EEE302 期末试题

大学学习

2014-15

Q1

a)

Channel capacity is defined as the maximum data rate over a communication channel, commonly denoted by C with a unit of bit(s)/sec or bps.

noiseless (Harley's Law)
$C=2B\log_2M$
AWGN (Shannon's Theorem)
$C=B\log_2(1+S/N)$

b)

(1)

$s(t)=\text{rect}(\dfrac{t}{\tau})$
$\text{rect}(\dfrac{t}{\tau})\iff \tau\text{sinc}(\tau f)$
$\tau f=\pi\Rightarrow f=\dfrac{\pi}{\tau}=10^6\pi$
$f_s=2\pi\ \text{MHz}$

$s(t)=\dfrac{1}{100\pi}100\pi\cdot\text{sinc}(100\pi t)$
$S(f)=\dfrac{1}{100\pi}\text{rect}(\dfrac{f}{100\pi})$
$B=50\pi$
$f_s=2B=100\pi$

(2)

c)

$\begin{aligned} p_o(-T)&=c_{-1}p_r(0)+c_0p_r(-T)+c_1p_r(-2T)\\ p_o(0)&=c_{-1}p_r(T)+c_0p_r(0)+c_1p_r(-T)\\ p_o(T)&=c_{-1}p_r(2T)+c_0p_r(T)+c_1p_r(0)\\ \end{aligned}$

$\begin{aligned} p_o(-T)&=c_{-1}-0.2c_0+0.1c_1\\ p_o(0)&=-0.2c_{-1}+c_0-0.2c_1\\ p_o(T)&=0.1c_{-1}-0.2c_0+c_1\\ \end{aligned}$

$c_{-1}=0.1961$, $c_0=1.0784$, $c_1=0.1961$

Q2

a)

b)

$N_o=kT=1.38\times10^{-20}\ \text{W/Hz}$
$P_n=N_oB=1.38\times10^{-14}\ \text{W}$

c)

(1)

$B_s=2/T_c=2R_c$
$\text{PG}=\dfrac{B_s}{B_m}=43.2\ \text{dB}$

(2)

$R_c=24\times10^7$

Q3

b)

(1)

(2)

Q4

a)

$\sigma=0.5$
$Q(\frac{\eta-\mu_0}{\sigma})\leq 0.8\%\Rightarrow\eta\geq1.205$

b)

$P_M=5.59\%$

c)

$P_F=0.798\%$
$P_M=5.54\%$

d)

$\eta=1.243$

e)

Appendix

2013-14

Q1

Q2

Q3

Q4