I am trying to understand some basic concepts with IFFT as represented here OFDM Modem page 50-53, and 108-114.

Some basics that are represented -

The base sampling rate is 8khz. The symbol period is 32ms, which would calculate to 250 total samples. The fft size is 64.
The bin separations are 32hz. The bandwidth used is from 1000-3000hz(1024-3072?)

I cannot get my mind around how to get from a ifft size of 64 to an output bandwidth of 1000-3000 and 32khz spacing. Is there some basic reason why I wouldn't use a much larger ifft size of something like 256, and then use only 32-96? That would calculate to 31.25hz per bin.

In short, this thesis is proposing that you use a fft size of 64 to cover a bandwidth of 2khz beginning at 1khz, 1k-3k. This isn't coming together for me at all, how this can get shifted.

If I have to be math'ed here, be gentle.

First attempt at upconversion using a spreadsheet:


Where 10 is the carrier freq, C(12) is a row containing a sine lookup value.

  • $\begingroup$ Hi and welcome to DSP.SE! You certainly don't expect us to read the whole master thesis. What page or section are you refering to and could you cite the most important facts so that the question remains understandable in case the link turns dead? $\endgroup$
    – Deve
    Apr 16, 2014 at 7:30
  • $\begingroup$ Right, I just wanted to have a reference. Page 50-53, and 108-114 $\endgroup$ Apr 16, 2014 at 10:22
  • $\begingroup$ I am unclear how much info I should include from the thesis, suggestions welcomed. $\endgroup$ Apr 16, 2014 at 10:29

1 Answer 1


To explain the supposed contradiction between the number of samples calculated by symbol length and sampling rate on the one hand and the IFFT size on the other hand, you have to take into account that the IFFT output is upsampled by a factor of 4. This means that for every output sample of the IFFT block there are four samples after upsampling. There's a cyclic prefix of 16 samples so that we arrive at a total number of samples of $N=4\cdot(64+16)=320$ as stated on page 53.

On the other hand we can calculate the number of samples by $N=Tf_\mathrm{S}=32\mathrm{ms}\cdot8\mathrm{kHz}=256$. The symbol duration $T$ has been taken from page 108. It think the guard interval has not been taken into account there and thus $N=64\cdot4=256$.

The process that is described as "upconversion" is exactly the frequency shift that you are observing (cf. figure 3-8). Citing from the report:

Once the baseband signal has been interpolated it is multiplied with the audio band carrier frequency to obtain a audioband representation.

To make this more clear using a formula, let $x(t)$ be the complex, analog OFDM signal (after digital-to-analog conversion (DAC)) and let $f_0$ be the desired carrier frequency. Then the radio frequency, upconverted signal $y(t)$ is $$ y(t) = x(t)\operatorname{exp}(j2\pi f_0 t) $$ This is a basic concept required in almost every radio system. For more information you might want to do some reserach on radio frequency upconversion and I/Q modulation.

The baseband bandwidth of an OFDM signal is approximately $f_\mathrm{S}/2$. But again, here we have to take the upsampling into account. If it is done correctly, it won't change the bandwidth of the signal. I.e. the sampling frequency before upsampling is actually $f_\mathrm{S}/4 = 2\mathrm{kHz}$ and we have to use this frequency for calcuating the OFDM signal bandwidth wich here is $B=1\mathrm{kHz}$. Finally, the band occupied by the radio frequency signal is $f_0 - B\ldots f_0+B$ which corresponds to 1kHz to 3kHz.

Concerning your suggestion to use a greater IFFT size and to left some subcarriers unmodulated to achieve upconversion - this is a valid approach in this specific example. Note however, that in general it requires a larger sampling frequency which is often undesirable.

  • $\begingroup$ This gets me started. I would like to find some C code and graphs of a short single sine upconversion, haven't found so far. $\endgroup$ Apr 17, 2014 at 12:01
  • $\begingroup$ @ErikFriesen The reason might be that RF upconversion is usually omitted in simulations/models of communications systems. Instead, the so-called equivalent baseband representation is used. $\endgroup$
    – Deve
    Apr 17, 2014 at 12:09
  • $\begingroup$ What do (t) and j represent? $\endgroup$ Apr 17, 2014 at 17:28
  • $\begingroup$ t is time and j is the imaginary unit, equal to the square root of -1 $\endgroup$
    – Deve
    Apr 17, 2014 at 17:30
  • $\begingroup$ I realize I need to take a math course here, obviously I am missing something, see above attempt in new edit. If this needs a new question I will start there. Otherwise feel free to ram home what I am not understanding here. $\endgroup$ Apr 17, 2014 at 18:01

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