# Continuous and discrete-time raised cosine filter properties in frequency domain

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

-ryan

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 $1/T_c$.
• $(5)$ 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 Sep 3 '16 at 6:01