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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 Fall '08
 Stark
 Normal Distribution, Univ. of Michigan

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