Sampling from a stochastic process

[Srumix]

Homework Statement

Given X(t) = cos(2$\pi$50t + ω),

where the stochastic variable ω is uniformly distributed between 0 and 2$\pi$.

Suppose the sampling frequency f_s is 30 Hz. What frequency interval is covered after the sampling?

Homework Equations

Normalized frequency when sampling can be made by changing the time units to t = k*d, where k is an integer and d is the sampling distance, so f_s = 1/d, which means d = 1/30 s.

The Attempt at a Solution

Performing the sampling we get

x(t) = cos(2$\pi$(50/30)*k + ω) = cos(2$\pi$(5/3)*k + ω) = cos(2$\pi$(2/3)*k + ω)

The normalized frequency is thus 2/3. Since the normalized frequency can be written as
v = f/f_s, this gives us a frequency interval as a point at 20 Hz ((2/3)*30 = 20)

Homework Statement

I think I should put here what my actual problem is. The problem is that the book give the answer as 10 Hz and I have no idea how they arrived at that. Could it be that it's wrong in the answers section of the book?

I would very much appreciate any feedback/pointers on what could be wrong here.

Thank in advance!

 

[lippyka]

The Attempt at a SolutionNo, the answer is not wrong. The normalized frequency is indeed 2/3, which gives a frequency interval at 20 Hz. However, since the stochastic variable ω is uniformly distributed between 0 and 2π, this means that the actual frequency range covered is 10 Hz (2/3*30-2/3*30 = 10 Hz).