ECE 567
STATISTICAL SIGNAL PROCESSING
FALL 2013
Homework Assignment #5
Solutions
1. The data model is
x(0)
.
.
=
.
x(N 1)
1
r1
.
.
.
1
r2
.
.
.
N
N
r1 1 r2 1
1
rp
.
.
.
N
rp 1
A1
A2
.
.
.
+
Ap
w(0)
.
.
.
w(N 1)
x
H
and then
= (H T H )1 H T x
C = 2 (H T H
ECE 567
STATISTICAL SIGNAL PROCESSING
Homework Assignment #1
Solutions
1. The decision rule is
d0 , (x) < 1
d1 , (x) > 1
d(x) =
with
e|xA|
e|x|
= e|x|xA|
A
x>A
e ,
2xA
e
, 0<x<A
=
A
e,
x<0
(x) =
Therefore we use
d0 , x > A/2
d1 , x < A/2
d(x) =
The detec
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ECE 567
STATISTICAL SIGNAL PROCESSING
FALL 2013
Final Exam
4 December 2013
Name:
This is an open-book, open-notes exam. The use of electronic calculators is permitted. The
exam lasts 120 minutes.
The exam consist of six true/false questions (lumped under
ECE 567
STATISTICAL SIGNAL PROCESSING
FALL 2016
Homework Assignment #1
Solutions
1. (a)
2
exp 12 x3
(x) =
x exp 21 x2
1
1 2
= exp
x .
3
3
x
3
Therefore, the ML detector is
1
exp
3
1 2
x
3
d1
>
<
d0
1.
(b) > is equivalent to x2 > 0 . For
d1
2 >
x <
d0
0
we
ECE 567
STATISTICAL SIGNAL PROCESSING
FALL 2016
Homework Assignment #1
Due: 6 September 2016
1. We observe x with Rayleigh PDF
(
p(x) =
2
exp 21 x2 , x 0
0,
x<0
x
2
with
H0 : 2 = 1
H1 : 2 = 3
In this situation, the event we wish to detect is an increase i
ECE 567
STATISTICAL SIGNAL PROCESSING
FALL 2016
Homework Assignment #2
Solutions
1. We have
2
N
0, N , under H0
1
x=
x(n)
N A, 2 , under H1
N n=0
N
N
1
X
PF A = P rcfw_|x| > | H0
p
p
= Q(/ 2 /N ) + 1 Q(/ 2 /N )
p
= 2Q(/ 2 /N )
p
1
PF A = Q(/ 2 /N )
2
ECE 567
STATISTICAL SIGNAL PROCESSING
FALL 2016
Homework Assignment #2
Due: 15 September 2016
1. For the parameter test
H0 : A = 0
H1 : A 6= 0
with measurements x(n) = A+w(n), n = 0, 1, . . . , N 1, where w(n) is white Gaussian
noise with (known) variance
EOE 56? STATISTICAL SIGNAL PROCESSING FALL 2013
Exam #1
3 October 2013
Name:
This is en openijook; open~11otes exam. The use of electronic calculators is permitted. The
exam lasts 75 minutes.
There are three questions on the exam. Do all your work
ECE 567
STATISTICAL SIGNAL PROCESSING
FALL 2013
Homework Assignment #9
Solutions
1. We have
= E [ | x]
=
p( | x)d
=
p(x | )p()
d
p(x | 1 )p(1 )d1
We have
N 1
p(x | ) = exp
x(n) + N u (mincfw_x(n) )
n=0
N 1
= exp(N ) exp
x(n) u (mincfw_x(n) )
n=0
N 1
p(
mm mg
Hr ~- ~anmwmmm~m ,uw- -
Whar . _ ..< m WMMMW WW I
Q; (x Mag-K: m? Effaimg %« C() wrMa (ilk) (x PE§§1\)t Ckyw gxcwkxf {NJ- QM]
3131 m1%
\m w) «r 3 am W i: 3? M E
w
la mug
$3, my ° _
WEB-m 3mm E mg: 3mm} M:
; Wm
: is
g mm%@
W;
ECE 567
STATISTICAL SIGNAL PROCESSING
FALL 2013
Homework Assignment #6
Solutions
1. (a)
x(n) = + w(n)
1
p(w(n) = exp (|w(n)|) ,
2
1
.
x = . + w
.
1
0
.
E [w ] = .
.
IID
0
VAR[w] = C = 2 I
The BLUE is
T
1
. 2
. Ix
.
1
1
= T
=
N
1
1
. 2 .
. I .
.
.
ECE 567
STATISTICAL SIGNAL PROCESSING
Homework Assignment #4
Solutions
1. We have x(n) N (0, 2 ) and
N 1
1 2
2 =
x (n)
N n=0
N 1
N 1
1 2
2
2 ] = 1
E [
E [x (n)] =
= 2
N n=0
N n=0
so its unbiased.
The variance is
[
]
var[ 2 ] = E ( 2 2 )2
[
]
= E ( 2 )2 2