GATE IT 2007 | Question: 61

Question

In the waveform (a) given below, a bit stream is encoded by Manchester encoding scheme. The same bit stream is encoded in a different coding scheme in wave form (b). The bit stream and the coding scheme are

• A. $1000010111$ and Differential Manchester respectively
• B. $0111101000$ and Differential Manchester respectively
• C. $1000010111$ and Integral Manchester respectively
• D. $0111101000$ and Integral Manchester respectively

Comments

did u get the explanation? if yes plz share

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https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Differential_Manchester_encoding.html

this link explains all

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How does IEEE format give a waveform like in figure b for the bitstream in option B? It can not be correct. The figure b gives a bitstream like in option A. So clearly option B is wrong, right?

Answer 1

THE AMBIGUOUS QUESTION WITHOUT ANY STANDARD MENTION !!

BOTH $A$ and $B$ is correct , it is just the convention which determine the correct answer.

see this from IIT KGP, they follow IEEE standard - 

http://nptel.ac.in/courses/Webcourse-contents/IIT%20Kharagpur/Computer%20networks/pdf/M2L4.pdf or archive

and this one from IITB , they follow G E Thomas version -- 

https://www.cse.iitb.ac.in/synerg/lib/exe/fetch.php?media=public:courses:cs348-spring08:slides:topic03-phy-encoding.pdf

As per $\text{IEEE}$ B is correct

As per $\text{G E Thomas}$ A is correct.

As we ususally follow IEEE thus B is correct 

Comments

@Pankaj Joshi:

I've had had the same concern you did:
Let us count the number of transitions for each bit period (starting and middle):
[?][2][2][2][2][1][2][1][1][1]
We can't tell for the first bit but the second one definitely has 2 transitions.

I'm guessing you're calling 2 transitions an inversion. And calling an inversion 0. So it should be:
"?000010111" as per you, matching option A. Most of the sources I've checked say the same.

Except this one:
https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Differential_Manchester_encoding.html

It says we can call 2 transitions a "0" and 1 transition a "1" or vice-versa.
This "vice-versa" has messed up everything for us, buddy!

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Integral Manchester?

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Sir, during transition in Differential manchester if next bit is 0 then there is a Inversion  , if next bit is 1 then there is NO inversion but , what representation will be used for 1st bit, since no bit precedes it ?

Answer 2

In this question option  A) is as per G E Thomas standard and option B)  is as per IEEE standard . 

Here Manchester code 1000010111  is based on  G E Thomas  standard and 0111101000 is based on  IEEE standard. Both are correct as there is no standard explicitly mentioned !!

But we generally follow IEEE standard thats why as per IEEE B is correct answer in this question .

Along with that, Rule for differential Manchester is, Let me post snap From Data Communication book ,Forouzen ->

Looking at this image, It is clear this book follow IEEE standard ,  And so this is differential Manchester.

AFAIK, there is no term as Integral Manchester, It was just put there for more confusion.

Comments

As per this snap (from Forouzen), transition from high to low is bit 0. But you are interpreteing 1st bit as 1. How?

Manchester encoding is also two types- https://en.wikipedia.org/wiki/Manchester_code

Here, optian A is as per G.E. Thomas 

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What will be used for initial bit in differential manchester encoding ?

Answer 3

according to G.E. Thomas' convention   '1' by high-to-low transition     and   '0' is expressed by a low-to-high transition 

In IEEE 802.3 '1' is represented by low-to-high and '0' represented by high-to-low 

From waveform (a) we got the bit sequence 

1.  1000010111 (G.E. THOMAS)

2.  0111101000 (IEEE 802.3)

As the standard is not given so we have to draw waveform of differential manchester encoding using both standard. And try to match with the given waveform (b)

 In Differential manchester if next bit is 0 then there is a Inversion  , if next bit is 1 then there is NO inversion

Only the the waveform following the standard G.E. Thomas is matching with the given waveform

So the correct option is  (A) .

Comments

@Deepak

Didn't get what exactly you tried to convey. Can you please make it more clear.

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@Deepak Agrawal 10 @Sumit Singh Chauhan @Chhotu @Anantk @Bikram @srestha 

Isn't the Bit stream should be 10000101111? Why the last '1' is neglected? 

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It should be there,they might have overlooked it

Answer 4

Manchester encoding(Used in Ethernet) -->

https://www.youtube.com/watch?v=XKtxxZ327UM. But There are two way to do Manchester encoding. 

1. In IEEE 802.3 '1' is represented by low-to-high and '0' represented by high-to-low.
2. G.E. Thomas' convention   '1' by high-to-low transition     and   '0' is expressed by a low-to-high transition.

Differential Manchester(Used in Token Ring) -->

https://www.youtube.com/watch?v=du_boiwX1yU

Now in mentioned question G.E. Thomas' convention is used and answer is (A) Part. 

Please notify if anything is not proper. Thank you @Shivam Chauhan ji and @Tuhin Dutta ji. 

PS:  Some additional info -->

Manchester encoding not only transmitting the actual data, but also the clock (meta data) due to its self clocking characteristic.

https://stackoverflow.com/questions/25834577/why-in-manchester-encoding-the-bit-rate-is-half-of-the-baud-rate

In both Manchester and Differential Manchester encoding scheme -->
Baud Rate =2*Bit Rate

Comments

Differential Manchester encoding means if next bit is zero then do transition.This is what i read in frouzen,

Based upon this a should be the answer.@joshi_nitish ,in your comment you said b is the answer.Can you tell hwo it can be differential Manchester encoding?