# ch09 - Chapter 9 Optimum Receivers and Signal Space...

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Chapter 9 Optimum Receivers and Signal Space Concepts 9.1 Problems Problem 9.1 a. Given H 1 , Z = N ,so f Z ( z | H 1 )= f N ( n ) | z = n =10 e 10 z u ( z ) Given H 2 , Z = S + N ,where S and N are independent. Thus, the resulting pdf under H 2 is the convolution of the separate pdfs of S and N : f Z ( z | H 2 Z −∞ f s ( z λ ) f N ( λ ) d λ =20 Z z 0 e 2 z e 8 λ d λ , λ > 0 which integrates to the result given in the problem statement. b. The likelihood ratio is Λ ( Z f Z ( Z | H 2 ) f Z ( Z | H 1 ) =0 . 25 ¡ e 8 Z 1 ¢ , Z> 0 c. The threshold is η = P 0 ( c 21 c 11 ) P 1 ( c 12 c 22 ) = 1 3 1

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2 CHAPTER 9. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS d. The likelihood ratio test becomes 0 . 25 ¡ e 8 Z 1 ¢ H 2 > < H 1 1 3 This can be reduced to Z H 2 > < H 1 ln (7 / 3) 8 =0 . 106 e. The probability of detection is P D = Z 0 . 106 2 . 5 ¡ e 2 z e 10 z ¢ dz . 925 The probability of false alarm is P F = Z 0 . 106 10 e 10 z dz . 346 Therefore, the risk is Risk = 1 4 (5) + 3 4 (5) (1 0 . 925) 1 4 (5) (1 0 . 346) = 0 . 714 f. Consider the threshold set at η .Th en P F = e 10 η η 0 . 921 to give P F 10 4 . Also, from part (e), P D =1 . 25 e 2 η 0 . 25 e 10 η For values of η 0 . 921 , P D is approximately equal to the & rst term of the above expression. For this range of η P D is strictly decreasing with η ,sothebestva lueo f P D occurs for η . 921 for P F 10 4 .There su l t ing P D is 0 . 198 . g. From part (f), P F = e 10 η and P D . 25 e 2 η 0 . 25 e 10 η
9.1. PROBLEMS 3 Figure 9.1: From part (d), the likelihood ratio test is Z H 2 > < H 1 ln (4 η +1) 8 = γ Ap lotof P D versus P F as a function of η or γ constitutes the operating characteristic of the test. It is shown in Figure 9.1. Problem 9.2 a. The likelihood ratio is Λ ( Z )= 1 2 e | Z | e 1 2 Z 2 2 π = r π 2 e 1 2 Z 2 | Z | b. The conditional pdfs under each hypothesis are plotted below. The decision regions R 1 and R 2 are indicated in Figure 9.2 for η =1 . Problem 9.3 De & ne A =( a 1 ,a 2 , a 3 ) and B b 1 ,b 2 3 ) .D e & ne scalar multiplication by α ( a 1 2 , a 3 ( α a 1 , α a 2 , α a 3 ) and vector addition by ( a 1 2 , a 3 )+ ( b 1 2 3 )=( a 1 + b 1 2 + b 2 3 + b 3 ) . The properties listed under &Structure of Signal Space& in the text then become:

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4 CHAPTER 9. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Figure 9.2: 1. α 1 A + α 2 B =( α 1 a 1 + α 2 b 1 , α 1 a 2 + α 2 b 2 , α 1 a 3 + α 2 b 3 ) ²R 3 ; 2. α ( A + B )= α ( a 1 + b 1 , a 2 + b 2 ,a 3 + b 3 ) α a 1 + α b 1 , α a 2 + α b 2 , α a 3 + α b 3 )=( α a 1 , α a 2 , α a 3 )+( α b 1 , α b 2 , α b 3 ) = α A + α B²R 3 ; 3. α 1 ( α 2 A α 1 α 2 A ) ²R 3 (follows by writing out in component form); 4. 1 & A = A (follows by writing out in component form); 5. The unique element 0 is (0 , 0 , 0) so that A +0= A ; 6. A a 1 , a 2 , a 3 ) so that A +( A )=0 . Problem 9.4 Consider ( x,y )= lim T →∞ Z T T x ( t ) y ( t ) dt Then ( T →∞ Z T T y ( t ) x ( t ) dt = lim T →∞ Z T T x ( t ) y ( t ) dt ) Also ( α ) = lim T →∞ Z T T α x ( t ) y ( t ) dt = α lim T →∞ Z T T x ( t ) y ( t ) dt = α ( )
9.1. PROBLEMS 5 and ( x + y,z ) = lim T →∞ Z T T [ x ( t )+ y ( t )] z ( t ) ( t ) dt = lim T →∞ Z T T x ( t ) z ( t ) dt +l im T →∞ Z T T y ( t ) z ( t ) dt =( x,z )+( ) Finally ( x,x ) = lim T →∞ Z T T x ( t ) x ( t ) dt

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## This note was uploaded on 04/28/2008 for the course EE ELCT 332 taught by Professor Cooper during the Spring '08 term at University of South Carolina Beaufort.

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ch09 - Chapter 9 Optimum Receivers and Signal Space...

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