According to me the Fundamental period is \$7/3\$ however is the signal periodic? I think it should satisfy this \$sin(6(pi/7)n + 1 ) = sin(6(pi/7)n + 1 + 7/3 )\$ , perform I need to attract the signal to check out if it is regular or is tbelow a formula I can multiply/add/divide some values of the equation and uncover out? To make it brief, how is a question choose this solved normally...

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

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

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

Therefore the period is \$frac73\$

You can inspect that means :

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

It works ! Signal is regular through duration 7, not 7/3! Due to the fact that we are talking around discrete situation not constant situation. Here "N" which is time period of discrete signal is always an integer.

Since "n" should be integer angular frequency(w) = 6π/7So time duration N = 2πm/(6π/7) = 7m/3But N is always an integer in discrete case, thus we would multiply it via m = 3. Therefore, N = 7"m" denoted the variety of cycles continuous signal has to repeat to set 1 period of discrete situation. A sinusoidal signal (such as \$sin\$) is indeed constantly routine.

The basic create of a sine wave is:

\$f(t) = Asin(2pi f t + phi)\$

Wright here \$phi\$ (greek letter phi) is the phase transition, \$f\$ is the frequency (in Hertz, or Hz \$=frac1s = s^-1\$) and also \$A\$ is the amplitude of the wave.

You deserve to usage the formula \$omega = 2pi f\$ where \$omega\$ is the angular frequency (in \$frac extrad exts\$), with the formula \$T = frac1f\$ wbelow \$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 suggests 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 counter by \$phi\$) however does not influence its period/frequency.