Is sin (n) the periodic signals ?

[wcz]

Is sin the periodic signals ?
If yes, please show me the way to show them.
Thanks.

 

[echo47]

Unclear question.
What is "n"?
"signals"? "them"? Are there more than one?

 

[aboozar.hamidipoor]

if n is a discrete quantity and (2π÷n) is an integer number then sin is a periodic function.

 

[wcz]

This question is related to discrete time signals.
The question is
X[n] = sin[n] a periodic signal ?
with n = is an integer.

I know Y[n] = cosine[n] is not a periodic signal in discrete time domain.
How about X[n]=sin[n] ?

 

[checkmate]

A periodic signal has to satisfy x=x(n+T) where T is the period.
A digital signal can only have real integer n, hence for a digital signal to be periodic, T must also be a real integer.

Now given x=sin, and T=2m*pi for some real integer m, is it possible to find a value for m in which T is a real integer? The answer is no since pi is not a rational number.

 

[tarkyss]

n is integer?
if yes, sin is not a periodic signals

 

[mryu]

I bet, f=sin is a period signal, because:

(1) Let sin(w0t) be the analog signal with frequency f0=2*pi/w0;
(2) so sin(w0t)=sin(2*pi*f0*t), if the sampling frequency is fs, t can be represented as t=n/fs
(3)then sin(w0t)=sin(2*pi*f0*t)=sin(2*pi*f0*n/fs)=sin(2*pi*f0/fs*n);
(4)if 2*pi*f0/fs = 1, sin(2*pi*f0/fs*n) = sin, n=integer;
(5) when 2*pi*f0/fs=1, fs=2*pi*f0>2*f0 meet the requirement of Nqyuist sampling theory.
and f0=fs/(2*pi)

you can use Matlab to confirm it.
N=16368;
n=1:N;
fs= 10e3;
s = sin;
plot(1/N*abs(fft(s)));

 

[checkmate]

You have only shown that an analog signal sin(wt) can be discretized under the appropriate sampling frequency fs into the digital form sin. But you have done nothing to show that sin is periodic for integer n.

If so, a periodic signal satisfies x=x(n+T). Please give your value of T here.

 

[swahlah]

I concede with checkmate,
Because sin can only be periodic if Sin = Sin(n+T)
And yes we need some T here.
This is the basic definition of any periodic signal

while to mryu, I think he is ignoring T here.

In short no arguments. As we know Sine funciton fulfils the property of a periodic signal so no matter what is the value of n, if n is an ineger there existss a T for which Sin=Sin(n+T).... and it is a periodic signal so i think it's a simple topic so let's keep it simple too.

-Regards

 

[mryu]

First, many thanks to Swahlah and checkmate. I do ignore the period.

A period analog signal, such as sinusoid, does not necessarily remain periodic when sampled at a given nqyuist rate,say,fs.
Assumed that x=sin(2*pi*f0*n*T)=sin(w0*n), where, w0=2*pi*f0*T=2*pi*f0/fs,
If x is periodic, that is, there must exist a N, and
x(N)=x(0) => sin(w0*N)=sin(0)=0 where requires that
w0*N=2*pi => 2*pi*f0/fs*N=2*pi => fs=N*f0;
With respect to signal sin, we can get that w0=1 =>fs = 2*pi*f0. Because 2*pi is not an integer, the sin will be never periodic.

Is above all right?

Thank you very much,

Best Regards!

 

[checkmate]

The condition for periodic signals is that x=x(n+T). We do not have to care about what is the original analog signal, nor about sampling frequencies as all these are not considered in the condition for periodic signals. mryu is simply thinking too much. The proof I have provided is sufficient, although I admit that I do not know how to prove that pi is irrational.

 

[swahlah]

Dear mryu,
I think u are trying to prove that sin is not a periodic function
Now on your derivation
You wrote

x(N)=x(0) => sin(w0*N)=sin(0)=0 where requires that
w0*N=2*pi => 2*pi*f0/fs*N=2*pi => fs=N*f0;
With respect to signal sin, we can get that w0=1 =>fs = 2*pi*f0. Because 2*pi is not an integer, the sin will be never periodic.

if w0*N = 0 then it means 2*pi*f0*n*T=0
and it doesnt mean w0*N=2*pi
it means w0*N=0.

- I think u better review your proof......

Otherwise Sin is a periodic function no matter you think or not because
Sin=Sin(n+T) and it can be seen in any book of DSP


-Regards

 

[ranjeeth]

sin cannot b periodic for n ==> integer
By definition we need 2 have sin=sin(n+N) for periodicity where N is an integer.
Now sin(n+K*pi)=sin where K is an integer.
therefore N must equal K*pi for sine function 2 be periodic,
K*pi can never be an integer.Therefore the argument of sin being periodic doesnot hold.

 

[1stfaruk]

https://en.wikipedia.org/wiki/Trigonometric_function

 

[swahlah]

i think Periodic Signal topic in the DSP book by proakis has good example about the SIN function as peroidic one............

-Regards

 

[emad1347]

If n is integer, No! sin is Not periodic:
For periodic signal we need to have: sin=sin(n+N)-> N=k*Π (That is impossible for N and k as periodic)
But for sin(Πn)=sin(Π(n+N))-> ΠN=Πk-> k=N: is periodic.

emad1347

[/img][/list]

 

[HMZ]

it is better to read first chapters(i think chapter 2) of Oppenheim's "signals and systems" then ccme here and read guy's comments.
there is some differences between discrete signals and continuos ones ,because they SHOULD itterate in INTEGER numbers but there is no multiplication of (dummy period=)2π/1 that be integer.
OK?


viva Persia

 

[purnapragna]

hi

actually sin cannot be derived from the multi plication of sin(t) with periodic impulse train, sin is a aperiodic signal

thnx

purna

 

[meisamz82]

I think sin , if n is integer , isn't periodic.

 

[haydaa]

if n represents an interval greater than 2*2pi; yes it is periodic.