1.
For 2 same waves traveling parallel to each other:
If one wave travel an extra distant as shown in the figure below. It infers a delay of arrival. And delay infers phase shift, vice versa: \boxed{ \text{distant } d \leftrightarrow \text{ delay } \tau \leftrightarrow \text{phase shift } \phi }
2.
Given a random variable:
\text{if: } & \Theta \sim \mathcal{U}([0,2\pi]) \\ \text{then: } & E[\cos(\Theta + c)] = 0Proof :
E[\cos(\Theta + c)] &= \int_{0}^{2\pi} \cos(\theta+c) f_{\Theta}(\theta) d\theta \\ (&\because f_{\Theta}(\theta) = \frac{1}{2\pi} )\\ &= \int_{0}^{2\pi} \cos(\theta+c) \frac{1}{2\pi} d\theta \\ &= \frac{1}{2\pi} \int_{0}^{2\pi} \cos(\theta+c) d\theta \\ &= \frac{1}{2\pi} [\sin(2\pi +c )-\sin(0+c)] \\ &= 03.
Given a random variable:
\text{if: } & \Theta \sim \mathcal{U}([0,2\pi]) \\ & t \text{ = time variable} \\ \text{then: } & E[\cos(c_0 t + \Theta +c_1 )] = 0Proof :
E[\cos(c_0 t + \Theta + c)] &= \int_{0}^{2\pi} \cos(c_0 t + \theta + c) f_{\Theta}(\theta) d\theta \\ (&\because f_{\Theta}(\theta) = \frac{1}{2\pi} )\\ &= \frac{1}{2\pi} \int_{0}^{2\pi} \cos(c_0 t + \theta + c) d\theta \\ \\ &\left(\begin{matrix} & t \text{ is treated as a constant}\\ &\text{make a new constant } c = c_0 t + c_1\\ \end{matrix}\right) \\ \\ &= \frac{1}{T} \int_{0}^{T}E\left[ \cos(\Theta + c)\right] dt \\ &(\text{ apply result 2. above}) \\ &= \frac{1}{T} \int_{0}^{T}0 dt \\ &= 0(time and frequency of the wave is irrelevant)
4.
Given two IID random variables:
\text{if: } &\Theta_0 \sim \mathcal{U}([0,2\pi]) \\ &\Theta_1 \sim \mathcal{U}([0,2\pi]) \\ \text{then: } & E[\cos( \pm\Theta_0 \pm \Theta_1 +c )] = 0Proof :
E[\cos(\Theta_0 + \Theta_1 + c)] &= (\frac{1}{2\pi}) ^2 \int_{0}^{2\pi} \int_{0}^{2\pi} \cos(\pm\theta_0\pm\theta_1+c) d\theta_0 d\theta_1 \\ &= (\frac{1}{2\pi}) ^2 \int_{0}^{2\pi} \big[\sin(\pm\theta_0\pm\theta_1+c)\big]^{2\pi}_{0} d\theta_1 \\ &= (\frac{1}{2\pi}) ^2 \int_{0}^{2\pi} \big[\sin(\pm2\pi\pm\theta_1+c) - \sin(\pm0\pm\theta_1+c)\big] d\theta_1 \\ &= (\frac{1}{2\pi}) ^2 \int_{0}^{2\pi} \big[\sin(\pm0\pm\theta_1+c) - \sin(\pm0\pm\theta_1+c)\big] d\theta_1 \\ &= (\frac{1}{2\pi}) ^2 \int_{0}^{2\pi} 0 d\theta_1 \\ &= 0