Harmonic Function (time averages) II

[jeff1evesque]

Statement:
$$A(t) = |A|cos(\omega t + \alpha), B(t) = |B|cos(\omega t + \beta)$$

The time average of the product is given by:
$$<A(t)B(t)> = \frac{1}{T} \int^{T}_{0}A(t)B(t)dt = \frac{|A||B|}{2T}\int^{T}_{0}cos(2\omega t + \alpha + \beta)dt + \int^{T}_{0}cos(\alpha - \beta)dt] = \frac{1}{2}|A||B|cos(\alpha - \beta)$$Relevant equations:
Note: using trigonometric identities:
$$A(t)B(t) = \frac{1}{2} |A||B|[cos(2\omega t + \alpha + \beta) + cos(\alpha - \beta)]$$Questions:
How is $$\int^{T}_{0}cos(2\omega t + \alpha + \beta)dt = 0?$$
When I worked it out I got $$\int^{T}_{0}cos(2\omega t + \alpha + \beta)dt = \frac{1}{(2w)} sin(u) |^{T}_{0} = \frac{1}{2w}sin(2\omega t + \alpha + \beta)^{T}_{0} \neq 0$$Is there a real-world application for taking the product of two harmonic functions (even an application for phasors)??

thanks

 

[cepheid]

Statement:

Questions:
How is $$\int^{T}_{0}cos(2\omega t + \alpha + \beta)dt = 0?$$
When I worked it out I got $$\int^{T}_{0}cos(2\omega t + \alpha + \beta)dt = \frac{1}{(2w)} sin(u) |^{T}_{0} = \frac{1}{2w}sin(2\omega t + \alpha + \beta)^{T}_{0} \neq 0$$

HOW is it not zero? By definition, ωT = 2π

$$\sin(2 \omega t + \alpha + \beta) |_0^T$$

$$= \sin(4 \pi + \alpha + \beta) - \sin(\alpha + \beta) = 0$$

​

because,

$$\sin(\theta + 2n\pi) = \sin(\theta)$$​

The integral of a cosine function over any whole number of periods is zero (that much is obvious just by looking at it).