# Convert Data from array

3 views (last 30 days)
David Jones on 22 Sep 2020
Commented: Star Strider on 24 Sep 2020
Hi
I have attached a Jpeg of my data file displayed in a plot,(0 to 80mS) it shows exactly the data I need how do I convert this data to a new array as is in the plot. The actual array use to create the plot has 16000 samples at 5uS per sample which is of no interest .I just need the 1s and 0s as the plot along with there timing and not 16000 1s and 0s
Thank you for any help
David

Star Strider on 22 Sep 2020
I am not certain what your data are, or what you want to do.
One option for reducing the size of the array and still getting the plot is to use the find function to detect the 1 and 0 levels, and simply store them. Use the stem function to plot them. Another (likely more robust) option is to use the Signal Processing Toolbox midcross function to detect the pulses.
The find function might return something like this:
idx = [1 5 6 9 21 23 30 35 40];
that you could then plot as:
figure
stem(idx, ones(size(idx)), 'Marker','none')
grid
.

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Star Strider on 23 Sep 2020
My pleasure!
David Jones on 24 Sep 2020
Hi
I think I have managed to convert the data and have attached a copy of the decoded file can you please show me how I can extract the bits as in the file (Kpnai.jpg) start bits 8,9 and then the bytes, any help would be greatly appreciated.
Thank You
Star Strider on 24 Sep 2020
I honestly have no idea how best to do that.
Try this:
y = D.manchesterDecoded;
st = [1 strfind(y, [0 1])];
en = strfind(y, [1 0]);
sten = sort([st en]);
dsten = diff([0 sten]); % <— This May Be What You Want
bitlen = min(dsten);
figure
histogram(dsten, 2*numel(st))
figure
plot(y, 'LineWidth',1.5)
grid
figure
plot(dsten)
grid
.

### More Answers (1)

Ameer Hamza on 22 Sep 2020
It seems that you want to reduce the number of samples. For your data, a good option seems to be decimate(): https://www.mathworks.com/help/signal/ref/decimate.html

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David Jones on 22 Sep 2020
Yes please see attached
Ameer Hamza on 22 Sep 2020
Check this code. It just keeps one value of 1s and 0s for each consecutive sequence along with the corresponding time
n = numel(sqwvx);
Ts = 5e-6;
t = 0:Ts:(n-1)*Ts;
idx = find(diff(sqwvx));
t_new = t(idx);
sqwvx_new = sqwvx(idx);
David Jones on 23 Sep 2020
Thank you for your help

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