Level Correction For Digital Down-Convertion?

Question

I am trying to understand a subtle (sorry if this is obvious) difference in the computed output power after a DDC process. Here is the story:

Suppose I generated a 10 MHz signal from a microwave source at -10 dBm and feed it to the a digitizer. This corresponds to a peak-to-peak voltage of 200 mV as measured by an oscilloscope for 50 Ohm lines and termination.

With the digitizer and by performing DDC, I am getting around +4 dBm as a result, 14 dBm above the expected -10 dBm of the source. Why is that?

I tried to go through the math step-by-step:

With the 200 mV input peak-to-peak signal, by multiplying this signal with a cos (for I) and -sin (for Q) with amplitude equals 1, I will obtain (suppose the signal has zero phase) I_DC = 50 mV and Q_DC = 0 mV after filtering out the high frequency part, just by trigonometry.

The power is then computed as P=I^2+Q^2= 2.5 mW. Taking log, we get P_log = 10*log(2.5mW) ~ -26 dB = 4 dBm. This is around 14 dBm higher than what I originally generated (-10 dBm).

If I wasn't missing something in the calculation, it seems I need a level correct factor of 14 dBm. But where did this gain come from? I guess by duplicating the signal to generate the I and Q there is a 6 dB gain. Then where is the remaining 8 dB from?

Appreciated if you can shine some light on it.

Edit1: (Answer?) Following the first reply, I might need to consider the 50 Ohm, so it will bring the power down by a factor 10*log(1/50), and it gives now P_dBm = 10*log(2.5e-3*1000/50) = -13.0103 dBm. I am now 3.0103 dB off, and it seems it comes from the trigonometry factor of half. By putting it back, I come back to the -10 dBm as desired.

Answer 1

I think you miscalculated the power generated by your signal source; note that you need to take the load impedance into account:

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

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

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 $-13\text{ dBm}$.

Answer 2

To add to Jason's correct answer on doing the math and including the 50 ohm impedance, the reason for the 3 dB difference is the DDC is a complex down-conversion, so only moves one of the two sidebands that represent the real sinusoidal 10 MHz signal to baseband.

See the first figure under the title "Downconversion" in this answer for a more detailed explanation. The properly designed DDC includes a low-pass filter to remove the higher frequency image that contains the other half of your signal power.

Frequency shifting of a quadrature mixed signal

(The reason the DDC output has "I" and "Q" is because it represents a complex signal. You need two real signals to define a single complex signal)

To help follow this explanation, here is a diagram from Wikipedia showing the "Digital Downconverter" structure. But personally I do not refer the the digital source as a DDS (Direct Digital Synthesizer) but a NCO (Numerically Controlled Oscillator). I reserve "DDS" to mean the combination of an NCO with a D/A converter such that it is a digitally controlled source with an analog output. I suspect but am not certain that the naming convention I follow for the DDS and NCO structures is standard practice. I also would not call the inputs/outputs "data" but rather "waveform".

This diagram (after reading my linked post this additional comment will be clearer) is a single complex multiplier with a real input (data in), and a complex input of a single tone (from the NCO) and a single complex output (represented as the I and Q real outputs).