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EEEN40060 Digital Communications Assignment 1 - Simulation of an 8-ary Solved

 

Assignment 1 - Simulation of an 8-ary 
Digital Communication System 
Digital Communications EEEN40060

Result: 

 

Modelling: 

1.       We first generated 100000 messages (a message consists of 3 bits). 

2.       On transmitter side, the reflected binary code (RBC), also known as Gray code is used to map the random input digital signal to symbols. Gray code is an ordering of the binary numeral system such that two successive values differ in only one bit (binary digit). Gray codes are widely used to prevent spurious output from electromechanical switches and to facilitate error correction in digital communications. In my code, I used Gray code and symbols in decimal form to make them more straight-forward. 

3.       AWGN was added to modulated vectors and demodulated the signal by calculating the Euclidean distance to each symbol and mapping the modulated signal (with the AWGN) to the nearest one. 

4.       Because of the equiprobable transmission, the metrics like SER and BER can be obtained by counting the number of error symbols. 

Analysis: 

l  Derivation of Theoretical SER 

 Recall the signal-space diagram for the 8-ary modulation scheme on the right.  

The overall SER equals to the sum of the SER of Red symbols and the SER of the Green (equiprobable transmission). 

 𝑅𝑅𝑅𝑅𝑅𝑅 𝑆𝑆𝑆𝑆𝑅𝑅 

                      𝑅𝑅                          𝑅𝑅                          𝑅𝑅                         𝑅𝑅

= 𝑃𝑃∅2 𝑛𝑛  

= 2𝑃𝑃𝑛𝑛≥𝑅𝑅 + 𝑃𝑃2𝑛𝑛≥ 𝑅𝑅 = 2𝑄𝑄 𝑅𝑅−𝑄𝑄2 𝑅𝑅 

                       2                             2                  2𝜎𝜎                2𝜎𝜎

Similarly, 

  𝐺𝐺𝐺𝐺𝑅𝑅𝑅𝑅𝑛𝑛 𝑆𝑆𝑆𝑆𝑅𝑅 

= 2𝑃𝑃𝑛𝑛≥𝑅𝑅 + 𝑃𝑃2𝑛𝑛≥ 𝑅𝑅 = 3𝑄𝑄 𝑅𝑅− 2𝑄𝑄2 𝑅𝑅 

                       2                             2                  2𝜎𝜎                   2𝜎𝜎

Average, 

𝑆𝑆𝑆𝑆𝑅𝑅 𝑅𝑅𝑅𝑅𝑅𝑅 𝑆𝑆𝑆𝑆𝑅𝑅 + 4 ∗𝐺𝐺𝐺𝐺𝑅𝑅𝑅𝑅𝑛𝑛 𝑆𝑆𝑆𝑆𝑅𝑅) 

     = 5 𝑄𝑄𝑅𝑅              3 2𝑅𝑅  

− 𝑄𝑄

          2       2𝜎𝜎        2         2𝜎𝜎

 The average energy to d, 𝑅𝑅 = 23𝐸𝐸𝑠𝑠, and 2𝜎𝜎2 = 2𝑁𝑁0 Substitute into the formula, we obtain: 

      𝑆𝑆𝑆𝑆𝑅𝑅 = 5         𝑆𝑆𝑠𝑠                 3 2 𝑆𝑆𝑠𝑠  

                     𝑄𝑄              − 𝑄𝑄

                 2           3𝑁𝑁0        2             3𝑁𝑁0

l  Value of Eb/N0 is greater 5.333 than where BER lies below e-4 

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