Plotted waveform frequency is way too low

I am using Quartus 12.1 sp1, vhdl and Altera Nios II programmed in C code for DE0-Nano Development. Basically, what I have is, Nios II system sends the data to serial port, Matlab access the serial port to real time plot the graph.

In my system, the rate the graph is plotted is much slower (11Hz) than the rate the processor speed (50MHZ).

Please refer to this image:

My question is, what should I change in order to get the waveform plotted with a period of 0.02 seconds instead of 0.16 minutes as I got now? What determine the period of the plotted waveform?

I posted the Matlab code here: EDIT:

   max_packets = 20000; %maximum number of packets to keep in memory (all are saved to log file)
    update_plot = 1; %how often we should update the plot
    plot_length = 12; %maximum length of plot
    figure('Position', [10, 50, 900, 900]); % new figure
    ax1 = subplot(3,1,1); % top subplot
    ax2 = subplot(3,1,2); % bottom subplot
    ax3 = subplot(3,1,3);
    drawnow;

    s = serial(port);
    cleanupObj = onCleanup(@()cleanupFunc(s)); %clean-up function
    set(s,'BaudRate',115200,'timeout',2); %timeout is in seconds 
    fopen(s);

    vref= NaN(12,max_packets);

    j = 0;
    tic; %start timer
    while 1; %do forever...
          if j < max_packets
          j = j+1;
      else
          fprintf('\n Buffer full, reset to 0\n');
          j = 1; %reset counter to stop memory overflow
      end    

        %get data line by line 
        line = fscanf(s);
        [c,vo] = disect_line(line);
        if c == 1 %start recording from set 1 

            vref(1,j) = vo; 

            for i = 2:12 %get set 2 to 12
                line = fscanf(s);
                [c,vo] = disect_line(line);
                if c == i

                    vref(i,j) = vo;

                else
                    fprintf('\nError\n');
                end
            end

           dlmwrite(logfile,[toc  vref(:,j).' ],'delimiter',',','-append'); %save data from all 12 sets to csv file
       update plot1 every update_plot packets, and after we have at least 2 data points (otherwise we get errors)

          if mod(j, update_plot)==0 && j>1  
              if j < plot_length
                  start = 1;
              else
                  start = j-plot_length+1;
              end    
           plot(ax1, start:j, vref(1,start:j).');

                for k = 1:1
                    text(j, vref(k, j), num2str(k), 'Parent', ax1);
                end
                xlim(ax1, [start j]);

                ylabel(ax1, 'Vref');

                title(ax1, ['Time elapsed = ' num2str(toc) ' s']);   

    end

    end

    function [c, vo]  = disect_line(line)

    c = str2double(line(1:2)); %set number
    vo = str2double(line(4:7));

    end

    function cleanupFunc(s)

    fprintf('\nClosed serial port.\n')
    fclose(s);
    delete(s);
    clear s;
    end

Answer 1

[Walter Roberson]

Your code does not define "start" or "j" and never changes them.

Remember too that start:j would increment by 1 and that if you want a different increment you need start:increment:j

[want2know]

Thank you for your reply, sorry for my mistake, some codes were left out while commenting the codes. My problem is, I can get the amplitude of the samples plotted correctly, means the data were sent correctly to serial port and Matlab can read the values correctly, the issue is with the frequency, but I dont know what actually determine the period of the plotted waveform in Matlab?

And, index j is used for the x axis where j represents the number of packet but does not represent a real time scale for the x axis, could this be the problem? Any idea how to build a time vector based on what is the sampling rate (time gap) between every point?

[Walter Roberson]

relative time = (sample_index - 1) * time_gap

[want2know]

Thank you for your reply. Could you please elaborate more? I am sorry, my understanding is, does the sample_index here represents 'j'? and time_gap refers to 0.02 seconds (1/50Hz)? Appreciate any resources or link so I can start working on it

[Walter Roberson]

No, "j" represents the sample index.

The time gap is 1/Fs

The first sample is at time 0

The second sample is at time 1/Fs

The third sample is at time 2/Fs

The fourth sample is at time 3/Fs

the N'th sample is at time (N-1)/Fs

[want2know]

Thank you for your reply. Forgive me, actually I don't understand what is the reason behind this? Is this the same as the common practice we used to generate a waveform :

For example, generating data with a 1000 Hz sample frequency, appropriate time vector is

t = (0:0.001:1)'; Do you mind to give me keywords or link of resources? much appreciated

[Walter Roberson]

Let Fs = 1000 then 1/Fs = 0.001 and that is what you used as your increment. So in 0:001:1, your first sample is at time 0, your second sample is at 1/Fs = 0.001, your third is at 2/Fs = 0.002, your fourth is at 3/Fs = 0.003