# What is the BER for multilevel modulated signal

Assume I want to transmit a low or high symbol which can be BPSK modulated. For ordinary BPSK, bit 0 can be mapped to 1+i0 and bit 1 mapped to -1+i0. In my question, each two bits are BPSK modulated and a level dependent so bit 01 is mapped to 2+0i and bits 00 is mapped to 1+0i. Also bits 10 is mapped to -1+0i and bits 11 is mapped to -2+0i. This is BPSK (not QPSK) signal since there are BPSK demodulator and level detector at the receiver in which each of them output one bit at symbol period.I want to know what demodulation scheme to use in order to detect the symbol level and hence calculating the BER.

As Dan Boschen has already pointed out, the question has very little to do with BPSK which acronym stands for Binary Phase-Shift Keying; this is a 4-ary system not a binary one, and as stated has very little to do with phase-shift keying either.

This is a baseband 4-ary communication system that uses pulse modulation. At the output of the demodulator, we observe one of $$4$$ signal levels ($$+2, +1, -1,-2$$) corrupted by noise which we will assume to be Gaussian with mean $$0$$ and variance $$\sigma^2$$. If $$b_0b_1$$ denotes the two bits being transmitted, then the conditional distribution of $$X$$, the demodulator output is given by \begin{align} X \sim N(+2, \sigma^2) &~~\text{if}~~ b_0b_1 = 01,\\ X \sim N(+1, \sigma^2) &~~\text{if}~~ b_0b_1 = 00,\\ X \sim N(-1, \sigma^2) &~~\text{if}~~ b_0b_1 = 10,\\ X \sim N(-2, \sigma^2) &~~\text{if}~~ b_0b_1 = 11. \end{align} The demodulator output is mapped to the nearest signal level and the decoded bits $$\hat{b}_0\hat{b}_1$$ are those corresponding to this signal level. That is, \begin{align}\text{If}~ X\in (1.5,\infty), &~\hat{b}_0\hat{b}_1 = 01,\\ \text{If}~ X\in (0, 1.5], &~\hat{b}_0\hat{b}_1 = 00,\\ \text{If}~ X\in (-1.5,0], &~\hat{b}_0\hat{b}_1 = 10,\\ \text{If}~ X\in (-\infty, -1.5], &~\hat{b}_0\hat{b}_1 = 11,\end{align} With $$P_e$$ meaning probability of bit error, we have that $$P_e(\hat{b}_0\neq b_0\mid b_0b_1 = 01) = P(N(+2, \sigma^2) < 0) = Q\left(\frac 2\sigma\right),\\ P_e(\hat{b}_0\neq b_0\mid b_0b_1 = 00) = P(N(+1, \sigma^2) < 0) = Q\left(\frac 1\sigma\right),\\ P_e(\hat{b}_0\neq b_0\mid b_0b_1 = 10) = P(N(-1, \sigma^2) > 0) = Q\left(\frac 1\sigma\right),\\ P_e(\hat{b}_0\neq b_0\mid b_0b_1 = 11) = P(N(-2, \sigma^2) > 0) = Q\left(\frac 2\sigma\right).$$ The average bit error probability for $$b_0$$ is thus $$P_e(\hat{b}_0\neq b_0) = \left.\left. \frac 12 \right[\displaystyle Q\left(\frac 1\sigma\right) + Q\left(\frac 2\sigma\right)\right].\tag{1}$$ On the other hand, the calculations for $$P_e(\hat{b_1}\neq b_1)$$ are a lot messier. We have that \begin{align}P_e(\hat{b}_1\neq b_1\mid b_0b_1 = 01) &= P(-1.5 < N(+2, \sigma^2) < 1.5)\\ &= Q\left(\frac{0.5}{\sigma}\right) - Q\left(\frac{3.5}{\sigma}\right),\\ P_e(\hat{b}_1\neq b_1\mid b_0b_1 = 00) &= P(N(+1, \sigma^2) < -1.5) + P(N(+1, \sigma^2) > 1.5)\\ &= Q\left(\frac{0.5}{\sigma}\right) + Q\left(\frac{2.5}{\sigma}\right),\\ P_e(\hat{b}_1\neq b_1\mid b_0b_1 = 10) &= P(N(-1, \sigma^2) < -1.5) + P(N(-1, \sigma^2) > 1.5)\\ &= Q\left(\frac{0.5}{\sigma}\right) + Q\left(\frac{2.5}{\sigma}\right),\\ P_e(\hat{b}_1\neq b_1\mid b_0b_1 = 11) &= P(-1.5 < N(-2, \sigma^2) < 1.5)\\ &= Q\left(\frac{0.5}{\sigma}\right) - Q\left(\frac{3.5}{\sigma}\right).\end{align} The average bit error probability for $$b_1$$ is thus $$P_e(\hat{b}_1\neq b_1) = Q\left(\frac{0.5}{\sigma}\right) + \left.\left. \frac 12 \right[\displaystyle Q\left(\frac{2.5}{\sigma}\right) - Q\left(\frac{3.5}{\sigma}\right)\right].\tag{2}$$ Thus the two bits being transmitted have significantly different average bit error probabilities with $$P_e(\hat{b}_1\neq b_1) >> P_e(\hat{b}_0\neq b_0)$$ for typical values of $$\sigma$$. $$Q(x)$$ is a very rapidly decreasing function of its argument and the dominant terms $$\displaystyle Q\left(\frac{0.5}{\sigma}\right)$$ and $$\displaystyle \frac 12 Q\left(\frac 1\sigma\right)$$ in $$(2)$$ and $$(1)$$ respectively can differ by an order of magnitude or more.

This is not BPSK! BPSK is binary phase shift keying with binary meaning 2 levels. What the OP is describing is Pulse Amplitude Modulation where multiple bits are mapped to multiple levels. In this case there are 4 levels so it would be 4-PAM. (As Dilip points out in the comments that the levels in 4-PAM would typically be equally spaced while here the inner levels are spaced further than the outer levels, complicating the BER computation slightly as detailed below but still PAM regardless).

The demodulation scheme would similar to BPSK demodulation in that you would multiply the received signal with the recovered carrier and low pass filter. Decision would be made on the output by setting decision thresholds half way between each of the levels (assuming equiprobable data) and then the derivation for BER follows similar to BPSK (solving for the tail probabilities under Gaussian distributions if the noise is assumed to be AWGN as is typically assumed, noting the different tail contributions involved since the levels aren't equally distributed as Dilip has pointed out - that said the thresholds at ±1.5 will likely dominate setting the error rate with significantly smaller error contribution from the threshold at 0). With equiprobable distribution of symbols and AWGN with standard deviation $$\sigma$$ this would result in the following Symbol Error Rate:

$$\text{SER} = \text{erfc}(0.5/(\sqrt{2}\sigma) + \frac{1}{2}\text{erfc}(1/(\sqrt{2}\sigma)$$

For smaller $$\sigma$$ values the second term would likely be insignificant. For larger $$\sigma$$ there would be additional tail contributions from the next adjacent threshold levels which would also significantly complicate the BER computation as detailed below. Thankfully under all practical applications these tails would certainly be insignificant and can be safely ignored.

To go from Symbol Error Rate to Bit Error Rate, consider how many bit errors occur when a symbol error occurs. Because the mapping done by the OP was properly Gray-coded (only one bit change between adjacent symbols), only one bit error would occur with each symbol error.

• Thanks Dan, but as i have written they are two separate blocks, one is Bpsk demodulator and one is level detector. – Riva11 May 11 '20 at 14:26
• @Riva11 If there are 4 amplitude levels in the signal, this is not a BPSK modulated signal. That said, a BPSK demodulator that multiplies the recieved signal with the carrier and low pass filters will still put out 4 levels, and to demodulate you would set the thresholds as I have indicated. Am I missing something else? – Dan Boschen May 11 '20 at 15:00
• What OP @Riva11 describes is not 4-PAM in the usual sense of the word. Standard 4-PAM would use equally-spaced levels $-3, -1, +1, +3$ instead of the $-2, -1, +1, +2$ used in the OP's system. Well, at least the dibits are assigned to the levels in Gray code order. Nonetheless, the first and second bits have different probabilities of error and so the BER calculation is messy too. – Dilip Sarwate May 11 '20 at 15:37
• @DilipSarwate Ah yes, good point! – Dan Boschen May 11 '20 at 16:01
• It is still called PAM not BPSK. PAM need not have equally spaced levels, but under AWGN would have the best BER if done as such. Here is an example of PAM with unequal levels to make up for non-linearity in the modulator. If the modulation is done by changing the amplitude of each pulse only, this is called PAM. – Dan Boschen May 12 '20 at 11:21