E&CE 358, Winter 2016: Solution #2. Prof. X. Shen

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1 E&CE 358, Winter 16: Solution # Prof. X. Shen xshen@bbcr.uwaterloo.ca

2 Prof. X. Shen E&CE 358, Winter 16 ( 1:3-:5 PM: Solution # Problem 1 Problem 1 The signal g(t = e t, t T is corrupted by additive white Gaussian noise with zero mean and power spectral density N / watts/hz. The optimum detector is a matched filter. (a Find and sketch the impulse response of the filter which is matched to the signal g(t. (b Derive an expression for the maximum output signal-to-noise ratio (SNR and express it in terms of the parameters, T and N. (c Consider the transmission of either s (t = g(t or s 1 (t = every T second interval, with equal probability, over a zero mean additive white Gaussian noise channel with power spectral density N / watts/hz. Sketch the pdf of the decision variable corresponding to s (t and s 1 (t sent, and derive the probability of error in making the decision as to whether s (t or s 1 (t was sent. ANSWER: g (t x(t t g t = e (, t T n(t N mn =, S n ( f = w / Hz a. Impulse response of the filter which is matched to the signal g(t is h(t = g(t t = { e (T t, t T, otherwise. g (t h(t e t e t b. f(t = g(t h(t = g(τh(t τdτ Problem 1 continued on next page... Page 1 of 9

3 Prof. X. Shen E&CE 358, Winter 16 ( 1:3-:5 PM: Solution # Problem 1 (continued g (t t T h( t τ τ τ f(t = t e (τ e [T (t τ] dτ = 1 [e (τ t (t+t e ] g (t T t T h( t τ τ τ f(t = t T For other values of t, f(t = e (τ e [T (t τ] dτ = 1 [e (t τ (3T e t] f (t E S t The average noise power at the output of the matched filter is E[N (t] = = N S n (fdf = where E S is Energy of the signal g(t. E s = G(f df = H(f df = N S n (f H(f df = g (tdt = G(f df = N E s N H(f df e t dt = 1 (1 e T Problem 1 continued on next page... Page of 9

4 Prof. X. Shen E&CE 358, Winter 16 ( 1:3-:5 PM: Solution # Problem 1 (continued n (t n h(t H ( f N S n ( f = S ( f = S ( f H ( n f n Therefore, the signal-to-noise ratio at the receiver output for t T is (SNR = f (t E[n (t] = Since f(t achieves its maximum value at t = T with f (t N E s f(t = 1 (1 e T = E s The (SNR,max achieves its maximum value at t = T, which is (SNR,max = f (T E[n (t] = c. Transmission system E s N = E s = 1 E s N N (1 e T = 1 e T N φ, S S i (t i =, 1 = g, t T 1 1, S =, t T 1 φ n(t h(t N R n ( τ = δ ( τ If S ϕ = g(t was sent, then the decision variable y(t will be i = S, i + n y, i y(t = S,ϕ (T + n (T = E s + n (T S t = T y(t = S DD * h( t n = n( t * h( t where n (T is a Gaussian R.V. with mean E[n (T ] = and variance E[n (t] = N E s Therefore, y(t is also Gaussian distributed with mean E[y(T ] = E s and variance So, E[(y(T E s ] = V ar[n (T ] = N E s y(t N(E s, N E s If S 1 (t = was sent, then the decision variable y(t will be y(t = S,1 (T + n (T = + n (T i Problem 1 continued on next page... Page 3 of 9

5 Prof. X. Shen E&CE 358, Winter 16 ( 1:3-:5 PM: Solution # Problem 1 (continued Therefore, y(t N(, N E s The conditional pdf of the decision variable y(t given that S ϕ (t = g(t was sent and the conditional pdf of the decision variable given that S 1 (t = was sent are shown below. Decision threshold: y(t = E s / Decision rule: If y(t E s /, then S ϕ (t = g(t was sent If y(t < E /, then S 1 (t = was sent Error event: y(t < E s /, given that S ϕ (t = g(t was sent y(t E s /, given that S 1 (t = was sent s 1 s φ ( y f y ( T S1 ( y f y ( T S E s Es So, the probability of error is P e = P (y(t < E s / S ϕ (t Pr(S ϕ (t sent + P (y(t E s / S 1 (t Pr(S 1 (t sent [ = 1 Es/ 1 exp [ (y E s ] ] ] 1 dy + exp [ y dy πes N E s N E s / πes N E s N = 1 1 E s N exp ( v dv + 1 exp ( u du π π Es N [ = 1 1 exp ( v dv + 1 ] exp ( u du π Es N π Es N [ ] = 1 Q( Es N + Q( Es N = Q( Es N Problem The purpose of a radar system is basically to detect the presence of a target, and to extract useful information about the target. Suppose that in such a system, hypothesis H is that there is no target present, so that the received signal x(t = w(t, where w(t is white Gaussian noise with power spectral density N /. For hypothesis H 1, a target is present, and x(t = w(t + s(t, where s(t is an echo produced by the target. Assumed that s(t is completely known. Problem continued on next page... Page 4 of 9

6 Prof. X. Shen E&CE 358, Winter 16 ( 1:3-:5 PM: Solution # Problem (continued (a Determine the structure of the optimal receiver. (b Evaluate the probability of false alarm defined as the probability that the receiver decides a target is present when it is not. (c Evaluate the probability of detection defined as the probability that the receiver decides a target is present when it it. (d Given that i the cost of a false alarm is the same as that of failing to detect a target when it is present, and ii the probability that there is a target is.1, determine the optimal decision threshold for the receiver. ANSWER: a. Optimal Receive Structure S1( t = S S = x(t h = S ( T t u = y(t t = T y(t S (t / w(t Let u = y(t be the decision variable The mean is The variance is where E s = S (τdτ = N E[u] = µ 1 = E[u S(t sent] = µ = E[u sent] = x(τ[s 1 (τ S (τ]dτ [S 1 (τ S (τ] dτ = N S(τ[S(τ ]dτ = E s [S(τ ]dτ = S (τdτ = N E s f u u ( ( u S f u S ES u Decision threshold: Let be the decision threshold, depends on E s and the probability that a target is present or not. Decision rule: if u, choose S(t Problem continued on next page... Page 5 of 9

7 Prof. X. Shen E&CE 358, Winter 16 ( 1:3-:5 PM: Solution # Problem (continued if u <, choose b. c. P F A = Pr{False Alarm} = Pr{ Receiver decides that a target is present when there is actually no target} = Pr{u no target} ] 1 (u = exp [ π du = 1 exp π P det = Pr{Detection} ] ( [ x x dx = Q = Q N E s = Pr{ Receiver decides that a target is present when there is a target} = Pr{u target present} 1 = exp [ (u E s ] π du = 1 exp π = 1 π Es exp ] ( [ x x dx = Q ( x dx = Q = Q E s N E s ( Es = Q E s N E s d. Find the decision threshold such that the average cost of the radar system is minimized. Let Y = Error event, X = no target, X = target present Then, Average Cost P (Y = P (Y XP (X + P (Y XP (X = P F A P (x + P fail P (x = (cost of false alarm P F A Pr{ no target} + (cost of failure to detect P fail Pr{targetpresent} where P F A = Pr{False Alarm} Pr{ target present} =.1 Pr{ no target} = 1.1 =.9 P fail = Pr{ Failing to detect a target} = 1 Pr{ Detection} Let cost of false alarm = cost of failure to detect = c Average cost = c P F A.9 + c P fail.1 Problem continued on next page... Page 6 of 9

8 Prof. X. Shen E&CE 358, Winter 16 ( 1:3-:5 PM: Solution # Problem (continued Let the decision threshold, as mentioned in part a be = βe s, where.5 β 1, because Pr{ no target} > Pr{ target present} From part b ( Es P F A = Q β From part c P fail = 1 P det = 1 Q N βe s E s N E s Therefore, the average cost is { ( Es Average cost = c.9 Q β N = Q ( (1 β E s N ( +.1 Q (1 β E } s N Find the value of β which minimizes the above cost function and use that value of β (let it be β min to calculate the optimal threshold (β min E s for the receiver. We cannot solve for β min, if we don t know the SNR E s /N for this system. Therefore, we consider solving for β min for a particular SNR. For example, for an SNR = 1 db The cost function is Method A : Trial and error 1 log E s N = 1, E s N = 1 cost =.9Q(β +.1Q((1 β =.9Q(4.5β +.1Q(4.5(1 β cost β=.5 =.9Q( Q(4.5.5 =.9Q(.5 +.1Q(4.5(.5 = Q(.5 =.1 cost β=1 =.9Q( Q(4.5 =.9Q( Q( = =.5 cost β=.75 =.9Q( Q( =.9Q( Q(1.15 = =.873 cost β=.6 =.9Q( Q( =.9Q(.7 +.1Q(1.8 = =.6716 Problem continued on next page... Page 7 of 9

9 Prof. X. Shen E&CE 358, Winter 16 ( 1:3-:5 PM: Solution # Problem (continued Therefore, Cost is minimum when β =.6 So, the optimum decision threshold at.6e s. Method B: %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % cost function for Es/N = 1 db % % Pr{no target} =.9 Pr{target} =.1 % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% cost β=.6 =.9Q( Q( =.9Q(.7 +.1Q(1.8 = =.6716 cost β=.8 =.9Q( Q(. 4.5 =.9Q( Q(.9 = =.1855 global s global ebno global a min_beta =.; max_beta =.; step =.5; B = []; dbebno = 1; ebno = 1^(dbebno/1; a = sqrt(*ebno; i = 1; for s=min_beta:step:max_beta B(i=.9*.5*erfc(a*s/sqrt(+.1*.5*erfc(a*(1-s/sqrt(; i = i+1; end s = min_beta:step:max_beta; plot(s,b; xlabel( Beta ; ylabel( Cost ; grid; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % cost function for Es/N = 1 db % % Pr{no target} =.5 Pr{target} =.5 % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% global s global ebno global a Problem continued on next page... Page 8 of 9

10 Prof. X. Shen E&CE 358, Winter 16 ( 1:3-:5 PM: Solution # Problem (continued min_beta =.; max_beta =.; step =.5; B = []; dbebno = 1; ebno = 1^(dbebno/1; a = sqrt(*ebno; i = 1; for s=min_beta:step:max_beta B(i=.5*.5*erfc(a*s/sqrt(+.5*.5*erfc(a*(1-s/sqrt(; i = i+1; end s = min_beta:step:max_beta; plot(s,b; xlabel( Beta ; ylabel( Cost ; grid; Cost.5 Cost Beta Beta (a cost function for Es/N = 1 db Pr{no target} =.9 Pr{target} =.1 (b cost function for Es/N = 1 db Pr{no target} =.5 Pr{target} =.5 Page 9 of 9