Seminar 1. Detectors and their errors

The correct answer is (c) ★

By Definition 1.3 of the lecture notes, the MAP detector chooses the symbol with the largest joint probability \Pr(x)W(y|x) for the observed output. For y=1 the two joint probabilities are \Pr(0)W(1|0)=\tfrac25\cdot\tfrac34=\tfrac{3}{10} and \Pr(1)W(1|1)=\tfrac35\cdot\tfrac12=\tfrac{3}{10}. They are equal, so there is a tie and the answer is (c). The likelihood favors 0 by a factor of \tfrac32, the prior favors 1 by the same factor, and the two effects compensate exactly. Either decision gives the same average error probability.

The correct answer is (a) ★

By Definition 1.3 of the lecture notes, the ML detector chooses the symbol with the largest W(y|x). For y=0 we have W(0|1)=\tfrac12>W(0|0)=\tfrac14, so it decides \hat x=1. For y=1 we have W(1|0)=\tfrac34>W(1|1)=\tfrac12, so it decides \hat x=0. The ML detector is not \hat x=y on this channel, because it decides the opposite bit. When x=0 is sent, the only wrong output is y=0, and by equation (1.6) the conditional error probability is P_{\rm e}(0)=W(0|0)=\tfrac14. The answer is (a). Option (b) is W(1|0), the error of the detector \hat x=y, which is not the ML detector here. Option (c) is the joint probability \Pr(0)W(0|0), which includes the prior, whereas the conditional error probability of a given detector does not depend on the input distribution.

The correct answer is (a) ★

By Definition 1.3 of the lecture notes, the MAP detector chooses the symbol with the largest \Pr(x)W(y|x). On the real line, the constellation is the three points -1, 0 and +1, and the observation y=\tfrac12 is the midpoint between 0 and +1. It is at the same distance \tfrac12 from 0 and from +1, so the channel law of Table 1.2 gives the same value for both, W(\tfrac12|0)=W(\tfrac12|{+1}). The prior of 0 is \tfrac12 and the prior of +1 is \tfrac14, so the product is twice as large for 0. The symbol -1 is farther, at distance \tfrac32, so it is less likely, and its prior is also \tfrac14. The MAP detector decides \hat x=0, whatever the noise variance, and the answer is (a). Option (c) is the outcome of the nearest-point detector, which ignores the prior and has its threshold exactly at y=\tfrac12.

The correct answer is (c) ★

The thresholds of the nearest-point detector of Proposition 1.5 are the midpoints -\tfrac12 and +\tfrac12, so the decision region of the symbol 0 is the interval -\tfrac12<y<\tfrac12. When 0 is sent the output is Y=Z, and an error occurs when Z<-\tfrac12 or when Z>\tfrac12. The two events are disjoint, so their probabilities add. By equation (1.23) of the lecture notes with the threshold \tfrac12 and \sqrt2\,\sigma=1, each of them is \tfrac12\erfc\bigl(\tfrac12\bigr). The sum of the two is P_{\rm e}(0)=\erfc\bigl(\tfrac12\bigr) and the answer is (c), as for the inner points of 4-PAM in Problem 1.12. Option (a) counts only one of the two sides. Option (b) uses the distance 1 to the neighbors instead of the distance \tfrac12 to the thresholds. The prior does not appear, because the nearest-point detector does not use it and the error probability is conditional.

The correct answer is (b) ★

Following Section 1.2 of the lecture notes, the energy of each symbol is \tfrac12+\tfrac12=1, because the real part and the imaginary part are both \pm\tfrac{\sqrt2}{2}. The average symbol energy is E_{\rm s}=1 and the signal-to-noise ratio is \mathrm{SNR}=E_{\rm s}/\sigma^2=1/\sigma^2. The answer is (b). Option (a) counts only one of the two components, and option (c) squares the sum \tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2} instead of adding the two squares.

The correct answer is (a) ★

The region of a symbol under the nearest-point detector is bounded by the perpendicular bisectors of the segments that join it to its neighbors, as explained in Section 1.5 of the lecture notes. For QPSK the bisectors are the real axis and the imaginary axis, so the four decision regions are the four quadrants, as in Figure 1.2(b) and Problem 1.13. The region of a symbol x is the quadrant where \re(y) has the sign of \re(x) and \im(y) has the sign of \im(x). In the complex plane, the constellation is the four corners of a square centered at the origin, one in each quadrant and all at distance 1 from the origin. An output with \re(y)>0 and \im(y)<0 is in the fourth quadrant, whose symbol is \tfrac{\sqrt2}{2}-j\tfrac{\sqrt2}{2}, so the detector decides \hat x=\tfrac{\sqrt2}{2}-j\tfrac{\sqrt2}{2}. The answer is (a). Option (b) has the two signs reversed, and it is the farthest symbol from this output. Option (c) is not correct, because the two signs determine the quadrant, whatever the magnitudes of \re(y) and \im(y).

The correct answer is (b) ★

By equation (1.19) of the lecture notes, the conditional error probability is the probability that the output falls outside the decision region of the symbol sent. The region of x=\tfrac{\sqrt2}{2}+j\tfrac{\sqrt2}{2} is the first quadrant, \re(y)>0 and \im(y)>0. The output is outside it when at least one of the two conditions fails, that is, when \re(y)<0 or \im(y)<0. The answer is (b). Option (a) requires the two conditions to fail together, which is only the probability of deciding the opposite symbol -\tfrac{\sqrt2}{2}-j\tfrac{\sqrt2}{2}. Option (c) places the thresholds at the symbol and not at the axes, and it gives \tfrac34 for every SNR. The value follows from the correct decision, as in Problem 1.13. The two parts of the noise are independent, and each part of the output stays positive with probability 1-\tfrac12\erfc\bigl(\tfrac{\sqrt2}{2\sigma}\bigr), so P_{\rm e}(x)=1-\bigl(1-\tfrac12\erfc\bigl(\tfrac{\sqrt2}{2\sigma}\bigr)\bigr)^2.