# Multi-path propagation

1. Feb 12, 2009

### raymond23

1. The problem statement, all variables and given/known data

Consider the multi-path propagation phenomenon discussed in class. Assume the signal from a sender takes 4 paths to arrive at the receiver, and the delay along each path is 3, 6, 9, 21 (in micro-seconds), respectively. Each symbol is 1 bit long. Two symbols can be successfully received/detected at the receiver if their received impulses are at least 1 micro-second apart. What is the maximum allowed transmission rate from the sender to the receiver?

2. Relevant equations

delay along each path is 3, 6, 9, 21 (in micro-seconds)

3. The attempt at a solution

no idea

2. Feb 12, 2009

### Delphi51

You could send at the maximum 1 bit per microsecond . . . for 3 microseconds. Then you are going to get interference from the path with delay 6. How long must you wait before things quiet down?

3. Feb 13, 2009

### raymond23

3+6=9 microseconds?? is it?

4. Feb 13, 2009

### Delphi51

The 2nd path message arrives 3 microseconds later.
You only get that long before you have to stop and wait.

5. Feb 16, 2009

### raymond23

there are four paths with delay 3, 6, 9, 21 (in micro-seconds)
for the first path require 3
first to second path require 3
second to third require 3
third to forth require 21-9=12
therefore 3+3+3+12 in total is it?
maximum allowed transmission rate from the sender to the receiver is 21?

6. Feb 16, 2009

### Delphi51

I don't understand your total.
I should say I don't know much about this - just interested!
I am thinking like this:
SEND time 0 to 2: Receive on path 1 from time 3 to 5 -> 3 bits delivered
Interference 6 to 8, 9 to 11 and 21 to 23

SEND time 9 to 11: Receive on path 1 from 12 to 14 - 3 bits delivered
Interference 15 to 17, 18 to 20, 30 to 32

So far 6 bits in 14 us. If you kept up the chart for a while, say 60 us, you would have a pretty good idea of the delivered bit rate.