# Average value of a wave

1. Jan 10, 2015

### ranju

1. The problem statement, all variables and given/known data
In the given fig. we have to find average value of the voltage wave . So , for that we should take limits from 0 to pi/2.!! But in solution it is taken as pi/2 to 3pi/2..! I did'nt get this..why so???

2. Relevant equations
average value of voltage = 1/pi limit integral Vmcos(theta)dtheta

3. The attempt at a solution
I am finding the average value by taking the limits fro 0 to pi/2..!!

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2. Jan 10, 2015

### Stephen Tashi

It would be best if you stated the problem exactly as it was worded.

As to relevant equations, state how you course materials define the average value of a wave. You stated a limit but didn't show what variable it applies to or what the variable approaches.

3. Jan 11, 2015

### ranju

A voltage wave is represented by a cosine function with 2 pi time period in given fig. as v=Vmcostheta . Find its average value.

4. Jan 11, 2015

### Stephen Tashi

You didn't clarify how your course materials define "average value" of a function. This is important.

In engineering, the important average associated with a perioid function is its "root mean square" value, which could be viewed as a type of average.

Taking the definition of average value as the area under the curve over one period divided by the length of the period, the average value of the cosine function would be zero due the cancellation of positive and negative areas.

Taking the definition of average to mean the area under the absolute value of the function over one period divided by the length of the period would give a nonzero answer..

If you integrate from 0 to pi/2 and divide the result by pi, I don't know what you are computing.

5. Jan 11, 2015

### ranju

in my course material average value for a sinusoidal wave is vav = 1/pi limit 0-pi integral Vm *sinwt*d(wt).. i.e. wave is integrated for half cycle..!!
And in the given problem , vav= 1/pi limit 0-pi/2 integral Vm cos theta d(theta)
And I got ans. as -Vm/pi... but in the book they have taken limits as pi/2 - 3 pi/2 instead of 0-pi/2.>!

6. Jan 11, 2015

### Staff: Mentor

We all know that the average value of a sine wave is zero, because it spends as much time above the x-axis as it does below the x-axis. So it is more likely that you are wanting to find the average value of one half-cycle of the sine wave. (This is appropriate once you rectify the sinewave; the period of the waveform halves once it is fullwave rectified.)

To find this average you can integrate over any half cycle you like, they are all identical. Just make sure the waveform starts from a particular level (say, 0V) and ends at that same level, and doesn't cross the x-axis anywhere between. That way you average over the rectified waveform's complete period.

The limits of the integration don't matter, provided they fall at the ends of one complete cycle (or half-cycle, here).

Last edited: Jan 11, 2015
7. Jan 11, 2015

### Stephen Tashi

The stated definition $av = \frac{1}{\pi} Lim_{? \rightarrow ? } \int_0^\pi sin(\omega t) d(\omega t)$ does not give complete information for the limit.

The usual definition of the average value of a function specifies the interval to be averaged.
The average value of $f(x)$ over the interval $[a,b]$ is defined as $\frac{1}{(b-a)} \int_a^b f(x) dx$.

If we apply that definition to the problem you stated, then, as Nacesent Oxygen pointed out, the answer is zero. If we don't want the answer to be zero, we have to speculate about what the problem meant to say.

8. Jan 11, 2015

### Staff: Mentor

For a periodic waveform, it's always averaged over one period. That goes without saying.

9. Jan 12, 2015

### leright

To clear up any confusion, RMS and average are two different but related quantities. For example, a sine wave with no offset has an average of zero but the RMS value is non-zero and positive.

The RMS value is simply the square root of the mean of the square of the waveform (hence 'RMS') . The RMS current (or voltage depending on your waveform) of a waveform is also equal to the DC current (or voltage) that produces the same amount of power dissipation as the waveform.

The average value of a PERIODIC signal is the integral (or area under) of the waveform over one period divided by the period. If the signal is aperiodic you need to be given the time period over which you average. The average of an APERIODIC signal is the area under the curve in the time period specified divided by the length of the time period.

It appears you are correct though. For a periodic waveform you integrate over one period and then divide by the period. They did not integrate over one period but only half a period.

However, it should be obvious that the average over one period is ZERO without even calculating anything.

Last edited: Jan 12, 2015
10. Jan 12, 2015

### leright

Oh, and you'd divided by 2pi, not pi. The period is 2pi.

11. Jan 12, 2015

### leright

Perhaps they want you to find the average of the cosine wave over only half a period. Please write exactly what the problem says.

Last edited: Jan 12, 2015
12. Jan 12, 2015

### leright

Unless the problem says to average over only half a period.

13. Jan 12, 2015

### Staff: Mentor

It would never say that as a complete goal, except where the average over half a period equals the average of the desired waveform over the whole period. (But there could be confusion in interpreting what is fundamentally going on.)

14. Jan 12, 2015

### leright

Why not? It wouldn't be the first time I've seen an unusual problem. It also wouldn't be the first time I've seen a mistake in a solutions manual.

I think the OP needs to post the question verbatim before we can help.

15. Jan 12, 2015

### ranju

I'hv written the complete problem.. my point is ..since for a complete cycle the average value of sine wave is zero..so we'll integrate for half cycle.. and I'hv taken the limits 0-pi/2 .but the ans coming is -Vm/pi.. but its wrong as per th textbook.. so what's my mistake..??

16. Jan 12, 2015

### Staff: Mentor

I think we are now waiting to see your working. If in your working you integrate from 0 to $\frac {\pi} 2$ then you will average that by dividing by $\frac \pi 2$.

17. Jan 12, 2015

### Staff: Mentor

0 .. pi/2 is a quarter cycle.

18. Jan 12, 2015

### ranju

I'hv shown my working in previous posts..
but then what limits we are going to take..!! if we integrate taking limits 0 to pi ..average value will be zero.>!!

19. Jan 12, 2015

### Staff: Mentor

It's handy when there exist multiple or conflicting interpretations of an ambiguously worded problem―because this allows you to choose whichever interpretation leads to the answer you think you should get.