Forouzan along with William stallings book follow IEEE 802.4 (token bus) and lower speed versions of IEEE 802.3 (Ethernet) standards . Where Tanenbaum follows G. E. Thomas standard.

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+25 votes

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

- $1000010111$ and Differential Manchester respectively
- $0111101000$ and Differential Manchester respectively
- $1000010111$ and Integral Manchester respectively
- $0111101000$ and Integral Manchester respectively

0

Forouzan along with William stallings book follow IEEE 802.4 (token bus) and lower speed versions of IEEE 802.3 (Ethernet) standards . Where Tanenbaum follows G. E. Thomas standard.

source

+12

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

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

+2

@joshi_nitish

As Bikram sir answer also gives a PDF for IEEE 802.3 which says about Differential Manchester Encoding as:

**The encoding of a 0 is represented by the presence of a transition both at the beginning and at the middle and 1 is represented by a transition only in the middle of the bit period.**

According to me Encoding will be

So how diagram b) is correct. Please help

0

If we take IEEE 802.3 convention then we get option B) bits but Bikram sir gave PDF has different encoding for Diifferential Manchester Encoding as i gave in above comment.

If we take GE Thomas convention then we get option A) bits. How does Diifferential Manchester Encoding done in this convention.

If we take GE Thomas convention then we get option A) bits. How does Diifferential Manchester Encoding done in this convention.

+24 votes

Best answer

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

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

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 **

+3

To know about Encoding plz read this pdf - http://nptel.ac.in/courses/Webcourse-contents/IIT%20Kharagpur/Computer%20networks/pdf/M2L4.pdf

In Manchester encoding we generally folow **IEEE 802.3** standard where *'1' is represented by low-to-high* and '0' represented by high-to-low .

And in Differential manchester **if next bit is 0 then there is a Inversion , **if next bit is 1 then there is NO inversion.

see that snapshot posted from Farauzan , it will be clear.

+1

answer can't be b because look at the second waveform if we select b as response it wont agree with the diffrential menchester encoding

+1

@pankaj joshi

**In IEEE 802.3** standard Manchester Encoding is where *'1' is represented by low-to-high* and '0' represented by high-to-low .

And in Differential manchester **if next bit is 0 then there is a Inversion , **if next bit is 1 then there is NO inversion.

now see the question and see for option B, it is correct as per IEEE standard.

+3

look at the differntial menchester the inversion in the second wave form are at 2,3,4,5 hence 0's at those positions

0

No, pankaj, plz read differential manchester

and see again, i follow IEEE standard, you missed something.

and see again, i follow IEEE standard, you missed something.

0

menchester has two conventions's but as far as I know differential menchester follows only one convention

as you said your self that inversion at 0s

now forget about wave a for a minute just see the wave b where are the inversions ?

as you said your self that inversion at 0s

now forget about wave a for a minute just see the wave b where are the inversions ?

0

@Pankaj Joshi

" menchester has two conventions's but as far as I know differential menchester follows only one convention" ??

why differential follow 1 conversion ? where you read ? need a link to verify...

read properly http://nptel.ac.in/courses/106105080/pdf/M2L4.pdf page 8-9

0

what are you trying to make me read I don't know

Differential menchester says that if next bit is 0 then invert otherwise don't

now can you provide a refrence that says otherwise because I haven't seen anything else

Differential menchester says that if next bit is 0 then invert otherwise don't

now can you provide a refrence that says otherwise because I haven't seen anything else

+1

@Pankaj josi who said that in differential manchester we invert if next bit 0 otherwise don't, you misunderstood this concept.

AFAIK in differential manchester if next bit same as current bit we will not invert if it is different then we will invert the waveform.So for waveform B can be done in differential manchester which simply means we can represent we start at one level if next bit is same invert if not stay there

AFAIK in differential manchester if next bit same as current bit we will not invert if it is different then we will invert the waveform.So for waveform B can be done in differential manchester which simply means we can represent we start at one level if next bit is same invert if not stay there

+2

In Differential manchester encoding presence or absence of transitions is required to represent a logical value. It has two variants one IEEE and another GE Thomas to which links are provided above in answer. So, here both (A) and (B) are correct.

PS :) In many pdf's and videos only one is mentioned ( the one author/faculty followed ), So it's likely to think any one of (A) or (B) answer.

PS :) In many pdf's and videos only one is mentioned ( the one author/faculty followed ), So it's likely to think any one of (A) or (B) answer.

0

As per the given pdf from IIT-KGP :

In the standard Manchester coding there is a transition at the middle of each bit period. A binary 1 corresponds to a low-to-high transition and a binary 0 to a high-to-low transition in the middle.

In Differential Manchester, inversion in the middle of each bit is used for synchronization. The encoding of a 0 is represented by the presence of a transition both at the beginning and at the middle and 1 is represented by a transition only in the middle of the bit period.

If the bit-stream in a) is 0111101000, aren't we supposed to have two transitions for a '0' and a single transition for '1'. But if the answer is B) the diagram shows just one transition for all '0' bits and two transitions for the '1' bits?

+1

everybody is talking about conventions which i got but my doubt is there should be **11 bits **in the answer or am i thinking wrong...somebody plz clear..!!

0

You are thinking wrongly here..

..there are 12 transitions and number of bits 10

see this 0111 1010 00 = 4+4+2 =10 bits in total

between each 2 transitions there is single bit exists.

0

still not getting..how you are counting 10 bits...but i am thinking like..between each dotted line there is one bit which is taking the wave up or down in manchester encoding (a) ....!!!

:(

:(

+2

Yes,

between each dotted line there is one bit which is taking the wave up or down in manchester encoding ,

yes , this is correct. see below answer , a snap from Farauzan is taken .

In this question they provide 10 bits only in options so we just not consider last transition ..

your asumtion is correct, option should have 11 bits .

+3

@Bikram sir i think @Pankaj joshi is right. answer should be a)

the link you said to read also says that in differential manchester 0 means transition and 1 means no transition. i studied differential manchester from nptel link u shared and according to it IEEE standard is not satisfying differential manchester. so answer should be a).

the link you said to read also says that in differential manchester 0 means transition and 1 means no transition. i studied differential manchester from nptel link u shared and according to it IEEE standard is not satisfying differential manchester. so answer should be a).

0

@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!

+10 votes

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.

+1

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

+2 votes

Manchester encoding(Used in Ethernet) -->

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

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

In both Manchester and Differential Manchester encoding scheme -->

Baud Rate =2*Bit Rate

+1

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 [email protected]_nitish ,in your comment you said b is the answer.Can you tell hwo it can be differential Manchester encoding?

Based upon this a should be the [email protected]_nitish ,in your comment you said b is the answer.Can you tell hwo it can be differential Manchester encoding?

+1 vote

0

Differential Manchester encoding is also not fixed like Manchester encoding. Either you can take 1 as transition or 0 as transition.

Because of this I think that thid question is ambiguous.

Because of this I think that thid question is ambiguous.

0

The encoding of a 0 is represented by the presence of a transition both at the beginning and at the middle and 1 is represented by a transition only in the middle of the bit period.

0

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

Isn't the Bit stream should be 1000010111**1? Why the last '1' is neglected? **

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