I was trying to figure out if the relationship between SNR and Es/N0 changes with roll-off factor.

  • $W$: RF bandwidth.
  • $B$: Base band bandwidth. $W = 2B$.
  • $E_s$: Symbol energy.
  • $P_s$: Symbol power.
  • $T_s$: Symbol period.
  • $P_n$: Noise power.
  • $N_0$: Noise spectral density.

Firstly, if we use sinc (Nyquist) pulses, we have $T_s·W=1$ and hence, $P_s/P_n = \frac{E_s/T_s}{N_0·W} = E_s/N_0$

Now, we use a raised cosine pulse with roll-off factor $\beta$. The channel bandwidth is increased into $W'=W(1+\beta)$. It also holds that $T_s·W'=(1+\beta)$.

Then, I would like to confirm that this is correct:
$P_s/P_n = \frac{E_s/T_s}{N_0·W'} = \frac{E_s·W'/(1+\beta)}{N_0·W'}= \frac{E_s}{N_0}\frac{1}{(1+\beta)}$
I would appreaciate if someone can confirm whether this reasoning is correct. Thanks.

  • $\begingroup$ After some reflection I feel it is quite logical, the symbol power is the same, but noise power has increased in a factor (1+$\beta$)... $\endgroup$ – MaLog Aug 15 '20 at 16:38
  • $\begingroup$ No it is not. Presuming this is root raised cosine (RRC) pulse shaping filter and you use match filter to demodulate, and your sinc and RRC filters are normalized, the SNR and EbNo should be intact with respect to the rolloff factor. Differently speaking, the noise at the output of your RRC is unchanged by changing $\beta$. Nonetheless, if your RRC filter is designed to output more than one samples per symbol, the SNR is reduced accordingly. Check this example fr.mathworks.com/help/comm/ug/… $\endgroup$ – AlexTP Aug 15 '20 at 17:09
  • $\begingroup$ There is not relationship between $β$ and the noise at the output of the RRC, but your last equations should be justified correctly. $\endgroup$ – Zeyad_Zeyad Aug 16 '20 at 8:16
  • $\begingroup$ Thanks for your comments, you both agree that the noise is unchanged but the bandwith is increased by a factor (1+beta) so noise power is increased by that factor... $\endgroup$ – MaLog Aug 16 '20 at 19:39
  • $\begingroup$ @MaLog Bandwidth increasing by rolloff factor does not mean that the power of matched filter output when the input is thermal noise increases. $\endgroup$ – AlexTP Aug 18 '20 at 8:32

Following the comments by AlexTP (thank you), there is one clear mistake in my reasoning. Using a roll-off factor indeed increases the bandwith but it colours noise too. The mistake is $P_n = W'N_0$ because that equation is assuming flat filter response. The intregration of the squared reception filter (RRC, and so, raised cosine) over all bandwidth is 1, so noise power is not modified by the roll-off and $P_s/P_n = E_s/N_0$.


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