RMS (root mean square) of sums of functions


by lynx1245
Tags: functions, root, square, sums
lynx1245
lynx1245 is offline
#1
Jan24-07, 01:31 AM
P: 2
1. The problem statement, all variables and given/known data

The voltage across a resistor is given by:
[tex]
v(t) = 5 + 3 \cos{(t + 10^o)} + \cos{(2 t + 30^o)} V
[/tex]
Find the RMS value of the voltage

2. Relevant equations
For a periodic function, [tex]f(t)[/tex], the rms value is given by:
[tex]
f_{rms} (t) = \sqrt{\frac{1}{T} \int_{0}^{T} f(t)^2 dt}
[/tex]
Where T is the period.

3. The attempt at a solution
I know that the solution is given by:
[tex]
v_{rms} (t) = \sqrt{5^2 + (\frac{3}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2} V
[/tex]

It seems that you take the sum of the squares of the respective RMS value of each piece of the original voltage. I can't figure out why you do this though. I don't think applying the equation given will easily give you this answer. It's hard to even find a period to integrate over from the original voltage equation. Any insight into why the sum of squares works would be helpful.
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SGT
#2
Jan24-07, 09:29 AM
P: n/a
To calculate the integral, you must express all your angles in radians and use the formula for the sum of arcs. So
[tex]3cos(t + 10^o) = 3cos(t + 10\pi /180) = 3[cos t cos\pi/18 - sin t sin \pi/18][/tex]
The period of integration is [tex]2 \pi[/tex], the period for the lower frequency 1 rad/s.
antonantal
antonantal is offline
#3
Jan24-07, 10:06 AM
P: 219
Quote Quote by lynx1245 View Post
It seems that you take the sum of the squares of the respective RMS value of each piece of the original voltage. I can't figure out why you do this though. I don't think applying the equation given will easily give you this answer. It's hard to even find a period to integrate over from the original voltage equation. Any insight into why the sum of squares works would be helpful.
First, you should take a look at why the RMS value of a cosinusoid is [tex]\frac{amplitude}{\sqrt{2}}[/tex] :http://en.wikipedia.org/wiki/Root_mean_square

Now, in your particular case you have a voltage composed of a DC component and two cosinusoids. In the formula of the RMS value,
[tex]f_{rms} (t) = \sqrt{\frac{1}{T} \int_{0}^{T} f(t)^2 dt}[/tex] you will have the integral of the sum of this components, which, because integration is a linear operation, equals the sum of the integral of each component. The integral of each component equals its RMS squared (and multiplied by T), and so the formula works.

The period of integration must be a common multiple of the periods of the individual components, because, any multiple of the main period of a signal is also a period. So you can take the period [tex]2 \pi[/tex] as SGT said. Also note that the RMS value of the DC component is it's actual value no matter what period you take since it's value doesn't vary over time.

EDIT: Also, I should have mentioned that the integral of the product of any two different components from your voltage over a period equals zero. The components are said to be orthogonal. You can verify this graphically, knowing that the integral is the sum of the signed areas between the function's graph and the abscise.

lynx1245
lynx1245 is offline
#4
Jan24-07, 07:43 PM
P: 2

RMS (root mean square) of sums of functions


Thanks for the help guys. Here's how I did it.

I expanded [tex]v_{rms}(t)^2[/tex] to:
[tex]v_{rms}(t)^2 = 5^2 + 30 \cos{(t+10^o)} + 10 \cos{(2t+30^o)} +9 \cos^2{(t+10^o)} + 6 \cos{(t+10^o)} \cos{(2t+30^o)} + \cos^2{(t+10^o)}
[/tex]

and using the product to sum identity:

[tex]
\cos{x} \cos{y} = \frac{1}{2} (\cos{(x-y)} + \cos{(x+y)})
[/tex]

Its easy to see that all the lone cosine terms and the product of the cosines will go away through the integration and all that will be left is the RMS of each respective term in the original equation.


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