Let's take a look at the signal waveforms:

You are right there is a diode voltage drop, let's assume for all intents and purposes the diode forward voltage drop is \$0.635V\$.

To compute the RMS voltage:

$$ V_{rms} = \sqrt{\frac{1}{p} \int_0^p V(t)^2 dt} $$

where \$p\$ is the period (in this case 1ms).

What is the output voltage?

Let's assume for a second that when \$V_{IN} < V_{DIODE}\$, \$V_{OUT} = 0\$. This isn't quite true, but should get us close to the correct answer.

So our output voltage for one period is:

\begin{equation}
V_{OUT} = \left\{
\begin{array}{lr}
0 & : 20\mu s < t\\
5 \sin(1000 \cdot 2 \pi t) - 0.635 & : \text{otherwise}\\
0 & : t > 480\mu s
\end{array}\right.
\end{equation}

plugging into the \$V_{rms}\$ calculation,

\begin{equation}
V_{rms} = \sqrt{\frac{1}{1 ms}\int_{20\mu s}^{480 \mu s}(5 \sin(1000 \cdot 2 \pi t) - 0.635)^2 dt} \approx 2.1V
\end{equation}

The minor difference in calculated values here and your measured values are due to the assumptions I made about diode behavior (constant diode voltage drop, \$V_{OUT}\$ behavior when diode isn't saturated), as well as component behavior not being ideal, nor having exactly the same characteristics as those I chose for the calculations.

Ok, what was the average voltage across the same time period?

\begin{equation}
V_{avg} = \frac{1}{p} \int_0^p V(t) dt\\
V_{avg} = \frac{1}{1 ms}\int_{20\mu s}^{480 \mu s}(5 \sin(1000 \cdot 2 \pi t) - 0.635) dt \approx 1.287V
\end{equation}

Frequency is measured by how frequently the period is completed in one second. The output signal completes a period twice as fast as the input frequency, as you can see in the diagram.

This is because the input wave is symmetrical, half positive and half negative. Making all the negative components into positive ones *doubles* the positive components. The positive component is only half of the original period. Thus the period is halved and the frequency is doubled.

Image from Electrapk.com

## Best Answer

The \$\boxed{\text{average of }|x|}\$ is \$\color{red}{\text{not mathematically the same as the}}\$ \$\sqrt{\text{average of } |x|^2}\$

One computes an average value of a signal but the other computes the power associated with that signal.