this is from Digital Signal Processing, 4th ed, Sanjit K Mitra, problem 2.39b.
The question is: Determine the fundamental period of
$x[n] = cos(0.6n\pi + 0.3\pi)$
Since x[n], is in square brackets, this means discrete time, not continuos time, thus the answer must be an integer.
The test is: $cos(0.6n\pi + 0.3%pi)$ = $cos(0.6(n+Nk)\pi + 0.3\pi)$
Where N is a positive integer (and the value we seek)
and k is any integer.
Where N is the answer.
Step 1: Inverse cos() both sides
$0.6n\pi + 0.3\pi$ = $0.6(n+Nk)\pi + 0.3\pi$
step 2: Simplify, remove common term $0.3\pi$ from both sides
And expand the () on the LHS
$0.6n\pi$ = $0.6n\pi + 0.6Nk\pi$
step 3: Remove $0.6\pi$, it is a common common terms:
$0$ = $0.6Nk\pi$
Step 4: The 0 on the LHS is really problematic, but we can also write it as $2\pi$ because $0$ equals $2\pi$
Thus $2\pi$ = $0.6Nk\pi$
step 5: Remove $\pi$ from both sides
$2$ = $0.6Nk$
$Nk$= $2/0.6$, or $3.33333$ repeating
One of these two (N or k) must be a non-integer.
Or is this problem misleading, in that this signal is periodic in the continuous domain, and aperiodic in the discrete domain.
Best Answer
I don't know why you make it so complicated. For any sinusoid of the form $\cos( \omega t + a )$ the fundamental period is $T= 2\pi /\omega$ . And any integer multiple of this is also a period, i.e $ k \, 2 \pi / \omega$
For $cos(0.6 \pi n+0.3 \pi)$ we have $T=k \, 2/0.6= k \, 10/3$. This is not an integer for $k=1$, but it is for $k=3$ Then the fundamental discrete period is $10$