Weighted-noise threshold based channel estimation for OFDM systems

Abstract

Orthogonal frequency division multiplexing (OFDM) technology is the key to evolving telecommunication standards including 3GPP-LTE Advanced and WiMAX. Reliability of any OFDM system increases with improved mean square error performance (MSE) of its channel estimator (CE). Particularly, a least squares (LS) based CE incorporating a time-domain denoising threshold, enables better MSE performance, while avoiding the need for a-priori knowledge of channel statistics (KCS). Existing optimal time-domain thresholds exhibit suboptimal behavior for completely unavailable KCS environments. This is because they involve consistent estimation of one or more KCS parameters, and corresponding estimation errors introduce severe degradation in MSE performance of the CE. To overcome the MSE degradation, this paper proposes a weighted-noise threshold, by introducing a modified hypothesis-testing-problem (HTP) interpretation. Derivation of resulting analytical MSE expression is also provided. Results of OFDM system simulations carried out in rayleigh faded ITU-TU6 and WiMAX-SUI4 channel environments with U-shaped power spectral densities, are presented. The performance results show that, compared to many of the existing thresholds, the proposed threshold renders better MSE performance to the CE and higher reliability to the OFDM system in terms of better bit error rate (BER) performance.

Appendix: proof of conditional expectations (24) and (25)

With the modified hypothesis shown in (16), the modified event III, namely ${{\left ({\left |{\hat h_{i,n}}\right |^{2} > \vartheta }\right )}/{\left ({\left | {h_{i,n}} \right |^{2} \le \left | {v_{i,n}} \right |^{2}}\right )}}$ occurs with a probability of P _{F A} . In this case, though actual tap is not present, the threshold detects as tap is existing, thus causing a squared error of $\left | \bar v_{i,n} + v_{i,n} \right |^{2}$. The mean of this squared error is given in (29) and (30).

$$ E\left\{{{{\left| {e_{i,n}} \right|^{2}}/{III}}} \right\} = E\left\{{{{\left| {\bar v_{i,n} + v_{i,n}} \right|^{2}}/{III}}} \right\} = E\left\{{{x/{III}}} \right\} $$

(29)

$$ E\left\{{{x/{III}}} \right\} = \int\limits_{ - \infty }^{\infty} {xf\left({{x /{III}}} \right)} dx. $$

(30)

Using Baye’s theorem, the conditional density function is given in (31).

$$ f\left({{x /{III}}}\right) ={\frac{{f(x,III)}}{{f\left({III}\right) = \int\limits_{- \infty }^{\infty} {f\left({x,III}\right)} dx}}}. $$

(31)

Simplifying (31), we get (32), which is used in (30), to finally obtain (24).

$$ \begin{array}{l} f\left({{x /{III}}}\right) = \frac{{\frac{1}{{\left({q + 1}\right){\sigma_{v}^{2}}}}e^{{{- x}/{\left({\left({q + 1} \right){\sigma_{v}^{2}}}\right)}}}}}{{\int\limits_{\vartheta}^{\infty} {\frac{1}{{\left({q + 1} \right){\sigma_{v}^{2}}}}e^{{{- x}/{\left({\left({q + 1}\right){\sigma_{v}^{2}}}\right)}}} dx}}},\vartheta \le x < \infty \\ \begin{array}{*{20}c} {} & = \end{array}\frac{{\frac{1}{{\left({q + 1}\right){\sigma_{v}^{2}}}}e^{{{- x}/{\left({\left({q + 1}\right){\sigma_{v}^{2}}}\right)}}}}}{{e^{{{- \vartheta }/{\left({\left({q + 1}\right){\sigma_{v}^{2}}}\right)}}}}}. \end{array} $$

(32)

On similar lines as above, modified event IV, ${{\left ({\left | {\hat h_{i,n}} \right |^{2} < \vartheta }\right )}/{\left ({\left | {h_{i,n}} \right |^{2} \le \left | {v_{i,n}} \right |^{2}}\right )}}$ has a probability of $\left (1-P_{FA} \right )$. In this case, the actual tap is not present, and the threshold also detects as no tap. Though the threshold identified correct, due to modified H ₀ in (16), a squared error of $\left | \bar v_{i,n} \right |^{2}$ exists. The mean of this squared error is simplified similar to that on event III to finally obtain (25). The conditional density function used in deriving (25) is (33).

$$ \begin{array}{l} f\left({{x/{IV}}}\right) = \frac{{\frac{1}{{q{\sigma_{v}^{2}}}}e^{{{- x}/{\left({q{\sigma_{v}^{2}}}\right)}}}}}{{\int\limits_{0}^{\vartheta} {\frac{1}{{q{\sigma_{v}^{2}}}}e^{{{- x}/{\left({q{\sigma_{v}^{2}}}\right)}}} dx}}},0 \le x \le \vartheta \\ = \frac{{\frac{1}{{q{\sigma_{v}^{2}}}}e^{{{- x}/{\left({q{\sigma_{v}^{2}}}\right)}}}}}{{1 - e^{{{- \vartheta}/{\left({q{\sigma_{v}^{2}}}\right)}}}}}. \end{array} $$

(33)