index error for find function

I am trying to find the value of Signal_fft_hilbert at the values of newf. But i get the error:

??? Attempted to access Signal_fft_hilbert(14.6484); index must be a positive integer or logical.

I want to find the value at exactly 14.6484. How can I find the value?

           newf=aa1(:)'
          for s=1:length(newf)
          ppks=find(Signal_fft_hilbert(newf(s)))
          end

[Dishant Arora]

You might have to go for interpolation to get the value at non- integral coordinates.

Answer 1

[Iain]

Ok, it looks like you've got a vector "Signal_fft_hilbert", which I'm going to call "v".

If want v(14.6484). The only way you can get v(14. ... ) is if you interpolate.

 doc interp1 % will tell you all about one function for interpolation.

Alternatively, if you want the v that corresponds to something having a value you could do something like:

 v( time == 14.684324)

or

 v(find(time > 14.00))

[ankita]

but 14.6484 isn't the only value. I have an array of such values in aa1. i want the v for all these values.

[Iain]

You haven't told us everything we need. Let me use an analogy that tells you the method:

 time = [0 0.1 0.2 0.3 0.4 .... ]; % or frequency, or number of bananas eaten, or "x"
 voltage = [1 2 3 4.5 .... ]; % or current, or power, or apes fed, or "y".

Now, if you want to know y, when x = 0.05, then you need to "interpolate" or fill in the blanks.

 interp1 is a simple method for automatically calculating the in-between parts, which needs to know, "x", "y" and the "x" you want to know y for.

If, instead, you have a list of times, which have known voltage, you can do something like:

 times_i_want = [0 0.2 0.4 ... ];
 for i = 1:numel(times_i_want)
    voltage_at_time_i_want(i) = voltage(times_i_want(i) == time);
 end

or...

    voltage_at_time_i_want(i) = voltage(find(times_i_want(i) == time));

[ankita]

Thanks alot Lain. I had 1 more doubt. This is my code

           for k=1:length(peaks)
             top1{k}=freq_s(locs(k))
              for s1=1:length(peaks)
                  xx{s1}=top1{k}*s1;

              end
              [xx1]=[xx(:)]';

          xx2=cell2mat(xx1);

          aa1(k,:)=xx2;

In the above answers you can see my output aa1. Is there a method where I can get output like for each top1 with their subscripts included like top11 for i=1,top12 for i=2..... see my basic idea is to create different arrays of top1,which is a frequency, and show the values of corresponding Signal_fft_hilbert. The user can then pick which top1 values suit them and they will get the final Signal_fft_hilbert values. Any suggestions?

[Chandrasekhar]

hope this works replace

top1{k}=freq_s(locs(k)) with 

evalin('base',['top1' num2str(k) '=freq_s(locs(k))';])

this will append the 'k' value to top1 every time and creates a new variable top1k in the workspace.

Answer 2

[Chandrasekhar]

k = find(newf == 14.6484);
ppks = Signal_fft_hilbert(k);

[ankita]

aa1 has a set of values. I want to find the Signal_fft_hilbert of all those values.

[Chandrasekhar]

ok, but the index should be a scalar value. is there any relation between newf and Signal_fft_hilber?

[ankita]

newf is the frequency and Signal_fft_hilbert gives the amplitude. I want to find the amplitude of these particular frequencies.

[Chandrasekhar]

does frequency at index1 of newf correspond to amplitude at index 1 of signal_fft_hilbert?

[ankita]

I tried this.

          k= find(newf==14.6484)
         abs(Signal_fft_hilbert(k))

Output was

k =

   Empty matrix: 1-by-0

ans =

   Empty matrix: 1-by-0

[Chandrasekhar]

k= find(newf==14.6484) this statement should give k = ,because the first element of newf is 14.6484

[ankita]

ya but it is giving me empty matrix 1-by-0

[Chandrasekhar]

Please find the data attached that i am using

[ankita]

Ah. I am not able to open this file

[Chandrasekhar]

copy the file and load it into your matlab workspace. load('Data.mat'); have you done the same?

[ankita]

ya i loaded your file and got index as 1. but it shows empty matrix in my program.

[Chandrasekhar]

Ok, you can use this data now. It is the same data that you had posted. what is the next step you have to do once you get the K value?

[ankita]

I have to find the corresponding Signal_fft_hilbert values for the corresponding frequencies. But why is the output coming for your data and not mine?

[Chandrasekhar]

can you save your data as .mat file and attach it. I will check it. Also let me know do you have data for Signal_fft_hilbert or it is only a graph that you have.

[ankita]

anki.m is my code and hilbertData.m is the Signal_fft_hilbert values

[Chandrasekhar]

the code(anki.m) doesnt execute. it is giving errors.

[ankita]

maybe because of the file in the first 2-3 lines. you can check the code from line 136. there the Signal_fft_hilbert is calculated.

Answer 3

[Niklas Nylén]

I misunderstood the question, removed my response but kept the answer to not lose the comments below.

[ankita]

I tried this already.Gives an error: ??? Subscript indices must either be real positive integers or logicals.

[Chandrasekhar]

can you share the data of aa1 and Signal_fft_hilbert

[ankita]

aa1 =

   14.6484   29.2969   43.9453   58.5938   73.2422   87.8906  102.5391
   51.2695  102.5391  153.8086  205.0781  256.3477  307.6172  358.8867
   65.9180  131.8359  197.7539  263.6719  329.5898  395.5078  461.4258
   80.5664  161.1328  241.6992  322.2656  402.8320  483.3984  563.9648
   95.2148  190.4297  285.6445  380.8594  476.0742  571.2891  666.5039
  109.8633  219.7266  329.5898  439.4531  549.3164  659.1797  769.0430
  124.5117  249.0234  373.5352  498.0469  622.5586  747.0703  871.5820

newf =

Columns 1 through 7

   14.6484   51.2695   65.9180   80.5664   95.2148  109.8633  124.5117

Columns 8 through 14

   29.2969  102.5391  131.8359  161.1328  190.4297  219.7266  249.0234

Columns 15 through 21

   43.9453  153.8086  197.7539  241.6992  285.6445  329.5898  373.5352

Columns 22 through 28

   58.5938  205.0781  263.6719  322.2656  380.8594  439.4531  498.0469

Columns 29 through 35

   73.2422  256.3477  329.5898  402.8320  476.0742  549.3164  622.5586

Columns 36 through 42

   87.8906  307.6172  395.5078  483.3984  571.2891  659.1797  747.0703

Columns 43 through 49

102.5391  358.8867  461.4258  563.9648  666.5039  769.0430  871.5820

this is the plot of Signal_fft_hilbert