shannon capacity qpsk
I know from my studies that QPSK transmit 2 bits per symbol wheras FSK can transmit 1 bit per symbol. This means that QPSK can send the same amount of bits as FSK using half as much bandwidth that FSK uses or QPSK can send twice as many bits as FSK using the same amount of BW that FSK uses.
Using the Channel Capacity formula derived by Shannon:
[BW]*(LOGbase2(1+S/N))
along with Eb/No charts for FSK and QPSK (see attached files for more insight)t o find my S/N for a given BER.
and an BW design of 1MHZ
my channel capacity for FSK is:
3.56 Mbps @ 10^-3 BER
3.71 Mbps @ 10^-4 BER
3.82 Mbps @ 10^-5 BER
3.89 Mbps @ 10^-6 BER
my channel capacity for QPSK is:
3.36 Mbps @ 10^-3 BER
3.63 Mbps @ 10^-4 BER
3.75 Mbps @ 10^-5 BER
3.84 Mbps @ 10^-6 BER
I was expecting QPSK channel capacity to be more than FSK's channel capacity. Is this a valid assumption to think that QPSK will have greater channel capacity than FSK? What am I doing wrong?
Well FSK has a much larger S/N ratio than QPSK.
So for the same signal rate QPSK needs a smaller bandwidth but has a much larger error probability, and a much smaller S/N ratio.
And your results seam fine.
p.s I'm browsing this with lynx and I haven't seen the images.
first Same MEAN POWER (1/2 of ones's and 1/2 of zeros'), Amplitude equal for all waveforms (ASK, FSK, PSK, etc)
then compare things:
ASK detector have one branch. Noise in only one branch. Mean Power
FSK detector have two branches. Noise in two branches. 2xMean Power
PSK detector only one branch. Noise in only one branch. 2xMean Power
QPSK(four points constellation, (I,Q)) detector have two branches. 2xMean Power
The probability of error is M-ary for QPSK. At first glance seems that FSK and QPSK have the same SNR. But the probability of error is what matters. The way of calculation is different. 1point have 3 other points for error decision.
The capacity of channel, Shannon law = B*log(1+SNR).
Search in a Book like COUCH or HAYKIN.
I hope that helped YOU.