Level Correction For Digital Down-Convertion?
I think you miscalculated the power generated by your signal source; note that you need to take the load impedance into account:

$$
\begin{align}
P = \frac{V_P^2}{2R} &amp;= \frac{(100\ \text{mV})^2}{100\ \Omega} \\
&amp;= \frac{0.01\ \text{V}^2}{100\ \Omega} \\
&amp;= 0.1\ \text{mW} \\ 
\\
\end{align}
$$

$$
\begin{align}
P_{\text{dBm}} &amp;= 10 \log_{10}\left(\frac{P}{1\ \text{mW}}\right) \\
&amp;= 10 \log_{10}\left(\frac{0.1\ \text{mW}}{1\ \text{mW}}\right) \\
&amp;= -10.0\text{ dBm}
\end{align}
$$

Likewise, if your DDC output signal has a peak amplitude of $50\text{ mV}$, then you can repeat the above calculations to yield a power level of $-13\text{ dBm}$.