Communications Test 6

The generator matrix for a rate 1/n repetition code has one column and ‘n’ rows with all elements unity. So, we have the generator matrix for rate 1/3 repetition code as:

\(\left[ G \right] = \left[ {\begin{array}{*{20}{c}} 1\\ 1\\ 1 \end{array}} \right]\)

In FDMA each user is assigned a particular frequeny Which cannot be used by other users.

In TDMA all users transmit using same carrier frequency but in different time slots.

We have the parity check matrix as:

\(\left[ H \right] = \left[ {\begin{array}{*{20}{c}}1&1&0&1&1&0&0\\1&1&1&0&0&1&0\\0&1&1&1&0&0&1\end{array}} \right]\) 

This can be written as:

\(\left[ {\begin{array}{*{20}{c}}{{h_{11}}}&{{h_{12}}}&{{h_{13}}}&{{h_{14}}}&1&0&0\\{{h_{21}}}&{{h_{22}}}&{{h_{23}}}&{{h_{24}}}&0&1&0\\{{h_{31}}}&{{h_{32}}}&{{h_{33}}}&{{h_{44}}}&0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1&0&1&1&0&0\\1&1&1&0&0&1&0\\0&1&1&1&0&0&1\end{array}} \right]\) 

Thus, the generator matrix for the parity check matrix is obtained as:

\(\left[ G \right] = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\{{h_{11}}}&{{h_{12}}}&{{h_{13}}}&{{h_{14}}}\\{{h_{21}}}&{{h_{22}}}&{{h_{23}}}&{{h_{24}}}\\{{h_{31}}}&{{h_{32}}}&{{h_{33}}}&{{h_{34}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\1&1&0&1\\1&1&1&0\\0&1&1&1\end{array}} \right]\) 

d_min=m

for error correction

d_min ≥ 2t+1

\(t \le \frac{{m - 1}}{2}\)

t≤ 2

so maximum correctable errors are 2

The number of channels is FDMA is given by

\(N = \frac{{{B_t} - 2Bguard}}{{{B_c}}}\) 

\( = \frac{{12.5 \times {{10}^6} - 2\left( {10 \times {{10}^3}} \right)}}{{30 \times {{10}^3}}}\) 

= 416.

Concept:

The error-correcting and detecting capability is given by the hamming distance

Error detection ≤  d_min - 1

Error correction ≤ \(\frac{{\left( {{d_{min}} - 1} \right)}}{2}\)

Application:

Parity matrix

\({P^T} = \left[ {\begin{array}{*{20}{c}} 1&1&0\\ 1&1&1\\ 1&0&1 \end{array}} \right]\)

Generator matrix

G = [I_K P^T]

\(G = \left[ {\begin{array}{*{20}{c}} 1&0&0&1&1&0\\ 0&1&0&1&1&1\\ 0&0&1&1&0&1 \end{array}} \right]\)

Code words

C = dG

All possible code words

Minimum weight of non – zero words = 3

d_min = 3

Error detection ≤  3 – 1

Error detection ≤  2

Error correction ≤  \(\frac{{{d_{min}} - 1}}{2}\)

\(≤ \frac{{2 - 1}}{2}\)

≤ 1

For CDMA \(\frac{{{E_b}}}{{{N_o}}} = \frac{{W/R}}{{{N_u} - 1}}\)

\(10\log (\frac{{{E_b}}}{{{N_o}}}) = \;10.5dB\) 

E_b/N_o = 11.3

\({N_u} = \frac{{W/R}}{{Eb/{N_o}}} + 1 = \frac{{1000}}{{11.3}} + 1\) 

= 89

Number of users m = 8

 Total spectrum (Bw) B_tot = 25 MHz

Guard Band, B_guard = 0

Number of users in GSM:

\(N = \frac{{m\left( {{B_{tot}} - 2{B_{guard}}} \right)}}{{{B_c}}}\) 

\(N = \frac{{8\left( {25 \times {{10}^6}} \right)}}{{200 \times {{10}^3}}}\) 

N = 1000

We have the (15, 11) single error-correcting codes, i.e. the word length is:

n = 15

and the length of the information symbol is:

k = 11

So, we get the length of the parity check symbol as:

r = n – k = 15 – 11 = 4

The parity check matrix for error-correcting codes is defined as:

\(\left[ H \right] = \left[ {\begin{array}{*{20}{c}}{{h_{11}}}&{{h_{12}}}& \cdots &{{h_{1K}}}&1&0& \cdots &0\\{{h_{21}}}&{{h_{22}}}& \cdots &{{h_{2K}}}&0&1& \cdots &0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\{{h_{r1}}}&{{h_{r2}}}& \cdots &{{h_{rK}}}&0&0& \cdots &1\end{array}} \right]\) 

Thus, the number of rows in the parity check matrix will be:

r = 4, and the number of columns will be:

r + k = 15