# COM347J1 Things to remember by dfhrf555fcg

VIEWS: 3 PAGES: 1

• pg 1
```									                                      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
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

```
To top