I wish to plot the EMI noise generated by high frequency switching circuits in simulink. The spectrum scope shows a plot in dB vs frequency. I want to plot this in log frequency scale.
If I understand correctly, you have data containing Frequency -> dB and you want to plot it using the log of the frequency?
For example if you have the following:
Freq dB
63 70
125 75
250 71
500 75
1000 65
2000 77
4000 79
8000 74
You want the log of the freq:
Freq dB Log(Freq)
63 70 1.8
125 75 2.1
250 71 2.4
500 75 2.7
1000 65 3.0
2000 77 3.3
4000 79 3.6
8000 74 3.9
And plot it in a graph:
80 |
- | x
76 | x
- | x x x
72 |
- | x x
68 |
- | x
64 |
|----|----|----|----|----|----|
1.0 1.5 2.0 2.5 3.0 3.5 4.0
If so, you can just take the data and convert each frequency to the log of the frequency.
Related
for example : I have a table as follows
id math science english history
1 80 90 90 90
2 70 60 81 78
3 69 50 45 80
4 30 40 10 80
i only want to find the maximum value in column math and science.
Is it possible?
Simply use this :
select max(science),max(math) from your_table
I have a set of 80 students and I need to sort them into 20 groups of 4.
I have their previous exam scores from a prerequisite module and I want to ensure that the average of the sorted group members scores is as close as possible to the overall average of the previous exam scores.
Sorry, if that isn't particularly clear.
Here's a snapshot of the problem:
Student Score
AA 50
AB 45
AC 80
AD 70
AE 45
AF 55
AG 65
AH 90
So the average of the scores here is 62.5. How would I best go about sorting these eight students into two groups of four such that, for both groups, the average of their combined exam scores is as close as possible to 62.5.
My problem is exactly this but with 80 data points (20 groups) rather than 8 (2 groups).
The more I think about this problem the harder it seems.
Does anyone have any ideas?
Thanks
One Possible Solution:
I would try going with a greedy algorithm that starts by pairing each student with another student that gets you closest to your target average. After the initial pairing you should then be able to make subsequent pairs out of the first pairs using the same approach.
After the first round of pairing, this approach leverages taking the average of two averages and comparing that to the target mean to create subsequent groups. You can read more about why that will work for this problem here.
However,
This will not necessarily give you the optimal solution, but is rather a heuristic technique to solve the problem. One noted example below is when one low value must be offset by three high values to reach the targeted mean. These types of groupings will not be accounted for by this technique. However, if you know you have a relatively normal distribution centered around your targeted mean then I think this approach should give a decent approximation.
First sort the goup by score. So it becomes:
AH 90
AC 80
.....
AB 45
AE 45
Then start combinning the first with the last:
(AE, AH, 67.5)
(AB, AC, 62.5)
(AD, AA, 60)
(AG, AF, 60)
And so on in the other case you will combine the two by two. First two with the last two.
Another way:
1. Find all the possible groups by 4 students.
2. Then for every combination of groups find the abs deviation from the average score and SUM it up for the combination of groups.
3. Choose the combination of groups with the lowest sum.
Initially, I did think about the top-bottom match option.
However, as John has highlighted, the results certainly aren't optimal:
Scores Students Avg.
40 94 40 94 'AE' 'DA' 'AI' 'AR' 67
40 90 40 88 'AK' 'CI' 'AM' 'BP' 64.5
40 85 40 80 'AQ' 'AW' 'AT' 'BD' 61.25
40 79 40 77 'AU' 'BC' 'AV' 'AB' 59
40 76 40 75 'AX' 'CG' 'AZ' 'CQ' 57.75
40 75 40 75 'BF' 'CB' 'BN' 'BQ' 57.5
40 75 40 74 'BR' 'BI' 'CF' 'CZ' 57.25
40 74 40 74 'CK' 'CO' 'CP' 'AL' 57
40 72 41 71 'DB' 'CN' 'AG' 'BO' 56
41 71 42 70 'CD' 'BM' 'AH' 'BS' 56
42 70 42 69 'BG' 'BL' 'CU' 'CX' 55.75
43 68 44 67 'BK' 'CY' 'AD' 'CE' 55.5
44 64 44 64 'BJ' 'CR' 'BZ' 'BY' 54
45 64 45 63 'BW' 'BV' 'CS' 'BE' 54.25
45 62 47 60 'CV' 'CH' 'AC' 'CM' 53.5
47 59 47 58 'BT' 'AY' 'CL' 'AP' 52.75
47 57 48 57 'CT' 'BA' 'BX' 'AS' 52.25
48 56 49 56 'CA' 'AJ' 'AN' 'AA' 52.25
50 55 50 54 'BB' 'AF' 'CJ' 'AO' 52.25
51 52 51 52 'CC' 'BU' 'CW' 'BH' 51.5
Suppose we have a 2D (5x5) matrix:
test =
39 13 90 5 71
60 78 38 4 11
87 92 46 45 35
40 96 61 17 1
90 50 46 89 63
And a second 2D (5x2) matrix:
tidx =
1 3
2 4
2 3
2 4
4 5
And now we want to use tidx as an idex into test, so that we get the following output:
out =
39 90
78 4
92 46
96 17
89 63
One way to do this is with a for loop...
for i=1:size(test,1)
out(i,:) = test(i,tidx(i,:));
end
Question:
Is there a way to vectorize this so the same output is generated without a for loop?
Here is one way:
test(repmat([1:rows(test)]',1,columns(tidx)) + (tidx-1)*rows(test))
What you describe is an index problem. When you place a matrix all in one dimension, you get
test(:) =
39
60
87
40
90
13
78
92
96
50
90
38
46
61
46
5
4
45
17
89
71
11
35
1
63
This can be indexed using a single number. Here is how you figure out how to transform tidx into the correct format.
First, I use the above reference to figure out the index numbers which are:
outinx =
1 11
7 17
8 13
9 19
20 25
Then I start trying to figure out the pattern. This calculation gives a clue:
(tidx-1)*rows(test) =
0 10
5 15
5 10
5 15
15 20
This will move the index count to the correct column of test. Now I just need the correct row.
outinx-(tidx-1)*rows(test) =
1 1
2 2
3 3
4 4
5 5
This pattern is created by the for loop. I created that matrix with:
[1:rows(test)]' * ones(1,columns(tidx))
*EDIT: This does the same thing with a built in function.
repmat([1:rows(test)]',1,columns(tidx))
I then add the 2 together and use them as the index for test.
I have a (symmetric) matrix M that represents the distance between each pair of nodes. For example,
A B C D E F G H I J K L
A 0 20 20 20 40 60 60 60 100 120 120 120
B 20 0 20 20 60 80 80 80 120 140 140 140
C 20 20 0 20 60 80 80 80 120 140 140 140
D 20 20 20 0 60 80 80 80 120 140 140 140
E 40 60 60 60 0 20 20 20 60 80 80 80
F 60 80 80 80 20 0 20 20 40 60 60 60
G 60 80 80 80 20 20 0 20 60 80 80 80
H 60 80 80 80 20 20 20 0 60 80 80 80
I 100 120 120 120 60 40 60 60 0 20 20 20
J 120 140 140 140 80 60 80 80 20 0 20 20
K 120 140 140 140 80 60 80 80 20 20 0 20
L 120 140 140 140 80 60 80 80 20 20 20 0
Is there any method to extract clusters from M (if needed, the number of clusters can be fixed), such that each cluster contains nodes with small distances between them. In the example, the clusters would be (A, B, C, D), (E, F, G, H) and (I, J, K, L).
Thanks a lot :)
Hierarchical clustering works directly with the distance matrix instead of the actual observations. If you know the number of clusters, you will already know your stopping criterion (stop when there are k clusters). The main trick here will be to choose an appropriate linkage method. Also, this paper(pdf) gives an excellent overview of all kinds of clustering methods.
One more possible way is using Partitioning Around Medoids which often called K-Medoids. If you look at R-clustering package you will see pam function which recieves distance matrix as input data.
Well, It is possible to perform K-means clustering on a given similarity matrix, at first you need to center the matrix and then take the eigenvalues of the matrix. The final and the most important step is multiplying the first two set of eigenvectors to the square root of diagonals of the eigenvalues to get the vectors and then move on with K-means . Below the code shows how to do it. You can change similarity matrix. fpdist is the similarity matrix.
mds.tau <- function(H)
{
n <- nrow(H)
P <- diag(n) - 1/n
return(-0.5 * P %*% H %*% P)
}
B<-mds.tau(fpdist)
eig <- eigen(B, symmetric = TRUE)
v <- eig$values[1:2]
#convert negative values to 0.
v[v < 0] <- 0
X <- eig$vectors[, 1:2] %*% diag(sqrt(v))
library(vegan)
km <- kmeans(X,centers= 5, iter.max=1000, nstart=10000) .
#embedding using MDS
cmd<-cmdscale(fpdist)
I have a text file containing RGB data for an image, how can I draw the image using this data in matlab?
data sample :
Red Green Blue
80 97 117
83 100 120
74 91 111
81 96 115
81 96 115
77 90 107
84 97 114
78 91 108
79 95 110
91 104 120
94 108 121
85 99 112
The IMAGE command takes an MxNx3 matrix and displays it as an RGB image. You can use LOAD and RESHAPE to get the data into the right format. Finally, IMAGE wants either integers between 0 and 255 or doubles between 0 and 1.0, so you need to cast or rescale your numbers. The following code snippet should show you how to put it all together.
x = load('rgbdata.txt'); % makes a 12x3 matrix
x = reshape(x, 2, 6, 3); % reshape pulls columnwise, assume 6x2 image
x = x/255; %scale the data to be between 0 and 1
image(x);