CN1805414A - SNR estimation method and system and channel compensation method and system - Google Patents

Abstract

The invention relates to a signal/noise ratio evaluation method and a channel compensation method, and relative systems. The invention uses high-order statistic quantity to evaluate the signal/noise ratio parallel distribution condition of wireless channel to process the optimized channel compensation on the Turbo code. The method comprises following steps: receiving the signal with several bytes from the channel; calculating the mean square value and the average valve of biquadrate of said signal and the ration between the average valve of biquadrate and the square of mean square value; measuring the channel parameter; calculating the signal/noise ratio of signal according to the channel parameter and the ratio. The invention can fully utilize the evaluated signal/noise ratio in two conditions as channel decline evaluation and blind evaluation, to provide the optimized channel compensation algorisms on the Turbo code.

Description

Signal-noise ratio estimation method and system and channel compensating method and system

Technical field

The present invention relates to the signal-to-noise ratio (SNR) estimation and the channel compensation of communications field chnnel coding, relate in particular to a kind of signal-noise ratio estimation method and system, and the method and system that utilizes above-mentioned signal-noise ratio estimation method that channel is compensated.

Background technology

Nakagami-m distributes and can describe the statistical property of wireless channel well ^[8,9], and because the simplicity of its formula and general applicability have obtained concern widely, thereby become the good mathematical model of generally acknowledging of wireless channel simulation.When m was 1, the Nakagami channel became Rayleigh (Rayleigh) channel.And when m was tending towards infinite, the Nakagami channel became additive white Gaussian channel (Additive White Gaussian Noise).These all are the channels in the practical communication environment.Signal-to-noise ratio (SNR) Algorithm under the Nakagami fading channel is a very interesting challenging technology.Estimation for Nakagami decline parameter m has report in document [8] [10].In practice, the estimation algorithm of signal to noise ratio all is essential in a lot of optimized signal processing technologies.For example the optimum decoding algorithm-MAP or the Log_MAP algorithm of at present known turbo sign indicating number all need to know signal to noise ratio, otherwise can bring certain snr loss.Certainly need not know channel status for Max_Log_MAP and the such suboptimum decoding algorithm of SOVA, but they there is the snr loss of 0.5dB than MAP algorithm.

Summers based on the quantitative analysis of low order statistics, defines in document [1]

$Z\≡\frac{E\left(r^2\right)}{E^2\left(\right|r\left|\right)}$

. under awgn channel, can derive $Z=f\left(\mathrm{\β}\right)=\frac{1+\mathrm{\β}}{\{\sqrt{\frac{2}{\mathrm{\π}}}\exp (-\frac{\mathrm{\β}}{2})+\sqrt{\mathrm{\β}}\mathrm{erf}{\left(\sqrt{\frac{\mathrm{\β}}{2}}\right)\}}^2}$ To try to achieve the β value by the Z value,, can't try to achieve analytic solutions because following formula is too loaded down with trivial details, so proposed an empirical formula in [1] based on the binomial match:

β≈-34.0516z ²+65.9548z-23.6184

Simultaneously, [1] points out that effectively the scope of the β of prediction is 0 to 6dB, thereby its estimation range exists significant limitation.

Ramesh adds up quantitative analysis based on low order equally in document [2], z under the Nakagami channel of further having derived and signal to noise ratio γ (being equal to β herein) relational expression:

$z=\frac{1+2\mathrm{\γ}}{{(\sqrt{\frac{2}{\mathrm{\π}}}\left(\frac{m}{m+\mathrm{\γ}}\right)m+\sqrt{\frac{2\mathrm{\γ}}{m}}\frac{\mathrm{\Γ}(m+\frac{1}{2})}{\mathrm{\Γ}\left(m\right)}(1-\frac{2I\left(m\right)}{\mathrm{\π}}))}^2}$

As seen following formula is also very numerous and diverse, can't obtain with the Z value is the analytical function of the β of dependent variable, thereby Ramesh also is the approximate solution that obtains the signal to noise ratio valuation with the method for fitting of a polynomial in document [2], not only method is loaded down with trivial details, and its estimation range exists significant limitation, all has than mistake at high s/n ratio and low signal-to-noise ratio.

Ramesh has provided the valuation formula under the Rayleigh channel (m=1):

$z_{\mathrm{rayleigh}}=\frac{\mathrm{\π}}{2}\frac{1+2\mathrm{\γ}}{{[1+\sqrt{\mathrm{\γ}}(\frac{\mathrm{\π}}{2}-{\cos }^{-1}\left(\sqrt{\frac{\mathrm{\γ}}{1+\mathrm{\γ}}}\right))]}^2}$

Following formula can't obtain the analytic solutions of gamma function, so Ramesh is similar to quadrinomial: γ=10971.3670Z-64731.6368Z ³+ 143212.2372Z ²-140825.8014Z+51838.6459

Document [8] [10] has obtained method of estimation to Nakagami channel fading parameters m based on high-order statistic, but the signal-noise ratio estimation method under the Nakagami channel of not deriving.

$\hat{m}=\frac{{\widehat{\mathrm{\μ}}}_2^2}{{\widehat{\mathrm{\μ}}}_4-{\widehat{\mathrm{\μ}}}_2^2}$

Wherein, use Represent the sample valuation of its k rank distance.

The patent aspect is not seen the report aspect the relevant signal-to-noise ratio (SNR) estimation under the Nakagami channel as yet.

Summary of the invention

Main purpose of the present invention provides a kind of real-time reliable the signal to noise ratio of wireless channel is carried out estimation approach and system, and to the method and system that channel compensates, makes it possible in the error correction coding such such as the turbo sign indicating number, improves error-correcting performance.

In one aspect of the invention, propose a kind of signal-noise ratio estimation method, comprised step: received the signal that comprises a plurality of bits from channel; Calculate the mean-square value and the biquadratic average of described signal, and described biquadratic average and described mean-square value square between ratio; The measured channel parameter value; And according to the signal to noise ratio of described channel parameter values and the described signal of described ratio calculation.

In another aspect of this invention, proposed a kind of signal-to-noise ratio (SNR) estimation system, having comprised: the importation is used for receiving the signal that comprises a plurality of bits from channel; The statistic calculating section, be used to calculate described signal biquadratic average, mean-square value and described biquadratic average and mean-square value square between ratio; The measuring parameters in channels part is used for from described channel measurement channel parameter; The snr computation part is used for calculating according to described ratio and described channel parameter the signal to noise ratio of described signal.

Another aspect of the present invention is a kind of channel compensating method, comprises step: receive the signal that comprises a plurality of bits from channel; Calculate the mean-square value and the biquadratic average of described signal, and described biquadratic average and described mean-square value square between ratio; The measured channel parameter value; Signal to noise ratio according to described channel parameter values and the described signal of described ratio calculation; According to described signal to noise ratio and described mean-square calculation channel compensation value; Utilize described channel compensation value that described channel is compensated.

In still another aspect of the invention, proposed a kind of channel compensation system, having comprised: the importation is used for receiving the signal that comprises a plurality of bits from channel; The statistic calculating section, be used to calculate described signal biquadratic average, mean-square value and described biquadratic average and mean-square value square between ratio; The measuring parameters in channels part is used for from described channel measurement channel parameter; The snr computation part is used for calculating according to described ratio and described channel parameter the signal to noise ratio of described signal; Channel compensation value calculating section is used for according to described signal to noise ratio and described mean-square calculation channel compensation value; The channel compensation part utilizes described channel compensation value that described channel is compensated.

Utilize said method and system, make it possible in the error correction coding such, improve the signal to noise ratio of the signal that receives, thereby improve error-correcting performance such as the turbo sign indicating number.

Description of drawings

The model that the method for the present invention that shows Fig. 1 adopts;

Fig. 2 is one group of asynchronous Z-beta curve of m;

Fig. 3 is the curve of function g (m);

Fig. 4 is the block diagram according to the signal-to-noise ratio (SNR) estimation system of first embodiment of the invention;

Fig. 5 is the concrete formation block diagram of the statistic calculating section in the signal-to-noise ratio (SNR) estimation system shown in Figure 4;

Fig. 6 is the block diagram according to the channel compensation value estimating system of second embodiment of the invention:

Fig. 7 is the block diagram according to the channel compensation system of third embodiment of the invention;

Fig. 8 is the flow chart of the course of work that is used for illustrating the signal-to-noise ratio (SNR) estimation system of first embodiment of the invention;

Fig. 9 is the flow chart of the course of work that is used for illustrating the channel compensation system of third embodiment of the invention;

Figure 10 is the signal to noise ratio valuation emulation testing figure as a result under the Gaussian channel; And

Figure 11 is the signal to noise ratio valuation emulation testing figure as a result under the Rayleigh channel.

Embodiment

At first this patent proposes and has proved two theorems, initial point rank theorem and the probability density distribution theorem of Nakagami decline stochastic variable.With these two theorems is Fundamentals of Mathematics, quantitative analysis has drawn the analytical expression accurately of signal to noise ratio valuation to this patent according to higher order statistical first, the scope that can accurately predict has been expanded greatly, carried out the sensitivity emulation contrast test of signal to noise ratio valuation simultaneously, proved the validity of new algorithm the decoding precision.

At first, we study some characteristics of Nakagami channel, on this basis, and by certain mathematical analysis with derive and to have drawn the formula of Analysis signal-to-noise ratio (SNR) accurately under the Nakagami channel, and, concrete implementation has been proposed channel compensation in conjunction with turbo decoding.

For the convenience of representing, we at first are defined as follows symbol:

U: through the ambipolar transmission bit of coding and BPSK modulation, u ∈ 1 ,+1},

μ: the interference of the large scale property taken advantage of, constant

A: the interference of the small scale property taken advantage of, all normalized Nakagami stochastic variable in side

The parameter that m:Nakagami distributes

E _s: receive the energy of bit, E _s=μ ²

N: additive white Gaussian noise

σ: the standard deviation of additive white Gaussian noise n, E (n ²)=σ ²

R: the receiving sequence of exporting under the Nakagami fading channel, compound stochastic variable (being called Nakagami decline stochastic variable in the literary composition)

L _c: the channel compensation value

β: the signal to noise ratio of estimation, this paper definition $\mathrm{\β}\≡\frac{{\mathrm{\μ}}^2}{2{\mathrm{\σ}}^2}$

Z: the ratio of high-order statistic, this paper definition $Z\≡\frac{E\left(r^4\right)}{E^2\left(r^2\right)}$

Pdf: probability density function

As shown in Figure 1, we suppose that at first the u sequence is through the general bipolarity bit (u ∈ { 1 ,+1) independently of the grade of BPSK modulation after the chnnel coding), it has experienced the property the taken advantage of interference μ of large scale and the property the taken advantage of interference a of small scale, also has additivity to disturb n; Suppose that channel is the memoryless smooth Nakagami fading channel through fully interweaving, μ, independent distribution between a and the n, the characteristic of channel is by the ternary vector as can be known $\stackrel{\→}{c}=(\mathrm{\μ},m,\mathrm{\σ})$ Describe; The r sequence is to obtain after receiving terminal has adopted the coherent demodulation of precise synchronization, the r sequence is carried out statistical analysis obtain the channel compensation value And both are multiplied each other send in the channel decoder, r=a μ u+n then.For the purpose of brief, in the literary composition r is called Nakagami decline stochastic variable.μ (as encoding block) in quite long timing statistics can think constant, and a is that all the normalized parameter in side is the stochastic variable of the Nakagami distribution of m, and n is that average is the stochastic variable of the Gaussian Profile of σ for the zero standard difference, and then signal to noise ratio is:

$\frac{\mathrm{Es}}{\mathrm{No}}=\frac{{\mathrm{\μ}}^{2}E\left(a^2\right)}{2{\mathrm{\σ}}^2}=\frac{{\mathrm{\μ}}^2}{2{\mathrm{\σ}}^2}$

The probability density function of a is:

$\mathrm{pdf}\left(a\right)=\frac{{2m}^2{a}^{2m-1}}{\mathrm{\Γ}\left(m\right)}\exp \left({-\mathrm{ma}}^2\right)---m>0.5,a>0---\left(1.1\right)$

Rayleigh channel is the Nakagami channel at m is 1 o'clock special case, and Gaussian channel is the special case of Nakagami channel when m is tending towards infinite.

The k rank square of a is:

$E\left(a^k\right)=\frac{\mathrm{\Γ}(m+\frac{k}{2})}{\mathrm{\Γ}{\left(m\right)m}^{\frac{k}{2}}}---\left(1.2\right)$

The probability density function of n is:

$\mathrm{pdf}\left(n\right)=\frac{\exp (-\frac{{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">n</figure-callout>}}^2}{2{\mathrm{\&sigma;}}^2})}{\sqrt{2\mathrm{\&pi;}}\mathrm{\&sigma;}},n\&Element;R---\left(1.3\right)$

The k rank square of n is

$E\left(n^k\right)=\left\{\begin{array}{cc}0& k=2t+1\\ {\mathrm{\&sigma;}}^k\underset{i=1}{\overset{t}{\mathrm{\&Pi;}}}(2i-1)={\mathrm{\&sigma;}}^k(k-1)!!& k=2t\end{array}\right.---\left(1.4\right)$

The probability-distribution function of u is:

$P\left(u\right)=\left\{\begin{array}{cc}-1& 1\\ 0.5& 0.5\end{array}\right\}---\left(1.5\right)$

The k rank of u are apart from being:

$E\left(u^k\right)=\left\{\begin{array}{cc}0& k=2t+1\\ 1& k=2t\end{array}\right.---\left(1.6\right)$

By above basis, we propose and proof Nakagami fading channel under the k rank of stochastic variable r apart from theorem and probability density function theorem.

The k rank of stochastic variable r under the theorem one Nakagami fading channel are apart from theorem

$E\left(r^k\right)=$

$\left\{\begin{array}{cc}0& k=2t+1\\ {\mathrm{\&mu;}}^k\underset{i=0}{\overset{t-1}{\mathrm{\&Pi;}}}(1+\frac{i}{m})+{\mathrm{\&sigma;}}^k(k-1)!!+\underset{i=1}{\overset{t-1}{\mathrm{\&Sigma;}}}\frac{k!}{\left(2i\right)!(k-2i)!!}{\mathrm{\&mu;}}^{2i}{\mathrm{\&sigma;}}^{k-2i}\underset{j=0}{\overset{i-1}{\mathrm{\&Pi;}}}(1+\frac{j}{m})& k=2t\end{array}\right.$

Proof:

(a)

Easily know by binomial expansion:

$E\left(r^{2t+1}\right)=\underset{i=0}{\overset{2t+1}{\mathrm{\&Sigma;}}}{\mathrm{\&mu;}}^i{C}_{2t+1}^{i}E\left(u^i\right)E\left(a^i\right)E\left(n^{2t+1-i}\right)$

Because i and 2t+1-i must have an odd number, and the odd ordered moment of u and n all is zero,

So i E (u ⁱ) E (n ^2t+1-i)=0

Hence one can see that:

E(r ^2t+1)＝0

(b)

Easily know by binomial expansion:

$E\left(r^{2t}\right)=\underset{i=0}{\overset{2t}{\mathrm{\&Sigma;}}}{\mathrm{\&mu;}}^i{C}_{2t}^{i}E\left(u^i\right)E\left(n^{2t-1}\right)E\left(a^i\right)$

$=\underset{i=0}{\overset{t}{\mathrm{\&Sigma;}}}{\mathrm{\&mu;}}^{2i}{C}_{2t}^{2i}E\left(u^{2i}\right)E\left(n^{2t-2i}\right)E\left(a^{2i}\right)+\underset{i=0}{\overset{t-1}{\mathrm{\&Sigma;}}}{\mathrm{\&mu;}}^{2i+1}{C}_{2t}^{2i+1}E\left(u^{2i+1}\right)E\left(n^{2t-2i-1}\right)E\left(a^{2i+1}\right)$

(odd even is split item)

Because the odd ordered moment of u and n all is zero,

So i E (u ²ⁱ⁺¹) E (n ^2t-2i-1)=0

Push away it thus:

$E\left(r^{2t}\right)=\underset{i=0}{\overset{2t}{\mathrm{\&Sigma;}}}{\mathrm{\&mu;}}^{2i}{C}_{2t}^{2i}E\left(u^{2i}\right)E\left(n^{2t-2i}\right)E\left(a^{2t}\right)$

$=\underset{i=0}{\overset{t}{\mathrm{\&Sigma;}}}{\mathrm{\&mu;}}^{2i}{C}_{2t}^{2i}E\left(n^{2t-2i}\right)E\left(a^{2i}\right)$

$={\mathrm{\&mu;}}^{k}E\left(a^k\right)+E\left(n^k\right)+\underset{i=0}{\overset{t-1}{\mathrm{\&Sigma;}}}{\mathrm{\&mu;}}^{2i}{C}_{2t}^{2i}E\left(u^{2t-2i}\right)E\left(a^{2i}\right)$ (by 1.6)

By (1.4), as can be known:

E(n ^2t-2i)＝σ ^2t-2i(2t-2i-1)！！

E(n ^2t)＝σ ^k(k-1)！！

According to character Γ (n+1)=n Γ (n) and (1.2) of Euler integral of the second kind, as can be known:

$E\left(a^{2i}\right)=\frac{\mathrm{\&Gamma;}(m+i)}{\mathrm{\&Gamma;}\left(m\right)m^i}=\underset{j=0}{\overset{i}{\mathrm{\&Pi;}}}(1+\frac{j}{m})$

$C_{2t}^{2i}=\frac{k!}{\left(2i\right)!(k-2i)!}$

The above all formulas of simultaneous, as can be known:

$E\left(r^{2t}\right)=$

${\mathrm{\&mu;}}^k\underset{i=0}{\overset{t-1}{\mathrm{\&Pi;}}}(1+\frac{i}{m})+{\mathrm{\&sigma;}}^k(k-1)!!+k!\underset{i=1}{\overset{t-1}{\mathrm{\&Sigma;}}}\frac{(k-2i-1)!!}{\left(2i\right)!(k-2i)!}{\mathrm{\&mu;}}^{2i}{\mathrm{\&sigma;}}^{k-2i}\underset{j=0}{\overset{t-1}{\mathrm{\&Pi;}}}(1+\frac{j}{m})$

$={\mathrm{\&mu;}}^k\underset{i=0}{\overset{t-1}{\mathrm{\&Pi;}}}(1+\frac{i}{m})+{\mathrm{\&sigma;}}^k(k-1)!!+\underset{i=1}{\overset{t-1}{\mathrm{\&Sigma;}}}\frac{k!}{\left(2i\right)!(k-2i)!!}{\mathrm{\&mu;}}^{2i}{\mathrm{\&sigma;}}^{k-2i}\underset{j=0}{\overset{i-1}{\mathrm{\&Pi;}}}(1+\frac{j}{m})k=2t$

Card is finished.

The probability density function theorem of stochastic variable r under the theorem 2:Nakagami fading channel

Setting m is Integral multiple, pdf (r) then)=

$\frac{m^m\exp (-\frac{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">m</figure-callout>}}{{2\mathrm{\&sigma;}}^{2}m+{\mathrm{\&mu;}}^2}{r}^2)}{\mathrm{\&Gamma;}\left(m\right)}\{\frac{{\left(\mathrm{\&mu;r}\right)}^{2m-1}}{{({\mathrm{\&mu;}}^2+2m{\mathrm{\&sigma;}}^2)}^{2m-\frac{1}{2}}}$

$+\underset{i=1}{\overset{[m-\frac{1}{2}]}{\mathrm{\&Sigma;}}}\frac{(2m-1)!}{\left(2i\right)!(2m-2i-1)!}\frac{\underset{k=0}{\overset{i-1}{\mathrm{\&Pi;}}}(k+\frac{1}{2})}{{({\mathrm{\&mu;}}^2+{2\mathrm{m\&sigma;}}^2)}^{2m-i-\frac{1}{2}}}{2}^i{\mathrm{\&sigma;}}^{2i}{\left(\mathrm{\&mu;r}\right)}^{2m-2i-1})$

Proof:

Make Z=μ a, then:

${\mathrm{pdf}}_z\left(Z\right)=\frac{{\mathrm{pdf}}_a\left(\frac{Z}{\mathrm{\&mu;}}\right)}{Z^{\&prime;}}=\frac{{\mathrm{pdf}}_a\left(\frac{Z}{\mathrm{\&mu;}}\right)}{\mathrm{\&mu;}}$

By (1.1), as can be known

$\mathrm{pdf}\left(Z\right)=\frac{{2m}^m{Z}^{2m-1}}{\mathrm{\&Gamma;}\left(m\right){\mathrm{\&mu;}}^{2m}}\exp (-\frac{{\mathrm{<figure-callout\; id="2"\; label="mZ"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">mZ</figure-callout>}}^2}{{\mathrm{\&mu;}}^2})---m>0.5,Z>0$

Make x=uZ

Because u is the ambipolar Bei Nuli stochastic variable of symmetry, and Z be monolateral on the occasion of the nakagami stochastic variable, and u and Z independent distribution, so by symmetrical analysis as can be known Z be the nakagami random distribution of bilateral even symmetry, and half of the corresponding probability density that its probability density on the occasion of part is Z, that is:

$\mathrm{Pdf}\left(x\right)=\frac{m^m{x}^{2m-1}}{\mathrm{\&Gamma;}\left(m\right){\mathrm{\&mu;}}^{2m}}\exp (-\frac{{\mathrm{<figure-callout\; id="2"\; label="mx"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">mx</figure-callout>}}^2}{{\mathrm{\&mu;}}^2})---m>0.5,x\&Element;R$

R=x+n  pdf _r=pdf _x pdf _n( represents convolution)

${\mathrm{pdf}}_r\left(r\right)=\underset{-\&infin;}{\overset{+\&infin;}{\&Integral;}}\frac{m^m{x}^{2m-1}}{\mathrm{\&Gamma;}\left(m\right){\mathrm{\&mu;}}^{2m}}\exp (-\frac{{\mathrm{<figure-callout\; id="2"\; label="mx"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">mx</figure-callout>}}^2}{{\mathrm{\&mu;}}^2})\frac{\exp (-\frac{{(r-x)}^2}{2{\mathrm{\&sigma;}}^2})}{\sqrt{2\mathrm{\&pi;}}\mathrm{\&sigma;}}\mathrm{dx}$

Order $c=\frac{r}{(1+\frac{{2\mathrm{m\&sigma;}}^2}{{\mathrm{\&mu;}}^2})}---a=\sqrt{\frac{m}{{\mathrm{\&mu;}}^2}+\frac{1}{2{\mathrm{\&sigma;}}^2}}---d=\frac{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">m</figure-callout>}}{2{\mathrm{\&sigma;}}^{2}m+{\mathrm{\&mu;}}^2}{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">r</figure-callout>}}^2$

And make y=x-c, then

${\mathrm{pdf}}_r\left(r\right)=$

$\frac{m^m}{\mathrm{\&Gamma;}\left(m\right){\mathrm{\&mu;}}^{2\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">m</figure-callout>}}\sqrt{2\mathrm{\&pi;}}\mathrm{\&sigma;}}\exp (-d)\underset{-\&infin;}{\overset{+\&infin;}{\&Integral;}}{(y+c)}^{2m-1}\exp \left({-a}^2{y}^2\right)\mathrm{dy}$

$=\frac{m^m}{\mathrm{\&Gamma;}\left(m\right){\mathrm{\&mu;}}^{2\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">m</figure-callout>}}\sqrt{2\mathrm{\&pi;}}\mathrm{\&sigma;}}\exp (-d)\underset{j=0}{\overset{2m-1}{\mathrm{\&Sigma;}}}{C}_{2m-1}^i{C}^{2m-1-i}\underset{-\&infin;}{\overset{+\&infin;}{\&Integral;}}{y}^i\exp \left({-a}^2{y}^2\right)\mathrm{dy}$

(binomial expansion)

Because $\&ForAll;i\underset{-\&infin;}{\overset{+\&infin;}{\&Integral;}}{y}^{2i+1}\exp (-a^2{y}^2)\mathrm{dy}=0$ (by odd symmetry)

So

${\mathrm{pdf}}_r\left(r\right)=$

$\frac{m^m}{\mathrm{\&Gamma;}\left(m\right){\mathrm{\&mu;}}^{2\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">m</figure-callout>}}\sqrt{2\mathrm{\&pi;}}\mathrm{\&sigma;}}\exp (-d)\underset{-\&infin;}{\overset{+\&infin;}{\&Integral;}}{(y+c)}^{2m-1}\exp \left({-a}^2{y}^2\right)\mathrm{dy}$

$=\frac{{2m}^m}{\mathrm{\&Gamma;}\left(m\right){\mathrm{\&mu;}}^{2\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">m</figure-callout>}}\sqrt{2\mathrm{\&pi;}}\mathrm{\&sigma;}}\exp (-d)\underset{i=0}{\overset{[m-\frac{1}{2}]}{\mathrm{\&Sigma;}}}{C}_{2m-1}^{2i}{C}^{2m-1-i}\underset{-\&infin;}{\overset{+\&infin;}{\&Integral;}}{y}^i\exp \left({-a}^2{y}^2\right)\mathrm{dy}$

Utilize formula $\underset{0}{\overset{\&infin;}{\&Integral;}}{x}^n\exp (-r^2{x}^2)\mathrm{dx}=\frac{\mathrm{\&Gamma;}[(n+1)/2]}{{2r}^{n+1}},$ Yi Zhi:

${\mathrm{pdf}}_r\left(r\right)=\frac{m^m}{\mathrm{\&Gamma;}\left(m\right){\mathrm{\&mu;}}^{2\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">m</figure-callout>}}\sqrt{2\mathrm{\&pi;}}\mathrm{\&sigma;}}\exp (-d)\underset{i=0}{\overset{[m-\frac{1}{2}]}{\mathrm{\&Sigma;}}}{C}_{2m-1}^{2i}{C}^{2m-1-2i}\frac{\mathrm{\&Gamma;}(i+\frac{1}{2})}{a^{2i+1}}$

Utilize the character of Euler integral of the second kind $\mathrm{\&Gamma;}(i+\frac{1}{2})=\sqrt{\mathrm{\&pi;}}\underset{k=0}{\overset{i-1}{\mathrm{\&Pi;}}}(k+\frac{1}{2})$ (i 〉=1) and $\mathrm{\&Gamma;}\left(\frac{1}{2}\right)=\sqrt{\mathrm{\&pi;}},$ Yi Zhi:

pdf(r)＝

$\frac{m^m\exp (-\frac{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">m</figure-callout>}}{{2\mathrm{\&sigma;}}^{2}m+{\mathrm{\&mu;}}^2}{r}^2)}{\sqrt{2}\mathrm{\&Gamma;}\left(m\right)\mathrm{\&sigma;}{\mathrm{\&mu;}}^{2m}}\{\sqrt{\frac{1}{\frac{m}{{\mathrm{\&mu;}}^2}+\frac{1}{2{\mathrm{\&sigma;}}^2}}}\frac{r^{2m-1}}{{(1+\frac{{2\mathrm{m\&sigma;}}^2}{{\mathrm{\&mu;}}^2})}^{2m-1}}$

$+\underset{i=1}{\overset{[m-\frac{1}{2}]}{\mathrm{\&Sigma;}}}\frac{(2m-1)!}{\left(2i\right)!(2m-2i-1)!}\frac{\underset{k=0}{\overset{i-1}{\mathrm{\&Pi;}}}(k+\frac{1}{2})}{{(\frac{m}{{\mathrm{\&mu;}}^2}+\frac{1}{{2\mathrm{\&sigma;}}^2})}^{i+\frac{1}{2}}}\frac{r^{2m-2i-1}}{{(1+\frac{{2\mathrm{m\&sigma;}}^2}{{\mathrm{\&mu;}}^2})}^{2m-2i-1}}\}$

$=\frac{m^m\exp (-\frac{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">m</figure-callout>}}{{2\mathrm{\&sigma;}}^{2}m+{\mathrm{\&mu;}}^2}{r}^2)}{\mathrm{\&Gamma;}\left(m\right)}\{\frac{{\left(\mathrm{\&mu;r}\right)}^{2m-1}}{{({\mathrm{\&mu;}}^2+2m{\mathrm{\&sigma;}}^2)}^{2m-\frac{1}{2}}}.$

$+\underset{i=1}{\overset{[m-\frac{1}{2}]}{\mathrm{\&Sigma;}}}\frac{(2m-1)!}{\left(2i\right)!(2m-2i-1)!}\frac{\underset{k=0}{\overset{i-1}{\mathrm{\&Pi;}}}(k+\frac{1}{2})}{{({\mathrm{\&mu;}}^2+{2\mathrm{m\&sigma;}}^2)}^{2m-\mathrm{<figure-callout\; id="1"\; label="i\; +"\; filenames="A20051000436200332.png,A20051000436200341.png"\; state="\{\{state\}\}">i</figure-callout>}+\frac{1}{2}}}{2}^i{\mathrm{\&sigma;}}^{2i}{\left(\mathrm{\&mu;r}\right)}^{2m-2i-1}\}$

Card is finished.

These two theorems are the strong tool of mathematical analysis and the theoretical foundation of Nakagami decline stochastic variable being carried out higher order statistic analysis.

The main purpose of signal to noise ratio estimation algorithm is by the analysis to the statistical property of the reception bit r sequence of certain-length, the signal to noise ratio β that obtains estimating, and then obtain channel compensation value L _cSaid above that Nakagami fading channel characteristic was by the ternary vector $\stackrel{\&RightArrow;}{c}=(\mathrm{\&mu;},m,\mathrm{\&sigma;})$ Describe, this paper is valuation μ under the known condition of hypothesis m, and σ adopts the square method of estimation in the classical theory of probability, and two unknown parameters need two of simultaneous apart from equation, and for for simplicity, this patent is with 2 rank distances and 4 rank moment equation simultaneous solutions.Based on Nakagami decline stochastic variable theorem above, this patent proposes and proves SNR signal to noise ratio valuation theorem under the Nakagami channel of analyzing based on High Order Moment first.

SNR signal to noise ratio valuation theorem under the theorem 3 Nakagami channels:

Definition β is the signal to noise ratio of estimation, $\mathrm{\&beta;}\&equiv;\frac{{\mathrm{\&mu;}}^2}{2{\mathrm{\&sigma;}}^2};$ Z is the ratio of high-order statistic, $z\&equiv;\frac{E\left(r^4\right)}{E^2\left(r^2\right)}$

Then:

$\left\{\begin{array}{ccc}& z=\frac{4(1+\frac{1}{m}){\mathrm{\&beta;}}^2+12\mathrm{\&beta;}+3}{4{\mathrm{\&beta;}}^2+4\mathrm{\&beta;}+1}& \mathrm{\&beta;}>0\\ \mathrm{\&beta;}=\frac{c}{2(1-c)}& c=\sqrt{\frac{3-z}{2-\frac{1}{m}}}& z\&Element;(1-\frac{1}{m},3)\end{array}\right.$

Proof:

Easily know by theorem 1:

$\left\{\begin{array}{c}E\left(r^4\right)=(1+\frac{1}{m}){\mathrm{\&mu;}}^4+6{\mathrm{\&mu;}}^2{\mathrm{\&sigma;}}^2+3{\mathrm{\&sigma;}}^4\\ E\left(r^2\right)={\mathrm{\&mu;}}^2+{\mathrm{\&sigma;}}^2\end{array}\right.$

Hence one can see that:

$z\&equiv;\frac{E\left(r^4\right)}{E^2\left(r^2\right)}=\frac{(1+\frac{1}{m}){\mathrm{\&mu;}}^4+6{\mathrm{\&mu;}}^2{\mathrm{\&sigma;}}^2+3{\mathrm{\&sigma;}}^4}{{\mathrm{\&mu;}}^4+2{\mathrm{\&mu;}}^2{\mathrm{\&sigma;}}^2+{\mathrm{\&sigma;}}^4}=\frac{(1+\frac{1}{m}){\left(\frac{\mathrm{\&mu;}}{\mathrm{\&sigma;}}\right)}^4+6{\left(\frac{\mathrm{\&mu;}}{\mathrm{\&sigma;}}\right)}^2+3}{{\left(\frac{\mathrm{\&mu;}}{\mathrm{\&sigma;}}\right)}^4+2{\left(\frac{\mathrm{\&mu;}}{\mathrm{\&sigma;}}\right)}^2+1}$

$=\frac{4(1+\frac{1}{m}){\mathrm{\&beta;}}^2+12\mathrm{\&beta;}+3}{4{\mathrm{\&beta;}}^2+4\mathrm{\&beta;}+1}$ β＞0

$3E^2\left(r^2\right)-E\left(r^4\right)=(2-\frac{1}{m}){\mathrm{\&mu;}}^4\&DoubleRightArrow;{\mathrm{\&mu;}}^2=\sqrt{\frac{3E^2\left(r^2\right)-E\left(r^4\right)}{2-\frac{1}{m}}}$

$\&DoubleRightArrow;{\mathrm{\&sigma;}}^2=E\left(r^2\right)-{\mathrm{\&mu;}}^2=E\left(r^2\right)-\sqrt{\frac{3E^2\left(r^2\right)-E\left(r^4\right)}{2-\frac{1}{m}}}$

$\&DoubleRightArrow;\mathrm{\&beta;}\&equiv;\frac{{\mathrm{\&mu;}}^2}{2{\mathrm{\&sigma;}}^2}=\frac{c}{2(1-c)}---c=\sqrt{\frac{3-z}{2-\frac{1}{m}}}$

Card is finished.

Discussion to SNR signal to noise ratio valuation theorem under the Nakagami channel

Inference 1:z is the subtraction function of β, and $\underset{\mathrm{\&beta;}\&RightArrow;0}{\lim }Z=3,$ $\underset{\mathrm{\&beta;}\&RightArrow;\&infin;}{\lim }Z=1+\frac{1}{m}$

Proof:

Order $f\left(\mathrm{\&beta;}\right)=4(1+\frac{1}{m}){\mathrm{\&beta;}}^2+12\mathrm{\&beta;}+3,$ g(β)＝4β ²+4β+1

By theorem 3 as can be known

$z=\frac{f\left(\mathrm{\&beta;}\right)}{g\left(\mathrm{\&beta;}\right)}\&DoubleRightArrow;z^{\&prime;}=\frac{{\mathrm{gf}}^{\&prime;}-{\mathrm{fg}}^{\&prime;}}{{\mathrm{<figure-callout\; id="2"\; label="g"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">g</figure-callout>}}^2}=\frac{8\mathrm{\&beta;}(\frac{1}{m}-2)(2\mathrm{\&beta;}+1)}{g^2\left(\mathrm{\&beta;}\right)}$

∵m＞0.5

∴ $\frac{1}{m}-2<0$

∴z′＜0

∴ z is the subtraction function of β

$\underset{\mathrm{\&beta;}\&RightArrow;0}{\lim }Z=\frac{\underset{\mathrm{\&beta;}\&RightArrow;0}{\lim }f\left(\mathrm{\&beta;}\right)}{\underset{\mathrm{\&beta;}\&RightArrow;0}{\lim }g\left(\mathrm{\&beta;}\right)}=\frac{3}{1}=3$

$\underset{\mathrm{\&beta;}\&RightArrow;\&infin;}{\lim }Z=\frac{\underset{\mathrm{\&beta;}\&RightArrow;\&infin;}{\lim }f\left(\mathrm{\&beta;}\right){\mathrm{\&beta;}}^{-2}}{\underset{\mathrm{\&beta;}\&RightArrow;\&infin;}{\lim }g\left(\mathrm{\&beta;}\right){\mathrm{\&beta;}}^{-2}}=\frac{4(1+\frac{1}{m})}{4}=1+\frac{1}{m}$

Card is finished.

Inference 2: when $m=\frac{1}{2}$ The time, Nakagami decline stochastic variable r is degenerated to the stochastic variable of normal distribution, thereby channel parameter μ, and σ can not valuation.

Proof:

By theorem 2, as can be known:

When $m=\frac{1}{2}$ The time, $\mathrm{pdf}\left(r\right)=\frac{\exp (-\frac{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">r</figure-callout>}}^2}{2({\mathrm{\&sigma;}}^2+{\mathrm{\&mu;}}^2)})}{\sqrt{2\mathrm{\&pi;}({\mathrm{\&sigma;}}^2+{\mathrm{\&mu;}}^2)}}$

So r is an average is that zero variance is σ ²+ μ ²The stochastic variable of normal distribution.As long as variance is certain, the probability density function of r just determines that also its each rank square is also definite thereupon, and satisfies the certain (σ of variance ², μ ²) obviously have infinite group to separate, thereby channel parameter μ, σ can not valuation.Work as in the analytical expression of β in the theorem 3 $m=\frac{1}{2}$ In time, singular solution occurs and obtain Z by the expression formula of Z being constantly equal to 3, all reflected this problem.

Card is finished.

Inference 3: know easily that by theorem three there are following relational expression in Z and β for awgn channel (m=∞)

$\left\{\begin{array}{ccc}& z=\frac{4{\mathrm{\&beta;}}^2+12\mathrm{\&beta;}+3}{4{\mathrm{\&beta;}}^2+4\mathrm{\&beta;}+1}& \mathrm{\&beta;}>0\\ \mathrm{\&beta;}=\frac{c}{2(1-c)}& c=\sqrt{\frac{3-z}{2}}& z\&Element;(1,3)\end{array}\right.$

Inference 4: know easily that by theorem three there are following relational expression in Z and β for the smooth Rayleigh channel (m=1) that fully interweaves

$\left\{\begin{array}{ccc}& z=\frac{8{\mathrm{\&beta;}}^2+12\mathrm{\&beta;}+3}{4{\mathrm{\&beta;}}^2+4\mathrm{\&beta;}+1}& \mathrm{\&beta;}>0\\ \mathrm{\&beta;}=\frac{c}{2(1-c)}& c=\sqrt{3-z}& z\&Element;(2,3)\end{array}\right.$

In order to obtain perceptual knowledge, Fig. 2 has drawn one group of asynchronous Z-beta curve of m.

We know the optimal decoding algorithm-MAP algorithm of Turbo decoding, need know channel information, and how research utilizes signal-to-noise ratio estimation algorithm to improve the performance of Turbo decoding below.At first suppose in the bipolarity code word of k turbo encoder output constantly to be $C_k\&equiv;(u_k,x_k^p)$ (u _kBe system bits, x _k ^pBe check digit), through the reception code word after the Nakagami fading channel be $Y_k\&equiv;(y_k^s,y_k^p)$ (y _k ^sBe the system bits that receives, y _k ^pBe the check digit that receives), then:

$y_k^s=a_k^s\mathrm{\&mu;}{u}_k+n_k^s$

$y_k^p=a_k^p\mathrm{\&mu;}{x}_k^p+n_k^p$

Wherein, a _k ^s, a _k ^pBe respectively the property the taken advantage of interference of the Nakagami distribution of system bits and check digit experience, n _k ^s, n _k ^pBe to be respectively the additivity interference of the normal distribution of system bits and check digit experience, these interference are separate.Now, we define by the transition probability γ of k-1 s ' state transitions constantly to k S state constantly _k(s ', s) ≡ P (Y _k, S _k=S/S _K-1=s '), then:

${\mathrm{\&gamma;}}_k(s^{\&prime;},s)\&equiv;P(Y_k,S_k=S/S_{k-1}=s^{\&prime;})$

$=P(s/S^{\&prime;})P(Y_k/s,S^{\&prime;})$

$=P\left(u_k\right)P(Y_k/C_k)$

(chain type decomposition)

With u _kThe priori likelihood ratio be defined as $L\left(u_k\right)\&equiv;\ln \frac{P(u_k=1)}{P(u_k=-1)},$ Then:

P(u _k)∝exp(u _kL(u _k)/2)

$P(Y_k/C_k)=P(y_k^s/u_k)P(y_k^p/x_k^p)$ (by the memoryless property and the independence of channel) supposes a _k ^s, a _k ^pKnown, then:

$P(y_k^s/u_k)P(y_k^p/x_k^p)=P(y_k^s/(u_k,a_k^s))P(y_k^p/(x_k^p,a_k^p))$

$=\frac{1}{\sqrt{2\mathrm{\&pi;}}\mathrm{\&sigma;}}\exp (-\frac{(y_k^s-a_k^s{\mathrm{\&mu;u}}_k)}{2{\mathrm{\&sigma;}}^2})\frac{1}{\sqrt{2\mathrm{\&pi;}}\mathrm{\&sigma;}}\exp (-\frac{(y_k^p-a_k^p\mathrm{\&mu;}{x}_k^p)}{2{\mathrm{\&sigma;}}^2})\&DoubleRightArrow;$

$P(Y_k/C_k)\&Proportional;\exp \left(\frac{\mathrm{\&mu;}(u_k{a}_k^s+x_k^p{a}_k^p)}{{\mathrm{\&sigma;}}^2}\right)$

Order $L_c\&equiv;\frac{2\mathrm{\&mu;}}{{\mathrm{\&sigma;}}^2}$ Then

$\mathrm{\&gamma;}(s^{\&prime;},s)\&Proportional;\exp (u_{k}L\left(u_k\right)/2)\exp \left(\frac{L_c}{2}(u_k{a}_k^s+x_k^p{a}_k^p)\right)$

Definition transfering sheet C (s ', s)=In γ (s ', s), then:

The property taken advantage of interference a in known Nakagami distribution _k ^s, a _k ^pCondition under,

$C(s^{\&prime;},s,a_k^s,a_k^p)\&Proportional;u_{k}L\left(u_k\right)/2+\frac{L_c}{2}(u_k{a}_k^s+x_k^p{a}_k^p)$

At this moment

$L_c\&equiv;\frac{2\mathrm{\&mu;}}{{\mathrm{\&sigma;}}^2},$

Again by E (r ²)=μ ²+ σ ², easily know:

$L_c=2\sqrt{\frac{2\mathrm{\&beta;}(1+2\mathrm{\&beta;})}{E\left(r^2\right)}}$

The property taken advantage of interference a in unknown Nakagami distribution _k ^s, a _k ^pCondition under

$C(s^{\&prime;},s)\&Proportional;\&Integral;\&Integral;C(s^{\&prime;},s,a_k^s,a_k^p)\mathrm{pdf}\left(a_k^s\right)\mathrm{pdf}\left(a_k^p\right)d\left(a_k^s\right)d\left(a_k^p\right)$

$=u_{k}L\left(u_k\right)/2+\frac{{\mathrm{\&mu;u}}_k}{{\mathrm{\&sigma;}}^2}\&Integral;a_k^s\mathrm{pdf}\left(a_k^s\right)d\left(a_k^s\right)+\frac{{\mathrm{\&mu;x}}_k^p}{{\mathrm{\&sigma;}}^2}\&Integral;a_k^p\mathrm{pdf}\left(a_k^p\right)d\left(a_k^p\right)$

$=u_{k}L\left(u_k\right)/2+\frac{\mathrm{\&mu;E}\left(a\right)}{{\mathrm{\&sigma;}}^2}(u_k+x_k^p)$

$=u_{k}L\left(u_k\right)/2+\frac{\mathrm{\&mu;g}\left(m\right)}{{\mathrm{\&sigma;}}^2}(u_k+x_k^p)$

Herein, $g\left(m\right)=\frac{\mathrm{\&Gamma;}(m+0.5)}{\sqrt{m}\mathrm{\&Gamma;}\left(m\right)}$ , its function curve sees Fig. 3, and visible g (m) is the monotonically increasing bounded function, and $\underset{m\&RightArrow;\&infin;}{\lim }g\left(m\right)=1,$ (proof is omitted).

So, at the property the taken advantage of interference a of unknown Nakagami distribution _k ^s, a _k ^pCondition under,

$L_c=2g\left(m\right)\sqrt{\frac{2\mathrm{\&beta;}(1+2\mathrm{\&beta;})}{E\left(r^2\right)}}$

To sum up, following channel compensation theorem is arranged:

The channel compensation theorem of theorem 4 turbo sign indicating numbers under the Nakagami fading channel

Based on above-mentioned mathematical theory, the present invention proposes the technical scheme of signal-to-noise ratio (SNR) estimation and channel compensation.Specify its process below in conjunction with Fig. 4-9.

Fig. 4 is the block diagram according to the signal-to-noise ratio (SNR) estimation system of first embodiment of the invention.Fig. 5 is the concrete formation block diagram of the statistic calculating section in the signal-to-noise ratio (SNR) estimation system shown in Figure 4.

Importation 110 at first receives data from channel, is that unit puts in the buffer area (not shown) with each encoding block.Suppose that the data that receive are accurately synchronous and process coherent demodulation, establish each encoding block and comprise the N bit, with { r ₁, r ₂..., r _NExpression.

Then, statistic calculating section 120 is used for estimating { r ₁, r ₂..., r _NThe ratio z of high-order statistic.As shown in Figure 5, statistic calculating section 120 comprises: biquadratic mean value computation unit 1201 is used to calculate input signal { r ₁, r ₂..., r _NThe biquadratic average, $\stackrel{\&OverBar;}{{\mathrm{<figure-callout\; id="4"\; label="r"\; filenames="A20051000436200341.png,A20051000436200392.png"\; state="\{\{state\}\}">r</figure-callout>}}^4}=\frac{\underset{i=1}{\overset{N}{\mathrm{\&Sigma;}}}{r}_i^4}{N};$ Mean-square calculation unit 1202 is used to calculate input signal { r ₁, r ₂..., r _NMean-square value, $\stackrel{\&OverBar;}{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">r</figure-callout>}}^2}=\frac{\underset{i=1}{\overset{N}{\mathrm{\&Sigma;}}}{r}_i^2}{N};$ With ratio calculation unit 1203, be used for from the biquadratic average of biquadratic mean value computation unit 1201 with from square the comparing of the mean-square value of mean-square calculation unit 1202, obtain ratio, $z=\frac{\stackrel{\&OverBar;}{{\mathrm{<figure-callout\; id="4"\; label="r"\; filenames="A20051000436200341.png,A20051000436200392.png"\; state="\{\{state\}\}">r</figure-callout>}}^4}}{{\stackrel{\&OverBar;}{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">r</figure-callout>}}^2}}^2}.$

Channel parameter estimation part 140 records corresponding m value from the channel circumstance of concrete application, the method for specifically measuring the m value has a lot, and referring to list of references [8], [9] and [10], above-mentioned list of references is incorporated herein by reference.Then, signal to noise ratio is found the solution according to the m value of measuring parameters in channels part 140 measurements and the ratio z of statistic calculating section 120 calculating in snr computation unit 130 $\mathrm{\&beta;}=\frac{\sqrt{\frac{3-z}{2-\frac{1}{m}}}}{2(1-\sqrt{\frac{3-z}{2-\frac{1}{m}}})},$ Overflow for preventing to calculate, when z was not less than 3, β got threshold value 0.1.

Fig. 6 is the block diagram according to the channel compensation value estimating system of second embodiment of the invention.

Channel compensation value estimating system shown in Figure 6 adopts the signal-to-noise ratio (SNR) estimation system according to first embodiment of the invention, the difference of itself and first embodiment has been to increase a channel compensation value calculating section 150 that is arranged on snr computation part 130 downstreams, and it is according to signal to noise ratio β and and the mean-square value of signal

The calculating channel offset, concrete operations are as follows: at decline amplitude a _iUnder the known situation, $\mathrm{Lc}=2\sqrt{\frac{2\mathrm{\&beta;}(1+\mathrm{\&beta;})}{\stackrel{\&OverBar;}{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">r</figure-callout>}}^2}}}.$ This situation is applicable to that front end receiver has had Rake receiver or equalizer can accurately estimate the occasion of channel fading; At decline amplitude a _lUnder the condition of unknown, $\mathrm{Lc}=2g\left(m\right)\sqrt{\frac{2\mathrm{\&beta;}(1+\mathrm{\&beta;})}{\stackrel{\&OverBar;}{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">r</figure-callout>}}^2}}},g\left(m\right)=\frac{\mathrm{\&Gamma;}\left(m\right)}{\sqrt{m}\mathrm{\&Gamma;}\left(m\right)},$ This situation is applicable to the occasion of blind estimation.

Fig. 7 is the block diagram according to the channel compensation system of third embodiment of the invention.

The difference of the channel compensation value estimating system of the channel compensation system of Fig. 7 and second embodiment of the invention is to have increased channel compensation part 160 and optional amplitude of fading estimating part 170.Therefore, only two above-mentioned parts are elaborated below, have omitted the part that repeats with first and second embodiment.

At decline amplitude a _iUnder the condition of unknown, 160 pairs of channel compensation parts are sent into the i moment bit r of Turbo decoder _iPursue the bit compensation, $x_i=\frac{L_c}{2}{r}_i,$ This is the block-by-block compensation; And at decline amplitude a _iUnder the known situation, just amplitude of fading estimating part 170 can be measured amplitude of fading a from concrete channel _i, the amplitude of fading difference of each bit in the data block compensates the value of being compensated to the i moment bit of sending into the Turbo decoder $x_i=\frac{L_c}{2}{a}_i{r}_i,$ What carried out this moment is by the bit compensation.

Fig. 8 is the flow chart of the course of work that is used for illustrating the signal-to-noise ratio (SNR) estimation system of first embodiment of the invention.

At step S10, receive data from channel, be that unit puts in the buffer area (not shown) with each encoding block.Suppose that the data that receive are accurately synchronous and process coherent demodulation, establish each encoding block and comprise the N bit, with { r ₁, r ₂..., r _NExpression.

At step S20, estimated signal { r ₁, r ₂..., r _NThe ratio z of high-order statistic.Particularly, calculate input signal { r ₁, r ₂..., r _NThe biquadratic average, $\stackrel{\&OverBar;}{{\mathrm{<figure-callout\; id="4"\; label="r"\; filenames="A20051000436200341.png,A20051000436200392.png"\; state="\{\{state\}\}">r</figure-callout>}}^4}=\frac{\underset{i=1}{\overset{N}{\mathrm{\&Sigma;}}}{r}_i^4}{N};$ Calculate input signal { r ₁, r ₂..., r _NMean-square value, $\stackrel{\&OverBar;}{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">r</figure-callout>}}^2}=\frac{\underset{i=1}{\overset{N}{\mathrm{\&Sigma;}}}{r}_i^2}{N};$ And, square comparing of the biquadratic average of signal and mean-square value, obtain ratio, $z=\frac{\stackrel{\&OverBar;}{{\mathrm{<figure-callout\; id="4"\; label="r"\; filenames="A20051000436200341.png,A20051000436200392.png"\; state="\{\{state\}\}">r</figure-callout>}}^4}}{{\stackrel{\&OverBar;}{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">r</figure-callout>}}^2}}^2}.$

At step S30, from the channel circumstance of concrete application, record corresponding m value.

Then, at step S40,, find the solution signal to noise ratio from the m value of measuring and the ratio z of calculating $\mathrm{\&beta;}=\frac{\sqrt{\frac{3-z}{2-\frac{1}{m}}}}{2(1-\sqrt{\frac{3-z}{2-\frac{1}{m}}})},$ Overflow for preventing to calculate, when z was not less than 3, β got threshold value 0.1.At last, whole flow process finishes.

Fig. 9 is the flow chart of the course of work that is used for illustrating the channel compensation system of third embodiment of the invention.

The difference of the flow chart of Fig. 9 and the flow chart of Fig. 8 is to have increased step S50-S70, is used for the calculating channel offset, and utilizes this channel compensation value that signal is compensated.Therefore, no longer step S10-S40 is repeated in this description below.

At step S50, according to signal to noise ratio β and and the mean-square value of signal

The calculating channel offset, concrete operations are as follows: at amplitude of fading a _lUnder the known situation, $\mathrm{Lc}=2\sqrt{\frac{2\mathrm{\&beta;}(1+\mathrm{\&beta;})}{\stackrel{\&OverBar;}{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">r</figure-callout>}}^2}}}.$ This situation is applicable to that front end receiver has had Rake receiver or equalizer can accurately estimate the occasion of channel fading; At decline amplitude a _iUnder the condition of unknown, $\mathrm{Lc}=2g\left(m\right)\sqrt{\frac{2\mathrm{\&beta;}(1+\mathrm{\&beta;})}{\stackrel{\&OverBar;}{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="A20051000436200331.png,A20051000436200332.png"\; state="\{\{state\}\}">r</figure-callout>}}^2}}},g\left(m\right)=\frac{\mathrm{\&Gamma;}}{\sqrt{m}\mathrm{\&Gamma;}\left(m\right)},$ This situation is applicable to the occasion of blind estimation.

Under the known situation of decline amplitude al, just can estimate the amplitude of fading a of channel at step S60 _i, then the i moment bit of sending into the Turbo decoder is compensated the value of being compensated at step S70 $x_i=\frac{L_c}{2}{a}_i{r}_i,$ This is by the bit compensation method; And at decline amplitude a _iUnder the condition of unknown,, the i moment bit of sending into the Turbo decoder is compensated the value of being compensated at step S70 $x_i=\frac{L_c}{2}{r}_i,$ This is the block-by-block compensation method.Then, whole flow process finishes.

Figure 10 is the signal to noise ratio valuation emulation testing figure as a result under the Gaussian channel, and we get frame length 1156 and 196, has carried out not having valuation compensation (L respectively _c=1), the contrast test of valuation compensation of the present invention and known channel compensation.Article three, solid line is the test result of short frame length 196, and three test results that dotted line is long frame 1156.

Figure 11 is the signal to noise ratio valuation emulation testing figure as a result under the Rayleigh channel, get frame length 1156 and 196, carried out not having valuation compensation (that is a=1 and β=1) respectively, compensation (that is a=1) when new algorithm valuation compensation during the known amplitude of fading a that draws by theorem 4 and unknown amplitude of fading a, the known channel compensation (that is known a, contrast test β).It is to be noted that known channel is the method by the bit compensation $(L_c=\frac{2\mathrm{a\&mu;}}{{\mathrm{\&sigma;}}^2}),$ Wherein the property the taken advantage of Rayleigh fading value a of each bit is known.Article four, solid line is the test result of frame length 1156, and four test results that dotted line is a frame length 196.

From Figure 10 and Figure 11, as can be known: the ber curve when the valuation backoff algorithm of known amplitude of fading a and known channel state is quite approaching, especially under the condition that frame length is grown and m is bigger.And compare with the curve that does not have compensation 0.2 to 0.5 decibel gain is arranged through the curve of overcompensation.These have all proved the validity of valuation of the present invention and compensation method.

Though with reference to specific embodiment the present invention has been done detailed explanation, the present invention can have other not break away from the particular form of its marrow and essential characteristics.Therefore the present invention exemplarily and has considered all aspects without limitation, is equal to all changes among meaning and the scope so want to comprise invention scope that claim is determined but not above-mentioned explanation or embodiment determine and claim.

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Claims (28)

1, a kind of signal-noise ratio estimation method comprises step:

Receive the signal that comprises a plurality of bits from channel;

Calculate the mean-square value and the biquadratic average of described signal, and described biquadratic average and described mean-square value square between ratio;

The measured channel parameter value; And

Signal to noise ratio according to described channel parameter values and the described signal of described ratio calculation.

2, signal-noise ratio estimation method according to claim 1 is characterized in that, comprises according to following formula according to the step of described channel parameter values and described ratio calculation signal to noise ratio and calculates signal to noise ratio β:

$\mathrm{\&beta;}=\frac{\sqrt{\frac{3-z}{2-\frac{1}{m}}}}{2(1-\sqrt{\frac{3-z}{2-\frac{1}{m}}})}$

Wherein, m is a channel parameter values, z be the biquadratic average of signal and mean-square value square between ratio.

3, signal-noise ratio estimation method according to claim 2 is characterized in that, described signal comprises N bit r ₁, r ₂..., r _N, and according to the mean-square value of following formula signal calculated

With the biquadratic average

$\stackrel{\&OverBar;}{r^2}=\frac{\underset{i=1}{\overset{N}{\mathrm{\&Sigma;}}}{r_i}^2}{N}$

$\stackrel{\&OverBar;}{r^4}=\frac{\underset{i=1}{\overset{N}{\mathrm{\&Sigma;}}}{r_i}^4}{N}.$

4, signal-noise ratio estimation method according to claim 3 is characterized in that, calculates described ratio z according to following formula:

$z=\frac{\stackrel{\&OverBar;}{r^4}}{{\stackrel{\&OverBar;}{r^2}}^2}.$

5. signal-noise ratio estimation method according to claim 3 is characterized in that ratio z is not less than 3, and signal to noise ratio β gets 0.1.

6, a kind of signal-to-noise ratio (SNR) estimation system comprises:

The importation is used for receiving the signal that comprises a plurality of bits from channel;

The statistic calculating section, be used to calculate described signal biquadratic average, mean-square value and described biquadratic average and described mean-square value square between ratio;

The measuring parameters in channels part is used for from described channel measurement channel parameter;

The snr computation part is used for calculating according to described ratio and described channel parameter the signal to noise ratio of described signal.

7, signal-to-noise ratio (SNR) estimation according to claim 6 system is characterized in that, described snr computation part is calculated signal to noise ratio β according to following formula:

$\mathrm{\&beta;}=\frac{\sqrt{\frac{3-z}{2-\frac{1}{m}}}}{2(1-\sqrt{\frac{3-z}{2-\frac{1}{m}}})}$

Wherein, m is a channel parameter values, z be the biquadratic average of signal and mean-square value square between ratio.

8, signal-to-noise ratio (SNR) estimation according to claim 7 system is characterized in that described statistic calculating section comprises:

Biquadratic mean value computation unit is used to calculate the biquadratic average of described signal;

The mean-square calculation unit is used to calculate the mean-square value of described signal; With

The ratio calculation unit, be used for from the biquadratic average of biquadratic computing unit with from square the comparing of the mean-square value of mean-square calculation unit, obtain ratio.

9, signal-to-noise ratio (SNR) estimation according to claim 8 system is characterized in that described signal comprises N bit r ₁, r ₂..., r _N, described biquadratic mean value computation unit is according to the biquadratic average of following formula signal calculated:

$\stackrel{\&OverBar;}{r^4}=\frac{\underset{i=1}{\overset{N}{\mathrm{\&Sigma;}}}{r_i}^4}{N}$

Described mean-square calculation unit is according to the mean-square value of following formula signal calculated:

$\stackrel{\&OverBar;}{r^2}=\frac{\underset{i=1}{\overset{N}{\mathrm{\&Sigma;}}}{r_i}^2}{N}.$

10, signal-to-noise ratio (SNR) estimation according to claim 9 system is characterized in that described ratio calculation unit calculates described ratio according to following formula:

$z=\frac{\stackrel{\&OverBar;}{r^4}}{{\stackrel{\&OverBar;}{r^2}}^2}.$

11, signal-to-noise ratio (SNR) estimation according to claim 7 system.It is characterized in that ratio z is not less than 3, signal to noise ratio β gets 0.1.

12, a kind of channel compensating method comprises step:

Receive the signal that comprises a plurality of bits from channel;

Calculate the mean-square value and the biquadratic average of described signal, and described biquadratic average and described mean-square value square between ratio;

The measured channel parameter value;

Signal to noise ratio according to described channel parameter values and the described signal of described ratio calculation;

According to described signal to noise ratio and described mean-square calculation channel compensation value;

Utilize described channel compensation value that described channel is compensated.

13, channel compensating method according to claim 12 is characterized in that, comprises according to following formula according to the step of described channel parameter values and described ratio calculation signal to noise ratio and calculates signal to noise ratio β:

$\mathrm{\&beta;}=\frac{\sqrt{\frac{3-z}{2-\frac{1}{m}}}}{2(1-\sqrt{\frac{3-z}{2-\frac{1}{m}}})}$

Wherein, m is a channel parameter values, z be the biquadratic average of signal and mean-square value square between ratio;

Under the known situation of channel fading amplitude, comprise according to following formula calculating channel offset Lc according to the step of described signal to noise ratio and described mean-square calculation channel compensation value:

$\mathrm{Lc}=2\sqrt{\frac{2\mathrm{\&beta;}(1+\mathrm{\&beta;})}{\stackrel{\&OverBar;}{r^2}}}.$

14, channel compensating method according to claim 12 is characterized in that, comprises according to following formula according to the step of described channel parameter values and described ratio calculation signal to noise ratio and calculates signal to noise ratio β:

$\mathrm{\&beta;}=\frac{\sqrt{\frac{3-z}{2-\frac{1}{m}}}}{2(1-\sqrt{\frac{3-z}{2-\frac{1}{m}}})}$

Wherein, m is a channel parameter values, z be the biquadratic average of signal and mean-square value square between ratio;

Under channel fading amplitude condition of unknown, comprise according to following formula calculating channel offset Lc according to the step of described signal to noise ratio and described mean-square calculation channel compensation value:

$\mathrm{Lc}=2g\left(m\right)\sqrt{\frac{2\mathrm{\&beta;}(1+\mathrm{\&beta;})}{\stackrel{\&OverBar;}{r^2}}},g\left(m\right)=\frac{\mathrm{\&Gamma;}\left(m\right)}{\sqrt{m}\mathrm{\&Gamma;}\left(m\right)}.$

According to claim 13 or 14 described channel compensating methods, it is characterized in that 15, described signal comprises N bit r ₁, r ₂..., r _N, and according to the mean-square value of following formula signal calculated With the biquadratic average

$\stackrel{\&OverBar;}{r^2}=\frac{\underset{i=1}{\overset{N}{\mathrm{\&Sigma;}}}{r_i}^2}{N}$

$\stackrel{\&OverBar;}{r^4}=\frac{\underset{i=1}{\overset{N}{\mathrm{\&Sigma;}}}{r_i}^4}{N}.$

16, channel compensating method according to claim 13 is characterized in that, also comprises from channel measurement channel fading amplitude alpha _iStep, and the step of utilizing described channel compensation value that described channel is compensated comprises according to the bit of following formula to signal and compensating:

$x_i=\frac{L_c}{2}{a}_i{r}_i.$

17, channel compensating method according to claim 14 is characterized in that, the step of utilizing described channel compensation value that described channel is compensated comprises according to the bit of following formula to signal and compensating:

$x_i=\frac{L_c}{2}{r}_i.$

18, according to claim 13 or 14 described channel compensating methods, it is characterized in that, it is characterized in that, calculate described ratio z according to following formula:

$z=\frac{\stackrel{\&OverBar;}{r^4}}{{\stackrel{\&OverBar;}{r^2}}^2}.$

19, according to claim 13 or 14 described channel compensating methods, it is characterized in that ratio z is not less than 3, signal to noise ratio β gets 0.1.

20, a kind of channel compensation system comprises:

The importation is used for receiving the signal that comprises a plurality of bits from channel;

The statistic calculating section, be used for signal calculated biquadratic average, mean-square value and described biquadratic average and described mean-square value square between ratio;

The measuring parameters in channels part is used for from described channel measurement channel parameter;

The snr computation part is used for calculating according to described ratio and described channel parameter the signal to noise ratio of described signal;

Channel compensation value calculating section is used for according to described signal to noise ratio and described mean-square calculation channel compensation value;

The channel compensation part utilizes described channel compensation value that described channel is compensated.

21, channel compensation according to claim 20 system, described snr computation part is calculated signal to noise ratio β according to following formula:

$\mathrm{\&beta;}=\frac{\sqrt{\frac{3-z}{2-\frac{1}{m}}}}{2(1-\sqrt{\frac{3-z}{2-\frac{1}{m}}})}$

Wherein, m is a channel parameter values, z be the biquadratic average of signal and mean-square value square between ratio;

Under the known situation of channel fading amplitude, described channel compensation value calculating section is according to following formula calculating channel offset Lc:

$\mathrm{Lc}=2\sqrt{\frac{2\mathrm{\&beta;}(1+\mathrm{\&beta;})}{\stackrel{\&OverBar;}{r^2}}}.$

22, channel compensation according to claim 20 system, described snr computation part is calculated signal to noise ratio β according to following formula:

$\mathrm{\&beta;}=\frac{\sqrt{\frac{3-z}{2-\frac{1}{m}}}}{2(1-\sqrt{\frac{3-z}{2-\frac{1}{m}}})}$

Wherein, m is a channel parameter values, z be the biquadratic average of signal and mean-square value square between ratio;

Under channel fading amplitude condition of unknown, described channel compensation value calculating section is according to following formula calculating channel offset Lc:

$\mathrm{Lc}=2g\left(m\right)\sqrt{\frac{2\mathrm{\&beta;}(1+\mathrm{\&beta;})}{\stackrel{\&OverBar;}{r^2}}},g\left(m\right)=\frac{\mathrm{\&Gamma;}\left(m\right)}{\sqrt{m}\mathrm{\&Gamma;}\left(m\right)}.$

23, according to claim 21 or 22 described channel compensation systems, it is characterized in that described statistic calculating section comprises:

Biquadratic mean value computation unit is used to calculate the biquadratic average of described signal;

The mean-square calculation unit is used to calculate the mean-square value of described signal; With

The ratio calculation unit, be used for from the biquadratic average of biquadratic computing unit with from square the comparing of the mean-square value of mean-square calculation unit, obtain ratio.

24, according to claim 21 or 22 described channel compensation systems, it is characterized in that described signal comprises N bit r ₁, r ₂..., r _N, described biquadratic mean value computation unit is according to the biquadratic average of following formula signal calculated:

$\stackrel{\&OverBar;}{r^4}=\frac{\underset{i=1}{\overset{N}{\mathrm{\&Sigma;}}}{r_i}^4}{N}$

Described mean-square calculation unit is according to the mean-square value of following formula signal calculated:

$\stackrel{\&OverBar;}{r^2}=\frac{\underset{i=1}{\overset{N}{\mathrm{\&Sigma;}}}{r_i}^2}{N}.$

25, channel compensation according to claim 21 system is characterized in that, also comprises amplitude of fading estimating part, is used for from described channel estimating channel fading amplitude alpha _i, and the step of utilizing described channel compensation value that described channel is compensated comprises according to the bit of following formula to signal and compensating:

$x_i=\frac{L_c}{2}{a}_i{r}_i.$

26, channel compensation according to claim 22 system is characterized in that, compensates according to the bit of following formula to signal:

$x_i=\frac{L_c}{2}{r}_i.$

27, according to claim 21 or 22 described channel compensation systems, it is characterized in that it is characterized in that, the ratio calculation unit calculates described ratio z according to following formula:

$z=\frac{\stackrel{\&OverBar;}{r^4}}{{\stackrel{\&OverBar;}{r^2}}^2}.$

28, according to claim 21 or 22 described channel compensation systems, it is characterized in that ratio z is not less than 3, signal to noise ratio β gets 0.1.

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