Hi all,
Why Es/No = SNR(Tsym/Tsamp)? It seems when sample rate increase, SNR will decrease?
Tsym is the period of symbol rate, Tsamp is over-sampled signal.
Any suggestions will be appreciated! Best regards, Davy
Es/No v.s. Over-sample SNR
Mar 21, 2006
5 Replies
It seems to me that by that equation, if sample rate increases, then Tsamp *decreases*, so SNR *increases*.
Oli
The more or less standard formula is Es/No = SNR * (bandwidth / symbol rate), where bandwidth is what was used to compute SNR. For a complex baseband system with white Gaussian noise, the bandwidth equals
1/Tsamp. So your formula is correct for this case.John
First, as mentioned, the equation indicates that SNR gets better as the sample rate increases.
Second, that equation would be correct if the noise of your system is dominated by white noise at the ADC -- in that case oversampling will allow less white noise to alias into your symbols. If you had a perfectly band limited signal and a noiseless ADC then that equation wouldn't be valid.
_Always_ understand the context of the problem you're working on, or your results will always be meaningless.
Tim Wescott
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Why?
SNR=(Es/No)*(Tsamp/Tsym), and Tsamp is proportional to SNR?
Davy said the following on 22/03/2006 13:49:
That's not the equation you first used...
Oli
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