Inverted Wrap-Around Limiting with Bussgang Noise Cancellation Receiver for OFDM Signals

Abstract

Orthogonal frequency division multiplexing (OFDM) is a widely used multicarrier modulation technique. High peak-to-average power ratio (PAPR) is a major drawback of OFDM systems which leads to power inefficiency and signal distortions. The simplest approach is to clip the high amplitudes at some predefined threshold before the signal is applied to the amplifier. The clipping causes in-band distortion as well as out-of-band radiation. Though the out-of-band radiation is reduced using filtering, it causes peak regrowth. The peak regrowth is taken care of using the repeated clipping and filtering. This paper presents the use of a different form of limiting, inverted wrap-around (IWrap). IWrap is presented and evaluated in terms of PAPR reduction and peak regrowth. The results show that IWrap limiting is much better for PAPR reduction as compared to conventional clipping technique. The performance of the proposed limiting technique is investigated with Bussgang noise cancellation iterative receiver for OFDM signals. The results of this iterative receiver with IWrap limiting are presented in terms of bit error probabilities.

Appendix

This appendix explains the derivation of nonlinear distortion factor $\alpha$ for IWrap limiting.

The limiting function f(z(n)) for IWrap could be written as follows:

$$\begin{aligned} f(z(n))={\left\{ \begin{array}{ll} z,&{} \text {if } 0\le \text {z}<A\\ 2A-z, &{} A\le \text {z}<\infty \end{array}\right. } \end{aligned}$$

(13)

According to central limit theorem, the OFDM signal may be considered Gaussian for large number of subcarriers. Hence the amplitude values of the signal follow a Rayleigh distribution. The probability density function (Pdf) of an OFDM symbol is:

$$\begin{aligned} P(z) = \frac{2z}{ \sigma ^{2}}\text {e}^{\frac{-z^{2}}{\sigma ^{2}}} \end{aligned}$$

(14)

The expectation from Eq. 9 could be written as:

$$\begin{aligned} \alpha =\frac{E\left\{ {z(n)f(z(n))}\right\} }{ \sigma ^{2}}=\frac{1}{\sigma ^{2}}\int _{0}^{\infty } zf(z)P(z)\hbox {d}z \end{aligned}$$

(15)

$$\begin{aligned} = \frac{1}{\sigma ^{2}}\left( \int _{0}^{A}zf(z)P(z)\hbox {d}z + \int _{A}^{\infty }zf(z)P(z)\hbox {d}z\right) \end{aligned}$$

(16)

solving Eq 16, taking first part

$$\begin{aligned} =\int _{0}^{A}zf(z)P(z)\hbox {d}z \end{aligned}$$

(17)

Putting in f(z) and P(z) into the equation from 13 and 14

$$\begin{aligned} =\int _{0}^{A}z z \frac{2z}{\sigma ^2}\text {e}^{\frac{-z^{2}}{\sigma ^{2}}}\hbox {d}z=\int _{0}^{A}\frac{z^2}{\sigma ^2}\text {e}^{\frac{-z^{2}}{\sigma ^{2}}}2z\hbox {d}z \end{aligned}$$

(18)

Evaluating the integral gives

$$\begin{aligned} \int _{0}^{A}zf(z)P(z)\hbox {d}z= & {} \left[ {\sigma ^{2}}\text {e}^t(t-1)\right] _{0}^{A} \end{aligned}$$

(19)

$$\begin{aligned}= & {} \left[ {\sigma ^{2}}\text {e}^{-\frac{{z}^2}{\sigma ^2}}\left( -\frac{z^2}{\sigma ^2}-1\right) \right] _{0}^{A} \end{aligned}$$

(20)

$$\begin{aligned}= & {} \left[ -\text {e}^{-\frac{z^2}{\sigma ^2}}\left( {z}^2+{\sigma ^2}\right) \right] _{0}^{A}\end{aligned}$$

(21)

$$\begin{aligned} \int _{0}^{A}zf(z)P(z)\hbox {d}z= & {} -\text {e}^{-\frac{A^2}{\sigma ^2}}\left( {A}^2+{\sigma ^2}\right) +\sigma ^{2} \end{aligned}$$

(22)

Solving the second part from 16

$$\begin{aligned} \int _{A}^{\infty }zf(z)P(z)\hbox {d}z= & {} \int _{A}^{\infty }z(2A-z)\frac{2z}{\sigma ^2}{e}^\frac{-z^2}{\sigma ^2}\hbox {d}z \end{aligned}$$

(23)

$$\begin{aligned}= & {} \int _{A}^{\infty }2Az\frac{2z}{\sigma ^2}{e}^\frac{-z^2}{\sigma ^2}\hbox {d}z - \int _{A}^{\infty }z^2\frac{2z}{\sigma ^2}{e}^\frac{-z^2}{\sigma ^2}\hbox {d}z \end{aligned}$$

(24)

Evaluating Eq. 24 gives

$$\begin{aligned} =2{A}^2 {e}^{\frac{-A^2}{\sigma ^2}} + {A\sigma }\sqrt{\pi } \hbox {erfc}\left( \frac{A}{\sigma }\right) - {e}^{\frac{-A^2}{\sigma ^2}}\left( A^2 + \sigma ^2\right) \end{aligned}$$

(25)

Using Eqs. 22 and 25 in Eq. 16

$$\begin{aligned} \alpha= & {} \frac{1}{\sigma ^2}\left( -\text {e}^{-\frac{A^2}{\sigma ^2}}\left( {A}^2+{\sigma ^2}\right) +\sigma ^{2} + 2{A}^2 {e}^{\frac{-A^2}{\sigma ^2}} \right. \nonumber \\&\left. + {A\sigma }\sqrt{\pi } \hbox {erfc}\left( \frac{A}{\sigma }\right) - {e}^{\frac{-A^2}{\sigma ^2}}\left( A^2 + \sigma ^2\right) \right) \end{aligned}$$

(26)

$$\begin{aligned} \alpha= & {} 1 -{2}\text {e}^{-\frac{A^2}{\sigma ^2}} + \frac{A}{\sigma }\sqrt{\pi } \hbox {erfc}\left( \frac{A}{\sigma }\right) \end{aligned}$$

(27)

Using clipping ratio $\gamma$ = A/$\sigma$,

$$\begin{aligned} \alpha =1 -{2}\text {e}^{-\gamma ^2} + \gamma \sqrt{\pi } \hbox {erfc}(\gamma ) \end{aligned}$$

(28)

Equation 28 is the distortion factor for IWrap.