Time Division Multiplexing - Synchronous & Statistical?

Can someone pls tell me what are they used for & whats the differnece between their characteristics? :nonod:

Thanks :happy:

 

Read here : http://en.wikipedia.org/wiki/Time-division_multiplexing :biggrin:

 

hey thanks for the link
you come to my rescue again


i having alot of problems in my assigments can you check if i did the following correctly?

Consider a transmission system using Cyclic Redundancy Check (CRC) technique with a
generator polynomial ​

​

​

​

10101. The receiver receives the following bit pattern:​

100111011000.

​

(a) Draw a digital logic circuit for a shift register implementation of CRC.
[8 marks]

so supposed i am supposed to place the XOR circuit to the right of any position for which the associated value for P(X) is one, i think it should be


A <---B + C <---D +


Can you help me out on this? thanks





​

 

I think you are right :smile:

Code:

  *----------------*------------------*
  |                |                  |
 \_/              \_/                 |
  *-- A -----> B --+--> C -----> D ---+-<--  Input

(* means a joint, + means XOR).

I hope it's correct :thinking:
I get the output as 10100101011

 

yeah... same ans as yours

how i wish you can go for the exams for me haha

 

:biggrin: .. LOL .. :rofl:

Best of luck for your exams.. :wink:

 

Hi Saswapadhi

i have one more question but its hard to state the whole question here

I attached my tutorial... its question 2... can you explain to me what is bit sequence? and what r they actually asking? :nonod::nonod:​

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​

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​

 

OK, I'll go through it and try to answer to the best of my knowledge :smile:

 

oh thanks my saviour! will wait for ya!

 

OK, I read the question .. it's pretty straight forward :

For each of the periods (1 to 8), just check the amplitude of the waveform, and its phase difference with respect to the previous period. Then write the 3-bit symbol from the table, corresponding to it.

E.g., for the first period, the symbol is 001, 'cuz amplitude is A2 and phase difference is 0.
for the second period, the symbol is 010, 'cuz amplitude is A1 and phase difference is 1/(4f).

I won't complete the answer 'cuz that would be a spoiler. Try it yourself and if you can't then I'll try to post complete answer.

 

period 3 A1
period 4 A2
period 5 A2
period 6 A1
period 7 A2
period 8 A1


but how do tell if the signals are 1/(4f)... 2/(4f).... 0 etc ??

 

Check the phase difference between the first point of a period and the last point of the previous period :wink:

Another hint : phase difference of period 3 is 1/(2f) = 2/(4f).

 

is there a phase where the difference between first point of a period and the last point of the previous period is = 3?

ok i give it a shot... so for

period 4 is 011 cuz amplitude is A2 and p.d is 1/4f
period 5 is 000 cuz amplitude is A1 and phase difference is 0
period 6 is 100 cuz amplitude is A1 AND P.D IS 1/2f (=2/4f)
period 7 is 101 cuz amplitude is A2 and p.d is 2/4f
period 8 is ?

 

Wrong answers !

I think you don't get it properly, try to follow my answer :
Period 1 is 001 [ Amplitude = A2 and PD = 0 ]
Period 2 is 010 [ Amplitude = A1 and PD = 1/(4f) ]
Period 3 is 100 [ Amplitude = A1 and PD = 2/(4f) ]
Period 4 is 011 [ Amplitude = A2 and PD = 1/(4f) ]
Period 5 is 101 [ Amplitude = A2 and PD = 2/(4f) ]
Period 6 is 000 [ Amplitude = A1 and PD = 0 ]
Period 7 is 011 [ Amplitude = A2 and PD = 1/(4f) ]
Period 8 is 110 [ Amplitude = A1 and PD = 3/(4f) ]

So, the bit sequence becomes : 001010100011101000011110

(I'm leaving, so I'll be able to read your reply after at least 2 hours)

 

mabbe u can explain why is period 3 1/2f and period 5 = 2/4f?
with your experience how can i tell?

 

haha i think i got it! thanks for all your help! ... u have been a great help already...

i was trying out this question and not sure if i did it right


1. A data source produces a 7-bit ASCII characters and send them over a 1kbps transmission line. Calculate the effective rate of the transmission for the following cases.

a) Ascynchronous transmission with a start bit, 1 stop bit and a parity bit

b) Synchronous transmission with a frame consisting of 48 control bits and 1024 information bits. The information firld contains a numbers of ASCII characters. Each character consists of 7 information bits and a parity bit.

(effective data rate = useful data bits excluding overhead/total time)


so for a... this is what i did

Total transmit time = 7/10(to the power of -3) = 0.007 seconds

Useful bits = 4
therefore Effective data rate = 4 / 0.007 = 571 bps


not sure abt b though shd i be using the same approach?

 

Equation for the waveforms of : (here p stands for pi)

(1) period 2 is x = A1 cos (2pft)
(2) period 3 is x = - A1 cos (2pft)
(3) period 4 is x = A2 sin (2pft)
(4) period 5 is x = - A2 sin (2pft)

So, phase difference between periods 2,3 and periods 4,5 is obviously half time-period = 2/(4f) !
And phase difference between periods 3,4 is quarter time-period = 1/(4f).

PS : You posted the last post while I was writing this reply.

 

Dude i really want to thank you for all your help... its really kind of you to render so much help to me, a complete stranger over the net...

i hope i didn't irritate you by askin so many questions...thanks once again!

 

Not at all, why do you think that way. You are not irritating me, you help me revise all these stuff.

Anyway, about your last question :

PART A:
Total data to be transmitted = 7 + 1 + 1 + 1 = 10 bits
Total transfer time = 10/1024 s
Effective transfer rate = 7*1024/10 = 716.8 bps

PART B: (not sure abt this one)
Total data to be transmitted = 48 + 7 + 1 = 56 bits
Total transfer time = 56/1024 s
Effective transfer rate = 7*1024/56 = 1024/8 = 128 bps

But synchronous transfer should be faster than asynchronous, right ?

 

yes synchronous shd be faster

http://www.computeruser.com/resources/dictionary/definition.html?lookup=4916

effective data rate = useful data bits excluding overhead/total time

shdn't it be 7/(10/1024)??