Adjusting my time axis(x axis) accordingly

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.

[Mohamed Jamal]

if you could help here at your free time, appreciated alot!

@Star Strider

Answer 1

[Kelly Kearney]

t = (0:(n-1))*(1/fs);
plot(t,y1)

[Mohamed Jamal]

n is the length of my signal vector? if so, then it should be (1:n)*1/fs no?

also (0:n-1)*(1/fs) it means the first I have 0, 1/fs , 2/fs , 3/fs .. yeah? but what I want is at time "n" it should be tn=t(n-1)+1/fs ..... thanks alot for your explanation.

[Kelly Kearney]

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.