00:01
All right, so this is a pretty fun problem.
00:03
We're told what the voltage is, and we're told first that the mean square of v is, by definition, the square root of the average value of v squared over the period.
00:19
So what does that mean? the period, we're told, is 1 over f.
00:24
So the interval is from 0 to 1 over f.
00:27
And the way that we find an average value is 1 over b minus a.
00:32
So that would be one over one over f.
00:35
The integral over that interval.
00:39
And then what we're integrating is the square of v.
00:42
So we have vp squared, sine squared of 2 pi, f t, d p.
00:49
So all this is is just the average value of v squared on this interval.
00:59
Okay, so let's clean this up a bit.
01:02
Note that vp squared is a constant, so i can factor that out.
01:06
And one over one over f is just f.
01:10
And all this is within the square root...