Finding the Mean and R.M.S. value

Finding the Mean and R.M.S. value Mean We begin with finding the mean value of a function using integration. The formula for this is given by Z b 1 Mean = y dt b−a a where a and b will be provided. Here we will go through an example. Example: Find the mean of y = cos (πt) between 0 and 0.5. Solution: Begin by writing down the formula Z b 1 y dt Mean = b−a a and substituting in what we know (y = cos(πt), a = 0 and b = 0.5) Z 0.5 1 Mean = cos (πt) dt 0.5 − 0 0 Z 0.5 1 = cos (πt) dt 0.5 0 Z 0.5 cos (πt) dt = 2 0

1 out front. Now we are ready to integrate 0.5 − 0 Z 0.5 Mean = 2 cos (πt) dt 0  0.5 1 = 2 sin (πt) (integrating cos to get sin) π 0     1 1 = 2 sin (π × 0.5) − sin (π × 0) π π   1 2 = 2 −0 = π π

simplifying the

1

R.M.S. Now for the R.M.S., the formula in this case is a little different. s Z b 1 R.M.S. = y 2 dt b−a a You will either be given a and b or need to work them out. If you need to work them out you need to remember that the R.M.S. is calculated by integrating over the period of y. Let us take an example. Example: Find the R.M.S. value of y = sin (3πt). Solution: We haven’t been given any limits for the integration so we need to calculate the period. 2π Period = ω where ω is the angular velocity which is the number multiplying t in our function, so 3π in this case. Therefore, Period =

2π 2 2π = = . ω 3π 3

Now we can write down our formula s R.M.S. =

1 b−a

Z

b

y 2 dt a

and substituting in what we know (y = sin (3πt), a = 0 and b = 23 ). {We integrate from 0 to 32 because all we know is the length of a period, it doesn’t matter where we start but starting at 0 will make it easier.} s Z 2/3 1 R.M.S. = [sin (3πt)]2 dt 2/3 − 0 0 s Z 2/3 1 = sin2 (3πt) dt 2/3 0 s Z 3 2/3 2 = sin (3πt) dt 2 0 Remember [sin(x)]2 = sin2 (x).

2

Now we don’t know a formula for integrating sin2 (3πt) so we need to replace it with something. We’ll use the formulae given to us sin2 (x) =

1 1 (1 − cos (2x)) or cos2 (x) = (1 + cos (2x)) 2 2

We’ll use the one for sin2 (x) and so 1 1 (1 − cos (2 × 3πt)) = (1 − cos (6πt)) 2 2 which we know how to integrate. Substituting this in and simplifying gives s Z 3 2/3 2 R.M.S. = sin (3πt) dt 2 0 s Z 3 2/3 1 (1 − cos (6πt)) dt = 2 0 2 s   Z 1 3 1 2/3 (1 − cos (6πt)) dt moving the × = 2 2 0 2 s Z 3 2/3 = (1 − cos (6πt)) dt ( simplifying ) 4 0 sin2 (3πt) =

Now we can integrate s Z 3 2/3 (1 − cos (6πt)) dt R.M.S. = 4 0 s  2/3 1 3 = sin (6πt) t− 4 6π 0 putting the limits in gives us s      1 2 3 2 1 = − sin 6π × sin (6π × 0) − 0− 4 3 6π 3 6π which can be evaluated to give s    3 2 = −0 −0 4 3 r 3 2 = × 4 3 r 1 = = 0.707 2 3