Here is the formula to compute magnitude of a raised-cosine filter: enter image description here

I know that 0 <= Beta <= 1. I am not sure what are the values for f (frequency) and Ts. I assume that some normalization will take place. I also know that Ts is symbol period and it is reciprocal of symbol rate. What are these parameters and relationship between them? Can someone provide an example of typical values of these parameters as used in the raised-cosine formula? Here is the graph from Wikipedia. How the normalization of frequency and magnitude is done?

enter image description here


The frequency $f$ is no parameter but it's the independent variable. The frequency response $Z(f)$ is a function of frequency. The only real parameter is $\beta$, defining the excess bandwidth, but you already know that $0\le \beta\le 1$. You also know that $T_s$ is the symbol period, so $1/T_s$ is indeed the baud rate, which is part of the system's specifications.

From the formula you can see that the minimum bandwidth is half the baud rate (for $\beta=0$, i.e. zero excess bandwidth). For $\beta=1$ (i.e., $100\%$ excess bandwidth), the bandwidth is twice as large, i.e. it equals the baud rate $1/T_s$.

  • $\begingroup$ Matt, I have looked at various graphs online. I have a sense that frequency is normalized as well as magnitude. I am confused with Fs = 1/(2Ts) on the graph. If you can shed light with example how and where normalization takes place (and for which parameters) it would be appreciated. Thank you. $\endgroup$ – Nebojsa Jun 22 '15 at 15:39
  • $\begingroup$ @Nebojsa: It doesn't matter how you plot it, you can normalize by whatever you like, but the basic formula remains the same (which is the one in your question), where frequency is simply in Hertz. Maybe you can link to an image (or add it to your question), so I understand what's troubling you. $\endgroup$ – Matt L. Jun 22 '15 at 16:03
  • $\begingroup$ Matt: I have updated my question with the graph. I want to find out relationship between frequency and Ts, and how the normalization of frequency is done. $\endgroup$ – Nebojsa Jun 22 '15 at 17:08
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    $\begingroup$ @Nebojsa: The frequency is not normalized. The characteristic points on the frequency axis are $f=1/2T_s$ and $f=1/T_s$. For $f>1/T_s$ the response is always zero (also for $\beta=1$), and for $\beta>0$ the response has a value of $1/2$ at $f=1/2T_s$, just as given by the formula. $\endgroup$ – Matt L. Jun 22 '15 at 18:38
  • $\begingroup$ ... "a value of $T_s/2$ at $f=1/2T_s$" (i.e. half the maximum value) I should have written in my previous comment. $\endgroup$ – Matt L. Jun 22 '15 at 18:55

The frequency $f$ is continuous and takes values between $-\infty$ and $\infty$.

The symbol period $T_s$ can take any value you want, depending on how fast you want to transmit. Note that, if you decrease $T_s$ (you transmit at a higher rate), then $Z(f)$ is non-zero for larger values of $f$; that is, its bandwidth increases.

Say you want to transmit 1,000 pulses per second, so that $T_s=1\times10^{-3}$ seconds. Then, the pulse's bandwidth is $$B=\frac{1+\beta}{.002}.$$ For $\beta=0$, the bandwidth is $B=500\,\textrm{Hz}$. For \beta=1$, the bandwidth is 1000 Hz.

When $\beta=0$, the pulse is a sinc, which has large sidelobes and decreases slowly. When $\beta=1$, the pulse has very small sidelobes and can be truncated in time, to give it a short duration, with little effect on its bandwidth. Which $\beta$ to use depends on the compromise you're willing to make on complexity (sinc pulses) and bandwidth (large $\beta$).


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