Suppose that you are asked to detect occurrences of +4 volt DC signal from a noisy observation Y and also mentioned that noise has Gaussian distribution N(0,σ

^{2}), what would you do to solve the problem ?

Figure 1 : DC voltage submerged in
noise

Observation
reveals that any sample of Y above +4v can be marked as occurrence of +4v DC. Samples of Y with magnitude below +4v need a
decision rule to compare Y against a “Threshold”
voltage either to agree or disagree on occurrence of +4v DC. Accuracy
of this decision rule, i.e. percentage of false detects or miss detects will be
dependent on “Threshold” voltage.

Problem Statement

Identify a decision rule Threshold for observation Y with a maximum false detection percentage α.

**Stochastic Theory**provides a detection methodology to solve the above stated problem. This problem is modeled as a two possible hypothesis testing, where H

_{1}and H

_{0}are the two hypothesis which can occur in observation Y with probability distributions P

_{1}and P

_{0}respectively.

H

_{1}: Y ~ P_{1}
H

_{0}: Y ~ P_{0}_{}

With
this introduction, we can formulate +A volt DC detection problem as below

H

_{1}: Y = +A + N(0,σ^{2})
H

_{0}: Y = N(0,σ^{2})
Where
Y is the observation, A is a deterministic signal (DC volts) and N(0,σ

^{2}) represent noise amplitude from a Gaussian distributed noise source with zero mean and variance σ^{2}. Probability distribution for hypothesis on Y is presented in Figure 2.
Figure 2 : Probability Distribution of
+A volt Hypothesis Testing

Objective
of this problem is to find a hypothesis decision rule on the observation Y

Before
writing the decision rule, let us make couple of assumptions here,

- Both the hypothesis are equally likely
- Cost for wrong decisions are equal (assume 1) and Cost for correct decisions are zero

With
the above assumptions, we can write likelihood decision rule δ(y) as,

δ(y) = 1 if
L(y) >= 1

δ(y) = 0 if
L(y) < 1

where
L(y) is likelihood ratio given as L(y) = p

_{1}(y)/p_{0}(y), p_{1}(y) is probability Y belongs to H_{1}and p_{0}(y) is probability Y belongs to H_{0}. So p_{1}(y) is N(A, σ^{2}) Gaussian distribution function and p_{0}(y) is N(0, σ^{2}) Gaussian distribution function.
Using
the Gaussian distribution function we get L(y) = p

_{1}(y)/p_{0}(y) = exp(-A^{2}+ 2yA), so decision rule can be re-written as
δ(y) = 1 if y
>= (A/2)

δ(y) = 0 if y
< (A/2)

So,
we compare observation Y against the threshold (A/2) and decide on the
resulting hypothesis.

Knowing
the decision Threshold, we are through half the initial problem now, so I will
try to find the false alarm probability P

_{F}for this decision rule. How to do that ?
P

_{F}(δ(y)) = P_{0}( L(y) >= 1)
False alarm probability is the probability
for the observation Y to fall in H

_{1}when it actually belong to H_{0}, which can be computed from the below expression. E is expectation function
P

_{F}(δ(y)) = E_{0}( y>(A/2) ) = 1 - ф( (A/2)/σ) )
Ф(x) =
P( z <= x) and z is a N(0,1) Gaussian distribution or in other terms Ф(x) is
cumulative distribution function of N(0,1) distributed random variable

*Till now we first made a decision rule and then computed its false alarm probability P*

_{F}, but our initial problem is to find decision rule with a desired false alarm probability P_{F}<= α. So to do this let us define a new decision rule δ_{F}(y) which satisfy P_{F}<= α condition
δ

_{F}(y) = 1 if y >= C
δ

_{F}(y) = 0 if y < C
Where
C is the desired threshold that gives us a decision rule with P

_{F}<= α, let us compute false alarm probability for this decision rule, with the objective to find C
P

_{F}(δ_{F}(y)) = E_{0}( y>C ) = 1 - ф( C/σ ) = α
So, decision rule threshold is function of
noise variance and P

_{F}. For a given P_{F}, Z = C/σ ratio is constant and Z can be estimated by fixing P_{F}. With the knowledge of σ decision rule can be formulated as
δ

_{F}(y) = 1 if y >= Z σ
δ

_{F}(y) = 0 if y < Z σIt can be observed that the decision rule conditioned on P_{F}, Threshold ‘C’ is independent of the mean of hypothesis H_{1}, i.e. independent of +A

Let us look at few variants of the initial
problem, which usually encounter in Communication Systems

**Problem Variant 1**

Hypothesis
testing problem to detect a “-A” volt DC is formulated as below

H

_{1}: Y = N(0,σ^{2})
H

_{0}: Y = -A + N(0,σ^{2})
Decision rule with no conditioning on false
alarm probability can be given as,

δ(y) = 1 if y
>= -A/2

δ(y) = 0 if y
< -A/2

Consider a hit on “–A” voltage occurrence, if
the observation satisfy the condition Y < -A/2 . Probability distribution
for hypothesis on Y is presented in Figure 3.

Figure 3 : Probability Distribution of
-A volt Hypothesis Testing

**Problem Variant 2**

Hypothesis
testing problem to detect a “|A|” volt DC is formulated as below

H

_{1}: Y = +A + N(0,σ^{2})
H

_{2}: Y = -A + N(0,σ^{2})
H

_{0}: Y = N(0,σ^{2})
Probability distribution for hypothesis on Y has
three different decision regions as presented in Figure 4

.

Figure 4 : Probability Distribution of
|A| volt Hypothesis Testing

Decision rule with no conditioning on false
alarm probability can be given as,

δ(y) = 1 if L

_{1}(y) >= 1
δ(y) = 0 if L

_{1}(y) < 1 and L_{2}(y) > 1
δ(y) = 2 if L

_{2}(y) <= 1
where
L

_{1}(y) = p1(y)/p0(y) and L_{2}(y) = p2(y)/p0(y),
p0(y)
= N(0,σ

^{2})
p1(y)
= N(+A,σ

^{2})
p2(y)
= N(-A,σ

^{2})
Using
these probability density functions, decision rule can be re-written as

δ(y) = 1 if y
>= A/2

δ(y) = 0 if A/2
> y < -A/2

δ(y) = 2 if y <= -A/2

Decision
rule can be simplified as

δ(y) = 1 if |y|
>= A/2

δ(y) = 0 if |y|
< A/2

This hypothesis testing problem is nothing but aPower Rise Detectionproblem often encountered in Communications systems.

Let us find a decision rule
conditioned on false alarm probability P

_{F}<= α
We can easily formulate the decision rule from
our working knowledge on initial problem

δ

_{F}(y) = 1 if y^{2}>= C^{2}
δ

_{F}(y) = 0 if y^{2}< C^{2}
Where P

_{F}(δ_{F}(y)) = E_{0}( |y|>C ) = 2[1 - ф( C/σ )]
If Z = C/σ, applying logarithm on both sides changes
to C dB = Z dB + σ dB. For a given false
alarm probability P

_{F}<= α, Power Rise Detection hypothesis is true if a Z dB rise in power over noise, is observed.**Problem Variant 3**

Hypothesis
testing problem to detect between “+A” and “-A” volt DC levels is formulated as
below

H

_{1}: Y = +A + N(0,σ^{2})
H

_{0}: Y = -A + N(0,σ^{2})
Probability distribution for hypothesis on Y has
two different decision regions as presented in Figure 5

Figure 5 : Probability Distribution of
two level [-A, +A] volt Hypothesis Testing

This hypothesis testing problem is akin toBPSK modulation symbol detectionproblem.

Decision rule can be given as,

δ(y) = 1 if L(y)
>= 1

δ(y) = 0 if L(y)
< 1

where
L(y) = p1(y)/p0(y)

p1(y)
= N(+A,σ

^{2})
p0(y)
= N(-A,σ

^{2})
Using
these probability density functions, decision rule can be re-written as

δ(y) = 1 if y
>= 0

δ(y) = 0 if y < 0

If
y > 0 detect symbol +A or else detect symbol –A and if y == 0 choose
arbitrarily between + A and -A. False
alarm probability will give us the Symbol error probability.

False alarm probabilities for both hypothesis
regions are given as below

P

P_{+A}(δ(y)) = E_{0}( y>0 ) = 1 - ф(A/σ)_{-A}(δ(y)) = E

_{1}( y<0 ) = 1 - ф(A/σ)

As we have two different symbols here, [+A
–A], and supposing equally likelihood of their occurrence

Error probability is given as P

[1] H. Vincent Poor, "An Introduction to Signal Detection and Estimation"

[2] Harry L. Van Trees, "Detection, Estimation and Modulation Theory"

_{E}= 0.5* P_{+A}(δ(y)) + 0.5* P_{-A}(δ(y))= 1 - ф(A/σ) .[1] H. Vincent Poor, "An Introduction to Signal Detection and Estimation"

[2] Harry L. Van Trees, "Detection, Estimation and Modulation Theory"

## No comments:

## Post a Comment