I came across the following solved question, but I have trouble understanding part of the solution. If anyone could shed some light into this I would be really grateful since this problem has been bugging me for a while.
We have the following power delay profile for a broadband bus:
relative delay (μs) || mean relative Power (dB)
0.0 || -1.0
0.5 || 0.0
1.0 || -3.0
1.5 || -6.0
2.5 || -7.0
4.5 || -11.0
There are a few questions such as: What's the mean excess delay (answer Tmean=0.79μs), RMS delay spread (answer Td=0.925μs) and the max excess delay, with Treff=-10dB (answer Tmax=2.5μs).
Up to this point I understand everything pretty well. But the final question is bugging me: "Can you tell that the bus is broadband, for a system that transmits with rate 25kbps and why?". The answer to this is:
The bus is narrow band, because Ts >> Td
I don't get it, at all. To begin with, how did he calculate Ts?
Best Answer
You would need to check this answer against the text since I see no definition of Ts in the question.
However, on the assumption that by Ts is meant symbol time, he is distinguishing between broadband and narrowband by comparing the symbol time with the delay. This is reasonable : if Ts >> Td, BW(s) << BW(channel) therefore this is a narrowband channel.
As for the question how did he calculate Ts? Trivially, by defining the datarate as 25kbps. Assuming a 2 level code (on/off, 1 bit per baud) Ts=1/25000, or 40 us.