Classic Detection Theory Problems in Communication Systems

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,σ²), 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 H₁ and H₀ are the two hypothesis which can occur in observation Y with probability distributions P₁ and P₀ respectively.

H₁ : Y ~ P₁

H₀ : Y ~ P₀

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

 H₁ : Y = +A + N(0,σ²)

 H₀ : Y = N(0,σ²)

Where Y is the observation, A is a deterministic signal (DC volts) and N(0,σ²) represent noise amplitude from a Gaussian distributed noise source with zero mean and variance σ². 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,

1. Both the hypothesis are equally likely
2. 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) = p₁(y)/p₀(y), p₁(y) is probability Y belongs to H₁ and p₀(y) is probability Y belongs to H₀. So p₁(y) is N(A, σ²) Gaussian distribution function and p₀(y) is N(0, σ²) Gaussian distribution function.

Using the Gaussian distribution function we get L(y) = p₁(y)/p₀(y) = exp(-A² + 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)) = P₀( L(y) >= 1)

False alarm probability is the probability for the observation Y to fall in H₁ when it actually belong to H₀, which can be computed from the below expression. E is expectation function

P_F(δ(y)) = E₀( 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)) = E₀( 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 σ

It can be observed that the decision rule conditioned on P_F, Threshold ‘C’ is independent of the  mean of hypothesis H₁, 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

H₁ : Y = N(0,σ²)

H₀ : Y = -A + N(0,σ²)

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

H₁ : Y = +A + N(0,σ²)

H₂ : Y = -A + N(0,σ²)

H₀ : Y = N(0,σ²)

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 L₁(y) >= 1

δ(y) = 0 if L₁(y) < 1 and L₂(y) > 1

δ(y) = 2 if L₂(y) <= 1

where L₁(y) = p1(y)/p0(y) and L₂(y) = p2(y)/p0(y),

p0(y) = N(0,σ²)

p1(y) = N(+A,σ²)

p2(y) = N(-A,σ²)

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 a Power Rise Detection problem 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 y² >= C²

δ_F (y) = 0 if y² < C²

Where P_F(δ_F(y)) = E₀( |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

H₁ : Y = +A + N(0,σ²)

H₀ : Y = -A + N(0,σ²)

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 to BPSK modulation symbol detection problem.

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,σ²)

p0(y) = N(-A,σ²)

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_+A(δ(y)) = E₀( y>0 ) = 1 - ф(A/σ) 

P_-A(δ(y)) = E₁( 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_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"