STEM Problem Detail

Problem Statement

Prompt

Please provide me step by step solution for the given Example - 6.12

Solution Rendering...

The Ricean probability density function is given by

\[
p(r) = \frac{r}{\sigma^{2}} \exp\!\left(-\frac{r^{2}+A^{2}}{2\sigma^{2}}\right)
I_{0}\!\left(\frac{Ar}{\sigma^{2}}\right)
\qquad \text{for } A \ge 0,\ r \ge 0
\tag{E6.12.1}
\]

where \( r \) is the Ricean amplitude and \( A \) is the specular amplitude.
By suitable transformation, the Ricean distribution can be expressed in terms of \( X \)
as

\[
p(X) = \frac{1+K}{\Gamma}
\exp\!\left(-\frac{X(1+K)+K}{\Gamma}\right)
I_{0}\!\left(\sqrt{\frac{4(1+K)KX}{\Gamma}}\right)
\tag{E6.12.2}
\]

where \( K = A^{2}/2\sigma^{2} \) is the specular-to-random ratio of the Ricean distribution.
The probability of error of DPSK and noncoherent orthogonal FSK in an AWGN channel
can be expressed as

\[
P_{e}(X,K_{1},K_{2}) = K_{1}\exp(-K_{2}X)
\tag{E6.12.3}
\]

where \( K_{1} = K_{2} = 1/2 \) for FSK and \( K_{1} = 1/2,\ K_{2} = 1 \) for DPSK.

To obtain the probability of error in a slow flat-fading channel we need to evaluate
the expression

\[
P_{e} = \int_{0}^{\infty} P_{e}(X)p(X)dX
\tag{E6.12.4}
\]

Substituting Equations (E6.12.2) and (E6.12.3) in Equation (E6.12.4) and solving the
integration we get the probability of error in a Ricean, slow flat-fading channel as

\[
P_{e} =
\frac{K_{1}(1+K)}{K_{2}\Gamma + 1 + K}
\exp\!\left(
-\frac{K_{2}K\Gamma}{K_{2}\Gamma + 1 + K}
\right)
\]

Substituting \( K_{1} = K_{2} = 1/2 \), for FSK, the probability of error is given by

\[
P_{e,\mathrm{NCFSK}} =
\frac{1+K}{\Gamma + 2 + 2K}
\exp\!\left(
-\frac{K\Gamma}{\Gamma + 2 + 2K}
\right)
\]

Similarly, for DPSK, substituting \( K_{1} = 1/2,\ K_{2} = 1 \), we get

\[
P_{e,\mathrm{DPSK}} =
\frac{1+K}{2(\Gamma + 1 + K)}
\exp\!\left(
-\frac{K\Gamma}{\Gamma + 1 + K}
\right)
\]

Loading solution...

Problem #97