Question

x[n]=cos(2*pi*n)

x[n]=cos(0.2*pi*n)

x[n]=cos(0.25*pi*n)

x[n]=cos(0.26*pi*n)

x[n]=cos(10*pi*n)

x[n]=cos((8/3)*pi*n)

those signals are periodic??

and for those signals in Part (a) that are periodic, determine the number of samples
per period

Answer 1

For a discrete signal,

,

rn = acos(wen)

In this signal,

If  $N_o= \frac{2\pi}{w_o}$ is a rational number then, signal is periodic and it's fundamental time period is No.

when No is irrataional number then signal is aperiodic.

Now for finding the number of samples per second 'N' for periodic discrete signal.

It is No when No is an integer other wise we multiply k with No to get least integer possible then N=k*No.

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Now In this question we have,

a)

  

rin) = cos(27)

comparing it we can see that here,

$w_o= 2\pi$

27 N.

since No is a rational number hence it is periodic signal.

and No is also an integer hence there are 'N=1' samples per period in the signal.

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b)

Here

$x(n)= cos(0.2\pi n)$

and

$w_o= 0.2\pi$

$N_o=\frac{2\pi}{w_o}=10$

since No is a rational number hence it is periodic signal.

and No is also an integer hence there are 'N=10' samples per period in the signal.

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c)

Here

$x(n)= cos(0.25\:\pi n)$

and

$w_o= 0.25\pi$

$N_o=\frac{2\pi}{w_o}=8$

since No is a rational number hence it is periodic signal.

and No is also an integer hence there are 'N=8' samples per second in a period in the signal.

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d)

Here

$x(n)= cos(0.26\: \pi n)$

and

$w_o= 0.26\pi$

$N_o=\frac{2\pi}{w_o}=\frac{100}{13}$

since No is a rational number hence it is periodic signal.

and No is not an integer hence we have to multiply with k taking k=13, we get minimum value of integer kNo then we he have N=100.

Hence there are 100 samples per period.

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e)

Here

$x(n)= cos(10\: \pi n)$

and

$w_o= 10\pi$

$N_o=\frac{2\pi}{w_o}=\frac{1}{5}$

since No is a rational number hence it is periodic signal.

and No is not an integer hence we have to multiply with k taking k=5, we get minimum value of integer kNo then we he have N=1.

Hence there are 1 samples per period.

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f)

Here

$x(n)= cos(\frac{8}{3}\: \pi n)$

and

$w_o= \frac{8}{3}\pi$

$N_o=\frac{2\pi}{w_o}=\frac{3}{4}$

since No is a rational number hence it is periodic signal.

and No is not an integer hence we have to multiply with k taking k=4, we get minimum value of integer kNo then we he have N=3.

Hence there are 3 samples per period.

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