megabits per second, information transfer equation? - Page 1
Forum section: electronics / beginners

author: CopperCone
author_rank: Super Contributor
timestamp: August 28, 2017, 07:01:14 pm
content: Very simple question, I think I am right butIf you have some kind of uplink capable of some speed measured in unit-bits per second,the equation is unit-links * time to see how many bits of information can be transferred right?So, if there are transmitters capable of 100MBPS, and the uplink is up for 400uS,it would be 100,000,000*0.0004, or 40,000 bits?

author: IanB
author_rank: Super Contributor
timestamp: August 28, 2017, 07:03:49 pm
content: In theory, yes. In practice, protocol, contention and timing factors will result in less than that (sometimes much less).For instance, compare the (theoretical) speed of USB 3.0 to what is actually achieved in practice.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: August 28, 2017, 07:38:00 pm
content: Also, make sure you follow the proper abbreviations for units. bits are written as 'b', bytes are written as 'B'. "100MBPS" means 100 megabytes per second. If you mean bits, you need to use a lower case b or spell it out. The "PS" really shouldn't be capitalized either, as that leads to confusion...did the person really mean megabytes, or was their caps lock stuck on? Same goes for "mbps".

author: CopperCone
author_rank: Super Contributor
timestamp: August 28, 2017, 08:38:05 pm
content: i meant bits.is there a table showing overhead for various protocols? like how many bits of information per packet?

author: IanB
author_rank: Super Contributor
timestamp: August 28, 2017, 08:51:51 pm
content: Quote from: CopperCone on August 28, 2017, 08:38:05 pmI don't think it's that simple. The "speed" defined by the link is the fastest electrical speed that the link is capable of. However, with fast links there are many system and environmental factors that affect the actual communication rate achieved. Not least of which is that the sending and receiving devices may not be able to keep up. So I think this is a "how long is a piece of string" question.

author: Kalvin
author_rank: Super Contributor
timestamp: August 28, 2017, 08:59:39 pm
content: There are also coding schemes using multiple amplitude levels for the symbols, each symbol containing a number of bits. In that case you will be talking about symbol rate in the cable and the bit rate = (number of bits per symbol) * symbol rate.

author: electrolust
author_rank: Unknown
timestamp: August 28, 2017, 09:15:07 pm
content: Quote from: CopperCone on August 28, 2017, 07:01:14 pmMinus all your typos, yes you are correct. And your typos are in fact critical errors, not easily overlooked typos. If you want to do this stuff, you are going to have to be a lot more precise in your note taking and what not.Two minor points:1. Protocol overhead. The data rate of a line is given in raw data. There will be protocol overhead on top of that, as well as bus contention for many types of links. So the calculation you have offered is the absolute maximum, theoretical rate, which can never be achieved.2. The data rate is expressed in powers of ten, as you have done. But memory is generally organized around powers of 2. Just something to watch out for.