5 Problems with Solution of Digital Communication - Exam 2 | EE 7620, Exams of Digital Communication Systems

Download 5 Problems with Solution of Digital Communication - Exam 2 | EE 7620 and more Exams Digital Communication Systems in PDF only on Docsity! EE 7620 Solutions to Exam 2 Nov. 10, 2008 Problem 1 1. An O.N. spanning set for the signal set is given by φ1(t) = √ 2 T cos(2π f0t), and φ2(t) = √ 2 T sin(2π f0t), 0≤ t ≤ T Then s0(t) = √ Eφ1(t) and thus s0 = ( √ E,0). Also sincd s1(t) = √ E cos(θ)φ1(t)− √ E sin(θ)φ2(t) we have s1 = ( √ E cos(θ), √ E sin(θ). A block diagram or the optimal receiver is shown below. r ( t ) t=T t=T Compare r 1 r 2 + _ Z ̂ 1 φ (t-T) 2 φ (t-T) cos( ) θ sin( ) θ Note that we have ignored the term √ E in the multipliers. 2. We have P(E) = Q ( d√ 2N0 ) , where d2 = ||s1− s1||2 = [ √ E− √ E cos(θ)]2 +[ √ E sin(θ)]2 = 2E[1− cos(θ)] To minimize P(E) we need to maximize d. From the above we see that d is maximized for θ = π which results in d24E. For this value of θ the signal set is BPSK. We need EbN0 ≥ 9.6dB and W ≥ Rb. Problem 2 1. EbN0 = P RN0 = 150,0005,000 = 30. Thus 10log Eb N0 = 14.7 dB. Thus the available SNR is 14.77 dB. The available R/W = 5,0002α = 5/8. Our system must require Eb N0 ≤ 14.7 dB and must have Rb/W ≥ 5/8. The simplest system with these requirements is BPSK. s0(t) = √ 2E T cos(2π f0t), s1(t) = √ 2E T cos(2π f0t +π), 0≤ t ≤ T where E = Eb = 30N0 and T = Tb = 1/R = .2 msec. 1 2. EbN0 = P RN0 = 1,250,0005,000 = 250. Thus 10log Eb N0 ≈ 24 dB. Thus the available SNR is 24 dB. The available R/W = 5,0002α = 4.54. Our system must require Eb N0 ≤ 24 dB and must have Rb/W ≥ 4.54. The simplest system with these requirements is 32PSK. si(t) = √ 2E T cos ( 2π f0t + 2π(i−1) 32 ) , i = 1,2, · · · ,32, 0≤ t ≤ T where E = Eb log2 32 = 5Eb = 5× 250N0 = 1250N0 and T = Tb log2 32 = 5Tb = 5/R = 1 msec. 3. EbN0 = P RN0 = 320,0005,000 = 64. Thus 10log Eb N0 ≈ 18.06 dB. Thus the available SNR is 18.06 dB. The available R/W = 5,0002α = 2.94. Our system must require Eb N0 ≤ 18.06 dB and must have Rb/W ≥ 2.94. The simplest system with these requirements is 8ASK. si(t) = ai √ 2 T cos(2π f0t), i = 1,2, · · · ,8, 0≤ t ≤ T where ai ∈ { ±7A 2 ,±5A 2 ,±3A 2 ,±A 2 } With this selection Eave = 1 8 ×2× [ 49A2 4 + 25A2 4 + 9A2 4 + A2 4 ] = 21A2 4 Now Eave = Eb log2 8 = 3Eb = 192N0. Thus A √ 4Eave 21 = 16 √ N0 7 Also T = Tb log2 8 = 3Tb = 3/R = .6 msec. Problem 3 1. We have W = 2.25−2 = .25 MHz. Therefore, Rb/W = 4 bits/sec/Hz. The system we choose must have Rb/W ≥ 4. (a) We can use q-ary PSK for q≥ 16. 16PSK has the smallest power requirement for which Eb N0 = 17.4 dB or 55. Thus Eb N0 = P RbN0 = 55, =⇒ P = 55×Rb×N0 = 110 W (b) No FSK system can provide Rb/W ≥ 4. (c) We can use q-ary ASK for q ≥ 16. 16ASK has the smallest power requirement for which EbN0 = 23.2 dB or 207. Thus P = 207×Rb×N0 = 414 W. 2 somes 717470,) binant ft a Bers ola wen yen aleuate y -— elk-$ I, > cheose | sles. [> me aitt) 2 inh 5. Using the union bound we get P(E)≤ 1 8 8 ∑ i=1 ∑ j 6=i Q ( di j√ 2N0 ) ≤ 7Q ( dmin√ 2N0 ) = 7Q   √ a2 N0   since dmin = a √ 2. Problem 5 1. Let y(t) denote the output of the filter. Then R = y(T ) and y(t) = r(t)∗h(t) = ∫ t 0 r(τ)h(t− τ) dτ = ∫ t 0 s(τ)e−a(t−τ) dτ+ ∫ t 0 NW (τ)e−a(t−τ) dτ Then R = y(T ) = e−aT ∫ T 0 s(τ)eaτ dτ+N = s+N where N = e−aT ∫ T 0 NW (τ)eaτ) dτ For s0(t) we have s = 0 and for s1(t) we get s = 1a √ E T (1− e−aT ). N is Gaussian and we need to find its mean and variance. Now E(N) = 0 and E[N2] = e−2aT ∫ T 0 ∫ T 0 E[NW (τ1)NW (τ2)]eaτ1)eaτ2) dτ1dτ2 = e−2aT ∫ T 0 ∫ T 0 N0/2δ(τ1− τ2)eaτ1eaτ2) dτ1dτ2 = e−2aT N0/2 ∫ T 0 e2aτ1 dτ1 = N0 4a [ 1− e−2aT ] Therefore σ2N = N0 4a [ 1− e−2aT ]. Now P(E) = Q ( d 2σN ) . Thus P(E) = Q [ √ E(1− e−aT )√ aT N0 (1− e−2aT ) ] 2. We need to maximize the argument of the Q function above. Let f (a) = √ E(1− e−aT )√ aT N0 (1− e−2aT ) 5 Taking the derivative of f (a) with respect to a we see that a = 0 sets the derivative to zero. For a = 0 h(t) = { 1 0≤ t 0 otherwise. But since we sample at t = T , this is the same as h(t) = { 1 0≤ t ≤ T 0 otherwise. which is the matched filter. No surprize. 3. For the optimal matched filter receiver we have P(E) = Q (√ ET 2N0 ) 6