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

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
MRCMaximal-ratio combining
EGCEqual-gain combining
Power of the fading amplitude r$\Omega =E\left[{r}^{2}\right]$, where $E\left[\cdot \right]$ 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

`${M}_{{\gamma }_{l}}\left(s\right)=\frac{1}{1-s{\overline{\gamma }}_{l}}$`

`${M}_{{\gamma }_{l}}\left(s\right)=\frac{1+K}{1+K-s{\overline{\gamma }}_{l}}{e}^{\left[\frac{Ks{\overline{\gamma }}_{l}}{\left(1+K\right)-s{\overline{\gamma }}_{l}}\right]}$`

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],

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

From [4] and [2],

`${P}_{b}=\frac{1}{k}\left(\sum _{i=1}^{M/2}\left({w}_{i}^{\text{'}}\right){\overline{P}}_{i}\right)$`

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,

`$\begin{array}{l}{\overline{P}}_{i}=\frac{1}{2\pi }\underset{0}{\overset{\pi \left(1-\left(2i-1\right)/M\right)}{\int }}\prod _{l=1}^{L}{M}_{{\gamma }_{l}}\left(-\frac{1}{{\mathrm{sin}}^{2}\theta }{\mathrm{sin}}^{2}\frac{\left(2i-1\right)\pi }{M}\right)d\theta \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\frac{1}{2\pi }\underset{0}{\overset{\pi \left(1-\left(2i+1\right)/M\right)}{\int }}\prod _{l=1}^{L}{M}_{{\gamma }_{l}}\left(-\frac{1}{{\mathrm{sin}}^{2}\theta }{\mathrm{sin}}^{2}\frac{\left(2i+1\right)\pi }{M}\right)d\theta \end{array}$`

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

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

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]),

`${P}_{s}={P}_{b}=\frac{2}{\pi }\underset{0}{\overset{\pi /2}{\int }}\prod _{l=1}^{L}{M}_{{\gamma }_{l}}\left(-\frac{1}{{\mathrm{sin}}^{2}\theta }\right)d\theta -\frac{2}{\pi }\underset{0}{\overset{\pi /4}{\int }}\prod _{l=1}^{L}{M}_{{\gamma }_{l}}\left(-\frac{1}{{\mathrm{sin}}^{2}\theta }\right)d\theta$`

### M-PAM with MRC

From equation 9.19 in [2],

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

From [5] and [2],

### M-QAM with MRC

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

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

From [5] and [2]:

For rectangular (nonsquare) M-QAM, $k={\mathrm{log}}_{2}M$ is odd, $M=I×J$, $I={2}^{\frac{k-1}{2}}$, $J={2}^{\frac{k+1}{2}}$, ${\overline{\gamma }}_{l}={\Omega }_{l}{\mathrm{log}}_{2}\left(IJ\right)\frac{{E}_{b}}{{N}_{0}}$,

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

From [5] and [2],

`$\begin{array}{c}{P}_{b}=\frac{1}{{\mathrm{log}}_{2}\left(IJ\right)}\left(\sum _{k=1}^{{\mathrm{log}}_{2}I}{P}_{I}\left(k\right)+\sum _{l=1}^{{\mathrm{log}}_{2}J}{P}_{J}\left(l\right)\right)\\ {P}_{I}\left(k\right)=\frac{2}{I\pi }\sum _{i=0}^{\left(1-{2}^{-k}\right)I-1}\left\{{\left(-1\right)}^{⌊\frac{i{2}^{k-1}}{I}⌋}\left({2}^{k-1}-⌊\frac{i{2}^{k-1}}{I}+\frac{1}{2}⌋\right)\underset{0}{\overset{\pi /2}{\int }}\prod _{l=1}^{L}{M}_{{\gamma }_{l}}\left(-\frac{{\left(2i+1\right)}^{2}3/\left({I}^{2}+{J}^{2}-2\right)}{{\mathrm{sin}}^{2}\theta }\right)d\theta \right\}\\ {P}_{J}\left(k\right)=\frac{2}{J\pi }\sum _{j=0}^{\left(1-{2}^{-l}\right)J-1}\left\{{\left(-1\right)}^{⌊\frac{j{2}^{l-1}}{J}⌋}\left({2}^{l-1}-⌊\frac{j{2}^{l-1}}{J}+\frac{1}{2}⌋\right)\underset{0}{\overset{\pi /2}{\int }}\prod _{l=1}^{L}{M}_{{\gamma }_{l}}\left(-\frac{{\left(2j+1\right)}^{2}3/\left({I}^{2}+{J}^{2}-2\right)}{{\mathrm{sin}}^{2}\theta }\right)d\theta \right\}\end{array}$`

### M-DPSK with Postdetection EGC

From equation 8.165 in [2],

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

From [4] and [2],

`${P}_{b}=\frac{1}{k}\left(\sum _{i=1}^{M/2}\left({w}_{i}^{\text{'}}\right){\overline{A}}_{i}\right)$`

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,

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

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

`${P}_{b}=\frac{1}{2\left(1+\overline{\gamma }\right)}$`

### Orthogonal 2-FSK, Coherent Detection with MRC

From equation 9.11 in [2],

`${P}_{s}={P}_{b}=\frac{1}{\pi }\underset{0}{\overset{\pi /2}{\int }}\prod _{l=1}^{L}{M}_{{\gamma }_{l}}\left(-\frac{1/2}{{\mathrm{sin}}^{2}\theta }\right)d\theta$`

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

`${P}_{s}={P}_{b}\text{​}=\frac{1}{{2}^{L}}{\left(1-\sqrt{\frac{\overline{\gamma }}{2+\overline{\gamma }}}\right)}^{L}\sum _{k=0}^{L-1}\left(\begin{array}{c}L-1+k\\ k\end{array}\right)\frac{1}{{2}^{k}}{\left(1+\sqrt{\frac{\overline{\gamma }}{2+\overline{\gamma }}}\right)}^{k}$`

### Nonorthogonal 2-FSK, Coherent Detection with MRC

From equations 9.11 and 8.44 in [2],

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

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

`${P}_{s}={P}_{b}=\frac{1}{2}\left[1-\sqrt{\frac{\overline{\gamma }\left(1-\mathrm{Re}\left[\rho \right]\right)}{2+\overline{\gamma }\left(1-\mathrm{Re}\left[\rho \right]\right)}}\right]$`

### Orthogonal M-FSK, Noncoherent Detection with EGC

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

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

For Rician fading from equation 41 in [8],

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

where

`$\begin{array}{c}{\overline{\gamma }}_{r}=\frac{1}{1+K}\overline{\gamma }\\ {\beta }_{nr}=\sum _{i=n-\left(L-1\right)}^{n}\frac{{\beta }_{i\left(r-1\right)}}{\left(n-i\right)!}{I}_{\left[0,\text{\hspace{0.17em}}\left(r-1\right)\left(L-1\right)\right]}\left(i\right)\\ {\beta }_{00}={\beta }_{0r}=1\\ {\beta }_{n1}=1/n!\\ {\beta }_{1r}=r\end{array}$`

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

### Nonorthogonal 2-FSK, Noncoherent Detection with No Diversity

From equation 8.163 in [2],

`${P}_{s}={P}_{b}=\frac{1}{4\pi }\underset{-\pi }{\overset{\pi }{\int }}\frac{1-{\varsigma }^{2}}{1+2\varsigma \mathrm{sin}\theta +{\varsigma }^{2}}{M}_{\gamma }\left(-\frac{1}{4}\left(1+\sqrt{1-{\rho }^{2}}\right)\left(1+2\varsigma \mathrm{sin}\theta +{\varsigma }^{2}\right)\right)d\theta$`

where

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