CORE1240 HW5

Q1

(a)

(i)

For channel 1

$$\begin{array}{rl}P(2)& =\left(\begin{array}{c}12\\ 2\end{array}\right)(0.25{)}^2(1-0.25{)}^{12-2}\\ & =0.2323\end{array}$$

For channel 2

$$\begin{array}{rl}P(2)& =\left(\begin{array}{c}12\\ 2\end{array}\right)(0.35{)}^2(1-0.35{)}^{12-2}\\ & =0.1088\end{array}$$

(ii)

For channel 1

$$\begin{array}{rl}P(3)& =\left(\begin{array}{c}12\\ 3\end{array}\right)(0.25{)}^3(1-0.25{)}^{12-3}\\ & =0.2581\end{array}$$

For channel 2

$$\begin{array}{rl}P(3)& =\left(\begin{array}{c}12\\ 3\end{array}\right)(0.35{)}^3(1-0.35{)}^{12-3}\\ & =0.1954\end{array}$$

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(b)

For channel 1

$$\begin{array}{rl}& \text{Maximum number of information bits per second}\\ & =(1+0.25{\log }_{2}0.25+(1-0.25){\log }_2(1-0.25))(120)\\ & =22.6466kbit/s\end{array}$$

For channel 2

$$\begin{array}{rl}& \text{Maximum number of information bits per second}\\ & =(1+0.35{\log }_{2}0.35+(1-0.35){\log }_2(1-0.35))(120)\\ & =7.9118kbit/s\end{array}$$

Q2

(a)

$$\begin{array}{rl}\text{Total bandwidth}& =(3)(39+39)(250\times 10^3)\\ & =58.5\mathrm{MHz}\end{array}$$

(b)

$\text{Let }C\text{ be the number of cells}$

$$\begin{array}{rl}C\text{ cells}\cdot \frac{39}{3}\frac{\text{channels}}{\text{cell}\cdot \text{provider}}\cdot 3\text{ providers}\cdot 4\frac{\text{time slots}}{\text{channel}}& =904800\text{ subscribers}\\ C& =5800\end{array}$$

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Q3

(a)

$$\begin{array}{rl}{\text{PN}}_A\cdot {\text{PN}}_B& =(+1-1-1+1-1+1+1-1)\cdot (+1+1-1-1-1-1+1+1)\\ & =+1-1+1-1+1-1+1-1\\ & =0\end{array}$$

Yes, they are orthogonal

(b)

$$\begin{array}{rl}{\text{Encoded}}_A=& (+1-1-1+1)\times (+1-1-1+1-1+1+1-1)\\ =& +1-1-1+1-1+1+1-1\\ & -1+1+1-1+1-1-1+1\\ & -1+1+1-1+1-1-1+1\\ & +1-1-1+1-1+1+1-1\end{array}$$ $$\begin{array}{rl}{\text{Encoded}}_B=& (+1+1+1-1)\times (+1+1-1-1-1-1+1+1)\\ =& +1+1-1-1-1-1+1+1\\ & +1+1-1-1-1-1+1+1\\ & +1+1-1-1-1-1+1+1\\ & -1-1+1+1+1+1-1-1\end{array}$$

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(c)

$$\begin{array}{rl}{\text{Encoded}}_A\times {\text{PN}}_B=& (+1-1-1+1-1+1+1-1\\ & -1+1+1-1+1-1-1+1\\ & -1+1+1-1+1-1-1+1\\ & +1-1-1+1-1+1+1-1)\\ & \times (+1+1-1-1-1-1+1+1)\\ =& +1-1+1-1+1-1+1-1\\ & -1+1-1+1-1+1-1+1\\ & -1+1-1+1-1+1-1+1\\ & +1-1+1-1+1-1+1-1\\ =& 0+0+0+0\end{array}$$

Nothing is output, signal is cancelled out

(d)

$$\begin{array}{rl}{\text{Encoded}}_A+{\text{Encoded}}_B=& (+1-1-1+1-1+1+1-1\\ & -1+1+1-1+1-1-1+1\\ & -1+1+1-1+1-1-1+1\\ & +1-1-1+1-1+1+1-1)\\ & +(+1+1-1-1-1-1+1+1\\ & +1+1-1-1-1-1+1+1\\ & +1+1-1-1-1-1+1+1\\ & -1-1+1+1+1+1-1-1)\\ =& +2+0-2+0-2+0+2+0\\ & +0+2+0-2+0-2+0+2\\ & +0+2+0-2+0-2+0+2\\ & +0-2+0+2+0+2+0-2\end{array}$$

(e)

$$\begin{array}{rl}({\text{Encoded}}_A+{\text{Encoded}}_B)\times {\text{PN}}_B=& (2+0-2+0-2+0+2+0\\ & +0+2+0-2+0-2+0+2\\ & +0+2+0-2+0-2+0+2\\ & +0-2+0+2+0+2+0-2)\\ & \times (+1+1-1-1-1-1+1+1)\\ =& +2+0+2+0+2+0+2+0\\ & +0+2+0+2+0+2+0+2\\ & +0+2+0+2+0+2+0+2\\ & +0-2+0-2+0-2+0-2\\ =& 8+8+8-8\end{array}$$

The output is 1110

Q4

(a)

$$\begin{array}{rl}\text{Required time}& =\frac{60\cdot 10^3\cdot 8}{30\cdot 10^6}\cdot 2\\ & =0.032\mathrm{s}\end{array}$$

(b)

$$\begin{array}{rl}\text{Required time}& =\frac{60\cdot 10^3\cdot 8}{30\cdot 10^6}+\frac{60\cdot 10^3\cdot 8}{15\cdot 10^6}\\ & =0.048\mathrm{s}\end{array}$$