Time-harmonic functions (Complex Vectors)

[jeff1evesque]

Statement:
<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. (#1)Relevant Question:
If we suppose v(t) is a complex vector, is the second equality above still true?Reasoning:
If v(t) is a complex vector, then v(t) = cos(\omega t)\hat{x} + sin(\omega t)\hat{y}.

But we also know (by sum to angle identity), 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)]

But is the following true:
V_0[cos(\omega t)cos(\theta)] = cos(\omega t)\hat{x} + sin(\omega t)\hat{y} ?
Because in the original equation (#1), if v(t) is a complex vector, I cannot see why v(t) = V_0cos(\omega t + \phi)?thanks,
Jeff

 

[cepheid]

I don't really see how v(t) ( as you have defined it ) is a complex vector in the first place.

 

[jeff1evesque]

What I wrote was really,
<br /> v(t) = cos(\omega t)\hat{x} + sin(\omega t)\hat{y} = A\hat{x} + jB\hat{y}<br />

the left equality is really "instantaneous form".

 

[cepheid]

The fact remains that v(t) is not a complex vector. Sure, as we've talked about before, you can express the cosine function or the sine function as phasors if you like. I just don't see what that has to do with anything. Perhaps you could clarify your question? If your question is:

What is \langle \textbf{v}(t) \rangle if \textbf{v}(t) is defined as:

\textbf{v}(t) \equiv \cos(\omega t) \hat{\textbf{x}} + \sin(\omega t) \hat{\textbf{y}}​

Then I think that the answer is:

\langle \textbf{v}(t) \rangle \equiv \frac{1}{T} \left( \hat{\textbf{x}} \int_0^T \cos(\omega t)\, dt + \hat{\textbf{y}} \int_0^T \sin(\omega t)\, dt \right)

= 0\hat{\textbf{x}} + 0\hat{\textbf{y}}​

which kind of makes sense when you think about it, since the vector goes around in a circle and ends up back where it started.

 

[jeff1evesque]

I am just curious because in my notes it says "The time-average of any time-harmonic function is always zero." That part makes sense, I think you answered that last time for me. But I don't understand why v(t) = V_0cos(\omega t + \phi)? In previous pages of my notes, we've defined v(t) as both complex vectors, and complex vectors in the instantaneous form (I wrote above).

 

[cepheid]

Without seeing your notes, it is difficult for me to comment further. The fact is,

\cos(\omega t) \hat{\textbf{x}} + \sin(\omega t) \hat{\textbf{y}} \neq V_0\cos(\omega t + \phi)​

The thing on the left hand side is a vector and the thing on the right hand side is a scalar. EDIT: As far as I can see, two different things are being talked about and are being referred to as "v(t)."

 

[jeff1evesque]

Since cosine and sine can be written as one another- by shifting the phase,

<br /> \frac{1}{T} \int^{T}_{0}v(t)dt = \frac{1}{T} \int^{T}_{0}V_0cos(\omega t + \phi)dt.<br />

Suppose \phi = 90\circ, then it follows

<br /> \frac{1}{T} \int^{T}_{0}v(t)dt = \frac{1}{T} \int^{T}_{0}V_0sin(\omega t)dt = - \frac{\omega V_0cos(\omega t)}{T}|^{T}_{0} \neq 0<br /> \Leftrightarrow T \neq 0

Doesn't that mean that a time-harmonic function is not always zero?

Thanks for reading my posts,JL

 

[cepheid]

\cos(\omega T) = \cos \left(\frac{2 \pi}{T}T\right) = \cos(2 \pi) = 1

\cos(0) = 1

\Rightarrow \cos(\omega T) - \cos(0) = 1 - 1 = 0​

Without doing any math, you can see that the integral of a sinusoidal function over any whole number of periods will be zero, because you will always be adding together an equal amount of "positive area" and "negative area" under the curve.

 

[jeff1evesque]

thanks

 

[Hootenanny]

thanks

As I said in the associate thread that I was helping you with, this only applies if T is in integer multiple of the period.

 

[jeff1evesque]

As I said in the associate thread that I was helping you with, this only applies if T is in integer multiple of the period.

For some reason in this example, I thought v(t) was some complex vector. Instead it shows that any sinusoid for harmonic functions have an time-average of zero, which is exactly what you said.

Thanks,JL

 

[cepheid]

As I said in the associate thread that I was helping you with, this only applies if T is in integer multiple of the period.

Oh yeah. I assumed T WAS the period. But I guess T is the time interval over which you are averaging. In any case, I already stated as well that the integral would be zero over some whole number of periods, so we both got the disclaimer in there.

 

[Hootenanny]

Oh yeah. I assumed T WAS the period. But I guess T is the time interval over which you are averaging. In any case, I already stated as well that the integral would be zero over some whole number of periods, so we both got the disclaimer in there.

No problem, I just thought that I would clarify since in the OP's previous question he stated that he didn't know what T was and it wasn't defined in his notes.