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From: | Ron Economos |
Subject: | Re: [Discuss-gnuradio] Power Line Communications - GNURadio |
Date: | Sun, 03 Apr 2016 07:05:51 -0700 |
User-agent: | Mozilla/5.0 (X11; Linux i686; rv:24.0) Gecko/20100101 Thunderbird/24.6.0 |
Carrier spacing = sample rate / IFFT size = 250,000 / 512 = 488.28125 HzOFDM symbol time = IFFT size / sample rate = 512 / 250,000 = 2.048 milliseconds.
However, only 97 carriers out of the possible 512 are used. 96 carry data and one is used as a pilot.
Bandwidth = active carriers * carrier spacing = 97 * 488.28125 Hz = 47363.28125 Hz. The signal occupies a little over 47 kHz from 42 to 89 kHz.
There is also a guard interval of 48 samples. So an OFDM symbol takes 2.048 + 0.192 = 2.24 milliseconds to transmit.
So you get to send 96 DBPSK, QQPSK or D8PSK symbols (1 per carrier) for each OFDM symbol every 2.24 milliseconds.
This gives an uncoded throughput of 1 / 0.00224 * 96 * 1 = 42847 bps for DBPSK
This gives an uncoded throughput of 1 / 0.00224 * 96 * 2 = 85714 bps for DQPSK
This gives an uncoded throughput of 1 / 0.00224 * 96 * 3 = 128571 bps for D8PSK
Ron On 04/03/2016 05:00 AM, Paul Creaser wrote:
With regards to OFDM I have one question. I have a 8 channel OFDM using 8PSK. To me a single symbol using 8PSK means 8 bits, hence one byte of data. Since I have 8 channels, that would mean 64 bytes. This would equate to a single OFDM symbol of 64 bytes? Going on further a payload of 16 OFDM symbols would mean 16*64 bytes of data. I'm not too sure of the relationship between bits, symbols and OFDM symbols even though it appears to be simple.
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