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Design >> RF Design >> NF requirement of the RX - Baseband
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Message started by aaron_do on Mar 14^th, 2009, 8:46am

Title: NF requirement of the RX - Baseband
Post by aaron_do on Mar 14^th, 2009, 8:46am

Hi all,

I'm trying to understand the NF requirement of the RX versus the spectral efficiency of the modulation.

Suppose we have an FSK signal. The bandwidth of the signal according to carson's rule is 2(Δf + fm). The SNR required is Eb/N0*bitrate/BW. This means that if we increase Δf, we reduce the required SNR for demodulation. However, the required NF of the receiver will stay roughly the same because the noise bandwidth will increase by about the same amount as the reduction in required SNR. So from this analysis, required NF of the receiver is independent of bandwidth once the data rate and modulation scheme have been chosen.

However, a matched filter can be used to maximize the SNR of the received FSK signal. In wideband FSK, the total noise BW instead of being 2(Δf + fM) will only be 2fm (i think). So therefore, the required NF of the receiver is reduced when a wide tone-separation is used.

So basically have one question...have i misunderstood something here?

thanks,
Aaron

EDIT: the place i think i might have gone wrong is: I assumed the BW in carson's rule = bandwidth used in expressing Eb/N0 in terms of SNR BW and Bit Rate. I guess the BW used in that expression should be approximately 2fm, not carson's ru BW...