SNR on amplifier

[Simon]

Well it's that time again of doing my now HND course assignment which as always is in monkey see monkey do fashion given the very selective information in the course module. This one is causing me a dilema around powers and voltages.

An amplifier has a bandwidth of 80-862 MHz
a gain of 20dB
Noise figure of 6 dB
Max output 85 dbµV
Input impedance 75 Ohms
Output Impedance 75 Ohms

If the signal level from the aerial is 5 dBmV and the input noise level is 20 dBµV, calculate the signal-to-noise ratio on the output of the amplifier.

So firstly as the source impedance has not been specified (this figures in the modules calculations) so I would assume also 75 Ohms. So do I assume the amplifier see's the full 5 dBmV and 20 dBµV or half due to the cable also having an impedance of 75 Ohms (I assume).

To convert from dBmV to dBµV I am unsure if I add 30 or 60 dB. I gather that 20 log V/µV is caveated by the impedances being the same which is standard.

If i assume 60 dB then that does make my noise ratio more plausible. So the 5 dBmV is 65 dBµV, this makes sense as knowing how these people operate they like to make things all tie together to help you see if you are getting it right at 65 dBµV is the answers toe tthe previous question about the same amplifier asking for the maximum permissible input given the specs.

Does the SNR change whether expressed in dBmV or dBmW ? I guess being a ratio no as all things being equal the ratio is the same.

[pwlps]

Input SNR is 65dBµV-20dBµV=45dB, if NF=6dB the output SNR is 45+6=51dB
edit: oops! 45-6=39dB 

Does the SNR change whether expressed in dBmV or dBmW ? I guess being a ratio no as all things being equal the ratio is the same.

SNR is expressed only in dB, not dBmV or dBmW.  dBmV or dBmW express the power level as a ratio (thus expressed in dB) to some fixed level (mV, mW, uV etc.).  SNR is already a ratio, so it's in dB.  Hence the strange arithmetics rules : dBmV-dBmV=dB

[Simon]

They want the output

I thought about doing it like that but the 6dB I assume is on power so can you add/subtract dBW and dBV like that? I came up with 39dB of SNR after converting to voltage,

so 85 dBµV - 6 - (20+20) = 85-46 = 39 dB which was my result.

[pwlps]

Yes, I corrected my error.  Doesn't matter if you convert to V,W, mV,mW etc., it's always a power ratio.

[Simon]

Yea I realised this is why with volts the dB doubles to make it equivalent to power which is proportional to the square of the voltage. In log math a power operation becomes a multiplication.