- #1
jeff1evesque
- 312
- 0
Statement:
[tex]<v(t)> = \frac{1}{T} \int^{T}_{0}v(t)dt = \frac{1}{T} \int^{T}_{0}V_0cos(\omega t + \phi)dt \equiv 0.[/tex] (#1)Relevant Question:
If we suppose [tex]v(t)[/tex] is a complex vector, is the second equality above still true?Reasoning:
If [tex]v(t)[/tex] is a complex vector, then [tex]v(t) = cos(\omega t)\hat{x} + sin(\omega t)\hat{y}.[/tex]
But we also know (by sum to angle identity), [tex]cos(\omega t + \theta) = cos(\omega t)cos(\theta) - sin(\omega t)sin(\theta) \Leftrightarrow V_0cos(\omega t + \theta) = V_0[cos(\omega t)cos(\theta) - sin(\omega t)sin(\theta)= V_0[cos(\omega t)cos(\theta)][/tex]
But is the following true:
[tex]V_0[cos(\omega t)cos(\theta)] = cos(\omega t)\hat{x} + sin(\omega t)\hat{y} ?[/tex]
Because in the original equation (#1), if v(t) is a complex vector, I cannot see why [tex]v(t) = V_0cos(\omega t + \phi)?[/tex]thanks,
Jeff
[tex]<v(t)> = \frac{1}{T} \int^{T}_{0}v(t)dt = \frac{1}{T} \int^{T}_{0}V_0cos(\omega t + \phi)dt \equiv 0.[/tex] (#1)Relevant Question:
If we suppose [tex]v(t)[/tex] is a complex vector, is the second equality above still true?Reasoning:
If [tex]v(t)[/tex] is a complex vector, then [tex]v(t) = cos(\omega t)\hat{x} + sin(\omega t)\hat{y}.[/tex]
But we also know (by sum to angle identity), [tex]cos(\omega t + \theta) = cos(\omega t)cos(\theta) - sin(\omega t)sin(\theta) \Leftrightarrow V_0cos(\omega t + \theta) = V_0[cos(\omega t)cos(\theta) - sin(\omega t)sin(\theta)= V_0[cos(\omega t)cos(\theta)][/tex]
But is the following true:
[tex]V_0[cos(\omega t)cos(\theta)] = cos(\omega t)\hat{x} + sin(\omega t)\hat{y} ?[/tex]
Because in the original equation (#1), if v(t) is a complex vector, I cannot see why [tex]v(t) = V_0cos(\omega t + \phi)?[/tex]thanks,
Jeff