# Homework Help: Harmonic Function (time averages)

1. Jul 18, 2009

### jeff1evesque

1. The problem statement, all variables and given/known data
The time-average of any time-harmonic function is always zero.

2. Relevant equations
$$<v(t)> = \frac{1}{T} \int_{0}^{T}v(t)dt = \frac{1}{T} \int_{0}^{T}V_{0}cos(\omega t + \phi)dt \equiv 0$$

3. The attempt at a solution
A harmonic function I think is defined as a sinusoidal function with some sort of fixed period, and a given phase. Here, I am just trying to understand notes I am reading. I don't understand why there is fraction of $$\frac{1}{T}$$ in the equation above. Could someone also
explain to me why the time averages is initially defined as $$<v(t)> = \frac{1}{T} \int_{0}^{T}v(t)dt$$, and why it is set to zero.

Thanks,

Jeffrey

2. Jul 18, 2009

### Hootenanny

Staff Emeritus
In general, we define the average value of a function, f, on some interval [a,b] by,

$$\bar{f} = \frac{1}{b-a}\int^{a}_{b}f\left(x\right)\; dx$$

Effectively what we are doing is summing the value of the function, f, at a set of equally spaced points along the interval [a,b], then dividing by the number of points. We then take the limit as the number of points approaches infinity to obtain the formula quoted above. See http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/applications/average.html" for more information.

Now in terms of your integral, T represents the time interval and therefore we set T = b - a and integrate along the segment [0,T].

Now, let's take a closer look at your example. Your integrand is most definitely a time-harmonic function, so let's compute the average explicitly:

$$\bar{v} = \frac{V_0}{T}\int^T_0 \cos\left(\omega t + \phi\right)\; dt$$

$$\bar{v} = \frac{V_0}{T\omega}\left.\sin\left(\omega t + \phi\right)\right|_0^T\; dt$$

Hence, $\bar{v}$ is zero iff,

$$\sin\left(\omega T + \phi\right) = \sin\left(\phi\right)$$

which reduces to

$$\omega T = 2n\pi\;\;\;,\;\;\; n\in\mathbb{Z}[/itex] Hence, the time-average of a time-harmonic function is zero iff, [tex]T = \frac{n}{f}$$

i.e. iff the time interval of integration is an integer multiple of the period.

Last edited by a moderator: Apr 24, 2017
3. Jul 18, 2009

### jeff1evesque

That makes sense, but why does my notes say it is always zero?

4. Jul 18, 2009

### Hootenanny

Staff Emeritus
Without seeing your notes, I couldn't possibly say. Do your notes give any conditions or definitions on T?

5. Jul 18, 2009

### jeff1evesque

It just says, The time-average of any time-harmonic function is always zero.

6. Jul 18, 2009

### Hootenanny

Staff Emeritus
Well that simply isn't true, as you can plainly see from explicitly computing the time-average of a general time-harmonic function. If T was the period, then I would agree with the statement, but in general it isn't true.

Are you using a textbook, or are these class notes?

7. Jul 18, 2009

### jeff1evesque

PDF notes from the lecturer. In this particular section of notes, we are discussing complex vectors and phasor notation to discuss sinusoidal varying vectors.