Question

Matlab help!

Answer 1

@phi

Answer 2

How to find max_L and max_C in the power matrix. I need to use find command.

Answer 3

Never mind got it figured out.

Answer 4

% 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-6:9.5e-5; % matrix C 2D matrix incr. from top to bottom

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

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

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

% defining new variable max PR
max_PR = max(max(P_R));

% find max L and C power
[max_L, max_C] = find(P_R == max_PR);

% 3D graph
surf(P_R)
xlabel('L'), ylabel('C'), title('Power (W)')

Answer 5

Got it.

Answer 6

First, I think you swapped the indices, and it should read
[max_C_ix, max_L_ix]= find(P_R == max_PR);

second, those are indices. the actual values are
max_C= C_o(max_C_ix)
or if you like: max_C= NewC_o(max_C_ix, max_L_ix);

similarly
max_L= L_o(max_L_ix)

Answer 7

Okay. I actually compared answers with another student and we got the same values.

Answer 8

what do you get for L and C ?

Answer 9

max_L = 0.075000
max_C = 9.5000e-05
?

Answer 10

I didn't get those numbers..

Answer 11

hold on @phi let me show you my ouput.

Answer 12

Did I also tell you I had the wrong increments for the C_o

Answer 13

i changed the indices

Answer 14

your "max values" are the index into the array where the max value is.
for example if you had
A= [ 1 3 1]
your "max" would be 2 (because A(2) is where the max value is)
if you want the actual value, you have to use the original array C_o or L_o

Answer 15

hold on

Answer 16

I am reading what you posted above and recoding.

Answer 17

[max_C, max_L] = find(P_R == max_PR) this doesn't read actual values just indices?

Answer 18

to find match up the indices, notice
P_R 49x10
has shape 49 x 10
the size 49 "goes with"
L_o 1x49

so you should write
[L_ix , C_ix ] = find(P_R == max_PR)
L_ix = 15
C_ix = 10

the actual values are
L_o(L_ix)
ans = 0.075000

C_o(C_ix)
ans = 9.5000e-05

Answer 19

as a check, do these values give the max power?

>> Lm= L_o(L_ix)
Lm = 0.075000
>> Cm= C_o(C_ix)
Cm = 9.5000e-05

>> II=(120)./(100 + j*2*pi*60*Lm - ((j)./(2*pi*60*Cm)));
>> 100*abs(II^2)
ans = 144.00

Answer 20

% indices of max_C and max_L
[max_C, max_L] = find(P_R == max_PR)

% actual max_C and max_L values
max_C = C_o(max_C)

max_L = L_o(max_L)


max_PR =

143.9982


max_C =

15


max_L =

10

Index exceeds matrix dimensions.

Error in lab03 (line 32)
max_C = C_o(max_C)

Answer 21

you should read my post up above. your indices are switched.

Answer 22

I just switched them. I did look at your post.

Answer 23

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

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

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

% defining new variable max PR
max_PR = max(max(P_R))

% indices of max_C and max_L
[max_C, max_L] = find(P_R == max_PR)

% actual max_C and max_L values
max_C = C_o(max_C)

max_L = L_o(max_L)

Answer 24

[L_ix , C_ix ] = find(P_R == max_PR)
L_ix = 15
C_ix = 10

the actual values are
L_o(L_ix)
ans = 0.075000

C_o(C_ix)
ans = 9.5000e-05

Answer 25

okay give me a sec.

Answer 26

[NewC_o, NewL_o] = meshgrid(C_o, L_o) L and C need to be swapped here?

Answer 27

I would not change that.
rather, when you use the find

Answer 28

In the lab it said that the inductor vector needed to increase from left to right and the cap had to increase from top to bottom. So when I use the meshgrid command does it matter where L and C are in ==> [M, I] = meshgrid()

Answer 29

max_L =

15


max_C =

10


ans =

0.0750


ans =

9.5000e-05

Answer 30

inductor vector needed to increase from left to right
that means L is going to be the rows
and C the columns
so I would do meshgrid(L,C)

Answer 31

[NewL_o, NewC_o] = meshgrid(L_o, C_o)

L_o(max_L)
C_o(max_C)

now it won't run it.

Answer 32

you can do the problem with either L or C "increasing left to right"
and you wrote your code to do the opposite of what your assignment asked.
It will still give the correct answer, but it's not in the shape they wanted you to use.

if you switch the order of the meshgrid, then you have to switch the order of the find.

Answer 33

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

The values for L will start at 0.005 and go to 0.245 in increments of 0.005. The values for C
will start at 5e-6 and go to 9.5e-5 in increments of 10e-6

so my vectors are correct right?

Answer 34

Yes those are correct.

Answer 35

My L is increasing from left to right? Is it not?

Answer 36

you can look at it in the command window

I think they are talking about the meshgrid

Answer 37

[NewL_o, NewC_o] = meshgrid(L_o, C_o) so L and C need to be swapped?

Answer 38

M = NewC_o and I = NewL_o

Answer 39

[M, I] = meshgrid()

Answer 40

clc; clear % clears command window and workspace

% 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-6:9.5e-5; % matrix C 2D matrix incr. from top to bottom

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

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

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

% defining new variable max PR
max_PR = max(max(P_R))

% find max L and C power
[C_ix, L_ix] = find(P_R == max_PR)
max_C= C_o(C_ix),
max_L= L_o(L_ix),

% test these give the max power value
II = (120)./(100 + j*2*pi*60*max_L - ((j)./(2*pi*60*max_C)));
100*abs(II^2),

Answer 41

okay it runs now, but why we switch L and C here...

[max_C, max_L] = find(P_R == max_PR)

Answer 42

figuring out how indexing works is one of the more complicated things in matlab

"find" returns two indices in some order. It turns out that is the order it uses.
I keep track of how it works by using a "toy example"
for example, using

>> A= [ 1 2 3; 4 5 6]
A =

1 2 3
4 5 6

>> [rr, cc]= find(A==6)
rr = 2
cc = 3


the first index is the row, and the 2nd index is the column

Answer 43

and
>> A(rr,cc)
ans = 6

returns the value

Answer 44

Thanks phi for pointing that out.