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Below is the signal :

$$ y[n] = \sin\left( \frac{6\pi}{7} n + 1 \right) $$

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...

Michael Hardy
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Anarkie
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  • Any signal $f(x)=\sin(\alpha x+\beta)$ is periodic – Hippalectryon Jun 29 '14 at 15:13
  • @Hippalectryon thanks but not all questions may come like this in front of me so I would like to understand the formulas and interactions between all the variables... Did I determine the fundamental period correct for example etc. – Anarkie Jun 29 '14 at 15:18

3 Answers3

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A $\sin$ signal has a period of $2\pi$.

Therefore you wanna know what $n$ has to be for $\frac{6\pi}7n$ to equal $2\pi.$

$\frac{6\pi}7n=2\pi\Leftrightarrow n=2\pi\frac{7}{6\pi}=\frac{14}6=\frac{7}3$

Therefore the period is $\frac{7}3$

You can check that way :

$y[n+\frac{7}3]=\sin\left(\frac{6\pi}7\cdot(n+\frac{7}3)+1\right)=\sin\left(\frac{6\pi}7n+2\pi+1\right)=\sin\left(\frac{6\pi}7n+1\right)=y[n]$

It works !

Hippalectryon
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  • Ok here we know that sin has 2pi as period, what if we didn't know? Another way to check? Or is my question non-sense =) – Anarkie Jun 29 '14 at 15:44
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    @Anarkie A function $f$ is periodic of period $\alpha$ when $f(x+\alpha)=f(x)$. Therefore if you find an $\alpha$ that works, then it is periodic :) – Hippalectryon Jun 29 '14 at 16:03
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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/7 So time period N = 2πm/(6π/7) = 7m/3 But 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.

Satyam Sharma
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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 $=\frac{1}{s} = s^{-1}$) and $A$ is the amplitude of the wave.

You can use the formula $\omega = 2\pi f$ where $\omega$ is the angular frequency (in $\frac{\text{rad}}{\text{s}}$), with the formula $T = \frac{1}{f}$ 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 $\omega = \frac{6\pi}{7}$, which means that the frequency is:

$f = \frac{\omega}{2\pi}=\frac{6\pi}{7}\frac{1}{2\pi}=\frac{3}{7}$.

Now you can find out the period as follows:

$T = \frac{1}{f} = \frac{1}{\frac{3}{7}}=\frac{7}{3}$

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.

aright
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  • I think I will understand better if I see some non periodic signals after calculating its period, would you be able to give some examples for non-periodic case? Because then I think everything inside a sin() should be periodic? Or are non-periodic signals are addition of multiple sines and cosines? – Anarkie Jun 30 '14 at 10:07