01/10/04 ian/modules/COM347J1/COM347J1_SUMS.ppt SUMS/1/21 COM347J1 Networks and Data Communications Ian McCrumRoom 5D03B Tel: 90.

01/10/04 www.eej.ulster.ac.uk/~ian/modules/COM347J1/COM347J1_SUMS.ppt SUMS/1/21

COM347J1Networks and Data Communications

Ian McCrum Room 5D03B

Tel: 90 366364 voice mail on 6th ring

Email: [email protected]

Web site: http://www.eej.ulst.ac.uk

Sums Lecture (and Tuts on ohp 15, 17 & 19)


Binary and Hex• Hex is just a shorthand way of writing binary• Binary is always written with b0 on the right• Hex always considers the binary is split into

groups of 4, starting at the right• For each group of 4 binary bits, use the table to

look up what single character to write down in place of the 4 bits

• E.g 01001001 is [0100][1001] is ‘4’ ‘9’• E.g 101110 is [oo10][1110] is ‘2’ ‘E’• Unfortunately data is transmitted bit zero first,

usually drawn with bit zero on left if drawing square waves, also RS232 uses –12V for ‘1’, +12v for ‘0’

• We could use “base-2” or “base-16” arithmetic, let the 4 bits have weights 8421

[0000] = ‘0’[0001] = ‘1’[0010] = ‘2’[0011] = ‘3’[0100] = ‘4’[0101] = ‘5’[0110] = ‘6’[0111] = ‘7’[1000] = ‘8’[1001] = ‘9’[1010] = ‘A’[1011] = ‘B’[1100] = ‘C’[1101] = ‘D’[1110] = ‘E’[1111] = ‘F’

So ‘A’ = 1010, 8+2 =10, one more than 9…


Basics

422 422 or

8222 or 823

162222 1624 or


Description

• In English we would say 2 to the power of 3 is 8. In other words 2 must be multiplied by itself 3 times in order that the answer 8 is achieved.

• What is 210 equal to?

• The answer is 1024

come on do it


More• Whenever you wish to work in reverse you

use a different way of describing the problem.

• One might say “to what power must I raise 2 to in order to produce the number 32?”

• And of course the answer is 5

• Another way of expressing that is by saying “what is the log of 32?” (really should say log to the base 2, or mathematically log2)


Mathematicians write that in the following way

5)32(2 Log

7)128(2 Log

6)64(2 Log

8)256(2 Log


But what if you are asked the following questionWhat is Log2 1000? - you know that it is less than 10 because Log2 1024 is equal to 10 but what is the exact

value?

)(

)()(

xLog

yLogyLog

z

zx

Here is a useful formula which shows you how to do it -

What that means in practice is - consider the log button on your calculator, it works to some base usually 10 but it

could have other values


You just take the log of the number you want the log2 (log to the base two) of, and divide it by the log

of the number 2.

Look at the formula again

)(

)()(

xLog

yLogyLog

z

zx

The x is the value 2 and the z is the base to which your calculator works (normally 10) so

)2(

)1000()1000(

10

102 Log

LogLog


So we were right, the answer is just less than 10

97.9301.0

0.3

)2(

)1000(

10

10 Log

Log

97.9693.0

91.6

)2(

)1000(

e

e

Log

Log

If your calculator has a ln button that is log to the base e where e=2.718

you can try it and you will get the same answer look!


When ever people have very big numbers or very small numbers, they might choose a different way to present them. This is also

true if they have some property they wish to highlight

One such case is power ratios in electronic circuits.

A person might wish to describe the ratio as1 : 100 or 1 : 1000000 or 1 : 100000000, you like me, might find it difficult to decide what

the actual ratio was when it is so written.


Now that you know about logs you might describe the ratios on the previous slide in

terms of log to the base 10.

1 : 100 as 2and 1 : 1000000 as 6

and 1 : 100000000 as 8

Now isn’t that more tidy and it also make you concentrate upon the important thing about the number, its magnitude, not its precise

value(these are Bells). - there’s more


However this can go too far and because of the nature of the value that we use the range

would be limited from 1.0 to 9.0 and we would need to include the decimal point to

give the required accuracy to our calculations.So we go one step further.

We multiply the log of the ratio by 10 and leave out the decimal point.

And we call them deciBells ie 1/10th part of a Bell. (dB)


So if you wish to find out the precise value of a ratio which is expressed in dB for example

40 dB

Take the 40 divide it by 10 so it is now in Bells thus 4 B and raise 10 to the power of the

number

Ie 104 = 10 x 10 x 10 x 10 or 10,000


Remember (or calculate) these

• 2^1 = 2 so log2(2)=1• 2^2 = 4 so log2(4)=2• 2^3 = 8 so log2(8)=3• 2^4 = 16 so log2(16)=4• 2^5 = 32 so log2(32)=5• 2^6 = 64 so log2(64)=6• 2^7 = 128 so log2(128)=7

• 10dB = 10:1• 20dB = 100:1• 30dB = 1000:1• 40dB = 10,000:1• 3dB = 2:1• 6dB = 4:1• 26dB = 400:1 • NB Add dB,

multiply the ratios


Example Tut Questions• Q1: Given the ASCII code for the letter ‘F’ is 0x46 draw the

binary pattern transmitted if the following data format is used– (a) One start bit, 8 data bits and one stop bit 8N1– (b) One start bit, 8 data bits, even parity and one stop bit 8E1– (c) One start bit, 7 data bits, odd parity and two stop bits 7O2

• Ans1: [0100][0110] so b0, bit zero, the least significant bit is on right. Remember unfortunately data is transmitted bit zero first, usually drawn with bit zero on left if drawing square waves, also RS232 uses –12V for ‘1’, +12v for ‘0’

• Note parity includes all data bits and the parity bit itself, so a data word of 0100 0110 would need a 1 added to give even parity, parity bits are added after bit 7 ( or bit 6)


• Hex 0x46 is 0100 0110 , we write binary bit zero on right

Transmitted data

XX__[0][1][1][0][0][0][1][0] XX

+12

-12

0

1

Start bit

‘0’ Stop bit of ‘1’


Now you can go and try Q1b and Q1cBe able to writing down the binary, calculating parity, adding stop and start bits and drawing the waveforms.

Related Questions:

• Q2: Calculate how long in milliseconds it takes to send a 8N1 character at 1200 baud

• Q3: How long to send “HELLO_WORLD” at 9600 baud assuming 8E2

• Q4: What is the overhead in sending a character in the format of 8N1?

• Q5, Book (ed 3) Q22, Q23, Pages 75 (see .pdf)


The formulae!

I H Vmax log2 2

I HS

Nmax log

2 1

Nyquist: When there is no noise… a perfect channel carrying ‘V’ levels

Shannon: When we have noise… a signal to noise level of s/n


Tut questions (3rd edition, p170-)• Q2-3 Analogue TV channels are 6MHz wide, how many

bits/sec can be sent if 4 level signals are used, ignoring noise

• Q2-4 A binary signal is sent over a 3kHz channel whose signal to noise ratio is 20dB, what is the maximum data rate possible

• Q6 If a modem uses 4 amplitude levels and 4 phase values how many different bits of information can be sent in a single data change.

• Q7 (a) If the signal to noise ratio in Q2-4 changes to 30dB what is the new maximum data rate? (b) for 36db?


Revision…I’ll repeat this Address the following areas in the recommended textbook

•general issues pp1-7

•network hardware ring v bus pp7-10

•wide area networks pp11-16

•network software - layers pp16-22

•connection oriented / connectionless pp22-27

•services, protocols peers pp27-28

•OSI reference model and layers pp28-34

•TCP/IP pp35-39

•Internet pp52-54

•Standards who and why? pp67-72


Revision…I’ll repeat this

•Syn/asych transfer

•Physical layer - bandwidth limited pp86-87

•baud v bits per second pp88-89

•Nyquist max data rate pp 89-90

•Shannon’s max data rate pp 90

•Media: (tape), UTP, Coaxial cable,

FOC, Tx light, Light sources, Wireless,

Radio,microwave,IR, pp 90-99

•Modems baud v bps pp125-127

•modulation methods pp127-130

01/10/04 ian/modules/COM347J1/COM347J1_SUMS.ppt SUMS/1/21 COM347J1 Networks and Data Communications Ian McCrumRoom 5D03B Tel: 90.

Documents

base e

log button

following way slide

bits e

binary bits

answer look

hex hex

sums221 binary