In communication systems, the transmitted signal is often passed through a root-raised cosine filter to avoid, in textbook language, intersymbol interference (ISI). An identical filter at the receiver end processes the signal so that the combined response of the two filters is a raised cosine filter.

The raised cosine filter has two properties: roll-off factor and oversampling rate.

I have a bipolar sequence ($\pm 1$ values) that I am supposed to process through a raised cosine filter. It is not clear to me what exactly is intersymbol interference here, and how can I visualize it. Here the symbol has only one bit. How should I choose the roll-off factor and the oversampling rate?


  • $\begingroup$ I think you should read this (free) textbook (linked at the bottom of the page): sethares.engr.wisc.edu/telebreak.html $\endgroup$
    – MBaz
    Commented Aug 15, 2016 at 14:21
  • $\begingroup$ Concerning oversampling, please refer to this question and its answers. $\endgroup$
    – Matt L.
    Commented Aug 15, 2016 at 18:49

1 Answer 1


Your bipolar sequence at one sample per symbol is equivalent to a perfectly synchronized sequence of symbols with no intersymbol interference (ISI). Intersymbol interference in this context refers to the tendency for symbols that precede and follow the current symbol to interfere at the receiver. That is, at the correct sampling instant for symbol $N$, the receiver also sees some of symbols $N-1$ and $N+1$ (and possibly more in each direction) mixed in. This causes an increase in symbol error rate.

In general, ISI is caused by some non-ideal channel impulse response between your transmitter and receiver. This can be caused by several phenomena, including:

  • A multipath communications channel (e.g. a wireless channel). In this case, the receiver observes a signal that can be modeled as your originally transmitted signal convolved with some impulse response.

  • The use of a non-ideal pulse shape. On the AWGN channel, the optimum receiver consists of a filter whose impulse response is matched to the transmitted pulse shape. The receiver therefore observes a stream of pulses where each has the shape of the originally transmitted pulse shape convolved with itself. The choice of the pulse shape is important to ensure that the receiver doesn't encounter any ISI after applying the matched filter.

This is why the root-raised cosine filter is often used. When convolved with itself, the result is a raised-cosine response, which has zero ISI (it is exactly zero for all multiples of the symbol period $T$, apart from $t=0$).

With that said, there are some practical considerations when designing your pulse shape. When you actually transmit it, you'll need to oversample it in order to meet the sampling theorem's reconstruction requirements. A typical oversampling rate might be 4.

The rolloff factor controls how much bandwidth the RRC-shaped signal will occupy. In general, the more excess bandwidth, the better you can expect your synchronization algorithm to perform (i.e. you would usually observe smaller error from your timing recovery subsystem with a larger rolloff factor). However, this comes at a cost of more bandwidth required to carry the signal.

  • $\begingroup$ Thanks. What is still unclear to me is why I need to oversample. I see in the problem specifications that my binary sequence needs to be oversampled by a factor of 1.5 (strange?). So, I need to design the RRC with an oversampling factor of 1.5. I understand the ISI and roll-off factor now. But the oversampling is still unclear. Any further insights? $\endgroup$
    – r2d2
    Commented Aug 15, 2016 at 15:38
  • $\begingroup$ If you're going to transmit a signal, in most cases you're going to need to reconstruct it to analog form at some point. In order to do that while respecting the sampling theorem, you'll need more than one sample per symbol (based on the bandwidth of the pulse-shaped signal). The number of samples per symbol that you do need will really depend on the rolloff factor (and therefore how much bandwidth the signal occupies). $\endgroup$
    – Jason R
    Commented Aug 16, 2016 at 13:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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