# How to find the maximum value of C that corresponds to a set of X,Y coordinates?

2 次查看（过去 30 天）
Brianna Miranda2022-7-29

I have a spectrogram plot that contains X,Y and C. I want to find the maximum values of C from that spectrogram and also the XY coordinates that they correspond to. I have this so far but I am not sure how to get the X,Y coordinates that each value of max C corresponds to.
imagesc(X,Y,10*log10(C))
maxC = max(C);

### 采纳的回答

Star Strider 2022-7-29
The spectrogram plot is actually a surf plot with the view set to display it looking from the top, so the ‘Z’ values will give you all the information you want.
One way of getting that information from the spectrogram is to use the medfreq funciton. See Track Chirps in Audio Signal for an example.
##### 2 个评论显示隐藏 1更早的评论
Star Strider 2022-7-29
With a spectrogram, the ‘maximum’ can be defined as the maximum value of the ‘Z’ matrix (the eqay solution), or the maximum frequency with respect to time (the solution that the link I posted could be useful to find).
The maximum value of the ‘Z’ matrix is simply:
[r,c] = find(Z == max(Z(:))
so for example —
[X,Y] = ndgrid(1:0.1:5);
Z = randn(size(X))
Z = 41×41
-0.0639 -2.4202 -0.5892 -0.3153 -1.2375 0.2582 0.5664 -0.0020 -0.4781 -0.4400 0.3104 -1.6870 2.3617 0.1904 0.4709 0.5604 0.9655 1.4496 0.8379 1.5481 -0.2459 0.1577 2.6907 1.9585 -0.7600 0.8633 -0.0593 -0.0278 -0.7414 0.2447 -0.8915 -1.3448 -0.9731 1.4001 -1.4113 0.0759 -0.3949 1.0819 -1.2504 0.2395 -0.5119 0.4586 -0.9246 0.8510 1.2941 1.2805 -0.4253 -0.0403 1.7856 -1.3657 1.5380 -0.8891 -0.9842 0.6527 -2.5084 -1.8022 0.3371 1.2054 -0.6248 -0.9365 0.0217 0.7413 0.9703 -1.3422 -0.6889 2.4894 -0.8549 -0.4740 -0.2225 -0.5694 -2.0222 1.1507 -0.2375 -2.5342 0.8704 0.9467 -1.1114 -2.4286 -0.3480 -1.4527 0.8740 -0.5818 -1.1733 0.6569 1.0697 0.2519 -1.4657 0.9207 -0.5496 0.5370 -0.3532 0.4664 -1.1207 -0.2304 0.0524 -0.0722 0.3842 -0.5589 -1.0253 0.6298 -0.3426 -0.8348 -2.4134 1.7251 -2.4607 -1.6548 -0.1585 -1.3724 1.6503 -0.1679 -0.6309 -0.2147 -0.6524 0.7183 -2.7147 1.3611 1.0592 -0.5393 1.2776 0.6410 0.7345 -0.1412 1.6778 -0.8870 0.1101 -0.2241 0.2429 1.8206 -0.6441 1.5836 -0.6047 1.9053 1.9305 0.3951 0.5758 -1.3763 0.2552 -1.7303 -0.0525 3.0846 -1.0426 0.0874 -0.3216 0.1369 0.2692 1.2349 0.5056 -1.2016 -0.8287 -0.7030 -1.6341 -0.6052 0.1854 -0.5525 -1.5532 -1.9856 0.6408 1.9057 -1.3372 1.1240 -0.4402 -1.4120 -0.1029 -0.5735 -0.2999 -0.0698 1.8575 0.2400 1.4729 0.6123 -0.0452 0.4546 0.5246 0.8361 2.2197 -0.6524 -0.1364 0.8422 0.4287 -0.7655 0.6935 0.9505 -0.1121 -1.8347 -1.2506 -1.0413 1.2313 -0.0750 -0.9505 0.1271 -0.5988 -1.3527 1.5817 0.4073 0.0950 -0.1533 -1.4091 1.1605 -1.0842 -0.9306 -0.3523 -1.8747 0.3412 1.2081 -2.6801 0.1715 0.0911 -0.1921 -1.0078 0.3955 0.1917 -0.5632 1.3781 1.4707 -0.7444 -1.0536 -0.4651 -0.4820 1.1504 -0.2833 1.1034 -0.0543 -0.1905 -0.2366 -2.0467 -0.1114 1.0882 -0.2817 0.5707 -0.7683 -1.1562 -0.3642 -1.0025 -0.7422 0.3502 -1.0852 -0.1734 1.7628 -0.7464 -0.8762 -1.6987 -0.3549 1.3508 -1.3206 -1.7158 0.6468 1.0099 -0.2371 0.2916 0.0334 0.3175 0.2897 -0.3463 1.5408 2.0720 0.1516 0.3823 -1.3584 1.1463 0.2069 1.2317 0.2429 1.4249 -0.0526 0.5708 -0.0585 0.1757 1.1555 -0.0869 -0.3246 1.1156 -0.5752 0.8764 -1.7859 -1.9668 -0.8118 0.0052 -1.3435 -0.5732 0.7530 -0.3858 0.0806 -0.7766 0.4792 2.5190 -0.6782 -1.3124 -0.4009 -1.7604 0.2375 -1.4506 0.9221 -0.9909 -0.3864 -0.5869 -0.0340 -1.1347 1.2365 -0.0547 -0.7845
[r,c] = find(Z == max(Z(:)))
r = 13
c = 9
maxZ = Z(r,c)
maxZ = 3.2838
figure
surf(X,Y,Z)
hold on
stem3(X(r,c),Y(r,c),Z(r,c), '^r', 'MarkerFaceColor','r')
hold off
grid on
.

### 更多回答（1 个）

Voss 2022-7-29
X = 1:3:31;
Y = 1:2:13;
C = rand(numel(Y),numel(X));
imagesc(X,Y,10*log10(C))
[maxC,idx] = max(C(:))
maxC = 0.9986
idx = 58
[yidx,xidx] = ind2sub(size(C),idx)
yidx = 2
xidx = 9
maxX = X(xidx)
maxX = 25
maxY = Y(yidx)
maxY = 3
line(maxX,maxY,'Marker','x','MarkerEdgeColor','r')

### 类别

Find more on Time-Frequency Analysis in Help Center and File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by