Signal Interference Noise Ratio Calculation

[Giowizzy]

Hi I'm a newby to the Electronic world and was given this problem to solve - please help by breaking down the problem

Transmitter TX1 sends data to Receiver RX without interference, the SNR is 10 db. Now suppose that TX2 starts transmitting resulting in interference. Suppose TX2 transmits at the same power as TX1 and the path loss from TX2 to RX is 5dB greater than the path loss from TX1 to RX. What's the SINR from TX1 to RX when TX2 is transmitting?

Thanks in advance

 

[chuckey]

SNR = signal +noise/noise, so in power terms signal = 9 units, noise = 1 unit ( 9+1 / 1 = 10 = 10dB). 5db down on TX1 = 9 X .31 = 2.79. New noise is therefore 1+2.79, so new equation is 9+3.79 /3.79 or 12.79/3.79 = 3.37 or 5.3 dB. Any better offers?
Frank

 

[Giowizzy]

Chuckey, I so appreciate it. Make sense to me now... how about this?

Suppose a receiver consists of an low noise amplier with a gain of 30 dB and noise gure of
2 dB, followed by a second stage of amplication of another 20 dB with a noise gure of 10
dB.
(a) What is the total noise gure and gain of the system?
(b) Suppose a 10 dB attenuator is placed at the input of the LNA. What is the resulting
overall gain and noise gure?
(c) What if the attenuator is placed at the output of the LNA?

 

[chuckey]

Similar way, NF = actual noise/ thermal noise. TH1 = total thermal noise from LNA1. O/P of LNA1 = 1000(30db)X ( TH1). Adding second amp O/P = 100(20 db)X (TH2)+100 X 1000X(TH1). So you can see that the thermal noise in the first stage is amplified by 100,000 and that due to the second stage, 100 fold. so it can be ignored in practise unless its a daft figure like 20db or so. So (a). NF = 2db. (b)1.gain = +30+20-10 = 40dB. Noise figure stays the same - no extra thermal noise (c) as (b). The signal to noise ration will change between (b) and (c), because the signal will be smaller in (b) and both the signal and noise will be reduced in (c).
Frank