What is the Maximum Transmission Rate for Multi-Path Propagation?

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Homework Help Overview

The discussion revolves around the maximum transmission rate for a signal experiencing multi-path propagation, where the signal takes multiple paths to reach the receiver, each with different delays. The problem involves understanding how these delays affect the ability to transmit bits without interference.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the implications of the delays on transmission rates, with some suggesting a maximum of 1 bit per microsecond based on the shortest delay. Others question how to account for interference from the different paths and the timing of received signals.

Discussion Status

The discussion is ongoing, with various interpretations of the timing and interference being explored. Some participants have attempted calculations based on the delays, while others express uncertainty about the total and the implications for transmission rates.

Contextual Notes

Participants note the requirement that received impulses must be at least 1 micro-second apart to avoid interference. There is also a mention of the original poster's lack of clarity on the problem, indicating potential gaps in understanding the multi-path propagation phenomenon.

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Homework Statement




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?

Homework Equations



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


The Attempt at a Solution



no idea
 
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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+6=9 microseconds?? is it?
 
The 2nd path message arrives 3 microseconds later.
You only get that long before you have to stop and wait.
 
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?
 
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.
 

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