A Blind Frequency Offset Estimator for Coherent M-PSK System in Autonomous Radio

Abstract

This paper presents a novel blind frequency offset estimator for coherent M-PSK systems in an autonomous radio. The proposed estimator is based on the spectrum of the signal’s argument. A data removal block is developed. We derive the distribution of the instantaneous phase, which is applied to indicate that the proposed estimator can be considered as a class of nonlinear least-squares estimator. We provide a method to analyze the asymptotic performance of the proposed estimator. This enable us to predict the mean-square error on frequency offset estimation for all signal-to-noise ratio (SNR) values. Computer simulations indicate that the proposed estimator achieves better performance than the original estimator. The performance of the proposed estimator as a blind estimator is also illustrated.

Appendix

We assume that

$$ \tan(z)=\frac{x+A\sin(\varphi)}{y+A\cos(\varphi)}, $$

(16)

where x and y are equal to q _R and q _I , respectively. The joint probability of x and y is given by

$$ p(x,y)=\frac{1}{2\pi\sigma^2}\exp \biggl\{-\frac{1}{2\sigma^2}\bigl(x^2+y^2 \bigr) \biggr\}. $$

(17)

In order to simplify the derivation, we divide the range of z into three parts. Thus, we can write the expression of p(z) as follows:

$$p(z)= \left \{ \begin{array}{l@{\quad}l} p_1(z), & -\pi<z<-\pi/2,\\ p_2(z), & -\pi/2\le z\le\pi/2,\\ p_3(z), & \pi/2 < z < \pi. \end{array} \right . $$

We derive the p ₂(z) and the other two use the same method. From (16), we obtain

$$ x=\tan(z)\bigl[y+A\cos(\varphi)\bigr]-A\sin(\varphi). $$

(18)

According to (17) and (18), p ₂(z) can be written as

$$\begin{aligned} p_2(z)&=\frac{1}{2\pi\sigma^2}\int_{-A\cos(\varphi)}^{\infty }\bigl \vert y+A\cos(\varphi)\bigr \vert \\ &\quad{}\times\exp \biggl\{ -\frac{1}{2\sigma^2} \bigl \{y^2+\bigl[\tan(z) \bigl(y+A\cos(\varphi )\bigr)-A\sin(\varphi) \bigr]^2 \bigr\} \biggr\}\,dy \\ &=\frac{1}{2\pi\sigma^2}\int_{-A\cos(\varphi)}^{\infty} \frac {1}{\cos^2(z)}\bigl[y+b+\bigl(A\cos (\varphi)-b\bigr)\bigr]\\ &\quad{}\times\exp \biggl\{- \frac{1+\tan^2(z)}{2\sigma^2}\bigl[(y+b)^2+a\bigr] \biggr\}\,dy \end{aligned} $$

where

Then, we get

$$p_2(z)=\frac{1}{2\pi}e^{-r}+\frac{1}{2}\sqrt{ \frac{r}{\pi}}\cos (z-\varphi)erfc\bigl(-\sqrt{r}\cos(z- \varphi) \bigr)e^{r\sin^2(z-\varphi)}. $$

Derive p ₁(z) and p ₃(z) following the same method, we get the same expression as p ₂(z). Thus,

(19)

Let

$$\left \{ \begin{array}{l} \mathbf{E}_\mathbf{1}=\{E\{z'(0)\},E\{z'(1)\},\ldots,E\{z'(N-1)\}\}\\[6pt] \mathbf{E}_\mathbf{2}=\{E\{z'^2(0)\},E\{z'^2(1)\},\ldots,E\{z'^2(N-1)\}\}\\[6pt] \mathbf{S}= \{0,\sin (\frac{2\pi k}{N} ),\ldots,\sin (\frac{2\pi k(N-1)}{N} ) \}\\[6pt] \mathbf{V}= \{0,\cos (\frac{2\pi k}{N} ),\ldots,\cos (\frac{2\pi k(N-1)}{N} ) \}; \end{array} \right . $$

z(n) is not the stationary process, and the values of E ₁ and E ₂ can be computed through p(z). Let m=2^D; we get

(20)

(21)

where

$$\lfloor x\rfloor= \left \{ \begin{array}{l@{\quad}l} -\pi, & x<-\pi,\\ \pi, & x>\pi,\\ x, & \mathrm{others}. \end{array} \right . $$

In order to compute the integrals in (20) and (21), we extend p(z) into Fourier series, which can be written as

$$ p(z)=\frac{1}{2\pi} \Biggl[1+2\sum_{i=1}^{\infty}b_i(r) \cos(iz) \Biggr], $$

(22)

with the Fourier series coefficients [10]

$$ b_i(r)=\frac{\sqrt{\pi r}}{2}e^{-r/2} \biggl[I_{\frac{i-1}{1}} \biggl(\frac{r}{2} \biggr)+I_{\frac{i+1}{1}} \biggl( \frac{r}{2} \biggr) \biggr], $$

(23)

where I _i (⋅) is the first kind modified Bessel function of order i. The integrals of E ₁ and E ₂ become integrable.

According to (16), we get

where sum(E ₂ ) is sum of all the elements in E ₂ .