The bandwidth of the raised cosine filter is w=0.5(1+r)Rs, where Rs is the symbol rate, and r is the roll-off factor. r=0 represents the Nyquist filter, for which Rs=2w. Any higher value for the roll-off factor will lead increasing the bandwidth, what about the symbol rate? is the symbol rate transmitted using the raised cosine filter higher or lower than the symbol rate transmitted using the Nyquist filter? What about the bit rate in the two cases? I know that more bandwidth can support more bit rate.


2 Answers 2


When comparing communications systems with different pulses, it's convenient to keep either the bandwidth $B$ or the pulse rate $R_s$ constant.

Say you have a channel with bandwidth $B$. Then, using sinc pulses you could transmit at rate $R_s = 2B$ baud. Using (square root) raised cosine pulses with excess bandwidth $\alpha$, you could transmit at rate $R_s = 2B/(1+\alpha)$ baud.

Now, say you want to transmit symbols at rate $R_s$. When using sinc pulses, you would need a channel with bandwidth $B=R_s/2$ hertz. On the other hand, if using square root raised cosine pulses, you would need $B=R_s(1+\alpha)/2$ hertz.

The bit rate is a separate issue; given a rate $R_s$, you can achieve the same bit rate regardless of the pulse used simply by using constellations with the same cardinality.

However, given a fixed bandwidth $B$, the pulse rate $R_s$ obtained with raised cosine pulses is lower. In order to match bit rates, the raised cosine system will need to use a larger constellation.


The symbol rate is not changed by the roll off factor. Consider the spectrum with no pulse shaping — the result in frequency is a Sinc function with the first null at $R$ where $R$ is the symbol rate (so the main lobe in frequency in that case has a null to bill BW equal to $2R$). The theoretical brick wall in frequency performance with an infinite duration pulse shaping filter (also as a Sinc but in time!) has a bandwidth in frequency equal to $R$. It is the implementation of realizable filters that leads to excess bandwidth over the theoretical brick wall limit and that doesn’t change anything about the symbol rate itself.


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