According to me the Fundamental period is $7/3$ but is the signal periodic? I think it should satisfy this $sin(6(pi/7)n + 1 ) = sin(6(pi/7)n + 1 + 7/3 )$ , do I have to draw the signal to see if it is periodic or is there a formula I can multiply/add/divide some values of the equation and find out? To make it short, how is a question like this solved normally...
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A $sin$ signal has a period of $2pi$.
Therefore you wanna know what $n$ has to be for $frac6pi7n$ to equal $2pi.$
Therefore the period is $frac73$
You can check that way :
It works !
Signal is periodic with period 7, not 7/3! Because we are talking about discrete case not continuous case. Here "N" which is time period of discrete signal is always an integer.
Because "n" should be integer angular frequency(w) = 6π/7So time period N = 2πm/(6π/7) = 7m/3But N is always an integer in discrete case, hence we would multiply it with m = 3. Therefore, N = 7"m" denoted the number of cycles continuous signal has to repeat to set 1 period of discrete case.
A sinusoidal signal (such as $sin$) is indeed always periodic.
The general form of a sine wave is:
$f(t) = Asin(2pi f t + phi)$
Where $phi$ (greek letter phi) is the phase shift, $f$ is the frequency (in Hertz, or Hz $=frac1s = s^-1$) and $A$ is the amplitude of the wave.
You can use the formula $omega = 2pi f$ where $omega$ is the angular frequency (in $frac extrad exts$), with the formula $T = frac1f$ where $T$ is the period of the signal (in seconds, $s$).
The (general form) formula then becomes:
$f(t) = Asin(omega t + phi)$
In your case (to make your formula fit the general form) you can say that $t=n$, $A=1$, $phi = 1$ and the angular frequency $omega = frac6pi7$, which means that the frequency is:
$f = fracomega2pi=frac6pi7frac12pi=frac37$.
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Now you can find out the period as follows:
$T = frac1f = frac1frac37=frac73$
It"s worthy to note that the term $phi$ in the equation only offsets the wave (i.e. if $phi eq 2pi k, k=<0,1,2,...>$ then the wave is offset by $phi$) but does not affect its period/frequency.