I came across this image in my notes and I fail to understand why BER seems to decrease as the modulation order increases and I was hoping for an answer

enter image description here

  • 6
    \$\begingroup\$ I think you're reading the graph incorrectly - the lower figure is better. So for example at 10dB BPSK is better than 16-PSK. \$\endgroup\$ – Kevin White Nov 20 '19 at 0:26

Put simply, for the same average energy per bit, the modulation scheme with more bits per symbol must have less distance between the points on its constellation diagram. Therefore a smaller noise excursion can produce a bit error.

| improve this answer | |

That is correct.

The channel with higher bit rate will suffer with a worse BER for the same SNR after converting CNR to SNR by demodulation.

Another way to understand this chart is the rise in SNR or Eb/No needed to keep the same BER with higher bandwidth from Shannon-Hartley Theorem. enter image description here

With more compression or bits per baud using more phases per bit, there is a need to increase the energy per bit to noise by a log of this bit ratio.

for \$ BER = 10^{-6}\$
2-PSK 10.5 dB (BPSK)
4-PSK 14.0 dB
8-PSK 18.5 dB

It appears to be ~ 1dB rise per bit/baud above the baseline for binary PSK at this BER threshold.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.