Hi,
Crutschow is correct.
I'm doing some math with 50 Ohms (just guessing)
-16dBm means -16dB × 1mW = 0.025mW
P = V × V / R --> V = sqrt(P × R) = sqrt (0.025mW × 50 Ohms) = 35.4 mV RMS.
Vpp = 2 × sqrt(2) × V_RMS = 100 mV
So your math seems to be correct.(for 50 Ohms)
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So what's happening?
The generator has a source impedance of 50 Ohms (to match the characteristic cable impedance and to match the termination resistance = load tesistance)
And it expects to drive a 50 Ohms load, thus it expects a voltage divider with 50 Ohms and 50 Ohms.
(Mind: the characteristic cable impedance does not need to be used, since it does not cause a voltage drop, just a delay in time)
The output voltage of this voltage divider is half of the input voltage
Or in other words the input voltage has to be twice of the (expected) output voltage.
So, by expecting a load impedance of 50 Ohms the generator drives 200mVpp into it's internal 50 Ohms source impedance.
But you did not connect (most probably) a 50 Ohms load, thus there is negligible current and thus negligible voltage drop across the source impedance.
--> You see the full generator voltage (200mVpp) at the output.
Klaus