Question

x[n]=x[n+r]

Now, if x[n]=A*cos[w*n+B] is Discrete Time periodic signal, then prove that

w*r=2*pi*q, where q is integer.

Answer 1

x[n] = x[n+r] =A*cos[w*n+B] =
A*cos[w*n + w*r +B] =
A*cos[w*n + w*r +B] =
A*cos[w*n + B + w*r] =
Then use sum of angles theorem for cosine...
A*cos[w*n + B]cos[w*r] - A*sin[w*n + B]sin[w*r] =
Sub in wr=2*pi*q ...
A*cos[w*n + B]cos[2*pi*q] - A*sin[w*n + B]sin[2*pi*q] =
and evaluate ...
A*cos[w*n + B]

Answer 2

It's more of a "Show" than a proof, but that's how

Answer 3

or u can simply equate the cos arguments:
x[n+r] = A*cos[w*n+w*r+B]
x[n]=A*cos[w*n+B] =A*cos[w*n+B+2*pi*q]<---because cos is periodic with 2pi
so u have
A*cos[w*n+w*r+B] = A*cos[w*n+B+2*pi*q]
equating the arguments of cos,
w*r = 2*pi*q

Answer 4

Yeah, but also not a proof

Answer 5

Thank you hartnn and Tonks!

May Allah Ta'ala guide you and us.