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 $2\pi$.

Therefore you wanna know what $n$ has to be for $\frac6\pi7n$ to equal $2\pi.$

$\frac6\pi7n=2\pi\Leftrightarrow n=2\pi\frac76\pi=\frac146=\frac73$

Therefore the period is $\frac73$

You can check that way :

$y=\sin\left(\frac6\pi7\cdot(n+\frac73)+1\right)=\sin\left(\frac6\pi7n+2\pi+1\right)=\sin\left(\frac6\pi7n+1\right)=y$

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:

$f(t) = A\sin(2\pi f t + \phi)$

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 $\omega = 2\pi f$ where $\omega$ is the angular frequency (in $\frac\textrad\texts$), with the formula $T = \frac1f$ where $T$ is the period of the signal (in seconds, $s$).

The (general form) formula then becomes:

$f(t) = A\sin(\omega t + \phi)$

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 $\omega = \frac6\pi7$, which means that the frequency is:

$f = \frac\omega2\pi=\frac6\pi7\frac12\pi=\frac37$.

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Now you can find out the period as follows:

$T = \frac1f = \frac1\frac37=\frac73$

It"s worthy to note that the term $\phi$ in the equation only offsets the wave (i.e. if $\phi \neq 2\pi k, k=<0,1,2,...>$ then the wave is offset by $\phi$) but does not affect its period/frequency.