Using the map decoding algorithm I got the result about BER vs Eb/NO. From this how can I calculate throughput? Is there any relationship between throughput and BER?
Throughput strictly depends on the source. If source doesn't fill all channel capacity you've not the maximum efficiency of the link. If you want to evaluate the throughput of correct received bits, under BER constraint, you can simply use:
You'd have to define things rather more tightly before you could get a useful formula.
eg the data getting through is D_sent x (1-BER)
(BER fractional rate eg 1 in 10^6 = 0.000001).
BUT ANY error either needs a resend or has forward correction overhead.
If any error occur at all then you need a resend and
throughput drops to D_sent - (packets with error(s) in them per second = pweitps) x (packet data content length )
And you also have the back channel overhead which may be part of your total bandwidth budget or may not.
To convert BER to pweitps you either assume every single error destroys a packet (as above), or you need to know statistical grouping of errors.
If each error bit loses one packet and if net error free data rate is D and if packet length is P then you lose BER x P bits/second due to errors so rate simplistically reduces to ~= (D - BER x P)/ D of previously.
he key challenges in bit-rate selection are determining which bit-rate provides the most throughput and knowing when to switch to another bit-rate that would provide more throughput. This thesis presents the SampleRate bit-rate selection algorithm. SampleRate sends most data packets at the bit-rate it believes will provide the highest throughput. SampleRate periodically sends a data packet at some other bit-rate in order to update a record of that bit-rate's loss rate. SampleRate switches to a different bit-rate if the throughput estimate based on the other bit-rate's recorded loss rate is higher than the current bit-rate's throughput. Measuring the loss rate of each supported bit-rate would be inefficient because sending packets at lower bit-rates could waste transmission time, and because successive unicast losses are time-consuming for bit-rates that do not work