信号检测与估计 第五章 答案

p(yi p(yi
|H1) |H0)
.
(18)
N i=1
ln
p(yi|H0)
=
N i=1
yi
=
0.
Therefore,
we
form
a
decision
variable
N
U = yi
i=1
(19)
If
U
=
0,
then
we
decide
H0.
If
U
=
0,
then
we
decide
H1.
The
minimum
cost
where µi = E{l|Hi} and σi2 = V {l|Hi} for i = 0, 1,
E{Lk|H1} = P11 ln τA + (1 − P11) ln τC ,
(24)
and
E{Lk|H0} = P10 ln τa + (1 − P10) ln τC
(25)
The unconditional variance is
∞
n
e(vi−θ0)dvi = enθ0 1 − e−θ1 n
θ1
(15)
Problem 5.13 (a) Because of the conditions, the threshold is τ = 1. Let
D1
L(yi)
=
p(yi|H1) p(yi|H0)
> <
1
(16)
D0
If
yi
=
1,
is
1 2
C01
P01
=
1 2
N +1 .
(d) The basic cost is
. 1
2N +1
For N
samples, there is an additional cost of
N 15
.
Thus, the total cost for N
samples
is
2−(N +1)
+
N 15
.
The
p0(v)
so we decide D1. If, on the other hand, any vi is less than θ1, then p(v) = 0 and L(v) = 0 so decide D0. The risk is the probability of choosing D1 when H0 is correct. This can be written as
Problem 5.16
Similar to the derivation in Section 5.11,
E{L2k|H1} = P11(ln τA)2 + (1 − P11)(ln τC )2
(20)
and
E{L2k|H0} = P10(ln τA)2 + (1 − P10)(ln τC )2
(21)
L(y) = p(y|H1) = p(y|H0)
N i=1
p(yi|H1)
N i=1
p(yi|H0)
(17)
and the log-likelihood ratio is If H0 is correct, all yi’s are zero, so
ln L(y) =
N i=1 N i=1
ln ln
1
τ − kµ
Pd
=
erfc 2
√ 2kσ2
(8)
and substituting in the value for τ found in part (a) results in the desired expression.
Problem 5.2
The measurement vector is
y = [yl, yl+i, ..., yl+k]
value
of
N
for
which
this
is
smallest
is
N
=
2.
Problem 5.15
This problem simply reiterates the fact that the decision variable U and the decision threshold are different concepts. If, in any of the examples in chapter 5, we had dealt with unequal a priori probabilities and/or non-uniform costs, the decision threshold would change as we learned in chapter 4. In this specific case of OOK, the decision variable can be determined as shown in Figure 5.11. The bias term would be included in the decision threshold.
√
τ = erfc−1 (2αf ) 2kσ2
(6)
(b) The probability of detection is defined as
∞
∞1
Pd =
τ
p1(yk)dyk =τ√Fra bibliotekexp
2πkσ2
−
(yk − kµ)2 2kσ2
dyk
(7)
√ Using the change of variable t = (yk − kµ)/ 2kσ2, it follows that
From Eq. F.8 in Appendix F,
V
{k|H1}
=
E {L2k |H1 }
−
σ12 E {Lk |H1 }/µ1 µ21
−
E 2 {Lk |H1 }
(22)
and
V
{k|H0}
=
E {L2k |H0 }
−
σ02 E {Lk |H0 }/µ0 µ20
−
E 2 {Lk |H0 }
(23)
(9)
starting at an arbitrary point l. The tap coefficient of the FIR filter are shown in Figure 5.4 and are
h∗i,k, h∗i,k−1, ..., h∗i,1, i = 0, 1
(10)
The filter output is
that
1 P (E) = erfc
E1
(12)
2
4N0
Problem 5.12
Let v = (v1, · · · , vn). This problem is a good one in using understanding and logic rather than just “chugand-plug” mathematical formulae. The a prior probabilities and costs assure that the decision threshold is unity. We can write the vector pdf as
Problem 5.10
Amending Figure 5.3 in the text, we can draw an OOK receiver structure as Figure 1. because for the
signal corresponding to “off”, u0 = 0. Equation 5.108 is still valid, and since ρr = 0 and E0 = 0, it follows
V {k} = π1V {k|H1} + π0V {k|H0}
(26)
p(v) =
enθ1 exp (− enθ0 exp (−
n in=1 i=1
vi) vi)
, ,
H1 H0
(13)
If all the individual vi are greater than or equal to θ1, it follows that
L(v) = p1(v) ≥ 1
(14)
(2)
and
1
p1 (yk )
=
√ 2πkσ2
exp
−
(yk − kµ)2 2kσ2
(3)
It follows that the probability of false alarm, specified by the Neyman-Pearson rule as αf , is
∞
αf =
p0 (yk )dyk
(4)
τ
√ Using the change of variable t = yk/ 2kσ2 inside the integral, we are left with
αf =
∞
1
√ τ / 2kσ2
√ π
exp
−t2
1 dt = erfc
2
τ √
2kσ2
(5)
and from this it follows that the threshold is
ylh∗i,1 + yl+1h∗i,2 + · · · + yl+kh∗i,k
k
=
yl−1+j h∗i,j