Aufgaben:Exercise 2.8: Code Comparison: Binary, AMI and 4B3T

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Eye diagrams of different codes

In the graphic three eye diagrams  (without noise)  are shown,  where in each case a rectangular NRZ basic transmission pulse and for the total system frequency response  (of transmitter, channel and decoder,  without encoder)  a cosine rolloff characteristic with rolloff factor  $r = 0.8$  are the basis.

For the individual eye diagrams it is furthermore assumed  (from top to bottom):

• the redundancy-free binary code,
• the AMI code (approx.  $37 \%$  redundancy),
• the 4B3T code (approx.  $16  \%$ redundancy).

Further,  the following conditions can be assumed:

• AWGN noise is present,  where holds:

$$10 \cdot {\rm lg}\hspace{0.1cm} ({s_0^2 \cdot T}/{N_0}) = 10\, {\rm dB}\hspace{0.05cm}.$$

• The detection noise power has the following value for the binary system  (due to the non-optimal receiver filter  $12 \%$ markup):

$$\sigma_d^2 = 1.12 \cdot {N_0}/({2 T})\hspace{0.05cm}.$$

• The symbol error probability of the binary system is:

$$p_{\rm S} = {\rm Q} \left( {s_0}/{ \sigma_d} \right) \hspace{0.05cm}.$$

• In contrast,  for the two redundant pseudo-ternary systems:

$$p_{\rm S} = {4}/{3} \cdot {\rm Q} \left( s_0/(2 \sigma_d) \right) \hspace{0.05cm}.$$

• It should be taken into account that the noise rms value  $\sigma_{d}$  may well change with respect to the redundancy-free binary system.

Notes:

• The exercise belongs to the chapter   "Symbolwise Coding with Pseudo-Ternary Codes".

• Reference is also made to the chapter  "Block Coding with 4B3T Codes".

• For numerical evaluation of the Q-function you can use the HTML5/JavaScript applet  "Complementary Gaussian Error Functions". 

Questions

1

Calculate the  (normalized)  noise root mean square for the  binary system.

$\sigma_{d}/s_{0} \ = \ $

2

What is the error probability of the binary system?

$\ p_{\rm S} \ = \ $ $\ \cdot 10^{-5}$

3

What is the noise rms value for the system with  AMI coding?

$\sigma_{d}/s_{0} \ = \ $

4

What is the error probability with AMI coding?

$\ p_{\rm S} \ = \ $ $\ \%$

5

What is the noise rms value when using the  4B3T code?

$\sigma_{d}/s_{0} \ = \ $

6

What is the error probability of the 4B3T code?

$\ p_{\rm S} \ = \ $ $\ \%$

Solution

Solution

(1)  From the given S/N ratio,  we obtain:

$$10 \cdot {\rm lg}\hspace{0.1cm}({s_0^2 \cdot T}/{N_0}) = 10\, {\rm dB} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{N_0} = { s_0^2 \cdot T}/{10}$$

$$ \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\sigma_d^2 = 1.12 \cdot {N_0}/({2 T}) = 0.056 \cdot s_0^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{ \sigma_d}/{s_0} \hspace{0.15cm}\underline { = 0.237}\hspace{0.05cm}.$$

(2)  From this,  it follows for the symbol error probability of the binary redundancy-free reference system:

$$p_{\rm S} = {\rm Q} \left( {s_0}/{ \sigma_d} \right)\approx {\rm Q}(4.22)\hspace{0.15cm}\underline { = 1.22 \cdot 10^{-5}} \hspace{0.05cm}.$$

(3)  The symbol duration  $T$  of the AMI encoded signal is equal to the bit duration  $T_{\rm B}$  of the binary signal.

• Therefore,  the bandwidth ratios do not change and the same noise rms value is obtained as calculated in point  (1):

$${ \sigma_d}/{s_0}\hspace{0.15cm}\underline { = 0.237} \hspace{0.05cm}.$$

(4)  Due to the ternary decision,  the argument of the Q-function is halved:

$$p_{\rm S} \approx{4}/{3}\cdot {\rm Q}(2.11)={4}/{3} \cdot 1.74 \cdot 10^{-2}\hspace{0.15cm}\underline { = 2.32 \cdot 10^{-2}} \hspace{0.05cm}.$$

• Here,  the factor  $4/3$  takes into account that the inner symbol  $0$  can be falsified in two directions.

(5)  When 4B3T coding is applied,  the symbol rate is reduced by  $25 \%$.

• By the same factor  $0.75$,  this makes the noise power smaller than calculated in  (1)  and  (3).  From this follows:

$${ \sigma_d}/{s_0} = \sqrt{0.75} \cdot 0.237 \hspace{0.15cm}\underline {\approx 0.205} \hspace{0.05cm}.$$

(6)  Due to the smaller noise rms value,  the error probability is now smaller than with the AMI code:

$$p_{\rm S} \approx {4}/{3} \cdot {\rm Q} \left( \frac{0.5}{ 0.205} \right) = {4}/{3} \cdot 0.833 \cdot 10^{-2}\hspace{0.15cm}\underline { = 1.11 \cdot 10^{-2}} \hspace{0.05cm}.$$

• But the 4B3T code cannot achieve the significantly smaller error probability of the redundancy-free binary code due to the ternary decision  (half eye opening).