Finding the mean

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Jakeboulter

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I need to find the mean voltage over a range of 0 less than or equal to t less than or equal to 3.6ms the function is v=220sin140*pie*t
The integration part I am ok with but I'm struggling to find the mean. Any help would be appreciated.
 
[MATH]\overline{V} = \displaystyle \dfrac 1 T \int_0^T v(t)~dt[/MATH]
 
Thank you so would it be right in saying I take the total voltage over 3.6ms and dividing it by 3.6.
 
if by "total voltage" you mean the integral of the voltage signal between 0 and 3.6ms then yes.
 
Jakeboulter said:
I need to find the mean voltage over a range of 0 less than or equal to t less than or equal to 3.6ms the function is v=220sin140*pie*t
The integration part I am ok with but I'm struggling to find the mean. Any help would be appreciated.
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Is this a problem from the math class or engineering/physics class? In engineering/physics, mean voltage may refer to RMS (root-mean-square) voltage.
Jakeboulter said:
Thank you so would it be right in saying I take the total voltage over 3.6ms and dividing it by 3.6.
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That statement is ambiguous at best. Why don't you perform the integration and then division. I am not sure - why do you have to say anything!
 
This is a maths problem. This is what I have worked out that is wrong, I am struggling to find out where I went wrong.
 

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Jakeboulter said:
This is a maths problem. This is what I have worked out that is wrong, I am struggling to find out where I went wrong.
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As I read your work, the integration is incorrect.

Hint:

\(\displaystyle \int \sin(\omega t) dt = - \frac{1}{\omega}\cos(\omega t) + C\)
 
You appear to have integrated [MATH]\sin(140)\pi t[/MATH], rather than, as I assume it is meant to be, [MATH]\sin(140\pi t)[/MATH].

The parentheses are important!
 
Jakeboulter said:
I need to find the mean voltage over a range of 0 less than or equal to t less than or equal to 3.6ms the function is v=220sin140*pie*t
The integration part I am ok with but I'm struggling to find the mean. Any help would be appreciated.
Click to expand...

Just as an aside, it's pi, or \(\displaystyle \pi\).
Pie is what you eat too much of after dinner.
 
Dr.Peterson said:
You appear to have integrated [MATH]\sin(140)\pi t[/MATH], rather than, as I assume it is meant to be, [MATH]\sin(140\pi t)[/MATH].

The parentheses are important!
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Thank you I did wonder about that as the question didn't come with parentheses.
 
Romsek said:
Just as an aside, it's pi, or \(\displaystyle \pi\).
Pie is what you eat too much of after dinner.
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Haha it is pi. I've not quite got the hang of the symbols on this forum yet, and should probably read through my text before a post.
 
Jakeboulter said:
Thank you I did wonder about that as the question didn't come with parentheses.
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If the problem was given without parentheses, they probably did still mean what I took it as; it is not uncommon to mean [MATH]\sin(140\pi t)[/MATH] when writing [MATH]\sin140\pi t[/MATH], though I consider that bad practice. Context would make it clearer what is intended; but I can't imagine any reason to mean [MATH]\sin(140)\pi t[/MATH].
 
Ok I am still stuck. I have tried a different way and am now getting negative answers. Surely voltage can't be negative.
 

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Cos(0) = 1 which is the maximum value of the cosine function. So how are you getting a negative answer? Please show your work.
You really should use equal signs! Your 3rd line does not equal your 2nd line! Your 5th line does not equal your 4th line! Why do you write your math so sloppy? if you want to do do side calculations, that is fine but do it on the side as opposed to right in the middle of your work?
 
This is how I worked it out roughly the very first time I looked at the question but I wouldn't have thought the answer would be 0. For some reason I just cant get this one.
 

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Jakeboulter said:
This is how I worked it out roughly the very first time I looked at the question but I wouldn't have thought the answer would be 0. For some reason I just cant get this one.
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Was the number"140" given to you - or was it derived by you from some other (larger) part of the problem?
 
Since [MATH]140\pi[/MATH] times 3.6 is an integer multiple of [MATH]2\pi[/MATH], you are integrating over a number of full cycles, so the mean voltage is indeed 0.

We need to go back to the question raised earlier: do you perhaps really need the RMS voltage, or some other kind of average?

This problem may have been assigned precisely to make you aware of these issues!

But it will help if you show us the exact, entire problem as given to you. The specific wording may be important.
 
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