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CASSPER Measurment 1

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MikeSundstrom

Electrical
Sep 18, 2008
3
I'm doing precompliance testing and I need to convert 37dBuV from a 30kHz BW measurment to a 120kHz BW measurment. I think it is a 10Log(BW2/BW1) x 37dBuV? Or will I need to convert the 37dBuV to uV and then convert? This is a narrow band signal. 50 ohm system. Any help would be greatly appreciated.

Mike

 
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UPDATE:
I have found that to convert dBuV to uV it is:

(dBuV/20)Log(inv)= uV

Sound about right? Still looking for the BW conversion formula.

Thanks Jeff for the posts.

Mike KB5UKT
 
Is the signal only noise? if so, the power will increase by 6 dB since you have 4x the bandwidth, but the voltage will increase only 3 dB, i.e. it'll double.

If it's an RF signal your measuring, changing bandwidth of a measurement device won't change any signal measurement (just the accuracy a bit since Signal is the same and noise is larger).

k

dB is always 10logX (log in base 10 of course), if you're converting volts to power it's 20logVoltage, but converting dBuV to uv would still be 10^37/10 or 10^3.7 or 5011 uv, or 5 millivolts. Pretty small voltage.
 
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