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As an non-specialist, this make little sense to me. Could sombody please put on a paragraph or two explaining in simple language what this is about. Thanks

Velela 14:40, 5 January 2006 (UTC)Reply

Would it be appropriate to merge this article with Bipolar encoding? Biscuittin 21:11, 7 August 2007 (UTC)Reply

Anon 21:13, 31 March 2008

It's all the same code. It is just a way to show all 0 in AMI code without disconect —Preceding unsigned comment added by 99.165.173.1 (talk) 02:14, 1 April 2008 (UTC)Reply

Hello, I think that there is a mistake at: "Example 3 of HDB3 encoding". the the first sequence should be B00V and not 000V.

thanks, Mike — Preceding unsigned comment added by 109.64.235.18 (talk) 20:23, 15 August 2011 (UTC)Reply

Example 2 of HDB3 encoding

The pattern of bits

" 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 0 0 "

encoded in HDB3 is " + 0 − B 0 0 V 0 + − B 0 0 V − + B 0 0 V 0 0 " which is:

" + 0 − + 0 0 + 0 + − + 0 0 + − + − 0 0 − 0 0 "

BUT there are now FOUR violation marks here and only THREE substitutions
+ 0 − [+ 0 0 +] 0 + − [+ 0 0 +] − + [− 0 0 −] 0 0
B 0 B [B 0 0 V] 0 V B [B 0 0 V] B B [B 0 0 V] 0 0
             ^    ^          ^             ^
             1    2          3             4

The violation mark 2 should be balance mark (B) not a violation mark (V). Violation marks only exist in the zero substitution patterns.

The correct sequence should be...
B 0 B [B 0 0 V] 0 B B [B 0 0 V] B B [B 0 0 V] 0 0
             ^               ^             ^
             1               2             3

Which gets encoded as...
+ 0 − [+ 0 0 +] 0 - + [- 0 0 -] + - [+ 0 0 +] 0 0
             ^               ^             ^
             1               2             3

The reason I know this to be true is because of examples presented in the book Digital Telephony: Third Edition written by John C. Bellamy.

In chapter 4: Digital Transmission and Multiplexing on page 177 they uses the following B3ZS example...
1 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 1

This shows where the zero sequences are...
1 0 1 [0 0 0] 1 1 [0 0 0] [0 0 0] 0 0 1 [0 0 0] 1
       even        even    even          odd

This should be pattern of marks
B 0 B [B 0 V] B B [B O V] [B 0 V] 0 0 B [0 0 V] B
       even        even    even          odd

The AMI with B3ZS should be...
+ 0 - [+ 0 +] - + [- 0 -] [+ 0 +] 0 0 - [0 0 -] +

— Preceding unsigned comment added by 1337CyberBob (talk • contribs) 19:49, 20 October 2013 (UTC)Reply