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Adjusting my time axis(x axis) accordingly

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Mohamed Jamal
Mohamed Jamal on 16 Jul 2020
Edited: Kelly Kearney on 16 Jul 2020
Hi guys, I have signal that I sampled called y1, it's implicitly vector of values (samples), what I want to do is this think:
my frequency sampling : fs=10KHZ. so (1/10kHz is in seconds)
I want my time (x axis) to be like this:
I want the first value in the vector y1 (actually y1 is my signal) to be at t0=0 , second value to be at t1=t0+1/fs , third value to be at t2=t1+1/fs , fourth value of my vector to be at t3=t2+1/fs, n value of my vector to b at time tn=t(n-1)+1/fs or tn=t0 + n*(1/fs) .
the time is actually x axis, and the values (amplitude) of my signal y1 (the values of vector y1) is actually y axis.
could anyone please help me how to implement that thing ? implement that my x axis(time axis) be according to what I said above (x axis(time axis) related to 1/fs) ..
Ofcourse assuming that first value of my signal y1 (of my vector y1) is started at t=0.
thanks alot.
  1 Comment
Mohamed Jamal
Mohamed Jamal on 16 Jul 2020
if you could help here at your free time, appreciated alot!

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Answers (1)

Kelly Kearney
Kelly Kearney on 16 Jul 2020
t = (0:(n-1))*(1/fs);
plot(t,y1)
  2 Comments
Kelly Kearney
Kelly Kearney on 16 Jul 2020
I'm not quite sure whether your use of "n" refers to the length of the signal vector, or the to the 0-based max index (i.e. length(y1)-1), or any generic index. In my example, it was the former. Whether you index from 1 or 0, both options match your requirement that the time at a given index is 1/fs greater than the previous index. You said you wanted the first point to correspond to t = 0, so that's what my example does.

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