This question is regarding digital modulation.
If 64 QAM always provides rates greater than BPSK or 16 QAM, then why do systems always use higher order modulation of QAM?
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$\begingroup$ Do you not want to achieve the greater rates promised by higher-order modulations? Conversely, if you are perfectly satisfied with the rate provided by BPSK on a channel, consider that using 64 QAM will allow you to get the same data rate on a channel with smaller bandwidth (which may save you, or your boss or your boss's boss, some money, and might even give you a pay raise). $\endgroup$– Dilip SarwateCommented May 17, 2015 at 16:12
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1 Answer
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There is a trade-off between modulation order, required energy and bit-error rate (BER). As you increase the modulation order, you need to increase the average energy per bit to keep the same BER.
As an easy example, consider BPSK vs 4-PAM. Say you have BPSK with distance between constellation points $d_{min}=1$; this works out to an average bit energy of 0.25 joules per bit.
Now, you want to double the bit rate by going to 4-PAM, but you want to keep roughly the same BER, which means having the same minimum distance $d_{min}$ between constellation neighbors. Now, your average symbol energy is $$\overline{E_s}=\frac{2\cdot1.5^2+2\cdot0.5^2}{4}=1.25$$ joules, or an average bit energy $\overline{E_b}=0.625$.
As you increase the modulation order, you'll see the energy required increase exponentially. Exactly the same thing happens when using QAM.
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$\begingroup$ Thanks, given the fact that higher order modulations have higher number of symbols, does this complicate the receiver side and might increase the probability of error in detection? Because points are closer to each other? $\endgroup$– TyroneCommented May 17, 2015 at 15:21
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$\begingroup$ Real-world transmission systems are noisy. The closer the points are to each other, the more likely it is that noise will corrupt the signal. This can only be avoided by increasing the power to increase the signal-to-noise ratio. $\endgroup$– Simon BCommented May 18, 2015 at 13:29
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$\begingroup$ @user253751 I don't understand your comment; could you clarify what your question is? $\endgroup$– MBazCommented Jul 5, 2020 at 1:03
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$\begingroup$ @user253751 Well, obviously that kind of energy per bit is not used in practice, especially in wireless systems. $\endgroup$– MBazCommented Jul 6, 2020 at 13:10