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...

You are watching: How to tell if a signal is periodic  A \$sin\$ signal has a period of \$2pi\$.

Therefore you wanna know what \$n\$ has to be for \$frac6pi7n\$ to equal \$2pi.\$

\$frac6pi7n=2piLeftrightarrow n=2pifrac76pi=frac146=frac73\$

Therefore the period is \$frac73\$

You can check that way :

\$y=sinleft(frac6pi7cdot(n+frac73)+1 ight)=sinleft(frac6pi7n+2pi+1 ight)=sinleft(frac6pi7n+1 ight)=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) = 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.