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I have been given a OFDM signal of crest factor of 12dB. The crest factor is the ratio of peak power to the average power of the signal. Does anyone have any idea how crest factor of 12dB is derived for an OFDM signal?(assumptions on peak power,average or RMS power etc). I would highly appreciate any input on this.

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  • $\begingroup$ You mean derived theoretically? Or for a given OFDM symbol? $\endgroup$ – Deve Oct 30 '15 at 12:33
  • $\begingroup$ For example, as per Nutaq , the crest factor of OFDM is equal to the number of subcarriers say K(where KxK is the peak value of signal and K is the RMS value of signal. Does this mean that the dynamic range of DAC is [-KxK, KxK]?(I wonder whether for large number of subcarriers the range of [-KxK, KxK] for a DAC will be practical or not!) $\endgroup$ – rmb Oct 30 '15 at 13:15
  • $\begingroup$ That's for the special case of all subcarriers having the same value. It's the worst case and very unlikely to ever happen. $\endgroup$ – Deve Oct 30 '15 at 13:30
  • $\begingroup$ I see! But in case all subcarriers don't have the same value, any idea on how to determine peak value of signal from its crest factor?I need to find the range of DAC from a given crest factor of OFDM! $\endgroup$ – rmb Oct 30 '15 at 13:37
  • $\begingroup$ As the crest factor is the ratio of peak and mean power, you have to know the mean power, if you'd like to calculate the peak power for a given crest factor. I can write up a longer answer on this when I find time later... $\endgroup$ – Deve Oct 30 '15 at 13:45
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In statistical case OFDM (without specific PAPR reduction precoding and with most of subcarriers used) signal could be modelled as white Gaussian process. So PDF of OFDM signal is well-known one for normal distribution:

$f(x) = 1/(\sigma\cdot\sqrt{2\cdot\pi}) \cdot \exp(-(x-\mu)^2/(2\sigma^2))$

For OFDM $\mu=0$. When we calculate crest factor we need to define the number of samples to be used for measurement. We say measurement for $1e+6$ samples is valid since crest factor doesn't increase significantly, when we take more values. In another words if we have e.g. $1e+2...1e+4$ samples, crest factor will vary from one measurement to another. But if we have a million of samples or more, crest factor of white noise or OFDM will converge to one specific value, it is about 13.5 dB. You can check it easily with MATLAB.

It can be derived analitically with aid of PDF of normal distribution. Look, we need to estimate how much peak value exceeds $\sigma$ to estimate crest factor. If for example $1e+6$ samples are used the probability of peak value is $1e-6$. It is rought approximation. For normal distribution large peaks have low probabilty and it decreases with peak amplitude increase (see plot of normal distr. PDF). We need to estimate maximum peak for million samples, so it is one peak for million and its probabilty is $1e-6$.

The equation to be solved is

$P(x)=\int_x^\infty f(x)dx = 1e-6$,

where $f(x)$ is presented above.

To avoid infinite integration we can roughtly simplify (forgive me, mathematicians!) this equation to something like this

$f(x) \le P(x) = 1e-6$.

By usign $f(x)$ instead of $P(x)$ and solving this equation for different $\sigma$ we can find out that $4.5\sigma \le x \le 5\sigma$ in general. The creast factor could be find as $20\cdot\log_{10}(x/\sigma)\approx 13...14$. The result is very similar to what we can observe in practice. It can be more accurate if someone suggest the way to approximate $\int_x^\infty f(x)dx$ term in the original equation (I believe it could be done since $f(x)$ is decreasing in the interval from $x$ to $\infty$).

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  • $\begingroup$ Very well explained! However, if I define my signal in MATLAB as signal = sqrt(0.5)*(randn(1,1e6) + 1i*randn(1,1e6)), can I find my peak value (without taking into account crest factor) as abs(max(signal)) ? $\endgroup$ – rmb Nov 2 '15 at 10:28
  • $\begingroup$ max(abs(signal)) will be better) $\endgroup$ – Serj Nov 2 '15 at 10:32
  • $\begingroup$ So if I take 13dB as the crest factor, will the peak amplitude be 1.91? (13 = 20.log(peak amplitude/1)). $\endgroup$ – rmb Nov 2 '15 at 10:36
  • $\begingroup$ It should be 4-5 times higher then &\sigma&. Use log10 in the equation. $\endgroup$ – Serj Nov 2 '15 at 10:40

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