3
$\begingroup$

In standard of 3GPP LTE physical layer(3GPP TS 36.211), the FFT size for baseband signal generation for both Dl and UL is 2048 and CP size is also determined. However there is a table in several resources said we have different FFT and CP size. E.g.

http://lteuniversity.com/ask_the_expert/f/59/t/405.aspx

How is it possible?

Improve this question
$\endgroup$
4
  • $\begingroup$ It's going to be hard to compare the two without a reference that you can point to that indicates that the size is always 2048. Do you have one? $\endgroup$
    – Jason R
    Commented Mar 2, 2012 at 1:50
  • $\begingroup$ Yes, please download the standard, then please search baseband signal generation. Then always N=2048. $\endgroup$
    – Hossein
    Commented Mar 2, 2012 at 1:59
  • 2
    $\begingroup$ @Hossein How about either quoting the relevant portions of the standard, or at the very least providing a link to where we can download it and references to the passages you are talking about. $\endgroup$
    – Jim Clay
    Commented Mar 2, 2012 at 15:25
  • 1
    $\begingroup$ The link to download standard:3gpp.org/ftp/Specs/archive/36_series/36.211 $\endgroup$
    – Hossein
    Commented Mar 2, 2012 at 22:18

2 Answers 2

Reset to default
6
$\begingroup$

Read the book 4G Lte Advanced by Stefan Parkvall and Eric Dahlman. According to the book, FFT size of 2048, with a corresponding sampling rate of 30.72 MHz, is suitable for the wider LTE carrier bandwidths, such as bandwidths of the order of 15 MHz and above. However, for smaller carrier bandwidths, a smaller FFT size and a correspondingly lower sampling rate can very well be used.(Lowering the sample rate simplifies the implementation in dual-mode HSPA/LTE terminals)

Also,nothing forbids the implementation of an OFDM modulator as a set of parallel modulators, Also, nothing prevents the use of a larger IFFT size, for example a size-2048 IFFT size, even in the case of a smaller number of OFDM subcarriers.

Tcp= 160·Ts and Tcpe=512.Ts. Change in CP induces change in the total number of REs(6X12, instead of 7X12). Note that #sub carriers remain 12 in both cases and FFT size is dependent on BW allocated which is expressed in #sub carriers.

Improve this answer
$\endgroup$
2
$\begingroup$

I think the answer vinay provides is correct although there are few more details that in my opinion provide a better answer to the fundamental question that Hossin is looking for.

The FFT size N=2048 sets the upper limit that can characterise all channel bandwidth from 1.4- 20 MHz (or 72-1200 occuppied subcarriers). Recall that resource blocks only have values on the occupied subcarriers and the rest are padded with zeros in order to fit the FFT size. So, for example, if you use 72 subcarriers (i.e 1.4 MHz CBW) and use an FFT size of 128 you will have to pad 56 subcarriers with zeros. The same can be said if you use an FFT size of 2048 where you will have to pad 1976 subcarriers. If you do it properly there should not be any difference in the spectrum between the two if you were to compute the FFT after you do the IFFT (both forms will show a total bandwidth of approx 1080 kHz = 15kHz*72).

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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