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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

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    \$\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 at 0:26
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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.

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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.

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