Peak to Average Power Ratio for OFDM
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Let us try to understand peak to average power ratio (PAPR) and its typical value in an OFDM system specified per IEEE 802.11a specifications.
What is PAPR?
The peak to average power ratio for a signal is defined as
, where
corresponds to the conjugate operator.
Expressing in deciBels,
.
PAPR of a single sine tone
Consider a sinusoidal signal
having the period
.
The peak value of the signal is
.
The mean square value of the signal is,
.
Given so, the PAPR of a single sine tone is,
.
Figure: Wave form of single sine tone
% Example Matlab/Octave script
clear all
close all
xt = sin(2*pi*1*[0:1/64:0.999]);
plot(xt,’b.-’,'LineWidth’,4)
grid on
xlabel(’sample number’)
ylabel(’amplitude’)
title(’sine wave’)
meanSquareValue = xt*xt’/length(xt)
peakValue = max(xt.*conj(xt))
PAPR of a complex sinusoidal
Consider a sinusoidal signal
having the period
.
The peak value of the signal is
.
The mean square value of the signal is,
.
Given so, the PAPR of a single complex sinusoidal tone is,
.
Figure: Waveform a single complex sinusoidal
close all
clear all
% defining a signal in frequency domain
% subcarrier +1 alone
xF = [zeros(1,6) zeros(1,26) 0 1 zeros(1,25) zeros(1,5) ];
xt = 64*ifft(fftshift(xF));
meanSquareValue = xt*xt’/length(xt)
peakValue = max(xt.*conj(xt))
plot(real(xt),’b',’LineWidth’,2)
hold on
plot(imag(xt),’g',’LineWidth’,2)
xlabel(’sample number’)
ylabel(’amplitude’)
title(’complex sinusoidal’)
legend(’real’, ‘imag’)
grid on
Maximum expected PAPR from an OFDM waveform
From the previous post (here), we have learned that an OFDM signal is the sum of multiple sinusoidals having frequency seperation where each sinusoidal gets modulated by independent information
. Mathematically, the transmit signal is,
For simplicity, let us assume that for all the subcarriers. In that scenario, the peak value of the signal is,
.
The mean square value of the signal is,
.
Given so, the peak to average power ratio for an OFDM system with subcarriers and all subcarriers are given the same modulation is,
.
It is reasonably intuitive that the above value corresponds to the maximum value of PAPR (when all the subcarriers are equally modulated, all the subcarriers align in phase and the peak value hits the maximum).
PAPR in IEEE 802.11a OFDM transmission
Per the IEEE 802.11a specifications, we have used subcarriers. Given so, the maximum expected PAPR is 52 (around 17dB!!). However, thanks to the scrambler, all the subcarriers in an OFDM symbol being equally modulated is unlikely.
Using a small script, the cumulative distribution of PAPR from each OFDM symbol, modulated by a random BPSK signal is obtained.
Click here to download.
Figure: Cumulative distribution (CDF) plot of PAPR from a random BPSK signal
As can be observed, the observed PAPR seems to be distributed from around +3.5dB to a maximum value of 10dB.
References
Hope this helps.
Krishna
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Comments
I am doing project on PAPR reduction in OFDM by selective tone reservation method can you help me to ind matlab program
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Hi …
Kindly upload a topic on the OFDM signal derivation equation…
thanx, i shall be thankful to you for help
@ Krishna Pillai.
Hi,
you are the man who has helped me in understanding the OFDMA concept and because of you I decided to do my project in it. But now i am stuck at a point and i want you to help/guide me through the problem. I have given you an personal email and i would request you to help me in understanding/solving the problem as well.
Regards
@Krishna Pillai
any of the way will be fine. but if it is a fresh post , please do send me the location.
Thanking you in anticipation,
umar
@umar:
On a closer look at the queries, I think each of the questions requires a blog-post or more to explain… 
Anyhow, let me try to provide my brief thoughts and some pointers.
Firstly, I do not have the Matlab code available for any of the simulation parameters listed below.
Trying to answer:
1)BER v.s. No of users (capacity)
[krishna] In an OFDMA system the number of users depends on the number of sub-carriers, each user being allocated a chunk of sub-carrriers. If the sub-carriers are orthogonal, the users do not interfere with each other. However, in the presence of frequency offset (and probably other impairments), there will be ICI (inter-carrier-interference) resulting in noise floor. Do you agree?
Where you looking for that?
2)BER v.s. distance
[krishna] This can be obtained from BER vs SNR plot.
If you get the path loss equations for the environment, for a given a transmit power, we can know the received power at a given distance. As the receiver noise bandwidth is fixed (typically equal to the signal bandwidth), the signal to noise ratio is defined. So, one can use path-loss to convert SNR to distance.
3)PDF v.s. CIR (dB)
[krishna] Probability distribution function (PDF?) of which event?
4)SNR vs outage probability
[krishna] Hopefully this link provides you pointers
http://tinyurl.com/4pvejj
5) No of users v.s rate sum capacity.
[krishna] No clue.
Further, you can look at Prof. Chockalingam’s publications listed in the URI: http://www.ece.iisc.ernet.in/~achockal/aboutme/tmp_node7.html
He has worked on schemes for cancellation of multiuser interference in OFDMA.
Hope this helps…
Firstly thanks for the responce and sorry for the delayed question for i was not able to get to the desk.
Thanks for the BER v.s. distance help, i think i will manage something from the guidance given.
But for the SNR vs outage probability, i am afraid the pointer you have provided only allows access to few pages and that too the unimportant one. I would request you to please help me in this regard and tell me how can i calculate required or threshold rate from the general capacity equation and how can i get the graph between snr and outage as we have to varry the capacity and form there get the snr. I am totally confused, need urgent help.
May be you can write a blog-post for this point only.
Regards
@Anatoliy: The best I have done to reduce the PAPR is to have a scrambler in place. The scrambler ensures that the long sequence of ones or zeros are avoided.
Some relevant links which I picked up from a quick Googling.
(a) Tone reservation : where some of the unused subcarriers in an OFDM system is modulated to reduce the PAPR.
(b) MERL has a paper on PAPR reduction by oversampling, soft clipping and filtering.
PAPR Reduction for WiMAX OFDM Systems
http://www.merl.com/projects/papr/
(c) Nice comp.dsp thread on this topic
Has The peak-to-average ratio problem of OFDM been solved?
http://tinyurl.com/6jn9ql
Sorry, I have not worked on any of the advanced methods. Hence unable to suggest one way or the other.
Good luck with you investigations. If you wish, kindly update us about your findings. It will be helpful to dsplog.com community
I have question about the MATLAB code. Why we need to take complex transpose while calculating mean square vale?
Sakib
Another interesting source for PAPR reduction I found is
http://to.swang.googlepages.com/peaktoaveragepowerratioreduction
http://to.swang.googlepages.com/C00-NTAH-20070827-008-LGE-PAPRReduct.pdf
Hi
I have one question… please help. I need your help.
My question is: could we use the reduction method PAPR in DVB-T not only IEE 802.11? For a example, if we use the compression method then the constellation diagram will be distorted and common home-receiver can’t correctly receive a signal.
For instance, in this document offer to use the algorithm u-Law.But then must be used specialized receiver.What algorithm possible to use in modulator with usual home receiver DVB-T?
just my 2 cents.
1) I feel like high PAPR is due to too many independent sub-signals to be added together. For example, 256-chip OFDM may have more chance of high PAPR than 64-chip OFDM.
2) for OFDM liked multi-carrier signals, if the signal is time-domain cyclic shift, it equals to frequency-domain phase rotation.
3) why not divide a, say, 256-chip OFDM signal into two, for example, 128-chip OFDM signals, and properly cyclic rotating one 128-chip OFDM tile the PAPR of the added signal is low enough.
any comments
Hi I am looking for the new 802.11n PHY specification. I will be grateful if some one can post it.
Sakib
Thanks. I know it is still in the draft stage. I know the fft size and other OFDM parameters but i am interested specially in the frequency band. Reports say they will be an extension of the current band (but 802.15.3 already have 3.1-10.6 Ghz). Also its not clear if they are going to use only 2.4 GHz and omit 5Ghz. If you can post some comment on the frequency band that will great.
Hi. Thanks for the prompt responses you provide to the questions. You know that for a high frequency, high bit rate OFDM signal-error vector magnitude (EVM) is an important parameter. Sometimes more important than PAPR-because even if the signal is clipped to a certain degree EVM can be very good.
Can you reflect on the relationship between EVM, SNR and BER of a QAM (QPSK is 4-QAM) OFDM signals? There is a paper titled -”On the extended relationships among EVM, BER and SNR as performance matrix.” But the equations there have problems. May be you can post some better references.
@Elibom: I tried to check the PAPR reduction by multiplying some subcarriers by j (as suggested in 802.11n 40MHz bandwidth). This operation does not reduce the PAPR for the random BPSK/QPSK case 
Hi, I have one very basic question, and hope you will explain it to me in basic way, i am a novice in MATLAB.
What does this function actually do. I have copied it from your code. paprSymboldB is a 193 bits vector whereas 0:0.5:15 is a 31 bit vector. What is actually stored in n and x.
[n x] = hist(paprSymboldB,[0:0.5:15]);
Please dont use words like one are found in MATLAB help, which i am copying down.
N = HIST(Y) bins the elements of Y into 10 equally spaced containers and returns the number of elements in each container. If Y is a matrix, HIST works down the columns.
@umar: Yeah, sometimes the help is not so helpful 
Ok, let me start:
hist(paprSymboldB,[0:0.5:15]) function computes the number of instances where paprSymboldB take the value 0dB, 0.5dB, 1dB, 1.5dB,…14.5dB, 15dB respectively. The number of elements is stored in n; and x is same as [0:0.5:15].
From n and the knowledge of the total number of events, one can get probability density function. Further, by doing a cumulative sum cumsum(), one can get the cumulative distribution function.
Does this help?
hi,
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