How to get a signal without NaN values?

4 ビュー (過去 30 日間)

Haya Ali 2023 年 3 月 12 日

0
リンク

この質問への直接リンク

https://jp.mathworks.com/matlabcentral/answers/1927200-how-to-get-a-signal-without-nan-values

コメント済み: Walter Roberson 2023 年 3 月 14 日

I have a signal and I have applied the Shannon's Nyquist Theorem to get the rate and the reconstructed signal by following formulas.

But I am getting NaN values in the Rate Signal. Is there any way that to not get the NaN values? Below is my code.

close all; clear all; clc;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
X=[11477.9
3
4
11413
9
1
9
1
3
9
4
7
4
2
2
6
4
1
4
5
3
1
9
2
3
8
5
8
8
8
9
5
6
11523
11516
1
9
3
9
1
7
5
11463
5
3
1
2
9
8
11456
5
6
1
11454
3
4
11501
8
2
2
8
6
3
5
4
9
6
3
6
2
5
6
6
2
5
11445
6
3
2
3
11456
6
7
8
6
4
11405
6
3
6
8
11405
7
2
5
6
8
4
8
9
11353
8
6
2
2
4
11341
3
8
3
1
7
4
9
6
2
3
5
3
5
7
6
11402
8
3
9
7
5
5
4
5
5
8
4
4
3
6
11409
3
6
3
6
5
6
6
5
6
5
11423
8
4
9
5
7
5
6
8
5
7
11395
8
5
8
3
4
3
11365
8
11404
6
9
5
3
2
1
6
1
3
4
6
5
7
5
3
1
3
6
9
8
9
3
6
9
5
11346
5
2
8
7
2
8
8
6
2
8
9
8
7
3
11295
2
8
1
2
7
5
6
4
3
11327
5
4
7
11367
1
2
7
2
9
3
6
11390
7
11429
1
3
11452
5
6
1
4
4
1
3
9
4
2
5
3
8
8
4
3
1
1
1
11355
8
7
7
9
2
2
5
3
3
8
2
6
9
5
9
8
5
2
7
8
1
6
2
4
7
8
8
6
5
6
7
3
11375
11398
4
2
3
11440
5
11438
1
1
6
1
6
1
8
7
3
7
5
11401
3
4
1
3
5
8
7
4
4
7
9
1
8
9
1
2
11530
11539
9
3
2
7
11490
9
1
9
6
8
9
4
1
11529
2
4
6
9
5
6
7
8
6
4
5
9
5
5
3
5
3
5
5
6
7
2
7
9
9
6
1
3
4
9
4
3
7
6
11355
8
7
9
4
5
11382
11370
6
1
11321
7
11294
5
6
3
9
7
9
2
11341
8
5
7
8
4
2
3
6
3
4
1
11465
5
1
6
7
11491
5
9
8
9
5
11463
3
9
5
3
8
11445
8
9
9
6
1
11402
8
7
11426
7
5
4
1
11423
11450
3
11486
3
1
8
5
5
8
7
2
6
3
7
8
8
11465
5
7
8
2
4
11508
9
9
6
1
4
4
3
8
11385
1
4
5
9
2
6
11323
3
3
5
6
5
4
4
5
11415
9
8
2
9
6
4
11411
2
2
8
7
9
4
9
4
9
4
3
4
11406
9
7
1
4
2
11472
3
7
3
8
1
3
6
9
8
1
5
1
1
6
8
7
5
2
4
5
5
8
4
4
4
9
11340
1
1
6
4
9
4
6
7
1
9
5
2
4
9
4
7
8
2
11746
5
11805
11714
6
9
2
9
11505
9
1
4
6
7
11450
1
1
6
11406
3
2
5
7
2
2
11267
5
4
3
6
1
5
9
8
1
2
2
8
8
1
4
6
8
9
8
11823
9
5
11550
1
9
8
9
11447
3
1
4
9
3
5
2
4
5
3
3
7
3
1
6
8
11395
2
8
3
11299
6
3
4
3
8
5
4
2
6
9
3
6
8
3
5
5
5
6
5
6
3
5
6
6
7
5
1
4
1
4
4
7
9
5
9
2
6
6
11476
11492
11438
4
6
6
6
9
4
1
9
7
11463
8
8
5
8
1
2
9
2
3
6
1
8
1
1
1
8
7
1
9
3
4
8
2
9
9
4
9
2
3
5
9
2
8
6
6
7
5
11439
2
3
11381
2
2
11391
9
8
11402
7
7
9
5
2
7
6
5
4
6
11321
2
6
8
11361
11363
9
1
11394
7
11394
1
4
2
11438
8
7
8
2
3
1
3
8
4
6
7
11317
3
11322
8
2
4
2
3
11325
7
3
11356
7
6
7
3
6
2
8
4
8
5
5
3
3
7
6
6
8
6
3
11380
9
6
4
2
7
11349
3
5
2
6
11385
8
5
5
6
9
3
3
3
2
7
11357
8
1
1
3
3
2
11311
8
1
1
3
4
9
1
1
4
3
5
9
7
5
4
6
3
2
9
6
5
1
5
5
9
2
9
5
4
1
9
5
8
3
3
2
2
7
4
3
8
8
6
2
3
6
5
7
9
2
5
11351
7
5
8
6
5
6
3
1
5
9
9
2
9
4
5
5
1
8
9
4
2
4
5
4
3
11332
9
2
5
11327
4
1
5
3
2
7
1
4
8
8
5
11366
11401
11366
2
6
8
1
3
9
9
9
2
9
1
8
8
8
8
9
6
11390
11363
6
8
5
1
4
5
7
5
2
5
2
8
9
8
3
8
9
2
9
3
7
11311
6
3
2
8
6
5
7
7
3
3
1
4
6
9
3
1
6
9
11360
8
11376
3
2
9
1
9
1
7
8
7
2
3
7
6
7
4
3
6
8
7
8
8
11368
7
6
1
3
3
3
1
4
9
7
2
2
6
2
5
5
7
1
7
6
8
6
7
5
11397
9
3
2
4
1
7
5
6
3
1
5
3
2
2
1
1
5
2
1
11380
3
8
6
3
7
6
7
9
7
11354
8
4
1
3
11370
6
9
5
3
6
11335
7
8
7
11345
7
4
6
6
11346
11325
1
8
5
5
6
2
2
4
4
11372
5
3
4
5
7
11356
2
5
6
7
8
4
3
9
7
8
8
2
2
9
1
11394
5
6
2
5
6
11414
3
4
6
5
6
3
9
2
6
4
7
1
6
8
7
4
5
9
5
6
3
6
1
3
2
4
6
7
5
1
6
2
6
7
1
6
2
7
11370
4
1
6
11336
5
4
5
9
9
5
5
2
5
7
7
4
3
2
2
8
11321
6
9
3
7
4
7
3
3
7
5
7
6
3
11361
9
9
8
];
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Addnum=20;%Number of additional data points in each time interval.
N=Addnum*length(X); %Length of the reconstructed signal(the continuous signal);
Rec=zeros([1,N]);
Rate=zeros([1,N]);
nx=zeros([1,N]);
NX=zeros([1,length(X)]);
zeroline=zeros([1,N]);
index=0;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for i=1:N
    if mod(i,Addnum)==1
        index=index+1;
        NX(index)=i;
    end
  nx(i)=i;  
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for i=1:N
Rec(i)=0;
Rate(i)=0;
for j=1:length(X)
    Rec(i)=Rec(i)+X(j)*sinc((1/Addnum)*(i-Addnum*(j-1)));
    Rate(i)=Rate(i)+X(j)*(cos((pi/Addnum)*(i-Addnum*(j-1)))/(i-Addnum*(j-1))-sin((pi/Addnum)*(i-Addnum*(j-1)))/((pi/Addnum)*(i-Addnum*(j-1))^2));
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
mean_X=mean(X);
X=X-mean_X;
mean_Rec=mean(Rec);
Rec=Rec-mean_Rec;
plot(nx,Rec,'sg');
hold on
plot(nx,zeroline,'k');
plot(nx,Rate,'*b');
plot(NX,X,'or')
axis([1000 10000 -400 600])
legend ('Reconstructed Signal','zeroline','Rate Signal','BOLD Signal X')

4 件のコメント
2 件の古いコメントを表示2 件の古いコメントを非表示

Dyuman Joshi 2023 年 3 月 12 日

What you are trying to do for the variable Rate? Do you have a reference formula?

What value it should be instead of NaN?

Haya Ali 2023 年 3 月 12 日

I have attached a formula for rate signal in the image. The value can be anything other than NaN

サインインしてコメントする。

サインインしてこの質問に回答する。

採用された回答

Image Analyst 2023 年 3 月 12 日

0
リンク

この回答への直接リンク

https://jp.mathworks.com/matlabcentral/answers/1927200-how-to-get-a-signal-without-nan-values#answer_1191180

MATLAB Online で開く

Try removing the nans or infs

badIndexes = isnan(Rate) | isinf(Rate);
nx2 = nx(~badIndexes);
Rate2 = Rate(~badIndexes);
plot(nx2, Rate2, '*b');

10 件のコメント
8 件の古いコメントを表示8 件の古いコメントを非表示

Image Analyst 2023 年 3 月 14 日

OK. Though it's probably more trouble to avoid/prevent the NaNs than to just fix them afterwards. But whatever you intuitively feel most comfortable with is fine.

Walter Roberson 2023 年 3 月 14 日

The code involves loop summing terms, some of which might be Nan. If you detect the potential for NaN at the term level, then you can omit adding that term to the summation, so you end up with a sum of non-NaN values that might contain useful information. If you wait until afterwards to detect the NaN then you lose the information in the non-NaN terms.

サインインしてコメントする。

その他の回答 (0 件)

サインインしてこの質問に回答する。

カテゴリ

Industries Medical Devices Cardiology

Help Center および File Exchange で Cardiology についてさらに検索

Community Treasure Hunt

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

Start Hunting!

Translated by