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as i sow in the specification of nfa n8975a its could be possible to measure nf of a device that its noise figure greater than the ENR of the noise source. as i know the formula of noise figure is NF(db) = ENR (db) -10log (y-1). as i understand the ENR OF THE NOISE SOURCE MUST BE GREATER THAN THE NOISE FIGURE OF THE DEVICE. CAN MODERATOR EXPLAIN ME THIS MATTER.
THANK YOU
as i sow in the specification of nfa n8975a its could be possible to measure nf of a device that its noise figure greater than the ENR of the noise source. as i know the formula of noise figure is NF(db) = ENR (db) -10log (y-1). as i understand the ENR OF THE NOISE SOURCE MUST BE GREATER THAN THE NOISE FIGURE OF THE DEVICE. CAN MODERATOR EXPLAIN ME THIS MATTER.
THANK YOU
When the noise figure of a device is higher than the ENR of the noise source, the device noise tends to mask the noise source output. In cases such as this, the Y-factor will be very close to 1. Accurate measurements of small ratios can be difficult. Generally the Y-factor method is not used when the noise figure is more than 10 dB above the ENR of the noise source. For this measurement challenge there are two suggested methods:
1) The signal generator twice-power method
2) Direct noise measurement method
See application note 57-1 from Agilent Technologies.
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