dB The exploration continues

[KM4FER]

About 1km from my house there is an AM broadcast station transmitting on 1.49Mhz.  Their signal gets into everything I try to do on a breadboard.  So,  just for fun I hooked up about 2m of antenna wire to one end of a 1k resistor and then attached my scope across the resistor with the probe on the wire/resistor connection and the ground clip on the opposite side of the resistor.  I get a beautiful eye pattern on the scope with a 35mV peak-to-peak voltage for the unmodulated carrier.  From this I can calculate the RMS voltage to be:

RMS = P2P/(2 X sqrt2) = 0.035/(2 X sqrt2) = 0.0124V = 12.4mV

The power dissipated in the resistor by this signal is:

P = E^2 / R = 0.0124^2 / 1000 = 154nW

Therefore the power loss in the resistor is:

dBm = 10 log P/1mW = 10 log( 154 x 10^-9 / 1 x 10^-3) = -38dBm

This makes sense since the power dissipated in the resistor is much smaller than the 1mW reference.


Now lets look at the voltage of the signal on the antenna:

dBmV = 20 log V/1mV = 20 log 0.0124 / 0.001 = 22dBmV

This also makes sense since the signal is a little over 10 times larger than the 1mV reference.
But if I reference the signal to 1V instead of 1mV then:

dBV = 20 log V/1V = 20 log 0.0124 / 1 = -38dBV

Again this makes sense since the signal is only about 1/100 the value of the reference voltage.


Fun stuff.

earl...