How to construct array with certain slope ?
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Dear Team,
I have a signal :
Signal = [ 0,10, 0, 10, 0, 10, -50, -10,-50,-10,-50, 30, 20, 30, 20] ;
Signal is y-axis values and i want to define x-axes values based on slope of incline and decline.
incline & decline slope to be 10%.
I would like to first scale the x-axes then interpolate 10 points in between them.
I want to plot such that its incline and declipne slope is 10 or any other user defined value ?
Can you please support in getting that slope part in signal ?
5 Comments
Paul Hoffrichter
on 5 May 2021
Edited: Paul Hoffrichter
on 5 May 2021
>> 0 to 10 it will take 1 secs
Sounds like 10 means 10 seconds, but elsewhere you refer to 10%. I still do not understand what the plot should look like. Are you looking to rotate the plot or just scale the x-axis.
Maybe clarify exactly what the axis should look like as a first step.
Answers (1)
Paul Hoffrichter
on 5 May 2021
Look at this rotation script example:
5 Comments
Paul Hoffrichter
on 6 May 2021
Edited: Paul Hoffrichter
on 6 May 2021
't' was my guess. Is 't' as written the exact values you want to compute? Or, are there some modifications that you would like to see to 't' ? After confirmed, I will see if I can derive 't' from the particular signal. Since signal values will vary, it would be helpful with another short signal, and then indicate what the value of 't' should be for that new signal. This may help to generalize the function to produce 't'.
Are all signal values multiple of 10? If not, this could affect the results.
t = [ 0 1 2 3 4 5 6 10 11 15 16 24 25 26 27];
stem(t)
Paul Hoffrichter
on 15 May 2021
NIKHIL,
Since the values of 't' in previous post is what you can use, here is a formulation of how to get those values.
Signal = [ 0,10, 0, 10, 0, 10, -50, -10,-50,-10,-50, 30, 20, 30, 20] ;
difSig = [0 diff(Signal)/10];
t = zeros(1, length(difSig));
for ii = 2:length(t)
if difSig(ii) < 0
t(ii) = t(ii-1) + 1;
else
t(ii) = difSig(ii) + t(ii-1);
end
end
plot(t,Signal), axis tight, grid on, grid minor
Hope this helps.
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