HELP!!!!!! Fundemental frequency

albbg said:

The period of a sinusoid is 2•Π that means it repeats every 2•Π. Matematically

sin(x)=sin(x+2•Π)

then considering two time values t1 and t2 we can write:

[(2n-1)/3]•Π•t2 = [(2n-1)/3]•Π•t1+2•Π

The period T is given by t2-t1. We can arbitrary set t1=0 so that T=t2, thus:

[(2n-1)/3]•Π•T = 2•Π rearranging:

T=(2•Π)/{[(2n-1)/3]•Π} ==> T=6/(2n-1)

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albbg said:

There should be something wrong. Since I saw the time dependency from 't' I expect that you are speaking of an analog signal and 'n' is a parameter to be set. In this case you will have a period that is T=6/(2n-1) function of the parameter 'n' and this is also the fundamental period.
If instead you are speaking of a discrete sequence, 't' should not appear or have to be a costant, then you have to find the fundamental period as the smallest integer over which the sequence repeats.

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