how to analysis the output snr when input noise much higher than thermal noise

[justtry114]

Hi all,
I have a question about how to caulculate the output cnr when input is signal mixed with AWGN noise? Acorrding to my understanding, NF of the receiver is defined when input noise equals to thermal noise.
we have sr5500 channel simulator in our lab, we want to find the snr threshold for QPSK demodulation of our receiver. we configure the SR5500 output signal level to -60dBm and output CNR to 2.5dB by adding AWGN noise interference to the signal, receiver can demodulated signal without error. Does that means our baseband demodulation threshold for QPSK is 2.5dB which can be used to estimated our receiver NF after measure the receiver sensitivity? By eaqution, NF = 174(thermal noise floor) + Sensitivity - 10*log10(BandWidth) - SNR(threshold of QPSK demodualtion: 2.5dB?)

Here's my understanding, Receiver noise figure is defined when input is thermal noise floor. But when input noise is much higer than thermal noise, like our case when use SR5500 to find the QPSK demodulation trheshold, to say, -57.5dBm noise in receiver bandwidth. Then input SNR is 2.5dB, what's output SNR if our receiver NF is 3dB? Here's my understanding, If input signal level is Psin = -60dBm and receiver gain is 50dB, then receiver output signal level is Psin + receiver_Gain = -10dBm, noise output is (-57.5dBm + receiver_Gain) + (-174 + receiver_Gain + NF) = -7.5dBm + -121dBm = -7.5XXdBm. That's output SNR is still equals to input SNR 2.5dB? Is that right?

Thank you for your help~

 

[tony_lth]

1st correction: SNR=(-60dBm)-(-57.5dBm)=-2.5dB, not a 2.5dB.
2nd: if the noise is -57.5dBm, it is not noise any more, it is called spurs. It can desense the receiver, you must use sw to calculate the mixer products, if any product falls into the working band, it will desense the receiver.

 

[jiripolivka]

Hi all,
I have a question about how to caulculate the output cnr when input is signal mixed with AWGN noise? Acorrding to my understanding, NF of the receiver is defined when input noise equals to thermal noise.
we have sr5500 channel simulator in our lab, we want to find the snr threshold for QPSK demodulation of our receiver. we configure the SR5500 output signal level to -60dBm and output CNR to 2.5dB by adding AWGN noise interference to the signal, receiver can demodulated signal without error. Does that means our baseband demodulation threshold for QPSK is 2.5dB which can be used to estimated our receiver NF after measure the receiver sensitivity? By eaqution, NF = 174(thermal noise floor) + Sensitivity - 10*log10(BandWidth) - SNR(threshold of QPSK demodualtion: 2.5dB?)

Here's my understanding, Receiver noise figure is defined when input is thermal noise floor. But when input noise is much higer than thermal noise, like our case when use SR5500 to find the QPSK demodulation trheshold, to say, -57.5dBm noise in receiver bandwidth. Then input SNR is 2.5dB, what's output SNR if our receiver NF is 3dB? Here's my understanding, If input signal level is Psin = -60dBm and receiver gain is 50dB, then receiver output signal level is Psin + receiver_Gain = -10dBm, noise output is (-57.5dBm + receiver_Gain) + (-174 + receiver_Gain + NF) = -7.5dBm + -121dBm = -7.5XXdBm. That's output SNR is still equals to input SNR 2.5dB? Is that right?

Thank you for your help~

You are right if the receiver chain from input to output is linear. Then the input S/N or C/N respecting noise figure and bandwidth will remain the same at the output.

System analyzers can either vary the input signal power while the input thermal noise is constant, or, now an easier situation is to leave the signal level constant and vary the noise input. If the receiver is linear , both methods generate equal results.