In communication systems, the raised cosine (RC) filter is split into root-raised cosine (RRC) filter at the transmitter and the receiver. The combined response of both RRC filters is the RC filter.

If the transmit RRC filter is analog (continuous-time or CT), how the combined response of both filters change if the receive RRC is placed (a) before digital sampling and (b) after?

For example, let $h_{\rm T_{RRC}}(t)$, $h_{\rm R_{RRC}}(t)$ and $h_{\rm RC}(t)$ be the impulse responses of the transmit RRC, receive RRC and RC filters, respectively. Let their frequency responses be $H_{\rm T_{RRC}}(f)$, $H_{\rm R_{RRC}}(f)$, and $H_{\rm RC}(f)$, respectively. Let the symbol period be $T_\rm c$.

  • Case 1: Receive RRC filter is continuous-time

    Here the system is as follows:

    Transmit RRC $\longrightarrow$ Receive RRC $\longrightarrow$ Sampling at rate $1/T_c$

    The combined response of the filters before sampling is $$H_{\rm T_{RRC}}(f)H_{\rm R_{RRC}}(f) = H_{\rm RC}(f) \tag{1}$$

    After sampling, the RC filter has the following desired property in time-domain $$h_{\rm RC}[n] = h_{\rm RC}(nT_\rm c) = \begin{cases} 1, \: n = 0 \\ 0, \: \text{otherwise}\end{cases}\tag{2}$$ $$\textrm{or,}\quad h_{\rm RC}(t)\cdot\sum\limits_{k=-\infty}^{\infty}\delta(t-kT_\rm c) = \delta(t)\tag{3}$$

    In frequency domain, this becomes

$$H_{\rm RC}(f)\star\frac{1}{T_\rm c}\sum\limits_{k=-\infty}^{+\infty}\delta\left(f - \frac{k}{T_\rm c}\right) = 1\tag{4}$$ $$\textrm{or,}\quad\frac{1}{T_\rm c}\sum\limits_{k=-\infty}^{+\infty}H_{\rm RC}\left(f - \frac{k}{T_\rm c}\right) = 1.\tag{5}$$

  • Case 2: Receive RRC filter is discrete-time

    Here the system is as follows:

    Transmit RRC $\longrightarrow$ Sampling at rate $1/T_\rm c$ $\longrightarrow$ Receive RRC

    Here, the sampled transmit RRC response is $$\frac{1}{T_\rm c}\sum\limits_{k=-\infty}^{+\infty}H_{\rm T_{RRC}}\left(f - \frac{k}{T_\rm c}\right).$$

Would the following hold true now: $$H_{\rm R_{RRC}}(f)\cdot\frac{1}{T_\rm c}\sum\limits_{k=-\infty}^{+\infty}H_{\rm T_{RRC}}\left(f - \frac{k}{T_\rm c}\right) = 1?\tag{6}$$



2 Answers 2


Root Raised Cosine has a bandwidth $B=\frac{1+\alpha}{2T_c}$ which is $(1+\alpha)$ times more than the minimum required bandwidth set by Nyquist no-ISI criterion. To satisfy the Nyquist sampling theorem now, one must sample at a rate of $2B$. In time domain, we can say that we have to have more samples/symbol than 1. In practice, and to facilitate the synchronization blocks down the chain, this sampling rate is selected such that we have an integer number of samples/symbol. The smallest integer greater than 1 is 2. In your eq (6), you are sampling at a rate of 1 sample/symbol which induces aliasing. Hence, the results of eq (5) and eq (6) will not be equivalent. For 2 or more samples/symbol for incoming signal and assuming no analog distortions, eq (5) and eq (6) will become equivalent after downsampling to symbol rate.

The best way to understand is to imagine a figure of RRC in frequency domain with aliases at integer multiples of symbol rate $1/T_c$.


how the combined response of both filters change if the receive RRC is placed (a) before digital sampling and (b) after

That depends on your sampling, but if you're sampling with a rate high enough to represent the full pre-filtered bandwidth, Nyquist says it doesn't matter.

  • $\begingroup$ But, mathematically, how does this check out? For example, how do equations 5 and 6 are equivalent? $\endgroup$
    – r2d2
    Sep 2, 2016 at 22:43
  • $\begingroup$ $(5)$ and $(6)$ aren't per se equivalent, but that's just because a RRC for rate $\frac1{T_C}$ has a bandwidth $> \frac1{T_C}$ and you're therefore in violation of Nyquist's sampling theorem $\endgroup$ Sep 3, 2016 at 6:01

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