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My question is kinda silly, I know it. But I cannot understand why root raised cosine filter or raised cosine filter can eliminate ISI. I googled before, but those explanation is not easy for me to understand. Maybe someone can help me this with some figures? Thank you so much for you help in advance.

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In addition to the frequency response that AlexTP showed, I think the time domain view of the impulse response gives good insight into the Zero-ISI operation of a Nyquist filter. Below is shown the result of transmitting three symbols, than each began as a single impulse. Each plot below is the "impulse response" for each symbol. Note the blue symbol was transmitted first and is maximum at time index 0 on the graph (but began much earlier than this in time). This position where the impulse response is maximum is our ideal sampling location to determine what was transmitted. One sample after the original impulse for the blue symbol, came an impulse for the red symbol. Notice at the ideal sampling location for red (where it is maximum), the response from all the other symbols is zero (meaning zero ISI).

Nyquist Filter

The pulses shown above have zero ISI at the sampling locations, but extend to infinity to achieve the maximimum conservation of bandwidth (1/(2T)). Of course, this is not feasible, so the raised cosine pulse shape applies a window to truncate the perfect Sinc function, resulting in a trade of bandwidth versus filter complexity, as well as trades with other characteristics related to performance for timing recovery.

Raised Cosine Pulse Shaping Filter

The filter is often split between transmitter and receiver as a "root-raised cosine filter" (RRC) on each side. Note that a RRC filter does NOT have zero-ISI until it is paired with the second RRC filter to form in cascade a raised cosine filter.

Below is an example eye diagram, showing the trajectory over 1 symbol for multiple pulses (each trajectory is based on the history of all previous symbols within the depth of the RRC filter and results in a different path when going from 0 to 1 or 0 to 0 etc). At time 0 is the ideal sampling instant to decide between 0 and 1, notice with the RRC filter that there is significant ISI, which is eliminated after passing through the second RRC filter in the receiver (assuming no other multipath effects have occurred). Also notice for the filter with alpha = 0.25 that there is a significant spread in the zero crossing locations, which would add more noise in timing recovery operations that are sensitive to zero crossings. Compare this to the eye diagram with alpha= 1 and notice the significant decrease in zero-crossing jitter, as well as a wider "eye-opening" which also means less sensitivity to jitter or other errors in the sampling clock location, but this benefit comes at the expense of less spectral efficiency.

QPSK Eye Diagram

enter image description here

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  • $\begingroup$ Dear Dan, first of all. thank you so much for your detail answer. What a helpful person you are!! However, i guess I am too native or "entry level" here. May I ask you, give me more idea about zero ISI and Nyquist theory? What I know about nyquist is the sample rate should be 2X high than original signal frequency. But I dont have any connection between Nyquist theory and ISI. I don't know you can attach some picture again or not, if not, do you think i need to post a new question? Thank you again, Dan. $\endgroup$ – Sunson29 Apr 11 '17 at 3:34
  • $\begingroup$ Thank you @Sunson29, I think between my answer and what AlexTP shows which is also very good is probably the best we could do to answer your question. I recommend finding a good book on DSP to help get you up to speed and I am sure you will have some very specific questions along the way (right now your question is much too broad to answer concisely, especially if you do not yet follow what we answered). There are some other posts you can search for where people have asked for good DSP books to get started with and I recommend that as your next step. $\endgroup$ – Dan Boschen Apr 11 '17 at 3:51
  • $\begingroup$ Dear Dan, I was studying this today. It's much better now. Please allow me to ask one question. It seems that Alpha = 1 is the best idea case, am I right? it seems that the eye is so clear. But because considering bandwidth efficiency, ppl more perfer to do 0.2 ~ 0.4 ? $\endgroup$ – Sunson29 Apr 12 '17 at 3:36
  • $\begingroup$ @Sunson29 If we did not care about spectral efficiency there would be no reason to do any raised cosine filter and we could simply transmit rectangular pulse shapes in time (which is a Sinc function in frequency). So you are correct from the point of view of accuracy in timing in the receiver, the larger alpha is better but from the point of view of spectral efficiency (which is quite a concern!), the smaller alpha is better. Hence there is a trade to be made. $\endgroup$ – Dan Boschen Apr 12 '17 at 12:05
  • $\begingroup$ Dear Dan, I am confused again. Because you said "If we did not care about spectral efficiency there would be no reason to do any raised cosine filter " . My question, I thought the main reason we have raised cosine filter is because of ISI elimination. Well, at least wiki said that. After "fix" the problem of ISI, then we talk about spectral efficiency. Please verify my understanding. Thank you, dan. $\endgroup$ – Sunson29 Apr 13 '17 at 0:36
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Because it satisfies Nyquist ISI criterion Nyquist_ISI_criterion.

The simplest way to apply this criterion to root raised cosine filter is looking at its frequency response of the composite filter "raised cosine" : Raised-cosine_filter

Raised cosine frequency resposne

You can see easily that the sum of $1/T$-shifted spectrum of raised cosine filter is constant.

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