1. TELE3113 Analogue and Digital
Communications –
Detection Theory (2)
Wei Zhang
w.zhang@unsw.edu.au
School of Electrical Engineering and Telecommunications
The University of New South Wales
7 Oct. 2009 TELE3113 1

2. Integrate-and-Dump detector
Integrate-and-Dump detector
r(t)=si(t)+n(t)  s (t ) = + A 0≤t ≤T for 1
si (t ) =  1
s2 (t ) = − A 0≤t ≤T for 0
t 0 +T
 a1 (t ) + no for 1
Output of the integrator: z (t ) = ∫ [si (t ) + n(t )]dt = 
t 0 +T
t0 a2 (t ) + no for 0
where a1 = ∫ Adt = AT
t0
t 0 +T
a2 = ∫ (− A)dt = − AT
t0
t 0 +T
no = ∫ n(t )dt
t0
7 Oct. 2009 TELE3113 2

3. Integrate-and-Dump detector
no is a zero-mean Gaussian random variable.
t0 +T
  t 0 +T

E{no } = E  ∫ n(t )dt  = ∫ E{n(t )}dt = 0
 t0
  t0

  t 0 +T  
2

{ }
σ no = Var{no } = E no = E  ∫ n(t )dt  
2 2 
  t0

 
 
t o +T t 0 +T
= ∫ ∫ E{n(t )n(ε )}dtdε
t0 t0
t o +T t 0 +T
η
= ∫ ∫ δ (t − ε )dtdε
t0 t0
2
t 0 +T
η ηT
= ∫
t0
2
dε =
2
1 1
pdf of no: f n (α ) =
−α 2 /( 2σ no )
2 2
e = e −α /(ηT )
o
2π σ no πηT
7 Oct. 2009 TELE3113 3

4. Integrate-and-Dump detector
 s1 (t ) = + A 0≤t ≤T for 1
As si (t ) = 
s 0 (t ) = − A 0≤t ≤T for 0 s0 s1
We choose the decision threshold to be 0. 0
− AT + AT
Two cases of detection error:
(a) +A is transmitted but (AT+no)<0 no<-AT
(b) -A is transmitted but (-AT+no)>0 no>+AT
Error probability:
Pe = P (no < − AT | A) P ( A) + P (no > AT | A) P (− A)
− AT 2 ∞ 2
e −α /(ηT )
e −α /(ηT )
= P ( A)
−∞
∫ πηT
dα + P (− A) ∫
AT πηT
dα
∞ 2
e −α /(ηT )
 2 A2T 
dα [P( A) + P (− A)]
∞ 2
e −u / 2
= ∫ πηT
Thus, Pe = Q

 Q Q(x ) = ∫ du
AT
 η   x 2π
∞ 2
e −u / 2 2α  2 Eb 
= ∫
T
du Qu =
2π ηT = Q
 η 
 Q Eb = ∫ A2 dt
2 A2T η
  0
7 Oct. 2009 TELE3113 4

5. Integrate-and-Dump detector
Consider two signal symbols s1 and s2.
Let Ed be the energy of the difference signal (s1- s2),
s2 s1
T
 Ed  signal symbol energy=si2
Ed = ∫ [s1 (t ) − s2 (t )] dt Pe = Q 
2
i.e.  2η 
0  
 s (t ) = + A 0≤t ≤T for 1
For example: If si (t ) =  1
s2 (t ) = − A 0≤t ≤T for 0
T T
Thus Ed = ∫ [s1 (t ) − s2 (t )] dt = ∫ [A − (− A)] dt = 4 A T
2 2
2
0 0
 4 A2T   2 A2T 
⇒ Pe = Q  = Q 
 2η   η 
   
7 Oct. 2009 TELE3113 5

6. Integrate-and-Dump detector
Example: In a binary system with bipolar binary signal which is a +A volt or –A volt pulse
during the interval (0,T), the sending of either +A or –A are equally probable.
The value of A is 10mV. The noise power spectral density is 10-9 W/Hz. The
transmission rate of data (bit rate) is 104 bit/s. An integrate-and-dump detector
is used.
(a) Find the probability of error, Pe.
(b) If the bit rate is increased to 105 bit/s what value of A is needed to
attain the same Pe, as in part (a).
η
Solution: (a) With = 10 −9 , P(+A)=P(-A)=0.5 , bit interval T=10-4 seconds, A=10mV
2
 2 A 2T   −4 
( )
−3 2
Pe = Q  = Q 2(10 × 10 ) (10 )  = Q 10 = 7.8 × 10 − 4
 η   2 × 10 −9 
   
 
( )
2
(b) Bit interval T=10 -5 seconds, for the same P , i.e. P = Q 2 A T  = Q 10
e e
 η 
 
2 A 2T
= 10 → A=
(
10 2 × 10 −9 )
= 31.62mV
η 2 10 −5 ( )
7 Oct. 2009 TELE3113 6

7. Optimal Detection Threshold
Pe = P (detect s 2 | s1 ) P( s1 ) + P (detect s1 | s 2 ) P( s 2 )
λ ∞
= P ( s1 ) ∫ f (r | s1 )dr + P( s 2 ) ∫ f (r | s 2 )dr
−∞ λ
λ
 λ 
= P ( s1 ) ∫ f (r | s1 )dr + P( s 2 ) 1 − ∫ f (r | s 2 )dr 
−∞  −∞ 
λ
= P( s 2 ) + ∫ [P(s ) f (r | s ) − P(s
−∞
1 1 2 ) f (r | s 2 )]dr
7 Oct. 2009 TELE3113 7

8. Optimal Detection Threshold
To find a threshold λo which minimizes Pe, we set dPe = 0
dλ
gives
Taking ln(.) on both sides, then
P( s1 ) f (λo | s1 ) = P ( s 2 ) f (λo | s 2 )
λo (s1 − s2 ) s12 − s2
2
P(s )
f (λo | s1 ) P ( s 2 ) − = ln 2
= σn 2
2σ n2
P(s1 )
f (λo | s 2 ) P( s1 )
o o
s +s σn 2
P(s )
− ( λo − s1 ) 2 /( 2σ no )
2 λo − 1 2 = o ln 2
f (λo | s1 ) e P( s 2 ) 2 s1 − s2 P(s1 )
= −( λ − s ) 2 /( 2σ 2 ) =
s1 + s2 σ no
2
f (λ o | s 2 ) e o 2 no P( s1 ) λo = +
P(s )
ln 2
2 s1 − s2 P(s1 )
2 2 2 2
λo ( s1 − s2 ) / σ no − ( s1 − s 2 ) /( 2σ no ) P( s 2 )
e = If P ( s1 ) = P ( s 2 ) ⇒ λo =
s1 + s 2
P ( s1 ) 2
7 Oct. 2009 TELE3113 8

9. Correlator Receiver
r r 2
Recall ML decision criterion: minimize r − si
r r
= ∫ [r (t ) − si (t )] dt = ∫ r 2 (t )dt + ∫ − 2 ∫ r (t ) si (t )dt
2
With r − si
2
si2 (t )dt
T T T T
1 24
4 3 1 24
4 3
constant energy of i - th signal
r r
∫ si2 (t )dt − 2 ∫ r (t ) si (t )dt
2
minimize r − si minimize
T T
Let ξ i denotes the energy of si(t).
ML decision criterion becomes
Detected
Find i to maximize signal
∫ r (t ) si (t )dt − 1 ∫ si2 (t )dt symbol
2
T T
1 24
4 3
ξi
Or simply: Find i to maximize ∫ r (t )s (t )dt
T
i
if all signal symbols have the same energy
Correlation Receiver
7 Oct. 2009 TELE3113 9

10. Matched Filter
The multiplying and integrating in correlation receiver can be reduced
to a linear filtering.
Consider the received signal r(t) passes through a filter hi(t):
i.e. r (t ) ∗ hi (t ) = ∫ r (τ )hi (t − τ )dτ
T
Let hi (τ ) = si (T − τ ) ⇒ ∫ r (τ )h (t − τ )dτ = ∫ r (τ )s (t − T + τ )dτ
i i for 0 ≤ t ≤ T
T T
Then we sample the filter output at t=T,
thus ∫ r (τ )h (t − τ )dτ
T
i = ∫ r (τ ) si (t − T + τ )dτ
T
= ∫ r (τ ) si (τ )dτ
T
(correlation)
t =T t =T
ML decision criterion:
∫ r (t ) s (t )dt − ∫ s
1 2
Find i to maximize i 2 i(t )dt
T T
1 24
4 3
becomes Find i to maximize ξi
Detected
signal
symbol
∫ r (τ )h (t − τ )dτ
T
i − 1 ∫ si2 (t )dt
2
T
t =T 1 24
4 3
ξi
7 Oct. 2009 TELE3113 10
Matched-Filter Receiver

11. Matched Filter
Consider the matched filter hi (t ) = si (T − t )
si(-t)
t t t
si(t) si(-t) hi(t)
0 T t -T 0 t 0 T t
Equivalence of matched filter and correlator:
Matched filter receiver
≤ ≤
Correlator receiver
7 Oct. 2009 TELE3113 11

12. Detection of PAM
Detection of M-ary PAM (one-dimension)
symbol s1 s2 sM/2-1 sM/2 sM/2+1 sM/2+2 sM-1 sM
… …
amplitude −3 E − E 0 E 3 E ( M − 1) E
2
Let si = E Ai where i = 1,2 ,...,M and Ai = (2i − 1 − M ) , energy of si is si
For equally probable signals, i.e. P( si ) = 1 / M
for i = 1,2,...,M
1 M 2 E M 2 E M E M (M 2 − 1)  M 2 − 1 
average energy ε av = ∑ si = ∑ Ai = ∑ (2i − 1 − M ) = =
 3 E
2
M i =1 M i =1 M i =1 M 3 
 
Received signal r = si + n = E Ai + n where n = 0, σ n = η 2
2
Average Prob(symbol error) Pe =
M −1
M
(
P r − si > E )
∞
M −1 2
πη ∫E
2
= e − x η dx
M
2( M − 1)  2 E 
∞
x 2 M −1 2
∫ Q 
2
let y = = e − y 2 dy =  η 
η M 2π 2 E /η
M  
7 Oct. 2009 TELE3113 12

13. Detection of QAM
…
Detection of M-ary QAM (two-dimension) : M=2k 2E
For one-dimension M -ary PAM:
2E
Average Prob(symbol error) Pe ( M − PAM ,1D )
… …
2( M − 1)  2 E 
= Q
 η 

M  
For two-dimension M-ary QAM:
Average Prob(correct decision) Pc ( M −QAM , 2 D ) = 1 − Pe ( ( ) 2
…
M − PAM ,1D )
Average Prob(symbol error) Pe ( M −QAM , 2 D ) = 1 − Pc ( M −QAM , 2 D )
 2E 
(
= 1 − 1 − Pe ( M − PAM ,1D )
)
2
For M=4, Pe ( 4 − PAM ,1D ) = Q
 η 

 
Average Prob(symbol error) Pe ( 4−QAM , 2 D ) = 1 − 1 − Pe ( ( 4 − PAM ,1D )
)
2
2
  2 E   2E   2 E 
= 1 − 1 − Q
 η 
  = Q
 η 
  2 − Q
 η 


      
7 Oct. 2009 TELE3113 13