You miss bits out, simple. Yes that means duplicate tags will app ere but if you say miss out every other code you will not see so many duplicates in practice. Anyway are you sure it is a 34 bit output.
You could always ask the seller on eBay what sort of tags to use and where to get them from.
Great to hear from you. So what you are saying is that the Wiegand output will simply be the last 26 bits of the code. In addition by doing this there is a greater likelihood of duplicate tags. If this is the case then do I simply treat the first and last bits as parity bits?
The reader can be set to either Wiegand 26 or 34.
I have asked the seller about tags. I have two options;
So what you are saying is that the Wiegand output will simply be the last 26 bits of the code.
No it is more likely that they will take selected bits so you don't get a run of duplicates. Like the first 20 and the last 6.
While it looks on the face of it that you will get 50% duplicates if you remove just one bit this only applies to a system with all the codes recognised or enrolled. This is impossible and so with a limited number of active codes in the system the probability of a duplicate code being seen is very low.
If this is the case then do I simply treat the first and last bits as parity bits?
That depends entirely on the reader, it could be but it doesn't have to be.
OK, so the 26 bits will be selected bits from the 64 bit ISO 18000-6C tag ID.
I didn't quite follow your second point. Do you mean that because I will be using only 26 bits (roughly half of the original bits) I therefore increase the likelihood of duplicates? Is that something to do with possibilities?