What does it mean when the propagation delay is given in bits?

Question

Studying ring latency for Ring topology, there were few mentions of propagation delay in bits. While I have learned to convert the time given in bits to seconds, what does the statement means when it says "the propagation delay of a channel is 10 bits"?

I came to a conclusion that for a bit to propagate the whole channel and reach at the end of the channel there must be 10 bits in the channel. Or in short, the capacity of the channel is 10 bits. 

Am I right? Please advice.

Answer 1

As you must know, the propagation delay (measured in time units) doesn't depend on the channel bandwidth and is a characteristic of the transmission media used. Given by t_p= l./v , where "l" is the length of the cable and "v" is the velocity of signal in that media.

Coming to your question, if the propagation delay is given in bits, then not only the media but also the bandwidth or data rate of the channel has a role to play.

The propagation delay in bits is the amount of data(bits) transmitted by the sender (A) in time t_p. You can observe that in this time, the first bit reaches the receiver and simultaneously the last bit is transmitted by the sender. So in other words you can say propagation delay in bits is the no. of bits required to fill up the channel. P.D.(bits)=t_p(s)×bw(bps)

So saying it is the capacity of the channel maybe misleading to some extent as we can have same propagational delay in bits for different pairs of channel bandwidth and media. Rather you can say it is the combined capacity of the channel+media.

But again in the specific cases such as Ethernet or Token ring where the media is more or less fixed (10base5 etc) you CAN say that it is the capacity of the channel only.

Hope this does the job. :)

Comments

Thanks :)