Eecs 455 univ of michigan fall 2012 october 3 2012 63

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Unformatted text preview: bove is constant for each j as is the last term. Thus finding the minimum is equivalent to finding the maximum of (sj , r ). EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 62 / 93 Lecture Notes 7 Example: M equal energy signals Thus the receiver should compute the inner product between the M different signals and find the largest such correlation. If the signals are all of duration T , i.e. zero outside the interval [0, T ] then this is also equivalent to filtering the received signal with a filter with impulse response sj (T − t ), sampling the output of the filter at time T and choosing the largest. EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 63 / 93 Lecture Notes 7 Demodulator (Equal Energy Case) ∗ s0 (t ) × ∗ (t ) s1 r (t ) ∗ r (t )s0 (t )dt × ∗ r (t )s1 (t )dt (r , s0 ) (r , s1 ) Find si with largest (r , si ) ∗ sM −1 (t ) × EECS 455 (Univ. of Michigan) ∗ r (t )sM −1 (t )dt Fall 2012 (r , sM −1 ) October 3, 2012 64 / 93 Lecture Notes 7 Notes about Optimum Receiver in AWGN Consider the case of equally likely signals (π0 = ... = πM −1 = 1/M ). The optimum receiver first maps the received signal into a N dimensional vector. (r (t ) → r ). The decision region is determined by the perpendicular bisectors of the signal points. Then the receiver finds which signal is closest (in Euclidean distance) to the received vector. (Find i for which r ∈ Ri ). EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 65 / 93 Lecture Notes 7 Example s0 (t ) T /3 2T /3 s2 (t ) T t T /3 2T /3 s1 (t ) T /3 2T /3 EECS 455 (Univ. of Michigan) T t T t s3 (t ) T t T /3 2T /3 Fall 2012 October 3, 2012 66 / 93 Lecture Notes 7 Orthonormal Basis Functions ϕ0 (t ) ϕ1 (t ) T T /3 t T t 2T /3 T t ϕ2 (t ) 2T /3 EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 67 / 93 Lecture Notes 7 Signal Vectors s0 = (+1, +1, +1) s1 = (−1, +1, −1) s2 = (+1, −1, −1) s3 = (−1, −1, +1) EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 68 / 93 Lecture Notes 7 Optimum Receiver 1 ϕ∗ (t ) 0 × r (t ) r (t )ϕ∗ (t )dt 0 r0 r (t )ϕ∗ (t )dt 1 r1 ϕ∗ (t ) 1 × smallest ||r − si ||2 ϕ∗ (t ) 2 × EECS 455 (Univ. of Michigan) Find si with r (t )ϕ∗ (t )dt 2 Fall 2012 r2 October 3, 2012 69 / 93 Lecture Notes 7 Optimum Receiver 2 ∗ s0 (t ) × ∗ (t ) s1 × ∗ r (t )s1 (t )dt × r (t ) ∗ r (t )s0 (t )dt ∗ r (t )s2 (t )dt ∗ s2 (t ) × ∗ r (t )s3 (t )dt (r , s0 ) (r , s1 ) Find si with (r , s2 ) largest (r , si ) (r , s3 ) ∗ s3 (t ) EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 70 / 93 Lecture Notes 7 Optimum Receiver 3 t = T /3, 2T /3, T r (t ) Find si with h(t ) = pT /3 (t ) largest Y (t ) (r , si ) T /3 r0 = Y (T /3) = r (t )dt r (t )ϕ0 (t )dt = 0 2T /3 r1 = Y (2T /3) = r (t )dt r (t )ϕ1 (t )dt = T /3 T r2 = Y (T ) = r (t )dt r (t )ϕ2 (t )dt = 2T /3 EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 71 / 93 Lecture Notes 7 Simplified Calculation (r , s0 ) = +r0 + r1 + r2 (r , s1 ) = −r0 + r1 − r2 (r , s2 ) = +r0 − r1 − r2 (r , s3 ) = −r0 − r1 + r2 First calculate x0 , x1 , x2 , x3 as follows x0 = +r0 x1 = −r0 x2 = r1 + r2 x3 = r1 − r2 EECS 455 (Univ. of Michigan) Fall 2012 Oc...
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This note was uploaded on 02/12/2014 for the course EECS 455 taught by Professor Stark during the Fall '08 term at University of Michigan.

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