Here is the processing of calculating the BER average

\begin{align} P_e &=\int^\infty_0Q(\sqrt{2uSNR})e^{-u}du \\ & =\frac{1}{\sqrt{2\pi}}\int^{\infty}_0\int^\infty_{\sqrt{2uSNR}}e^{-\frac{t^2}{2}}e^{-u}dtdu \\ & =\frac{1}{\sqrt{2\pi}}\int^{\infty}_0\int^\frac{t^2}{2SNR}_{0}e^{-\frac{t^2}{2}}e^{-u}dtdu \\ & =\frac{1}{\sqrt{2\pi}}\int^{\infty}_0(1-e^{-\frac{t^2}{2SNR}})e^{-\frac{t^2}{2}}dt \\ & =\frac{1}{2}(1-\sqrt{\frac{SNR}{1+SNR}}) \end{align}

channel state $h$ is a complex gaussian r.b.$CN(0,1)$,$SNR=E_s/N_0$,the BER of received symbols is $P_{e|h}=Q(\sqrt{2|h|^2SNR})$

I don't really understand these computation procedure step by step,for example.

Why is $\int^\infty_{\sqrt{2uSNR}}$ become $\int^\frac{t^2}{2SNR}_{0}$ ? Why is $\int^\frac{t^2}{2SNR}_{0}e^{-u}du$ become $(1-e^{-\frac{t^2}{2SNR}})$ ?

Why is $\frac{1}{\sqrt{2\pi}}\int^{\infty}_0(1-e^{-\frac{t^2}{2SNR}})e^{-\frac{t^2}{2}}dt=\frac{1}{2}(1-\sqrt{\frac{SNR}{1+SNR}})$ ?

Can anyone explain them and teach me?


This is pure mathematics. However, these are some hints.

1) In 2-D coordinates $t$ and $u$ the multiple integral scans the region limited by $t = f_u(u) = 0$ and $t = g_u(u) = \sqrt{2u\textrm{SNR}}$ for $t > 0$ and $u >0$. It is the same region limited by two functions $u = f_t(t) = 0$ and $u = g_t(t) = t^2/2\textrm{SNR}$ for $t > 0$ and $u >0$.

2) because $\int_a^b e^{-u} du = e^{-a} - e^{-b}$

3) use 2) with the identity Q(0) = 1/2.


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