附录D 式(7.5-18)和式(7.5-20)的推导

1. 式 (7.5-18) 的推导

令 \(t = \frac{h}{\sigma_{\mathrm{n}}} \sqrt{1 + n}\) , 则 \(t^2 = \frac{h^2}{\sigma_{\mathrm{n}}^2} (1 + n), \frac{h}{\sigma_{\mathrm{n}}^2} = \frac{t}{\sqrt{(1 + n) \sigma_{\mathrm{n}}}}, \mathrm{d}h = \frac{\sigma_{\mathrm{n}} \mathrm{d}t}{\sqrt{(1 + n)}}, \mathrm{d}t = \frac{\sqrt{1 + n}}{\sigma_{\mathrm{n}}} \mathrm{d}h,\) 并令 \(\alpha=\frac{A}{\sigma_{n}\sqrt{1+n}}\) ，于是 \(\alpha t=\frac{Ah}{\sigma_{n}^{2}}\) 将以上公式代入式

\[ \int_ {0} ^ {\infty} t I _ {0} (\alpha t) \mathrm{e} ^ {- (\alpha^ {2} + t ^ {2}) / 2} \mathrm{d} t = 1 \]

中，则得

\[ \int_ {0} ^ {\infty} \frac {h}{\sigma_ {\mathrm{n}}} \sqrt {1 + n} I _ {0} \left(\frac {A h}{\sigma_ {\mathrm{n}} ^ {2}}\right) \exp \left[ - \left(\frac {A ^ {2}}{(1 + n)} + h ^ {2} (1 + n)\right) / 2 \sigma_ {\mathrm{n}} ^ {2} \right] \left(\frac {\sqrt {1 + n}}{\sigma_ {\mathrm{n}}} \mathrm{d} h\right) = 1 \]

\[ \int_ {0} ^ {\infty} \frac {h}{\sigma_ {\mathrm{n}} ^ {2}} (1 + n) I _ {0} \left(\frac {A h}{\sigma_ {\mathrm{n}} ^ {2}}\right) \exp \left[ - \left(\frac {A ^ {2}}{(1 + n)} + h ^ {2} (1 + n)\right) / 2 \sigma_ {\mathrm{n}} ^ {2} \right] \mathrm{d} h = 1 \]

于是

\[ P _ {e} = \mathrm{e} ^ {- \frac {A ^ {2}}{2 \sigma_ {n} ^ {2}}} \sum_ {n = 1} ^ {M - 1} (- 1) ^ {n - 1} \binom {M - 1} {n} \int_ {0} ^ {\infty} \frac {h}{\sigma_ {n} ^ {2}} I _ {0} \left(\frac {A h}{\sigma_ {n} ^ {2}}\right) \mathrm{e} ^ {- (1 + n) h ^ {2} / 2 \sigma_ {n} ^ {2}} \mathrm{d} h \]

中的积分为

\[ \begin{array}{l} \int_ {0} ^ {\infty} \frac {h}{\sigma_ {\mathrm{n}} ^ {2}} I _ {0} \left(\frac {A h}{\sigma_ {\mathrm{n}} ^ {2}}\right) \exp \left[ - \frac {(1 + n) h ^ {2}}{2 \sigma_ {\mathrm{n}} ^ {2}} \right] \mathrm{d} h \\ = \frac {1}{1 + n} \exp \left(\frac {A ^ {2}}{2 (1 + n) \sigma_ {\mathrm{n}} ^ {2}}\right) \int_ {0} ^ {\infty} \frac {h}{\sigma_ {\mathrm{n}} ^ {2}} (1 + n) I _ {0} \left(\frac {A h}{\sigma_ {\mathrm{n}} ^ {2}}\right) \exp \left[ - \left(\frac {A ^ {2}}{(1 + n)} + h ^ {2} (1 + n)\right) / 2 \sigma_ {\mathrm{n}} ^ {2} \right] \mathrm{d} h \\ = \frac {1}{1 + n} \exp \left(\frac {A ^ {2}}{2 (1 + n) \sigma_ {\mathrm{n}} ^ {2}}\right) \\ \end{array} \]

因此，误码率 \(P_{e}\) 公式变成

\[ \begin{array}{l} P _ {\mathrm{c}} = \mathrm{e} ^ {- \frac {A 2}{2 \sigma_ {\mathrm{n}} ^ {2}}} \sum_ {n = 1} ^ {M - 1} (- 1) ^ {n - 1} \binom {M - 1} {n} \int_ {0} ^ {\infty} \frac {h}{\sigma_ {\mathrm{n}} ^ {2}} I _ {0} \left(\frac {A h}{\sigma_ {\mathrm{n}} ^ {2}}\right) \mathrm{e} ^ {- (1 + n) h ^ {2} / 2 \sigma_ {\mathrm{n}} ^ {2}} \mathrm{d} h \\ = \mathrm{e} ^ {- \frac {A ^ {2}}{2 \sigma_ {\mathrm{n}} ^ {2}}} \sum_ {n = 1} ^ {M - 1} (- 1) ^ {n - 1} \binom {M - 1} {n} \frac {1}{n + 1} \mathrm{e} ^ {A ^ {2} / 2 (n + 1) \sigma_ {\mathrm{n}} ^ {2}} \\ = \sum_ {n = 1} ^ {M - 1} (- 1) ^ {n - 1} \binom {M - 1} {n} \frac {1}{n + 1} \mathrm{e} ^ {- n A ^ {2} / 2 (n + 1) \sigma_ {n} ^ {2}} \\ \end{array} \]

2. 式 (7.5-20) 的推导

由代数式

\[ 1 - x ^ {n} = (1 - x) \left(1 + x + x ^ {2} + \dots + x ^ {n - 1}\right) \quad (\text {附D-1}) \]

可以将式 (7.5-15) 中的 \([1 - P(h)]\) 当作是上式中的 \(x\) ，把 \((M - 1)\) 当作上式中的 \(n\) ，则式 (7.5-15) 可以改写为

\[ \begin{array}{l} P _ {r} (h) = 1 - [ 1 - P (h) ] ^ {M - 1} = 1 - \left[ 1 - \mathrm{e} ^ {- h ^ {2} / 2 \sigma_ {\mathrm{n}} ^ {2}} \right] ^ {M - 1} \\ = \mathrm{e} ^ {- h ^ {2} / 2 \sigma_ {\mathrm{n}} ^ {2}} \sum_ {k = 0} ^ {M - 2} \left(1 - \mathrm{e} ^ {- h ^ {2} / 2 \sigma_ {\mathrm{n}} ^ {2}}\right) ^ {k} \tag {附D-2} \\ \end{array} \]

又由于上式中

\[ 1 - \mathrm{e} ^ {- h ^ {2} / 2 \sigma_ {\mathrm{n}} ^ {2}} \leqslant 1 \quad (\text {附D-3}) \]

即式 (附 D-2) 中求和的各项均不大于 1, 所以有

\[ P _ {e} (h) \leqslant (M - 1) \mathrm{e} ^ {- h ^ {2} / 2 \sigma_ {\mathrm{n}} ^ {2}} \quad (\text {附D-4}) \]

将上式和式 \((7.5-16)\) 代入式 \((7.5-17)\) ，得出

\[ \begin{array}{l} P _ {e} = \int_ {0} ^ {\infty} p (h) P _ {e} (h) \mathrm{d} h \leqslant \\ \int_ {0} ^ {\infty} (M - 1) \mathrm{e} ^ {- h ^ {2} / 2 \sigma_ {\mathrm{n}} ^ {2}} \frac {h}{\sigma_ {\mathrm{n}} ^ {2}} I _ {0} \left(\frac {A h}{\sigma_ {\mathrm{n}} ^ {2}}\right) \exp \left[ - \frac {1}{2 \sigma_ {\mathrm{n}} ^ {2}} \left(h ^ {2} + A ^ {2}\right) \right] \mathrm{d} h \\ \end{array} \]

(附 D - 5)

现在，令 \(t = \sqrt{2} h / \sigma_{\mathrm{n}},\alpha = A / \sqrt{2}\sigma_{\mathrm{n}}\) ，则

\[ \alpha t = A h / \sigma_ {\mathrm{n}} ^ {2}, \frac {\alpha^ {2} + t ^ {2}}{2} = \frac {1}{2} \left(\frac {A ^ {2}}{2 \sigma_ {\mathrm{n}} ^ {2}} + \frac {2 h ^ {2}}{\sigma_ {\mathrm{n}} ^ {2}}\right) = \frac {A ^ {2} / 2 + 2 h ^ {2}}{2 \sigma_ {\mathrm{n}} ^ {2}} \]

\[ \mathrm{d} t = (\sqrt {2} / \sigma_ {\mathrm{n}}) \mathrm{d} h, \mathrm{d} h = (\sigma_ {\mathrm{n}} / \sqrt {2}) \mathrm{d} t \]

将这些关系代入式 (附 D-5)，并考虑到式 (7.5-19)，即

\[ \int_ {0} ^ {\infty} t I _ {0} (\alpha t) \mathrm{e} ^ {- (\alpha^ {2} + t ^ {2}) / 2} \mathrm{d} t = 1 \]

得到式 \((7.5-20)\) ，即

\[ P _ {\mathrm{e}} \leqslant \frac {M - 1}{2} \mathrm{e} ^ {- A ^ {2} / 4 \sigma_ {\mathrm{n}} ^ {2}} \int_ {0} ^ {\infty} t I _ {0} (\alpha t) \exp \left(- \frac {\alpha^ {2} + t ^ {2}}{2}\right) \mathrm{d} t = \frac {M - 1}{2} \mathrm{e} ^ {- A ^ {2} / 4 \sigma_ {\mathrm{n}} ^ {2}} \]

附录 D 式 (7.5-18) 和式 (7.5-20) 的推导