I understand that the bit error performance of BPSK and QPSK are the same. Does this mean that if I fix the transmitter power (eg. 1W), at the receiver I would be able to have the same performance, no matter whether I am using BPSK or QPSK.
This is counter-intuitive, as QPSK uses half the bandwidth, and it looks like I’m would always be better off using a QPSK rather than BPSK.
@Anthony: Well, as you know, the symbol error rate(SER) vs Es/No curve of QPSK is worse than BPSK by around 3dB. QPSK and BPSK have identical BER vs Eb/No curves.
So, if we consider the Tx power as a constraint (Es/No), we can see that BPSK will be better off than QPSK by 3dB (and rightly so, since the decision regions became smaller – one quadrant for QPSK instead of 2 quadrants for BPSK).
The bandwidth argument might not be valid for the following reason:
If we consider passband transmission, minimum required bandwidth for BPSK is only fc to fc+1/2T Hz (we can shave of half the B/W as the spectrum is symmetric). For QPSK we need to send fc-1/2T Hz to fc+1/2T Hz.
Hope this helps.
Thank you for your quick response.
I’m confused about the last part of your answer. I thought in a typical BPSK receiver, I need to use fc-1/2T to fc+1/2T to transmit also. At least, that’s what I was taught in school…
So, if the bandwidth is halved, the noise in the passband is also reduced by 3dB (assuming AWGN). The sensitivity of the receiver is also reduced by 3dB. This will negate the 3dB that Es/No that BPSK have over QPSK.
@Anthony: Yes, bandwidth of BPSK is from fc-1/2T to fc+1/2T. However, since BPSK is a real signal, the spectrum is symmetric around fc. So, theoretically we need to send only half the bandwidth to enable reliable demodulation of the information.
Well, regarding the noise removal with halfband width, may I try to argue as follows:
even if we pass the full bandwidth, we will be ignoring the imaginary part of the noise for BPSK demodulation. Hence the performance should be same irrespective of full-half bandwidth.
Does that makes sense?
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