題組內容

4. Consider the BPSK system whose transmitted signal is given by

$$x(t) = \begin{cases} A \cos(2\pi fc t), & m = 0, \\ -A \cos(2\pi fc t), & m = 1, \quad 0 \le t \le T, \quad f_c >> \frac{1}{T}, \end{cases}$$

The message $m$ has the distribution of $\mathbb{P}(m=0) = 4/9$ and $\mathbb{P}(m=1) = 5/9$. The received signal is given by $y(t) = x(t) + n(t)$ where $n(t)$ is a zero mean white Gaussian noise with power spectral density $N_0/2$ for all $f$.

iii) (10pt) Derive the average probability of error based on $ \eta_{\mathrm{MAP}} $ in part ii).

Hint: $Q(x) \triangleq \intx^\infty \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \, \mathrm{d}y$ and $Q(-x) = \int{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \, \mathrm{d}y$.

商品廣告・Roblox Rivals 競爭者 3D 列印實體造型道具

Key 爪刀

12 cm・洞口約 1.7 cm；3D 列印實體造型道具，2–7 天寄出。

黑洞狙擊槍

24 cm・軟性材質・瞄準鏡造型；3D 列印實體造型道具，2–7 天寄出。

非遊戲虛寶。價格、規格及供貨情形以蝦皮商品頁為準。

題組內容

iii) (10pt) Derive the average probability of error based on \( \eta_{\mathrm{MAP}} \) in part ii).

Hint: $Q(x) \triangleq \intx^\infty \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \, \mathrm{d}y$ and $Q(-x) = \int{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \, \mathrm{d}y$.

相關申論題

相關試卷

題組內容

iii) (10pt) Derive the average probability of error based on \( \eta_{\mathrm{MAP}} \) in part ii).Hint: $Q(x) \triangleq \intx^\infty \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \, \mathrm{d}y$ and $Q(-x) = \int{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \, \mathrm{d}y$.

相關申論題

相關試卷

iii) (10pt) Derive the average probability of error based on \( \eta_{\mathrm{MAP}} \) in part ii).

Hint: $Q(x) \triangleq \intx^\infty \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \, \mathrm{d}y$ and $Q(-x) = \int{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \, \mathrm{d}y$.