Question

The random process \(X(t)=A\cos(\omega_0 t+\theta)\) is wide-sense stationary, where \(A\) and \(\omega_0\) are constants and \(\theta\) is a uniformly distributed random variable on the interval \((0,2\pi)\). Then the mean value is

• A. \(>0\)
• B. \(1\)
• C. \(\pi\)
• D. \(0\)

Hint:

A sinusoidal random process with uniformly distributed phase over \((0,2\pi)\) always has zero mean.

Answer 1

The Correct Option is D

Concept: The mean of a random process is \[ m_X(t)=E[X(t)] \] For a uniformly distributed phase variable \(\theta\), \[ f_\theta(\theta)=\frac1{2\pi}, \qquad 0<\theta<2\pi \]

Step 1: Write the expectation.
\[ m_X(t)=E[A\cos(\omega_0 t+\theta)] \] \[ =\frac{A}{2\pi}\int_0^{2\pi} \cos(\omega_0 t+\theta)\,d\theta \]

Step 2: Evaluate the integral.
\[ m_X(t) = \frac{A}{2\pi} \Big[\sin(\omega_0 t+\theta)\Big]_0^{2\pi} \] \[ = \frac{A}{2\pi} \left[ \sin(\omega_0 t+2\pi)-\sin(\omega_0 t) \right] \] \[ =0 \]

Step 3: Final Answer.
\[ \boxed{m_X(t)=0} \] Hence, \[ \boxed{\text{Correct Option (D)}} \]