When designing a root raised cosine filter on Matlab, there is an option to set its "roll-off factor". What does this mean? I have been looking around for a while but I can't seem to find a clear explanation. Any help will be appreciated.


Think of the root-raised cosine as a family of functions. The roll-off factor is a parameter that chooses one of the possible functions.

All root-raised cosine pulses look similar to the sinc pulse. The interesting thing about the roll-off factor is that it controls two features of the function:

  • The rate at which the function's lobes (or ripples) decrease. At one extreme, with roll-off set to 0, they decrease slowly (as $1/t$, with $t$ equal to time). At the other end, when the roll-off factor is 1, they decrease as $1/t^3$.

  • As the roll-off factor increases, so does the function's bandwith. Basically, if the function's bandwidth is $B$ when the roll-off is 0, then it becomes $2B$ when the roll-off is 1.

You can choose any value for the roll-off factor between 0 and 1, so you can select the function with the ripple and bandwidth that are best suited to your needs.

| improve this answer | |
  • $\begingroup$ So if the roll-off changes the bandwidth, then it also changes the cut-off frequency because the 3dB point would have moved right? I would have thought the bandwidth of the raised cosine filter would be determined solely by the value you define for it's cutoff frequency and the roll-off would determine the transition bandwidth because if you decide to set the rolloff factor then you do not set the transition bandwidth. $\endgroup$ – KillaKem Oct 24 '14 at 9:01
  • $\begingroup$ @KillaKem, in my answer I've used the absolute definition of bandwidth; that is, the function's spectrum becomes $0$ for $|f|>B$. If you define bandwidth as the cut-off frequency, then you're right: the roll-off factor will affect the transition bandwidth. If the roll-off is zero, then the transition bandwidth is essentially zero. If it's one, then the transition bandwidth is roughly equal to the pass band. $\endgroup$ – MBaz Oct 24 '14 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.