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)
and
γ0 =
N0T ln(1/PF A) 2
Note
that If X
∼ N (0, σ2),
Y
∼ N (0, σ2),
X
and Y
are independent,
then R
=
√ X2 + Y 2
is
Rayleigh-
distributed with pdf:
fR(R)
=
R σ2
exp
−R2 2σ2
+
cos
2φ)
T
A2 cos2(ω0t + θ)dt
0
=
A2T 2
+
A2 2
T 0
cos 2(ω0t + θ)dt ≈
A2T 2
≡ Es,
symbol energy
We have
Λ(x(t))
=
e−
A2 T 2N0
·
1 2π
π
exp
−π
2A N0
T
x(t) cos(ω0t + θ)dt
0
H1 dθ >< Λ0
sin ω0t
ec - ( )2 es - ( )2
? R2 ⊕ 6
- Th: γ02
-
2. Quadrature receiver in matched-filter implementation:
x(t)-
-⊗ 6
cos ω0t -⊗ -
6
sin ω0t
hi(t) hq (t)
ec - ( )2 es - ( )2
where the Marcum Q-function is defined as Q(a, b) ≡ ∞ x · e−(x+a)2/2 · I0(ax)dx
b
IX - 35
5 Noncoherent FSK with random phase
x(t) =
A sin(ω0t + φ) + n(t), H0 A sin(ω1t + θ) + n(t), H1
1 2π
π θ:−π
exp
−
T 0
[x(t)−A
cos(ω0
t+θ)]2
dt
N0
dθ
Λ(x(t)) =
exp
− T x2(t)dt
0
N0
=
1 2π
π
exp
−π
−
A2 N0
T 0
cos2(ω0t
+
θ)dt
+
2A N0
T
x(t) cos(ω0t + θ)dt
0
dθ
Note that
cos2 φ
=
1 2
(1
,R ≥ 0
IX - 34
4.2 PD
Under H1, x(t) = A cos(ω0t + θ) + n(t),
ec = es =
T
cos ω0t[A cos(ω0t + θ) + n(t)]dt
0
T
sin ω0t[A cos(ω0t + θ) + n(t)]dt
0
and
E[ec|H1, θ]
T
n(t) cos ω0tdt
0
T
n(t) sin ω0tdt
0
Therefore,
E[ec|H0] = E[es|H0] = 0
T
2
E[e2c |H0]
=
E
n(t) cos ω0tdt
0
=
N0 2
T
cos2 ω0tdt
0
=
N0T 4
= E[e2s|H0]
E[eces]
=
N0 2
T
cos ω0t sin ω0t dt ≈ 0,
0
uncorrelated/independent Gaussian rv’s
PF A = P [R ≥ γ0|H0]
=
R≥γ0
√ 2π
1 N0T /4
2
exp
−
(e2c + e2s) 2N0T /4
desdec
=
2 πT N0
2π 0
∞
exp−
2R2 N0 T
·R
dR
dφ
γ0
PF A
=
exp(−
2γ02 N0T
H0
2. Quadrature representation: Rewrite
T
x(t) cos(ω0t + θ)dt = ec · cos θ − es · sin θ
0
=
e2c + e2s
cos(θ + tan−1
es ec
)
= R cos(θ + ψ)
where
ec ≡ es ≡
T
x(t) cos ω0tdt,
Pe = P [H0|H1] = P [R0 > R1|H1]
∞
=
Pe|R1 · p1(R1) dR1
R1 :0
∞
∞
=
dR1 · p1(R1) ·
dR0 · p1(R0)
R1 :0
R0 :R1
Under H1, R1 is Rician-distributed and R0 is Rayleigh-distributed with pdf’s:
0
T
2
x(t) · sin ω0tdt
0
Bt choosing Λ0 = 1 and noting that I0(·) is monotonic, we have
H1 R1 >< R0
H0
5.2 Performance
Assume P [H0] = P [H1] = 1/2, |ω1 − ω0| is very large, R0 is independent of R1,
IX - 36
=
1 2π
π
f (ec, es|H1, θ)dθ
−π
=
2 N0πT
exp
−
2R2 N0
2AR N0
Finally,
PD = P [R ≥ γ0|H1]
=
2π ·
2 N0πT
·
∞
R · exp
γ0
−
2R2 N0T
· exp
−
A2T 2N0
· I0
2AR N0
dR
= Q 2Es/N0 , −2 lnPF A
? R2 ⊕ 6
- Th: γ02
-
where the impulse responses are
hi(t) = hq(t) =
cos ω0(T − t), 0 ≤ t ≤ T
0,
o.w.
sin ω0(T − t), 0 ≤ t ≤ T
0,
o.w.
IX - 33
4 Performance analysis
H1 ><
N0 2A
I0−1(γ)
≡
γ0
H0
H1 R2 ≡ e2c + e2s >< γ02
H0
3 Optimum noncoherent receiver
1. Envelope correlator implementation:
x(t)-
-⊗ 6
cos ω0t -⊗ -
6
T 0
(·)dt
T 0
(·)dt
ICM 5507 Detection & Estimation
Notes
IX. Detection of Signals with Random Phase
Fall 2002 S.F. Hsieh
1 Noncoherent OOK Model
Consider a noncoherent On-Off Keying(OOK) system:
=
AT
cos θ 2
=
µec|θ
E[es|H1, θ]
=
−AT sin θ 2
=
µes|θ
σe2c|θ
=
σe2s|θ
=
N0T 4
E[(ec − µec|θ) · (es − µes|θ)] ≈ 0
Thus, conditioned on H1 and θ, ec and es are independent Gaussian rv’s, and their (conditional) joint pdf is
about the phase. 2. An unstable transmitter oscillator with no control over phase. 3. A transmission medium(channel) that introduces an unknown phase. 4. A receiver that does NOT track the phase of the transmitted sinusoid.
Since n(t) is Gaussian, for a fixed θ,
ec = es =
T
x(t) cos ω0tdt
0
T
x(t) sin ω0tdt
0
are Gaussian random variables, too.
4.1 PF A
Under H0, x(t) = n(t),
