EDIT: There was a small error in my code when plotting the Squared RC. Below is the correct frequency response, and clearly RC² doesn't fulfill Nyquist's criterion.

Correct plot

---------------------------------- Original question (to discard)

So first let me start by asking if using a Raised Cosine filter at both the transmitter and receiver (matched filter) gives a response that is ISI free? (Assuming an ideal channel)

In other words is the SQUARED (not root-squared) frequency response of RC $H_{RC}(f)^2$ still fulfills Nyquist's ISI criterion?

I avoided a direct derivation since the mathematical expressions are quite intimidating so I turned to Matlab and quickly plotted the frequency response of both the RC & RC squared and tried to calculate: $\sum H(f-k/Ts)$ and check if it's constant.

The graphs show that both $H_{RC}(f)$ and $H_{RC}(f)^2$ add up to a constant after performing the previous operation. Moreover I performed an ifft on $H_{RC}(f)^2$ and plotted it and it seems to have zeroes at $kTs$ (for $k\neq 0$) just like the RC time domain function.

RC and RC SQUARED Frequency response and ISI criterion

So bets are, if my Matlab code is correct, that RC² fulfills Nyquist's criterion. Am I missing something here?

Now if the RC² is also ISI free, why wouldn't we use an RC at both the transmitter and receiver? Why do we turn to a root-raised cosine instead?

Why did we go from an RC to a Root-RC? Most textbooks say it's to split the filtering between TX & RX and to get an overall ISI free response but the same could be done with a RC filter.

Thanks in advance.


1 Answer 1


There must be something wrong in your code because it's quite straightforward to show that the squared frequency response of a raised cosine pulse does not satisfy the Nyquist criterion.

A raised cosine pulse satisfies the Nyquist criterion simply because a raised (co)sine function and an inverted raised (co)sine function add up to a constant:


where one of the two expressions corresponds to the rising edge of the frequency response whereas the other corresponds to the falling edge of a shifted version of the frequency response.

Now if you square the frequency response you get the following expression in the region where two shifted spectra overlap:

$$(1+\sin(x))^2+(1-\sin(x))^2=2(1+\sin^2(x))\neq const$$

which is not constant, and which proves that the squared frequency response of a raised-cosine pulse does not correspond to a Nyquist pulse.

  • $\begingroup$ There has been a silly mistake in my code when plotting the frequency response. I updated the post.Thanks for the explanation! $\endgroup$
    – br4him
    Feb 10, 2016 at 12:45
  • $\begingroup$ @br4him: Good to know that we arrive at the same result! :) $\endgroup$
    – Matt L.
    Feb 10, 2016 at 12:54

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