Impulse Noise Correction in OFDM Systems

Abstract

In orthogonal frequency division multiplexing (OFDM) systems, the impulse noise causes catastrophic accuracy degradation since the impulse noise affects all the subcarriers in a symbol due to the fast Fourier transform (FFT) operations at the receiver. Potential causes of impulse noise include erasure channel, power switching, and circuit failure in integrated circuits. In this paper, from a practical observation, a novel iterative impulse error correction scheme is proposed. This scheme is referred to as the impulse noise location and value search algorithm, which is based on the crucial observation of the relationship of the impulse noise and the symbol constellation. In a 512-FFT OFDM system at 25 dB additive white Gaussian noise signal-to-noise ratio, for quadrature amplitude modulation (QAM)-4 and QAM-8 modulation, simulation results show that our proposed novel scheme can effectively correct impulse errors that corrupt up to 20.7 % and 13.9 % of the received time-domain signal at known locations. In addition, without the knowledge of impulse noise location, the proposed scheme still can correct at least 9.96 % of the received time-domain signal for QAM-4 modulation.

Appendices

Appendix A: Proof of Probability of Correct Demodulation for PAM-2 Modulation

Figure 20 illustrates the area of correct demodulation for PAM-2 modulation. The transmitted symbol s _f is either 1 or -1. In Fig. 20, if s _f = 1, the area of correct demodulation Ar ea ₁ is

$$\begin{array}{rll} Area_{1} & = & \pi r^{2}-\pi r^{2}\left(\frac{\theta}{2\pi}\right)+\sqrt{r^{2}-1} \\ \\ \theta & = & \arccos\left(\frac{2-r^{2}}{r^{2}}\right), \end{array} $$

(13)

where x-axis and y-axis represent the real part and imaginary part of the reconstructed symbol, respectively. Then, the probability of correct demodulation for PAM-2 modulation p ₂ is

$$\begin{array}{rll} p_{2} & = & \frac{1}{2}\left[\frac{Area_{1}}{\pi r^{2}}+\frac{Area_{-1}}{\pi r^{2}}\right] \\ & = &\begin{cases} 1, \ r\leq1 \\ \frac{1}{\pi r^{2}}\left[\pi r^{2}\left(\frac{2\pi-\theta}{2\pi}\right)+\sqrt{r^{2}-1}\right], \ r>1 \\ \end{cases} \\ \theta &=& \arccos\left(\frac{2-r^{2}}{r^{2}}\right). \end{array} $$

(14)

This corresponds to (8).

Figure 20

Area of concern to calculate p ₂ for PAM-2 modulation.

Appendix B: Proof of Probability of Correct Demodulation for QAM-4 Modulation

Figure 21 illustrates the area of correct demodulation for QAM-4 modulation. The transmitted symbol s _f is randomly chosen from {±1±j} or {(±1,±1)}. For QAM-4, two cases $1< r\leq \sqrt {2}$ and $r>\sqrt {2}$ need to be considered separately. When $1< r\leq \sqrt {2}$ in Fig. 21(a), if s _f = 1+j = (1,1), the area of correct demodulation $Area_{s_{f}}$ is

$$ \begin{array}{rll} Area_{s_{f}} &=& \pi r^{2}-\pi r^{2}\left(\frac{2\theta_{1}}{2\pi}\right)+2\sqrt{r^{2}-1} \\ \theta_{1} &=& \arccos\left(\frac{2-r^{2}}{r^{2}}\right). \end{array} $$

(15)

Figure 21

Areas of concern to calculate p ₄ for QAM-4 modulation, including (a) $1< r\leq \sqrt {2}$ and (b) $r>\sqrt {2}$.

In addition, when $r>\sqrt {2}$ in Fig. 21b, if s _f = 1+j, the area of correct demodulation $Area_{s_{f}}$ is

$$\begin{array}{rll} Area_{s_{f}} & = & \pi r^{2}\left(\frac{\theta_{2}}{2\pi}\right)+1+\sqrt{r^{2}-1} \\ \\ \theta_{2} & = & \arccos\left(\frac{-2\sqrt{r^{2}-1}}{r^{2}}\right). \end{array} $$

(16)

Therefore, based on (15) and (16), the probability of correct demodulation for QAM-4 modulation p ₄ is

$$\begin{array}{rll} p_{4} &=& \sum\limits_{s_{f}\in\{\pm1\pm j\}}\frac{1}{4}\frac{Area_{s_{f}}}{\pi r^{2}} \\ &=& \begin{cases} 1, \ r\leq1 \\ \frac{1}{\pi r^{2}}\left[\pi r^{2}\left(\frac{2\pi-2\theta_{1}}{2\pi}\right)+2\sqrt{r^{2}-1}\right],{} \sqrt{2}\geq r>1 \\ \frac{1}{\pi r^{2}}\left[\pi r^{2}\left(\frac{\theta_{2}}{2\pi}\right){}+{}1{}+{}\sqrt{r^{2}-1}\right], r>\sqrt{2} \\ \end{cases} \\ \theta_{1} &=& \arccos\left(\frac{2-r^{2}}{r^{2}}\right) \\ \theta_{2} &=& \arccos\left(\frac{-2\sqrt{r^{2}-1}}{r^{2}}\right). \end{array} $$

(17)

This corresponds to (9).

Appendix C: Proof of Probability of Correct Demodulation for QAM-8 Modulation

Figure 22 illustrates the area of correct demodulation for QAM-8 modulation. The transmitted symbol s _f is randomly chosen from {±1±;j,±3±j} or {(±1,±1),(±3,±1)}. When $1< r\leq \sqrt {2}$ in Fig. 22a, if s _f ∈{±1±j}, the area of correct demodulation $Area_{s_{f}}$ is

$$\begin{array}{rll} Area_{s_{f}} & = & \pi r^{2}-\pi r^{2}\left(\frac{3\theta_{1}}{2\pi}\right)+3\sqrt{r^{2}-1} \\ \\ \theta_{1} &=& \arccos\left(\frac{2-r^{2}}{r^{2}}\right). \end{array} $$

(18)

If s _f ∈{±3±j}, the area of correct demodulation $Area_{s_{f}}$ is similar to that in QAM-4 modulation, and $Area_{s_{f}}$ is

$$\begin{array}{rll} Area_{s_{f}} & = & \pi r^{2}-\pi r^{2}\left(\frac{2\theta_{1}}{2\pi}\right)+2\sqrt{r^{2}-1} \\ \\ \theta_{1}& = &\arccos\left(\frac{2-r^{2}}{r^{2}}\right). \end{array} $$

(19)

Moreover, when $r>\sqrt {2}$ in Fig. 22b, if s _f ∈{±1±j}, the area of correct demodulation $Area_{s_{f}}$ is

$$\begin{array}{rll} Area_{s_{f}} & = & \pi r^{2}\left(\frac{\theta_{2}}{2\pi}\right)+\frac{3}{2}\left(1+\sqrt{r^{2}-1}\right) \\ \\ \theta_{2} & = & \arccos\left(\frac{r^{2}-2}{r^{2}}\right). \end{array} $$

(20)

If s _f ∈{±3±j}, the area of correct demodulation is also similar to that in QAM-4 modulation, and $Area_{s_{f}}$ is

$$\begin{array}{rll} Area_{s_{f}}& = &\pi r^{2}\left(\frac{\theta_{2}}{2\pi}\right)+1+\sqrt{r^{2}-1} \\ \\ \theta_{3}& = &\arccos\left(\frac{-2\sqrt{r^{2}-1}}{r^{2}}\right). \end{array} $$

(21)

From (18)–(21), the probability of correct demodulation for QAM-8 modulation p ₈ is

$$ \begin{array}{rll} p_{8} &=& \sum\limits_{s_{f}\in\{\pm1\pm j,\pm3\pm j\}}\frac{1}{8}\frac{Area_{s_{f}}}{\pi r^{2}} \\ &=& \begin{cases} 1, \ r\leq1 \\ \frac{1}{\pi r^{2}}\left[\pi r^{2}\left(\frac{2\pi-2\theta_{1}}{2\pi}\right)+2\sqrt{r^{2}-1}\right], \ \sqrt{2}\geq r>1 \\ \frac{1}{\pi r^{2}}\left[\pi r^{2}\left(\frac{\theta_{2}}{2\pi}\right)+1+\sqrt{r^{2}-1}\right], \ r>\sqrt{2} \\ \end{cases} \\ \end{array} $$

$$\begin{array}{rll} \theta_{1} &=& \arccos\left(\frac{2-r^{2}}{r^{2}}\right) \\ [6pt] \theta_{2} &=& \arccos\left(\frac{-2\sqrt{r^{2}-1}}{r^{2}}\right). \end{array} $$

(22)

This corresponds to (10).

Figure 22

Areas of concern to calculate p ₈ for QAM-8 modulation, including (a) $1< r\leq \sqrt {2}$ and (b) $r>\sqrt {2}$.