General interference analysis of <Emphasis Type="Italic">M</Emphasis>-QAM and <Emphasis Type="Italic">M</Emphasis>-PSK wireless communications

Albert Mráz; László Pap

doi:10.1007/s11276-012-0469-5

Wireless Networks

April 2013, Volume 19, Issue 3, pp 331–344 | Cite as

General interference analysis of M-QAM and M-PSK wireless communications

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Albert MrázEmail author
László Pap

Article

First Online: 26 June 2012

Abstract

This work includes an accurate and general technique, which has been developed to analyze the symbol-error-rate (SER) of coherent M-Quadrature Amplitude Modulation (QAM) and Phase Shift Keying (PSK) schemes for various conditions of the transmission. A calculation scheme has been presented dealing with Additive White Gaussian Noise and different fading effects influencing the analyzed M-ary QAM or PSK signals. Furthermore, the authors have considered the effects of multiple interference signals embedded into a stochastic model having numerous parameters, which can be assumed as random variables with adjustable distributions. The resulted M-QAM and M-PSK SER expressions have been extended for multi-carrier transmission based on the analytical calculation of the stochastic interference model.

Keywords

M-QAM M-PSK Symbol-error-rate (SER) Rayleigh-, Rice-, Nakagami-fading stochastic interference model

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Notes

Acknowledgments

Our work presented in this paper was supported from the KMOP-1.1.2-08/1-2008-0001 project by the BME-Infokom Innovátor Nonprofit Kft.

Appendix 1: Calculation of the equivalent Gaussian component

Interference component

The deduction of expression (12) will be detailed as

$$ {\mathbb{E}}\left[ {{\left( R_{m,k}^{\left( 2 \right)} \right)}^2} \right]={\mathbb{E}}\left[ \left( \hbox{Re}{{\left\{ \frac{2\sqrt{{s_0}{s_k}}}{{{T}_{\rm s}}}\int\limits_{-\infty }^{\infty }{{{\eta }_{k}}(t)\left( d_{m}^{*} \right){{g}^{*}}(t){{e}^{j\left( 2\pi \left( {{f}_{0}}-{f_k} \right)t+{{\Upphi }_{0}} \right)}}dt} \right\}}} \right)^2 \right]. $$

(50)

The ∑ _k=1 ^K sum will be neglected during this calculation, since the interference components arisen from different sources are independent. Furthermore, the sum components will be eliminated during the calculation of the expected value and cross-products will be zeros during the operation of raising to the second power.

$$ \begin{aligned} &{\mathbb{E}}\left[ {{\left( R_{m,k}^{\left( 2 \right)} \right)}^2}\right]\\ &=\frac{4{s_k}{s_0}}{4{{T}_{\rm s}}^2}{\mathbb{E}}\left[ {{\left( \int\limits_{-\infty }^{\infty }{\left[ {{\eta }_{k}}(t)d_{l}^{*}{{g}^{*}}(t){{e}^{j\left( 2\pi \Updelta {f_k}t+{{\Upphi }_{0}} \right)}}+{{\eta}_{k}}^{*}(t){d_l}g(t){{e}^{-j\left( 2\pi \Updelta {f_k}t+{{\Upphi}_{0}} \right)}} \right]}dt \right)}^2} \right]\\ &=\frac{4{s_0}{s_k}}{4{{T}_{\rm s}}^2}{\mathbb{E}}\left[ \left( \int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{\left[ {{\eta }_{k}}(t)d_{l}^{*}{{g}^{*}}(t){{e}^{j\left( 2\pi \Updelta {f_k}t+{{\Upphi }_{0}} \right)}}+\eta _{k}^{*}(t){d_l}g(t){{e}^{-j\left( 2\pi \Updelta {f_k}t+{{\Upphi }_{0}} \right)}} \right]}} \right.\right.\\ &\left. \cdot \left[ {{\eta }_{k}}\left( \rho \right)d_{l}^{*}{{g}^{*}}\left(\rho \right){{e}^{j\left( 2\pi \Updelta {f_k}\rho +{{\Upphi }_{0}} \right)}}+\eta _{k}^{*}\left( \rho \right){d_l}g\left( \rho \right){{e}^{-j\left( 2\pi \Updelta {f_k}\rho +{{\Upphi }_{0}} \right)}} \right]dtd\rho \right]\\ &=\frac{4{s_k}{s_0}}{4{{T}_{\rm s}}^2}\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{{{R}_{{{\eta }_{k}}}}\left( t-\rho \right){{\left| {d_l} \right|}^2}\left[ {{g}^{*}}(t)g\left( \rho \right){{e}^{j2\pi \Updelta {f_k}\left( t-\rho \right)}}+g(t){{g}^{*}}\left( \rho \right){{e}^{-j2\pi \Updelta {f_k}\left( t-\rho \right)}} \right]}}dtd\rho, \end{aligned} $$

(51)

since the correlation function has a symmetrical attribute, i.e. ${{{R}_{{{\eta }_{k}}}}\left( t-\rho \right)={{R}_{{{\eta }_{k}}}}\left( \rho -t \right)}$. Hereafter, the phase $\Upphi_0$ will be neglected, as η_k(t) also disposes of a random phase. In this way it has effect of no interest in final results’ point of view. Equal cross-products will remain, since in-phase and quadrature components are independent. After the introduction of new variables (co-ordinate transformation)

$$ \left(\begin{array}{l} \tau =t-\rho \\ \eta =t\\ \end{array}, \left | \begin{array}{ll}\frac{\partial \tau}{\partial t} & \frac{\partial \eta }{\partial t}\\ \frac{\partial \tau }{\partial \rho } & \frac{\partial \eta }{\partial \rho} \end{array} \right | = \left | \begin{array}{ll} 1 & 1\\ -1 & 0 \end{array} \right | = 1\right) $$

(52)

let us notice, that

$$ {\mathcal{F}}\left\{ {{R}_{{{\eta }_{k}}}}\left( \tau \right){{e}^{j2\pi \Updelta {f_k}\tau }} \right\}=\int\limits_{-\infty }^{\infty }{{{R}_{{{\eta }_{k}}}}\left( \tau \right){{e}^{j2\pi \Updelta {f_k}\tau }}{{e}^{-j2\pi f\tau }}}d\tau ={{S}_{{{\eta }_{k}}}}\left( f-\Updelta {f_k} \right), $$

(53)

where $\mathcal{F}$ denotes the Fourier-transform, and

$$ {\mathcal{F}}\left\{ {{g}^{*}}\left( \tau \right)g\left( t-\tau \right) \right\}=\int\limits_{-\infty }^{\infty }{{{g}^{*}}\left( \tau \right)g\left( t-\tau \right){{e}^{-j2\pi ft}}d \tau}={{\left| G\left( f \right) \right|}^2} $$

(54)

in terms of Parseval’s theorem [19]. After that

$$ \begin{aligned} &{\mathbb{E}}\left[ {{\left( R_{m,k}^{\left( 2 \right)} \right)}^2}\right] =\frac{4{s_k}{s_0}}{4{{T}_{\rm s}}^2}{{\left| {d_m} \right|}^2}\left[ \int\limits_{-\infty }^{\infty }{{{S}_{{{\eta }_{k}}}}\left( f-\Updelta {f_k} \right){{\left| G\left( f \right) \right|}^2}}df+\int\limits_{-\infty }^{\infty }{{{S}_{{{\eta }_{k}}}}\left( f+\Updelta {f_k} \right){{\left| G\left( f \right) \right|}^2}}df \right], \end{aligned} $$

and during the symmetry of ${{\left| G\left( f \right) \right|}^2}$, the results of the two integrals above will be equal, i.e.

$$ {\mathbb{E}}\left[ {{\left( R_{m,k}^{\left( 2 \right)} \right)}^2} \right]=2\frac{{4s_k}{s_0}}{{4{T}_{\rm s}}^2}{{\left| {d_m} \right|}^2}\int\limits_{-\infty }^{\infty }{{{S}_{{{\eta }_{k}}}}\left( f-\Updelta {f_k} \right){{\left| G\left( f \right) \right|}^2}}df. $$

(55)

The AWGN component

Let us consider the detailed calculation of expression (14).

$$ \begin{aligned} {\mathbb{E}}\left[ {{\left( R_{m}^{\left( 3 \right)} \right)}^2}\right]={\mathbb{E}}\left[ {{\left( \hbox{Re}\left\{ \frac{\sqrt{2{s_0}}}{{{T}_{\rm s}}}\int\limits_{-\infty }^{\infty }{n(t)\left( d_{m}^{*} \right){{g}^{*}}(t){{e}^{j\left( 2\pi {{f}_{0}}t+{{\Upphi }_{0}} \right)}}dt} \right\} \right)}^2}\right]\\ & =\frac{2{s_0}}{4{{T}^2}}{\mathbb{E}}\left[ \int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{\left[ n(t)d_{m}^{*}{{g}^{*}}(t)+{{n}^{*}}(t){d_m}g(t) \right]}}\left[ n\left( \rho \right)d_{m}^{*}{{g}^{*}}\left( \rho \right)+{{n}^{*}}\left( \rho \right){d_m}g\left( \rho \right) \right]dtd\rho \right]\\ & =\frac{2{s_0}}{4{{T}_{\rm s}}^2}\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{\left[ {\mathbb{E}}\left[ n(t)\cdot {{n}^{*}}\left( \rho \right) \right]\cdot {{\left| {d_m} \right|}^2}{{g}^{*}}(t)g\left( \rho \right)+{\mathbb{E}}\left[ {{n}^{*}}(t)\cdot n\left( \rho \right) \right]\cdot {{\left| {d_m} \right|}^2}g(t){{g}^{*}}\left( \rho \right) \right]dt}}d\rho \end{aligned} $$

(56)

In the addition above, the two components are equal, therefore

$$ \begin{aligned} {\mathbb{E}}\left[ {{\left( R_{m}^{\left( 3 \right)} \right)}^2} \right]&=\frac{2{s_0}}{4{{T}_{\rm s}}^2}2\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{2{N_0}\delta \left( t-\rho \right){{\left| {d_m} \right|}^2}{{g}^{*}}(t)g\left( \rho \right)}}dtd\rho\\ & =\frac{2{s_0}}{{{T}_{\rm s}}^2}{{\left| {d_m} \right|}^2}{N_0}\int\limits_{-\infty }^{\infty }{{{\left| g(t) \right|}^2}dt=\frac{2{s_0}}{{{T}_{\rm s}}}{{\left| {d_m} \right|}^2}{N_0}}, \end{aligned} $$

(57)

since $\int\limits_{-\infty }^{\infty }{{{\left| g(t) \right|}^2}dt}=1$. Only either of the noise components appears in the final result, since

$$ \begin{aligned} {\mathbb{E}}\left[ n(t)n\left( \rho \right) \right]&={\mathbb{E}}\left[ \left( {{n}_{\rm I}}(t)+j{{n}_{\rm Q}}(t) \right)\left( {{n}_{\rm I}}\left( \rho \right)+j{{n}_{\rm Q}}\left( \rho \right) \right)\right]\\ &={\mathbb{E}}\left[ {{n}_{\rm I}}(t){{n}_{\rm I}}\left(\rho \right) \right]-{\mathbb{E}}\left[ {{n}_{\rm Q}}(t){{n}_{\rm Q}}\left( \rho \right) \right]=0, \end{aligned} $$

(58)

through

$$ {\mathbb{E}}\left[ {{n}_{\rm I}}(t){{n}_{\rm Q}}\left( \rho \right) \right]={\mathbb{E}}\left[ {{n}_{\rm I}}\left( \rho \right){{n}_{\rm Q}}(t) \right]=0. $$

Appendix 2: Calculation of the spectral overlap for OFDM

The general expression of the spectral overlap has been defined in (16) as

$$ v\left( \Updelta {f_k} \right)=\frac{\int\limits_{-\infty }^{\infty }{{{S}_{{{\eta }_{k}}}}\left( f-\Updelta {f_k} \right)S\left( f \right)df}}{\int\limits_{-\infty }^{\infty }{S\left( f \right)df}}. $$

(59)

Let us calculate the $S\left( f \right)$ spectral power density function of an OFDM subcarrier. Assuming rectangular shapes for s(t) time-domain signals for the subcarriers

$$ s(t)=\left\{\begin{array}{l} \sqrt{{s_0}}\left[ \hbox{V} \right],\hbox{ if }t\in \left[ -\frac{T_{\rm s}}{2},\frac{T_{\rm s}}{2} \right) \\ 0,\hbox{ otherwise.} \\ \end{array} \right. $$

(60)

After that, the $S\left( f \right)$ spectral power density function of the s(t) useful signal can be expressed by applying the Fourier-transformation

$$ \begin{aligned} S\left(f\right)&=\frac{1}{T_{\rm s}}{{\left| \int\limits_{-\frac{T_{\rm s}}{2}}^{\frac{T_{\rm s}}{2}}{s(t){{e}^{-j2\pi ft}}dt} \right|}^2}=\frac{{s_0}}{T_{\rm s}}{{\left| \int\limits_{-\frac{T_{\rm s}}{2}}^{\frac{T_{\rm s}}{2}}{{{e}^{-j2\pi ft}}dt} \right|}^2}=\frac{{s_0}}{T_{\rm s}}{{\left| \frac{\sin \left( \pi f T_{\rm s} \right)}{\pi f} \right|}^2}\\ &={s_0}T_{\rm s}{{\left( \frac{\sin \left( \pi f T_{\rm s} \right)}{\pi f T_{\rm s} } \right)}^2} \left[ \frac{\hbox{W}}{\hbox{Hz}} \right]. \end{aligned} $$

(61)

Similarly, by assuming the same T _s symbol period for the interferers, the spectral power density function of interfering signal k can be given with as ${{S}_{{{\eta }_{k}}}}\left( f \right)={{T}_{\rm s}}{{\left( \frac{\sin \left( \pi f T_{\rm s} \right)}{\pi f T_{\rm s} } \right)}^2}$. Since R _η _k(0) = 1, the spectral overlap can be expressed as

$$ \begin{aligned} v\left( \Updelta {f_k} \right)&=\frac{\int\limits_{-\infty }^{\infty }{{{S}_{{{\eta }_{k}}}}\left( f-\Updelta {f_k} \right)S\left( f \right)df}}{\int\limits_{-\infty }^{\infty }{S\left( f \right)df}}=\frac{\int\limits_{-\infty }^{\infty }{T_{\rm s}{{\left( \frac{\sin \left( \pi \left( f-\Updelta {f_k} \right)T_{\rm s} \right)}{\pi \left( f-\Updelta {f_k} \right)T_{\rm s}} \right)}^2}{s_0}T_{\rm s}{{\left( \frac{\sin \left( \pi fT_{\rm s} \right)}{\pi fT_{\rm s}} \right)}^2}df}}{\int\limits_{-\infty }^{\infty }{{s_0}T_{\rm s}{{\left( \frac{\sin \left( \pi fT_{\rm s} \right)}{\pi fT_{\rm s}} \right)}^2}df}} \\ & =T_{\rm s}\frac{\int\limits_{-\infty }^{\infty }{{{\left( \frac{\sin \left( \pi \left( f-\Updelta {f_k} \right)T_{\rm s} \right)}{\pi \left( f-\Updelta {f_k} \right)T_{\rm s}} \right)}^2}{{\left( \frac{\sin \left( \pi fT_{\rm s} \right)}{\pi fT_{\rm s}} \right)}^2}df}}{\int\limits_{-\infty }^{\infty }{{{\left( \frac{\sin \left( \pi fT_{\rm s} \right)}{\pi fT_{\rm s}} \right)}^2}df}}. \end{aligned} $$

(62)

The integral in the denominator of (62) considering $T_{\rm s}=\frac{1}{\Updelta f_{\rm c}}$ can be given with

$$ \int\limits_{-\infty }^{\infty }{{{\left( \frac{\sin \left( \pi \frac{f}{\Updelta f_{\rm c}} \right)}{\pi \frac{f}{\Updelta f_{\rm c}}} \right)}^2}}df=\Updelta f_{\rm c}. $$

(63)

Overlap for integer multiple distances of the subcarrier spacing

Let us consider the calculation of expression (62) by substituting (63) for $\Updelta f_k=l\cdot \Updelta f_c$ frequency distances of the interferer, i.e. for ${l\in \mathbb{N}}$ multiples of the subcarrier spacing. In this case

$$ v\left( l\cdot \Updelta f_c \right)=\frac{T_{\rm s}}{\Updelta f_c}\int\limits_{-\infty }^{\infty }{{{\left( \frac{\sin \left(\pi \frac{ \left( f-l\cdot \Updelta f_c \right)}{\Updelta f_c} \right)}{\pi\frac{ \left( f-l\cdot \Updelta f_c \right)}{\Updelta f_c}} \right)}^2}{{\left( \frac{\sin \left(\pi \frac{ f}{\Updelta f_c} \right)}{\pi\frac{f}{\Updelta f_c}} \right)}^2}df}=\frac{T_{\rm s}}{\Updelta f_c}\frac{\Updelta f_c}{{l^2}{{\pi }^2}}=\frac{T_{\rm s}}{{l^2}{{\pi }^2}}. $$

(64)

We can calculate $v\left( l\cdot \Updelta f_c \right)$ also for l = 0 by the application of the integration with substitution, i.e. let $x=\pi f/\Updelta {f_c}\Rightarrow f=x\Updelta {f_c}/\pi $. Accordingly

$$ \begin{aligned} v\left(0\cdot \Updelta f_c \right)&=\frac{{{T}_{s}}}{\Updelta {f_c}}\int\limits_{-\infty }^{\infty }{\frac{{{\sin }^2}\left( x-0 \right)}{{{\left( x-0 \right)}^2}}\frac{{{\sin }^2}\left( x \right)}{{{x}^2}}\frac{df}{dx}dx}=\frac{{{T}_{s}}}{\Updelta {f_c}}\int\limits_{-\infty }^{\infty }{\frac{{{\sin }^{4}}\left( x \right)}{{{x}^{4}}}\frac{df}{dx}dx}\\ & =\frac{{{T}_{s}}}{\Updelta {f_c}}\frac{\Updelta {f_c}}{\pi }\int\limits_{-\infty }^{\infty }{\frac{{{\sin }^{4}}\left( x \right)}{{{x}^{4}}}dx}=\frac{{{T}_{s}}}{\pi }\cdot \frac{2}{3}\pi ={{T}_{s}}\frac{2}{3}=0,66\dot{6}\cdot{{T}_{s}}, \end{aligned} $$

(65)

since $df/dx=\Updelta {f_c}/\pi $. Finally, we get the general form of $v\left( l\cdot \Updelta f_c \right)$ as

$$ v_{\rm OFDM}\left( l\cdot \Updelta f \right)=\left\{\begin{array}{ll} \frac{1}{{l^2}{{\pi }^2}}T_{\rm s},&l\neq0 \\ \frac{2}{3}T_{\rm s}, & l=0. \end{array} \right. $$

(66)

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Authors and Affiliations

Albert Mráz
- 1
Email author
László Pap
- 1

1.Department of TelecommunicationsBudapest University of Technology and EconomicsBudapestHungary