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COM347J1 Things to remember

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					                                      COM347J1: Things to remember
        [0000]       =   ‘0’           2x2=4 = 22             NB count the zeros!                         Standard
        [0001]       =   ‘1’        2x2x2=8 = 23              50dB = 100,000 (5 Zeros)                    Baud rates
        [0010]       =   ‘2’        24=16                                                                 110
        [0011]       =   ‘3’        25=32                     10dB = 10:1                                 300
        [0100]       =   ‘4’        26=64                     20dB = 100:1                                600
        [0101]       =   ‘5’        27=128                    30dB = 1000:1                               1200
        [0110]       =   ‘6’        28=256                    40dB = 10,000:1                             4800
        [0111]       =   ‘7’        29=256                    3dB = 2:1                                   9600
        [1000]       =   ‘8’        210=1024                  6dB = 4:1                                   19.2 kBaud
        [1001]       =   ‘9’                                  26dB = 400:1                                38.4
        [1010]       =   ‘A’        220 = 210 x 2 10                                                      115
                                                              NB Add dB, multiply the ratios
        [1011]       =   ‘B’        Or 1024x1024
                                                              26db = 20dB+6dB = 100x4
        [1100]       =   ‘C’        ie 1M roughly
        [1101]       =   ‘D’
        [1110]       =   ‘E’                                                       If there is no noise NYQUIST
                                    to what power must I raise 2 to in
        [1111]       =   ‘F’                                                              Imax = 2 H log2(V)
                                    order to produce the number 32?” –
                                                                                          I=BITS PER SEC
                                    answer is 5. Ie LOG2(32)=5
                                                                                       H=BANDWIDTH (Hz)
                                                                                   V=number of changes or levels
        E.g      101110 is [oo10][1110] is ‘2’ ‘E’                                  So log2(V) is number of bits
                                                                                        needed to encode these
                                                                                       If V=1024, log2(V) = 10
        If noise is mentioned – use SHANNON
                            Imax = H log2(1+ S/N)
           Convert the S/N dB value to a ratio; e.g 26dB = 400:1 so                2 amplitude levels
                (1+ S/N) = 401 and note log2(401) needs 9 bits                     3 phase levels
        The bandwidth H is the difference between the lowest and                   6 possibilities
        highest frequencies (3500-500 = 3000) So ans =3000x9 BPS                   3 bits can represent
                                                                                   6 numbers
            Eg Humans write 0x82hex as binary like this
                                                                     4 amplitude levels
                                                                     4 phase levels
                                                                     16 possibiliites
       1 0 0 0 0 0 1 0
      msb
                                                                     - 4 bits can represent
                                                                      16 numbers
                                                       lsb
                                                   Eg But data is sent lsb first – and we usually         A format of 8E1
                                                   write down the earliest bits on the left – so          Means 8 data
                                                   we have to reverse the data bits, dropping
                01000001                           the msb if we only need 7 bits
                                                                                                          bits, Even parity
                                                                                                          and one stop bit
 Start bit
                                                           Stop bit, also same state as idle

              001000001 0 1                                     parity bit –add it to the data to make
                                                                total Odd or Even
                                                                Parity can be ‘E’, ‘O’, ‘N’, ‘M’ ‘S’
First to                                                        We won’t use the last two
arrive -        11 bits are sent in this case
start bit
then lsb       Note: Least significant bit sent first (after start), parity usually sent after MSB

				
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posted:2/26/2010
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