understand the general idea. I think what I am going to need to do is just break the matrix down.

Answer 49

btw, I would make these different variables
[L_o, C_o] = meshgrid(L_o,C_o);

the L_o on the output is a 2D matrix and the L_o on the input is a vector.
to retrieve your original values, it might be easier to use the vectors
example
L_o(ix2)
and C_o(ix(ix2))

Answer 50

% (d) Recall that both the max and min functions can return not only the
% maximum values in a column, but also the element number where those
% values occur. Use this capability to determine the values of times at
% which the maxima and minima occur in each column.

max(vector_thermo(:, 2:4));

Answer 51

%% 4.6
% The area of a triangle is, area = 1/2*base*height (see Figure P4.6).
% Find the area of a group of triangles whose base varies from 0 to 10 m
% and whose height varies from 2 to 6 m. Choose an appropriate spacing
% for your calculational variables. Your answer should be a
% two-dimensional matrix.

base = 0:2:10; % base vector in meters
height = 2:0.75:6; % height vector in meters

Answer 52

I am a little confused. So I am suppose to decide on increments so the matrix have same dim so I can multiply?

Answer 53

or do I use meshgrid() command?

Answer 54

you can use mesh grid
[ bb, hh] = meshgrid(base, height);

Area= 0.5.*bb.*hh

Answer 55

though your height values leave out the max value of 6 (because 0.75 does not divide evenly into 6)

Answer 56

that is funny I did that... But what got me was the 2d dim. That doesn't output as 2D

Answer 57

So I was trying to do is suppress all dim. except two columns. however every column (output side) is the same answer.

Answer 58

>> [bb,hh]= meshgrid(base,height)
bb =

0 2 4 6 8 10
0 2 4 6 8 10
0 2 4 6 8 10
0 2 4 6 8 10
0 2 4 6 8 10
0 2 4 6 8 10

hh =

2.0000 2.0000 2.0000 2.0000 2.0000 2.0000
2.7500 2.7500 2.7500 2.7500 2.7500 2.7500
3.5000 3.5000 3.5000 3.5000 3.5000 3.5000
4.2500 4.2500 4.2500 4.2500 4.2500 4.2500
5.0000 5.0000 5.0000 5.0000 5.0000 5.0000
5.7500 5.7500 5.7500 5.7500 5.7500 5.7500

>> Area= 0.5.*bb.*hh
Area =

0.00000 2.00000 4.00000 6.00000 8.00000 10.00000
0.00000 2.75000 5.50000 8.25000 11.00000 13.75000
0.00000 3.50000 7.00000 10.50000 14.00000 17.50000
0.00000 4.25000 8.50000 12.75000 17.00000 21.25000
0.00000 5.00000 10.00000 15.00000 20.00000 25.00000
0.00000 5.75000 11.50000 17.25000 23.00000 28.75000

Answer 59

remember you have to use .* (dot times)

Answer 60

I did. that isn't a two-dim matrix

Answer 61

that is a 6x6 matrix

Answer 62

whos Area
Variables in the current scope:

Attr Name Size Bytes Class
==== ==== ==== ===== =====
Area 6x6 288 double

Total is 36 elements using 288 bytes

Answer 63

6 x 6 is two-dimensional
1 x 6 or 6 x 1 is one dimensional

Answer 64

or, in other words, you need two indices to get at a value
Area(2,3)
for example

Answer 65

hold on i need to read something.

Answer 66

dimension refers to the vector living in the space? This is starting to come back. when we have a 2x2 matrix it lives in R^2 ?

Answer 67

R^2 = two-dimensional?

Answer 68

yes R^2 or R x R is two-dimensions

intuitively each dimension is perpendicular to all the other dimensions
point, line, plane, cube
it's hard to think beyond 3 D

Answer 69

Okay thanks. If I have any further questions I will post them here.

Answer 70

@jim_thompson5910

Answer 71

%% 4.10
% Create a 6x6 magic matrix.
V_mag = magic(6);

% (a) What is the sum of each of the rows?
sum(V_mag, 2);

% (b) What is the sum of each of the columns?
sum(V_mag);

% (c) What is the sum of each of the diagonals?
sum(diag(V_mag, 0));
sum(diag(V_mag, 1));
sum(diag(V_mag, 2));
sum(diag(V_mag, 3));
sum(diag(V_mag, 4));
sum(diag(V_mag, 5));
sum(diag(V_mag, -1));
sum(diag(V_mag, -2));
sum(diag(V_mag, -3));
sum(diag(V_mag, -4));
sum(diag(V_mag, -5));

Answer 72

is there another way to solve part c?

Answer 73

I wanted to find the sum of each diagonal outputting all at the same time

Answer 74

let me think

Answer 75

`diag(V_mag)` should return the elements along the diagonal

`sum(diag(V_mag))` will then add up the diagonal elements

Answer 76

yes I agree

Answer 77

I am thinking too

Answer 78

I need sum of each diagonal. There are 11 diagonals.

Answer 79

what is the output when you type in `diag(V_mag)` ?

Answer 80

>> diag(V_mag)

ans =

35
32
2
17
14
11

>>

Answer 81

oh nvm, you don't want just the main diagonal. I see what part (c) is asking now

Answer 82

V_mag =

35 1 6 26 19 24
3 32 7 21 23 25
31 9 2 22 27 20
8 28 33 17 10 15
30 5 34 12 14 16
4 36 29 13 18 11

Answer 83

ok how about `diag(V_mag,0)`

post both V_mag and the result of whatever `diag(V_mag,0)` spits out

Answer 84

>> diag(V_mag, 0)

ans =

35
32
2
17
14
11

Answer 85

ok it looks like 0 is the main diagonal

Answer 86

yes, I agree.

Answer 87

how about diag(V_mag, 1)

Answer 88

I'm guessing V_mag doesn't change each time you run the script?

Answer 89

that will display the next diagonal to the right of 0

Answer 90

>> diag(V_mag, 1)

ans =

1
7
22
10
16

Answer 91

>> diag(V_mag, -1)

ans =

3
9
33
12
18

Answer 92

that is the diagonal below 0

Answer 93

everything looks good

Answer 94

Okay thanks. I figured out my next problem, but I have a question about it.

Answer 95

%% 4.11
% Extract a 3x3 matrix from the upper left-hand corner of the magic
% matrix you created in Problem 4.9. Is this also a magic matrix?
V_mag(1:3, 1:3);

Answer 96

why is it 1:3, 1:3 not 0:3, 0:3 ?

Answer 97

I wonder if `diag(V_mag, :)` will work

instead of having a number in the second argument, have a colon

tell me what the output of that is