0
$\begingroup$

I would like to do know why do we need to do oversampling in a raise cosine filter. There is already a question like this but the answer is very confusing and not sufficient.

  • I need to know what this oversampling factor does?
  • The higher is better or the lower?
$\endgroup$
  • 1
    $\begingroup$ Note that you don't need to oversample; all you need to do is to meet Nyquist. Having said that, if you could be more precise about what you find confusing in the answer you linked to, I can try to clarify. $\endgroup$ – MBaz Jun 5 '16 at 15:06
2
$\begingroup$

A (baseband) pulse amplitude modulated (PAM) signal is given by

$$s(t)=\sum_{k}A_kp(t-kT)\tag{1}$$

where $A_k$ are the data symbols, $p(t)$ is the transmit pulse, and $1/T$ is the symbol rate. Note that $(1)$ describes a continuous-time signal. If you want to simulate a digital communication system, you need to use sampled versions of continuous-time functions, such as the transmit pulse $p(t)$. For this purpose, you need to define a sampling rate greater than the symbol rate. The ratio between the sampling frequency and the symbol rate is the oversampling factor.

Assume you choose a sampling rate an integer (oversampling) factor $m$ greater than the symbol rate:

$$\frac{1}{T_s}=\frac{m}{T}\tag{2}$$

From $(1)$, the sampled signal $s(nT_s)$ is then

$$\begin{align}s(nT_s)&=\sum_{k}A_kp(nT_s-kT)\\&=\sum_{k}A_kp\left(\left(n-k\frac{T}{T_s}\right)T_s\right)\\&=\sum_{k}A_kp((n-km)T_s)\end{align}\tag{3}$$

Or, in notation for discrete-time signals

$$s[n]=\sum_{k}A_kp[n-km]\tag{4}$$

$\endgroup$
0
$\begingroup$

The purpose of Oversampling in raised cosine filter is to narrow the spectral content of the transmitted sequence. An oversampling factor is chosen such that the available bandwidth becomes entirely occupied with the signal. The upsampled signal is then followed by a lowpass RRC filter to remove the multiple copies of the upsampled spectrum.

Refer to the book " Signal Processing for Communications" by prandoni and vetterli. [Chapter 12]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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