RMS of Fullwave rectified sine wave.

[pooface]

Homework Statement

Determine the RMS value of fullwave rectified sine wave.

Homework Equations

RMS = \sqrt{({1}/{b-a})\int^{b}_{a}[(fx)]^{2}dx}

The Attempt at a Solution

Notes: The Period of a full wave rectified sine wave is pi.

a=0
b=pi

Let's do square root at the end.

=1/pi \int^{pi}_{0}sin^{2}xdx

=1/pi [pi/2 - [sin(2pi)]/4] - 1/pi [pi/2 - [sin(2pi)]/4]

=1/2 - 1/2 ? ? ? ? ? ?

Where am i going wrong? Sorry I am not good with latex code, even with the reference.

 

[Mindscrape]

So you used the double angle identity.

sin^2(x) =\frac{1}{2}(1-cos(2x))

\int_0^{\pi} \frac{1}{2}(1-cos(2x)) = \frac{\pi}{2} - \left[sin(2x)/4 \right]^{x=\pi}_{x=0}

\int_0^{\pi} \frac{1}{2}(1-cos(2x)) = \frac{\pi}{2}-(0)

I'm not sure what you were doing.

 

[pooface]

I used integral of sin^2(u) du is = u/2 - [sin(2u)]/4 + C

When i sub pi in the term sin2u , then this becomes sin2pi which is 0.

 

[Mindscrape]

Right so you have

\sqrt{\frac{1}{\pi}\frac{\pi}{2}}

RMS = 1/√2