[SOLVED] Sum of 2 noise sources

[guetguet]

Hi everyone !

I would like to know the total noise power caused by two different noise sources. Let's assume the level of these noise sources to be respectively -90dBm and -107dBm (for example).

Hope you enjoy your day !

Regards.

 

[Russ Phillips]

I am assuming the impedance for both sources are the same and that they are being added together by an ideal adder. First, convert each power level from dBm to Watts. Next, calculate the RMS voltage of each source alone. Square each of these two voltages and add them, then take the square root. This gives the total RMS voltage. Calculate the power in Watts that corresponds to this and then convert back to dBm.

For the levels you specified for example, if the impedance is 50 Ohms:
P1 = 0.001*10^(-90/10) = 1 x 10^-12 Watts ==> V1 = SQRT(P1*50) = 7.07 x 10^-6 Volts
P2 = 0.001*10^(-107/10) = 19.95 x 10^-15 Watts ==> V2 = SQRT(P2*50) = 998.8 x 10^-9 Volts

VT = SQRT(V1^2 + V2^2) = SQRT( (7.07 x 10^-6)^2 + (998.8 x 10^-9)^2 ) = 7.141 x 10^-6 Volts
PT = (VT^2)/R = ((7.141 x 10^-6)^2)/50 = 1.01995 x 10^-12 Watts
Total Power in dBm = 10*log(PT/0.001) = 10*log(1.01995 x 10^-12/0.001) = -89.914 dBm

 

[albbg]

It's correct but a little bit more complex than needed. You don't need to convert to voltage, then back to power.

The uncorrelated noise sums in power (of course linear power that is W, mW, etc) then, if the two impedances are the same, regardless to the actual value:

Ptot(mW)=P1(mW)+P2(mW) ==> in dBm ==> Ptot(dBm)=10*log10[P1(mW)+P2(mW)]

since P(mW)=10^(dBm/10)

P1(mW)=10^(-90/10)
P2(mW)=10^(-107/10)

thus

Ptot(dBm)=10*log10[10^(-90/10)+10^(-107/10)] = -89.914 dBm

 

[guetguet]

Thank you ! I followed this idea in the morning and I am glad to know that I was right.
Thanks again and have a good day.

 

[Russ Phillips]

That is simpler. Thanks!