If I extend the length of the Cyclic Prefix (Guard Interval) in an OFDM signal, the loss in data rate is clear by means of the time domain, but I can't see it in the frequency domain as the BW is given by Nsc/T_useful.

What am I missing?

I think that maybe calculating BW out of Nsc/T_useful is only correct after removing the CP in the demodulation process while the transmitted signal won't look the same "on the air", is that true?


You are correct that the spectrum seen over-the-air is different depending on the length of the cyclic prefix (CP) used. Even so, it isn't going to change the occupied bandwidth significantly. The easiest way to see the change caused by the CP is to consider the spectrum of a single symbol for a single subcarrier and then extrapolate that to the bigger picture.

A common drawing that I'm sure you have seen before illustrates the orthogonality of each subcarrier spectrum. The spectrum of each subcarrier is a sinc function. The spectrum of each subcarrier reaches a null at exactly the center frequency of the other subcarriers. This drawing is not an accurate depiction of what the subcarrier signal spectra look like over-the-air.

We will assume a sampling period of $1$ second to simplify the math involved. Let's also define the subcarrier bandwidth to be the bandwidth from the first null left of the sinc function peak to the first null right of the sinc function peak. This definition ignores that the sinc is not bandlimited (i.e., it really has an infinite bandwidth), but this definition is convenient for us to gain some intuition about how the shape of each subcarrier's spectrum is changing with the CP length.

Over-the-air, the subcarrier bandwidth as defined above is $\frac{2}{N + N_{CP}}$ where $N$ is the FFT size and $N_{CP}$ is the length of the cyclic prefix that you are using. If you were to consider the entire symbol window, the subcarriers would not appear orthogonal. Applying the rectangular window at the receiver (i.e., CP removal) restores you to a window length where the adjacent subcarriers are orthogonal to one another (each one having an integer number of periods in the window). After this process, the subcarrier bandwidth looks like the typical pictures that people draw to illustrate the subcarrier orthogonality. The subcarrier bandwidth of this rectangular windowed version of the signal is $\frac{2}{N}$ as you would expect.

You will notice that the subcarrier bandwidth over-the-air is equal to $\frac{2}{N + N_{CP}}$, but we are overlapping the subcarriers at a spacing of $\frac{1}{N}$. So, the bandwidth required for transmission is nearly the same for the signal with or without the cyclic prefix. The only thing that changes, is at the edges of the occupied bandwidth. With the cyclic prefix, the first null associated with the leftmost subcarrier extends $\frac{1}{N + N_{CP}}$ Hz to the left of the main spectrum before hitting its first null. Similarly, the rightmost subcarrier spectrum extends $\frac{1}{N + N_{CP}}$ Hz to the right before hitting its first null.

Since all of the subcarriers are modulated by random data and added together, it is very hard to perceive this slight change in the spectrum profile. The easiest way to verify it is in the slope of the magnitude of the side lobes of the OFDM spectrum. For example, if we choose $N_{CP} = 0$ (not a likely choice) and observe the spectrum, the side lobes will "roll off" at a slower rate than if we choose $N_{CP} = N$. I'd recommend trying to plot the spectrum of these two choices against each other with your favorite numerical software package just to verify this behavior for yourself.

  • $\begingroup$ Very nice answer, more detailed than my earlier one. Good summary of the nuance involved with the OP's question. $\endgroup$ – Jason R Feb 14 '17 at 1:57

There isn't any obvious effect in the frequency domain. However, extending the duration of the cyclic prefix does make your signal more resistant to multipath channels with a large delay spread. As a rule of thumb, you need to size the cyclic prefix duration to cover the entire delay spread. If this condition is met, then the linear convolution that is imposed on your signal by the multipath channel will manifest itself as circular convolution on the cyclic-prefixed signal. This circular convolution can be equalized by simple pointwise multiplication in the frequency domain at the receiver.

This is why one of OFDM's advantages is often cited as the fact that it permits the use of a "one-tap equalizer." In single-carrier systems, equalization can be one of the harder things to get right, so anything that can simplify that portion of the receiver is a help.

  • $\begingroup$ Sorry but apart from your first sentence ("There isn't any obvious effect in the frequency domain"), this is a very detailed answer to a question I haven't asked. You haven't answered my question and just told me things I already know. Besides that, I don't think there is no effect in the frequency domain.Imagine that I take the cyclic prefix to be 2*T_useful or 0.25*T_useful. I'm pretty sure the spectrum won't look the same. $\endgroup$ – Gabizon Feb 12 '17 at 18:55
  • $\begingroup$ It doesn't help that you've edited your question to change it. You originally asked what effect the duration of the guard interval has in the frequency domain. As I mentioned, it doesn't really have an effect. You choose the duration of the cyclic prefix to trade off multipath resistance and information throughput. There is no information conveyed in the guard interval, so if you make it too long, your throughput suffers. $\endgroup$ – Jason R Feb 12 '17 at 19:04
  • $\begingroup$ I didn't edit the first part of my question, I only added the second passage as a possible solution and my question is still how does the duration affect the frequency domain, as you stated. Of course my throughput suffers if I make the cyclic prefix too long, but how is it possible that it doesn't affect the spectrum of the signal at all? Let's assume T_useful=0.8 ms, an OFDM signal with a 2 ms symbol length (T_useful+T_guard) - excessively large cyclic prefix, or 1 ms symbol length looks the same on the spectrum? $\endgroup$ – Gabizon Feb 12 '17 at 19:22
  • $\begingroup$ I was specifically addressing the criteria that you would use to choose an appropriate cyclic prefix length. Why would you expect there to be any difference in the bandwidth of the signal based on the cyclic prefix length? You're still transmitting on all of the same subcarrier frequencies during the guard interval, so it occupies the same bandwidth. $\endgroup$ – Jason R Feb 13 '17 at 1:02

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