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        COM 5140代做、c/c++程序設(shè)計(jì)代寫

        時(shí)間:2024-05-07  來源:合肥網(wǎng)hfw.cc  作者:hfw.cc 我要糾錯(cuò)



        COM 5140 Error-Correcting Codes 
        Spring 2024
        Project No. 1
        Due 4:00pm, May 16, 2024
        You are expected to produce a program to implement the Viterbi decoding algorithm for
        the widely used (2, 1, 6) convolutional code with generator matrix
        G(D) = (1 + D
        2 + D
        3 + D
        5 + D
        6
        1 + D + D
        2 + D
        3 + D
        6
        ).
        This code is assumed to be transmitted over an additive white Gaussian noise (AWGN)
        channel.
        The deliverable will consist of three parts:
        ? Part I, Demonstration. At the time of demonstration, we will let you know: the
        number of decoded bits N, the bit signal-to-noise ratio (SNR) Eb/N0 (in dB), the seed
        for the random number generator, and hard or (unquantized) soft decision. You should
        then report in each case the number of decoded bit errors made by your decoder and
        the corresponding bit error rate (BER). I want you to truncate your survivors at length
        **, outputting the oldest bit on the survivor with the best metric.
        ? Part II, Report. You should run experiments with your Viterbi decoder to produce
        performance curves showing the relationships between Eb/N0 (in dB) and the decoded
        BER (in logarithmic scale), with both hard-decision decoding, which corresponds to
        decoding on a binary symmetric channel (BSC), and unquantized soft-decision decod?ing, for Eb/N0 ranging from 1 dB to 6 dB for hard decision and 1 dB to 4 dB for
        unquantized soft decision, with increments of 0.5 dB. Please also include your simula?tion data in tabular form, listing for each data point: the bit SNR E0/N0, the number
        of decoded bits, the number of decoded bit errors, and the BER. (These detailed data
        are only required for the two mandatory performance curves.) Please hand in before
        the deadline a report (in a hard copy) which includes, among other things, performance
        curves, and (optional) discussions of issues like output decision alternatives (best-state,
        fixed-state, majority-vote), survivor truncation length, etc. Your computer program
        with comments should be attached at the end of the report.
        ? Part III, Program file. You also need to submit, before the deadline, your program
        file. Please put all of your programs into a single file with your registration number and
        proj1 as the file name, say, 105064851 proj1.c or 105064851 proj1.cpp. (If, after all
        kinds of attempts, you are still unable to put all of your programs in a single file, please
        compress your files into a single rar or zip file and use your registration number along
        with proj1 as the file name, say, 105064851 proj1.rar or 105064851 porj1.zip.)
        Upload your file to the eeclass system.
        Additional Details on Project No. 1
        1. Use the recursion
        ul+6 = ul+1    ul
        , for l    0
        with the initial conditions u0 = 1, u1 = u2 = u3 = u4 = u5 = 0 to generate the infor?mation bits. Ensure that the generated sequence is 100000100001 . . . and is periodic
        with period 63.
        2. Encode the information sequence using the generator matrix G(D).
        3. The encoder outputs 0  s and 1  s. However, the input to the AWGN channel is normal?ized to   1. Therefore, map 0  s to +1  s and 1  s to ?1  s.
        4. To simulate the AWGN channel with unquantized soft-decision decoding, add a normal
        (Gaussian) random variable of mean zero and variance   
        2
        to the   1  s generated at the
        previous step. For a binary code of rate R on the AWGN channel with antipodal
        signaling, the relationship between Eb/N0 and   
        2
        is given by
          
        2 =
         2R
        Eb
        N0
        
        ?1
        so for example for a R = 1/2 code, the relationship is simply
          
        2 =
        
        Eb
        N0
        
        ?1
        .
        Please remember that Eb/N0 is always quoted in   dBs,   which equals 10 log10(Eb/N0).
        Thus for example, a value of Eb/N0 of 4 dB for a R = 1/2 code corresponds to a value
        of   
        2 = 0.3981.
        5. Use the following segment of pseudo code to generate normal random variables of
        mean zero and variance   
        2
        . The procedure normal outputs two independent normal
        random variables, n1 and n2, and Ranq1 is a function which generates a random variable
        uniformly distributed in the interval (0, 1).
        unsigned long long SEED;
        // SEED must be an unsigned integer smaller than 4101842887655102017.
        unsigned long long RANV;
        int RANI = 0;
        main()
        {
                
                
                
        }
        2
        normal(n1, n2,   )
        {
        do{
        x1 = Ranq1();
        x2 = Ranq1();
        x1 = 2x1 ? 1;
        x2 = 2x2 ? 1;
        s = x
        2
        1 + x
        2
        2;
        } while (s    1.0)
        n1 =   x1
        q ?2 ln s/s;
        n2 =   x2
        q ?2 ln s/s;
        }
        double Ranq1()
        {
        if ( RANI == 0 ){
        RANV = SEED    4101842887655102017LL;
        RANV   = RANV >> 21;
        RANV   = RANV << 35;
        RANV   = RANV >> 4;
        RANV = RANV * 2685821657**6338717LL;
        RANI++;
        }
        RANV   = RANV >> 21;
        RANV   = RANV << 35;
        RANV   = RANV >> 4;
        return RANV * 2685821657**6338717LL * 5.42101086242752217E-20;
        }
        6. To get the output of the BSC, take the sign of the output of the AWGN channel and
        map +1  s to 0  s and ?1  s to 1  s.
        7. In your decoder, truncate the survivors to length ** and output the oldest bit on the
        survivor with the best metric. To decode N bits, generate N + 31 bits in (1). Finally
        compare the decoded information sequence with the original information sequence. If
        there are K bit errors, K/N will be a good estimate of the decoded BER.
        8. As a partial check, some typical values are listed below.
        Eb/N0 BER (BSC) Eb/N0 BER (AWGN)
        4.5 dB 2.1    10?3 2.5 dB 2.2    10?3
        5.0 dB 6.4    10?4 3.0 dB 5.3    10?4
        3
        Other Notes for Demonstration
        1. The survivor truncation length corresponds to the actual storage requirement of the
        survivors. For example, a survivor truncation length of ** for this code means that
        each survivor stores ** bits.
        2. For the illustration below, suppose a state is described as the content of the feed-forward
        shift register in the encoder s = (s1, s2, s3, s4, s5, s6), where the input information bit
        first fed to s1 and then shifted from left to right. In the trellis diagram, consider placing
        the states vertically from top to bottom in the order of (0 0 0 0 0 0), (1 0 0 0 0 0),
        (0 1 0 0 0 0), (1 1 0 0 0 0), (0 0 1 0 0 0), . . ., (1 1 1 1 1 1). What to do in case of tied
        metrics? In the   add-compare-select   step the two metrics could be equal. In this case,
        if 0  s and 1  s are equally probable to occur in the transmitted information sequence, in
        principle you can safely select either case, and it will not affect the decoder performance.
        Yet for the purpose of demonstration, always choose the upper branch as the survivor.
        If best-state output decision is employed, in case of tied metrics, in principle you can
        also safely select either case, but again for the purpose of demonstration, always choose
        the survivor of the uppermost state.
        3. Except in the procedure normal for generating noise, if a random number is needed in
        your program, use other random number generators instead of the function Ranq1, for
        the purpose of demonstration.
        4. Each call of the procedure normal can return two independent normal random vari?ables, n1 and n2. Please use both of them in your program. Specifically, since this is
        a (2, 1) code, each branch transition consists of two encoded bits, say x1 and x2. Add
        n1 and n2 to x1 and x2, respectively, to get the two channel outputs y1 and y2, i.e.,
        y1 = x1 + n1 and y2 = x2 + n2.

         

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