1

I am learning about OFDM and QAM.

Every explanation I have read illustrates how the separate carriers are sinc functions that are arranged to overlap in a nice orthogonal way so as to cancel out interference from neighbouring carriers.

The part I don't understand is how is it possible to generate these neat perfectly overlapping sinc functions, when they contain constantly changing symbols. Each transition from 1 symbol to the next, in the time domain is a random change in phase and amplitude. This must correspond to a random change in the sideband pattern of the carrier in the frequency domain, and therefore disrupt the orthogonal arrangement.

Marcus Müller
  • 88,280
  • 5
  • 131
  • 237
John Spence
  • 421
  • 1
  • 3
  • 13

2 Answers2

1

These sinc functions, as you've noticed, have zeros in a distance of the subcarrier spacing \$\Delta f\$.

Remember how these sincs come to be (the texts you've been reading most definitely mention that!): The sinc function is the Fourier transform of the rectangle function. Scaled to yield zeros in frequency domain every \$\Delta f\$, the width \$T\$ of that rectancgle must be \$T=\frac1{\Delta f}\$.

So, that answers your question: all your sincs are just the result of having a rectangle in time domain, and multiplying it by \$e^{j2\frac{n\cdot\Delta f}{f_\text{sample}}t}\$, so to shift it in frequency to yield the \$n\$th subcarrier. The QAM symbol is just a complex factor you multiply the result with – that is just a constant factor and doesn't change the shape, neither in time nor frequency domain.

Now, what's \$\Delta f\$, when you think about it? In OFDM, you use the \$N\$-point DFT to divide your Nyquist bandwidth (complex!) \$f_\text{sample}\$ into \$N\$ equally large subcarriers, so \$\Delta f = \frac{f_\text{sample}}{N}\$. Therefore, the width of the rectangle \$T=\frac1{\Delta f}=N\cdot\frac{1}{f_\text{sample}} = N\cdot T_\text{sample}\$. That very simply means that the sinc shapes are just the effect of turning on a (complex) oscillation of frequency \$n\cdot\frac{f_\text{sample}}N\$ for exactly \$N\$ samples.

Each transition from 1 symbol to the next, ...

Such a transition simply doesn't happen within one OFDM symbol: For the duration of one of these rectangles, the symbol for each subcarrier is constant. So, you use \$N\$ samples to send a single symbol, but you gain the ability to send \$N\$ symbols at once. So, nothing lost, nothing gained here!

Marcus Müller
  • 88,280
  • 5
  • 131
  • 237
  • That helps s lot. So in what way do the sync functions change visually to represent different QAM symbols? – John Spence May 28 '20 at 22:17
  • 1
    **sinc**, not **sync**. Please don't mix up these words! They are not visually different, because the QAM symbol is just a constant complex factor you multiply with the sinc function. – Marcus Müller May 29 '20 at 09:31
0

The FT of a rectangle is a Sinc. So basically any basis vectors with a constant (thus rectangular) envelope will produce a Sinc shaped spectrum. That’s just how the math of an FFT (all basis vectors orthogonal) works out.

Now the transition between FFT blocks won’t be constant envelope, thus will have spectral splatter. But a cyclic prefix, essentially a repeat of the data will the exact same spectrum is usually pre or postpended to each block. The length of this cyclic prefix is designed to be longer than the impulse response of the channel. Thus, a whole block of clean (unmodulated basis) data, all carriers with only a Sinc shape spectrum, will be left over after the cyclic prefix. And the random change due to block transitions has already completely died down after the impulse response of a linear time invariant channel. So assuming clean Sinc shaped carriers is a good assumption (barring other channel impairments.)

hotpaw2
  • 4,731
  • 4
  • 29
  • 44
  • It seems that the cyclic prefix will fill in the gaps between rectangular blocks and therefore there would not be a sync function associated with them. – John Spence May 29 '20 at 09:14
  • 1
    @JohnSpence it's a **sinc** function, really, not a **sync** function ;) – Marcus Müller May 29 '20 at 09:30
  • @JohnSpence your notion that there'd be "splatter" is however, not right: the sinc shape **is** the splatter! – Marcus Müller May 29 '20 at 09:32
  • So if the sinc function is the splatter, then what is producing them is the abrupt change as one symbol is spliced onto the next. These transitions are constantly changing. In some transitions there will be a stronger phase and amplitude shift than in others. Some transitions there will be no change and therefore no splatter to produce a sinc function. – John Spence May 29 '20 at 09:58
  • Not constantly changing. Only changing between blocks. After impulse response dies out (during the cyclic prefix and after the change), all the effects of the changes have faded out, leaving only a clean sinc shaped spectrum carriers. – hotpaw2 May 29 '20 at 17:54