In a receiver the input signal is 100 V, while the internal noise at the input is 10 V. With amplification the output signal is 2 V, while the output noise is 0.4 V. The noise figure of receiver is
NF = (100/10)/(2/0.4) or 2.
A receiver is operated at a temperature of 300 K. The transistor used in the receiver have an average output resistance of 1 k. The Johnson noise voltage for a receiver with a bandwidth of 200 kHz is
v2 = 4kBTR where symbols have their usual meanings, hence v = 1.8 µV.
A resistor R 1 k is maintained at 17C. The rms noise voltage generated in a bandwidth of 10 kHz is
v2 = 4kBTR where symbols have their usual meanings, hence v = 0.4 µV.
A mixer stage has a noise figure of 20 dB. This mixer stage is preceded by an amplifier which has a noise figure of 9 dB and an available power gain of 15 dB. The overall noise figure referred to the input is
The required answer is 7.94 + (100-1)/31.62 = 11.07.
A system has three stage cascaded amplifier each stage having a power gain of 10 dB and noise figure of 6 dB. the overall noise figure is
The gain at each stage is 10db, hence the required answer is 4 + (4-1)/10 + (4-1)/100 or 4.33.
(Q.6-Q.8) An amplifier when used with a source of average noise temperature 60 K, has an average operating noise figure of 5.6.
Q. The Te is
The required answer is 60 (5-1) or 240K.
If the amplifier is sold to engineering public, the noise figure that would be quoted in a catalog is
The required answer is 1 + (240/290).
What average operating noise figure results when the amplifier is used with an antenna of temperature 30 K?
The required answer is 1 + 240/30 = 9 or 9.54 db.
What is the maximum average effective input noise temperature that an amplifier can have if its average standard noise figure is to not exceed 1.7?
The required answer is 290 x (1.7-1) or 203K.
If a matched attenuator with a loss of 3.2 dB is placed between the source and the amplifier’s input, what is the operating spot noise figure of the attenuator amplifier cascade if the attenuator’s physical temperature is 290 K?
Here Te is 290[(20.89-1) + (2.089)(7.98-1)] or 4544.4K. Hence the required answer is 1 + 4544.4/290 = 212 or 13.3 db.