Analytical Expressions Used in berfading Function and Bit Error Rate Analysis App

R2026b

This section covers the main analytical expressions used in the berfading function and the Bit Error Rate Analysis app.

Notation

This table describes the additional notations used in analytical expressions in this section.

Description	Notation
MRC	Maximal-ratio combining
EGC	Equal-gain combining
Power of the fading amplitude r	$$\Omega =E\left[r^2\right]$$, where $$E[\cdot ]$$ denotes statistical expectation
Number of diversity branches	$$L$$
Signal to Noise Ratio (SNR) per symbol per branch	$${\overline{\gamma }}_l=\left({\Omega }_l\frac{E_s}{N_0}\right)/L=\left({\Omega }_l\frac{k{E}_b}{N_0}\right)/L$$ For identically-distributed diversity branches, $$\overline{\gamma }=\left(\Omega \frac{k{E}_b}{N_0}\right)/L$$
Moment generating functions for each diversity branch	For Rayleigh fading channels: $$M_{{\gamma }_l}\left(s\right)=\frac{1}{1-s{\overline{\gamma }}_l}$$ For Rician fading channels: $$M_{{\gamma }_l}\left(s\right)=\frac{1+K}{1+K-s{\overline{\gamma }}_l}{e}^{\left[\frac{Ks{\overline{\gamma }}_l}{(1+K)-s{\overline{\gamma }}_l}\right]}$$ K is the ratio of the energy in the specular component to the energy in the diffuse component (linear scale). For identically-distributed diversity branches,$$M_{{\gamma }_l}\left(s\right)=M_{\gamma }\left(s\right)$$ for all l.

M-PSK with MRC

From equation 9.15 in [2],

$$P_s=\frac{1}{\pi }{\displaystyle \underset{0}{\overset{(M-1)\pi /M}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{{\sin }^2(\pi /M)}{{\sin }^2\theta }\right)}d\theta }$$

From [4] and [2],

$$P_b=\frac{1}{k}\left({\displaystyle \sum _{i=1}^{M/2}(w_i^{\text{'}}){\overline{P}}_i}\right)$$

where $$w_i^{\text{'}}=w_i+w_{M-i}$$, $$w_{M/2}^{\text{'}}=w_{M/2}$$, $$w_i$$ is the Hamming weight of bits assigned to symbol i,

$$\begin{array}{l}{\overline{P}}_i=\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{\pi (1-(2i-1)/M)}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{1}{{\sin }^2\theta }{\sin }^2\frac{(2i-1)\pi }{M}\right)}d\theta }\\ -\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{\pi (1-(2i+1)/M)}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{1}{{\sin }^2\theta }{\sin }^2\frac{(2i+1)\pi }{M}\right)}d\theta }\end{array}$$

For the special case of Rayleigh fading with $$M=2$$ (from equations C-18 and C-21 and Table C-1 in [6]),

$$P_b=\frac{1}{2}\left[1-\mu {\displaystyle \sum _{i=0}^{L-1}\left(\begin{array}{c}2i\\ i\end{array}\right)}{\left(\frac{1-{\mu }^2}{4}\right)}^i\right]$$

where

$$\mu =\sqrt{\frac{\overline{\gamma }}{\overline{\gamma }+1}}$$

If $$L=1$$, then:

$$P_b=\frac{1}{2}\left[1-\sqrt{\frac{\overline{\gamma }}{\overline{\gamma }+1}}\right]$$

DE-M-PSK with MRC

For $$M=2$$ (from equations 8.37 and 9.8-9.11 in [2]),

$$P_s=P_b=\frac{2}{\pi }{\displaystyle \underset{0}{\overset{\pi /2}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{1}{{\sin }^2\theta }\right)}d\theta }-\frac{2}{\pi }{\displaystyle \underset{0}{\overset{\pi /4}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{1}{{\sin }^2\theta }\right)}d\theta }$$

M-PAM with MRC

From equation 9.19 in [2],

$$P_s=\frac{2(M-1)}{M\pi }{\displaystyle \underset{0}{\overset{\pi /2}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{3/(M^2-1)}{{\sin }^2\theta }\right)}d\theta }$$

From [5] and [2],

$$\begin{array}{c}{P}_b=\frac{2}{\pi M{\log }_{2}M}\\ \times {\displaystyle \sum _{k=1}^{{\log }_{2}M}{\displaystyle \sum _{i=0}^{(1-2^{-k})M-1}\left\{{(-1)}^{\lfloor \frac{i{2}^{k-1}}{M}\rfloor }\left(2^{k-1}-\lfloor \frac{i{2}^{k-1}}{M}+\frac{1}{2}\rfloor \right){\displaystyle \underset{0}{\overset{\pi /2}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{{(2i+1)}^{2}3/(M^2-1)}{{\sin }^2\theta }\right)}d\theta }\right\}}}\end{array}$$

M-QAM with MRC

For square M-QAM, $$k={\log }_{2}M$$ is even (equation 9.21 in [2]),

$$\begin{array}{c}{P}_s=\frac{4}{\pi }\left(1-\frac{1}{\sqrt{M}}\right){\displaystyle \underset{0}{\overset{\pi /2}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{3/(2(M-1))}{{\sin }^2\theta }\right)}d\theta }\\ -\frac{4}{\pi }{\left(1-\frac{1}{\sqrt{M}}\right)}^2{\displaystyle \underset{0}{\overset{\pi /4}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{3/(2(M-1))}{{\sin }^2\theta }\right)}d\theta }\end{array}$$

From [5] and [2]:

$$\begin{array}{c}{P}_b=\frac{2}{\pi \sqrt{M}{\log }_2\sqrt{M}}\\ \times {\displaystyle \sum _{k=1}^{{\log }_2\sqrt{M}}{\displaystyle \sum _{i=0}^{(1-2^{-k})\sqrt{M}-1}\left\{{(-1)}^{\lfloor \frac{i{2}^{k-1}}{\sqrt{M}}\rfloor }\left(2^{k-1}-\lfloor \frac{i{2}^{k-1}}{\sqrt{M}}+\frac{1}{2}\rfloor \right){\displaystyle \underset{0}{\overset{\pi /2}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{{(2i+1)}^{2}3/(2(M-1))}{{\sin }^2\theta }\right)}d\theta }\right\}}}\end{array}$$

For rectangular (nonsquare) M-QAM, $$k={\log }_{2}M$$ is odd, $$M=I\times J$$, $$I=2^{\frac{k-1}{2}}$$, $$J=2^{\frac{k+1}{2}}$$, $${\overline{\gamma }}_l={\Omega }_l{\log }_2(IJ)\frac{E_b}{N_0}$$,

$$\begin{array}{c}{P}_s=\frac{4IJ-2I-2J}{M\pi }{\displaystyle \underset{0}{\overset{\pi /2}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{3/(I^2+J^2-2)}{{\sin }^2\theta }\right)}d\theta }\\ -\frac{4}{M\pi }(1+IJ-I-J){\displaystyle \underset{0}{\overset{\pi /4}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{3/(I^2+J^2-2)}{{\sin }^2\theta }\right)}d\theta }\end{array}$$

From [5] and [2],

$$\begin{array}{c}{P}_b=\frac{1}{{\log }_2(IJ)}\left({\displaystyle \sum _{k=1}^{{\log }_{2}I}{P}_I(k)}+{\displaystyle \sum _{l=1}^{{\log }_{2}J}{P}_J(l)}\right)\\ P_I(k)=\frac{2}{I\pi }{\displaystyle \sum _{i=0}^{(1-2^{-k})I-1}\left\{{(-1)}^{\lfloor \frac{i{2}^{k-1}}{I}\rfloor }\left(2^{k-1}-\lfloor \frac{i{2}^{k-1}}{I}+\frac{1}{2}\rfloor \right){\displaystyle \underset{0}{\overset{\pi /2}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{{(2i+1)}^{2}3/(I^2+J^2-2)}{{\sin }^2\theta }\right)}d\theta }\right\}}\\ P_J(k)=\frac{2}{J\pi }{\displaystyle \sum _{j=0}^{(1-2^{-l})J-1}\left\{{(-1)}^{\lfloor \frac{j{2}^{l-1}}{J}\rfloor }\left(2^{l-1}-\lfloor \frac{j{2}^{l-1}}{J}+\frac{1}{2}\rfloor \right){\displaystyle \underset{0}{\overset{\pi /2}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{{(2j+1)}^{2}3/(I^2+J^2-2)}{{\sin }^2\theta }\right)}d\theta }\right\}}\end{array}$$

M-DPSK with Postdetection EGC

From equation 8.165 in [2],

$$P_s=\frac{\sin (\pi /M)}{2\pi }{\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{1}{\left[1-\cos (\pi /M)\cos \theta \right]}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\left[1-\cos (\pi /M)\cos \theta \right]\right)}d\theta }$$

From [4] and [2],

$$P_b=\frac{1}{k}\left({\displaystyle \sum _{i=1}^{M/2}(w_i^{\text{'}}){\overline{A}}_i}\right)$$

where $$w_i^{\text{'}}=w_i+w_{M-i}$$, $$w_{M/2}^{\text{'}}=w_{M/2}$$, $$w_i$$ is the Hamming weight of bits assigned to symbol i,

$$\begin{array}{l}{\overline{A}}_i=\overline{F}\left(\left(2i+1\right)\frac{\pi }{M}\right)-\overline{F}\left(\left(2i-1\right)\frac{\pi }{M}\right)\\ \overline{F}(\psi )=-\frac{\sin \psi }{4\pi }{\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{1}{\left(1-\cos \psi \cos t\right)}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\left(1-\cos \psi \cos t\right)\right)}dt}\end{array}$$

For the special case of Rayleigh fading with $$M=2$$ and $$L=1$$ (equation 8.173 from [2]),

$$P_b=\frac{1}{2(1+\overline{\gamma })}$$

Orthogonal 2-FSK, Coherent Detection with MRC

From equation 9.11 in [2],

$$P_s=P_b=\frac{1}{\pi }{\displaystyle \underset{0}{\overset{\pi /2}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{1/2}{{\sin }^2\theta }\right)}d\theta }$$

For the special case of Rayleigh fading (equations 14.4-15 and 14.4-21 in [1]),

$$P_s=P_b\text{}=\frac{1}{2^L}{\left(1-\sqrt{\frac{\overline{\gamma }}{2+\overline{\gamma }}}\right)}^L{\displaystyle \sum _{k=0}^{L-1}\left(\begin{array}{c}L-1+k\\ k\end{array}\right)\frac{1}{2^k}}{\left(1+\sqrt{\frac{\overline{\gamma }}{2+\overline{\gamma }}}\right)}^k$$

Nonorthogonal 2-FSK, Coherent Detection with MRC

From equations 9.11 and 8.44 in [2],

$$P_s=P_b=\frac{1}{\pi }{\displaystyle \underset{0}{\overset{\pi /2}{\int }}{\displaystyle \prod _{l=1}^L{M}_{{\gamma }_l}\left(-\frac{(1-\mathrm{Re}\left[\rho \right])/2}{{\sin }^2\theta }\right)}d\theta }$$

For the special case of Rayleigh fading with $$L=1$$ (equations 20 in [8] and 8.130 in [2]),

$$P_s=P_b=\frac{1}{2}\left[1-\sqrt{\frac{\overline{\gamma }(1-\mathrm{Re}[\rho ])}{2+\overline{\gamma }(1-\mathrm{Re}[\rho ])}}\right]$$

Orthogonal M-FSK, Noncoherent Detection with EGC

For Rayleigh fading, from equation 14.4-47 in [1],

$$\begin{array}{l}{P}_s\text{}=1-{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{{\left(1+\overline{\gamma }\right)}^L\left(L-1\right)!}{U}^{L-1}{e}^{-\frac{U}{1+\overline{\gamma }}}{\left(1-e^{-U}{\displaystyle \sum _{k=0}^{L-1}\frac{U^k}{k!}}\right)}^{M-1}dU}\\ P_b=\frac{1}{2}\frac{M}{M-1}{P}_s\end{array}$$

For Rician fading from equation 41 in [8],

$$\begin{array}{l}{P}_s={\displaystyle \sum _{r=1}^{M-1}\frac{{(-1)}^{r+1}{e}^{-LK{\overline{\gamma }}_r/(1+{\overline{\gamma }}_r)}}{{\left(r(1+{\overline{\gamma }}_r)+1\right)}^L}\left(\begin{array}{c}M-1\\ r\end{array}\right)}{\displaystyle \sum _{n=0}^{r(L-1)}{\beta }_{nr}\frac{\Gamma (L+n)}{\Gamma (L)}{\left[\frac{1+{\overline{\gamma }}_r}{r+1+r{\overline{\gamma }}_r}\right]}^n}{}_{1}F{}_1\left(L+n,L;\frac{LK{\overline{\gamma }}_r/(1+{\overline{\gamma }}_r)}{r(1+{\overline{\gamma }}_r)+1}\right)\\ P_b=\frac{1}{2}\frac{M}{M-1}{P}_s\end{array}$$

where

$$\begin{array}{c}{\overline{\gamma }}_r=\frac{1}{1+K}\overline{\gamma }\\ {\beta }_{nr}={\displaystyle \sum _{i=n-(L-1)}^n\frac{{\beta }_{i(r-1)}}{(n-i)!}{I}_{[0,(r-1)(L-1)]}(i)}\\ {\beta }_{00}={\beta }_{0r}=1\\ {\beta }_{n1}=1/n!\\ {\beta }_{1r}=r\end{array}$$

and $$I_{[a,b]}(i)=1$$ if $$a\le i\le b$$ and 0 otherwise.

Nonorthogonal 2-FSK, Noncoherent Detection with No Diversity

From equation 8.163 in [2],

$$P_s=P_b=\frac{1}{4\pi }{\displaystyle \underset{-\pi }{\overset{\pi }{\int }}\frac{1-{\varsigma }^2}{1+2\varsigma \sin \theta +{\varsigma }^2}{M}_{\gamma }\left(-\frac{1}{4}(1+\sqrt{1-{\rho }^2})(1+2\varsigma \sin \theta +{\varsigma }^2)\right)d\theta }$$

where

$$\varsigma =\sqrt{\frac{1-\sqrt{1-{\rho }^2}}{1+\sqrt{1-{\rho }^2}}}$$