Understanding Binary Transmission Over AWGN Channels
School
Ege Üniversitesi**We aren't endorsed by this school
Course
ELECTRICAL EEE400
Subject
Electrical Engineering
Date
Dec 12, 2024
Pages
3
Uploaded by ProfessorMusicElephant38
Problem 1 (50 points) The signal depicted in Fig. 1 is to be used in conjunction with binary transmission over an AWGN channel with power spectral density Np/2. The signaling format is polar and bits are assumed to be equally likely. Ans(t) +A T Figure 1: Signal used in binary transmission. (a) Find the value of A that results in unit energy, i.e., £, = 1. 2 1 E,=4A’=1 - A=< 2 E,=4A=1 — A=~ 8— = o - —2. (b) Sketch carefully the output v(t) of the matched filter for s(¢), when the input is s(f). For what value of t is the output maximum? What is the maximum value of v(f) equal to? y 2T I : > T |1 |2 3 B -1/2 4= t=1 A S(l“t) t=3 ‘;5(3‘1) v(3)=-1/4 2 ' | } - T l > T t=2 A s(2-1) t=4: p=ie v(2)=o v(4)=+l ‘ S | I l1 l2 I3 l4 ’ B Figure 2: Graphical computation of the convolution. The maximum output occurs, as expected, at t = T = 2. The maximum value equals the energy of the signal s(t), v(2) = E, = 1.
v(t) Figure 3: Output of matched filter. (¢) Find the probability of a bit error P, with an optimum receiver. Express your answer in terms of the Q-function and Nj. 2F 2 P —Q(\/m) ‘Q(\/T\‘G)' (d) Repeat part (c¢), assuming now that the output of the matched filter is erroneously sampled at t = 3. Estimate the loss (in dB) compared to part (¢) to achieve the same value of P.. At t = 3, the output of the matched filter is v(3/2) = —1/4. Since the sign of the output is negative, the decisions will be erroneous with large probability. To estimate F,, assume that +s(f) is transmitted. Then the output of the matched filter, V', is a Gaussian r.v. of mean -1/4 (instead of +1 with correct sampling) and variance: 9 B 4'\r0 - 1 - 2 20 (2 g == YT /0 (2) 8No The probability of error is the probability that Y < 0 which is given by 11 0+1/4 1 By i —). 7373 Q( oy ) Q( GNO) There are two bounds for P.. If Ny =~ 0 (noise tends to zero) then P, =~ 1. On the other hand, for large values of Ny, P. =~ 1/2. In other words, 1 -<Pe<1- 2 -— . and the probability of a bit error does not go to zero, regardless of the amount of signal power. Consequently, the loss, compared to the case with properly sampled matched filter output, is infinite! 3. Data is transmitted over a baseband AWGN channel with Ng = 1 x 107! W/Hz. The pulse shape p(t) shown in the figure below is employed and the signaling scheme is binary polar, that is, s(t) = {+p(t). bft = 1; —p(t), bit = 0.
p(t) » = 1 2 -A = ——m——————— (a) Draw a block diagram of the optimum receiver. Solution: Matched Decision filter device From h(t)=p(2-t) z:. I Estimated channel I bit t=2 f Threshold (b) Find the value of the amplitude A that gives a bit error probability Ple] = 2.3 x 10—, Solution: The energy of the signal is E, = 2A42. Therefore, oF. 142 » P[G]—Q( !VO)—Q( W —23)(10 " Table 1 (last page) yields, 4A2 10-11 from which A = 5.53 uV. Type B: A =6.71 uV. 3.5,