Correlated and Uncorrelated Signals
Problem: we have two signals x[n]
and y[n] . How “close” are they to each other?
Example: in a radar (or sonar) we transmit a pulse and we expect a return
1
0.8
0.6
Transmit
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
0
2
4
6
8 10 12 14 16 18 20
1.5
1
0.5
Receive
0
-0.5
-1
-1.5
-2
0
20
40
60
80
100
120
140
160
180
Example: Radar Return
Since we know what we are looking for, we keep comparing what we receive
with what we sent.
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
-0.2
-0.2
-0.4
-0.4
-0.6
-0.6
-0.8
-0.8
-1
0
2
4
6
8
10
12
14
16
18
20
-1
0
2
4
6
8
10
12
14
16
18
20
1.5
1
0.5
0
Receive
-0.5
-1
-1.5
-2
0
Similar?
20
NO!
40
60
80
100
120
140
Think so!
160
180
Inner Product between two Signals
We need a “measure” of how close two signals are to each other.
This leads to the concepts of
• Inner Product
• Correlation Coefficient
Inner Product
Problem: we have two signals x[n]
and y[n] . How “close” are they to each other?
Define: Inner Product between two signals of the same length
N 1
rxy   x[n] y *[n]
n 0
Properties:
N 1
N 1
rxx   x[n]x [n]   x[n]  0
*
n 0
2
n 0
2
rxy  rxx ryy
2
rxy  rxx ryy
if and only if
y[n]  C x[n] for some constant C
How we measure similarity (correlation coefficient)
Assume: zero mean
Compute:
 xy 
| rxy |
rxx ryy
Check the value:
0   xy  1
 xy  0
x,y uncorrelated
 xy  1
x,y strongly correlated
Back to the Example: with no return
x[n]
x[n] y[n]
y[n]
2
3
3
2
2
1.5
1
0.5
1
0
0
1
0
-0.5
-1
-1
-1
-2
-2
-1.5
-3
-2
-3
0
100
200
300
400
500
600
700
800
900
0
100
200
300
400
500
600
700
800
0
100
200
300
400
500
600
700
800
900
1000
900
rxy  2.27
rxx  500
ryy  982
 xy  0.003 NO Correlation!
Back to the Example: with return
x[n]
x[n] y[n]
y[n]
2
2.5
3
1.5
2
2
1
1.5
0
1
1
0.5
0
-0.5
-1
0.5
-1
-1.5
-2
0
-2
0
100
200
300
400
500
600
700
800
900
-3
0
100
200
300
400
500
600
700
800
900
-0.5
0
100
200
300
400
500
600
700
800
900
rxy  494
rxx  500
ryy  754
  0.8 Good Correlation!
1000
Inner Product in Matlab
Take two signals of the same length. Each one is a vector:
x  x(1) x(2)  x( N )
Row vector
y  y(1)
Row vector
y(2)  y( N )
Define: Inner Product between two vectors
 y * (1) 
 * 
N
y ( 2) 
*

rxy   x(n) y (n)  x(1) x(2)  x( N )
  
n 1
x
 *

 y ( N )
rxy  x * y'
y'
conjugate,
transpose
Example
Take two signals:
y
2.5
x
2
3
2
1.5
1
1
0.5
0
0
-0.5
-1
-1
-1.5
-2
-2
-2.5
0
50
100
150
200
Compute these:
rxy  x * y'  19.7
rxx  x * x'  218.8
ryy  y * y'  241.9
250
300
-3
0
50
100
150
200
250
300
Then:
 xy 
19.7
 0.0856 0
218.8  241.9
x,y are not correlated
Example
Take two signals:
Compute these:
3
x
rxy  x * y'  230.9
2
1
rxx  x * x'  229.6
0
-1
ryy  y * y'  234.3
-2
-3
y
0
50
100
150
200
250
300
Then:
3
2
 xy 
1
230.9
 0.9955 1
229.6  234.3
0
-1
x,y are strongly correlated
-2
-3
0
50
100
150
200
250
300
Example
Take two signals:
y
3
x
3
2
2
1
1
0
0
-1
-1
-2
-2
-3
0
50
100
150
200
Compute these:
rxy  x * y'  230.9
rxx  x * x'  229.6
ryy  y * y'  234.3
250
300
-3
0
50
100
150
200
250
300
Then:
230.9
 xy 
 0.9955 1
229.6  234.3
x,y are strongly correlated
Typical Application: Radar
Send a Pulse…
s[n]
n
N
… and receive it back with noise, distortion …
y[n]
n
n0
Problem: estimate the time delay
n0 , ie detect when we receive it.
Use Inner Product
“Slide” the pulse s[n] over the received signal and see when
the inner product is maximum:
N 1
rys [n]   y[n  ]s *[]
rys[n]  0, if n  n0
 0
n
y[]

n0
s[]

N
Use Inner Product
“Slide” the pulse x[n] over the received signal and see when
the inner product is maximum:
N 1
rys [n]   y[n  ]s*[]  MAX
 0
y[]
  n0

s[]

N
if
n  n0
Matched Filter
Take the expression
N 1
rys [n]   y[n  ]s*[]
n 0
 s*[ N  1] y[n  N  1]  ...  s*[1] y[n  1]  s*[0] y[n]
Compare this, with the output of the following FIR Filter
rˆ[n]  h[0] y[n]  ...  h[1] y[n  1]  h[ N  1] y[n  N  1]
Then
y[n]
h[n]
rˆ[n]  rys[n  N  1]
h[n]  s*[ N 1  n], n  0,...,N 1
Matched Filter
This Filter is called a Matched Filter
y[n]
rˆ[n]
h[n]
h[n]  s*[ N 1  n], n  0,...,N 1
The output is maximum when n  N  1  n0
i.e.
n  n0  N 1
rˆ[n]  rys[n  N  1]
Example
We transmit the pulse s[n], n  0,...,N  1 shown below, with
length N  20
1
0.8
0.6
0.4
s[n]
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
0
2
4
6
8
10
12
14
16
18
Max at n=119
20
 n0  119 20  1  100
Received signal:
12
y[n]
1.5
rˆ[n]
1
y[n]
0.5
0
8
6
h[n]
-0.5
10
4
2
0
-1
-2
-1.5
-4
-2
0
20
40
60
80
100
120
140
160
180
-6
0
20
40
h[n]  s*[ N 1  n], n  0,...,N 1
60
80
100
120
140
160
180
200
How do we choose a “good pulse”
We transmit the pulse s[n], n  0,...,N  1 and we receive
(ignore the noise for the time being)
y[n]  As[n  n0 ]
rˆ[n]
h[n]
rˆ[n]  rys[n  N  1]
h[n]  s*[ N 1  n], n  0,...,N 1
where
N 1
rys [n]  A s[n  n0  ]s*[]
n 0
 A rss [n  n0 ]
N 1
The term rss [n]   s[n  ]s*[]
 0
is called the “autocorrelation of s[n]”. This characterizes
the pulse.
Example: a square pulse
rss [n]
N
s[n]
1
0
N 1
N
n
N
n
N 1
rss [n]   s[n  ]s*[]  ?
 0
N 1
N 1
rss [0]   s[]s []   s[]  N
2
*
 0
N 2
See a few values:
 0
N 2
rss [1]   s[1  ]s []  1  N  1
*
 0
rss [k ] 
 0
N 1 k
rss [k ] 
N 1 k
 s[k  ]s []  1  N  k
*
 0
N 1 k
 0
N 1
*
s
[

k


]
s
[]  1  N  k

 0
k
Compute it in Matlab
s[n]
20
1
18
0
N 1
16
n
14
12
10
N=20;
% data length
s=ones(1,N);
% square pulse
8
6
4
2
rss=xcorr(s);
% autocorr
n=-N+1:N-1;
% indices for plot
stem(n,rss)
% plot
0
-20
-15
-10
-5
0
5
10
15
20
Example: Sinusoid
25
1
20
0.8
15
0.6
0.4
10
0.2
0
5
-0.2
-0.4
0
-0.6
-5
-0.8
-1
0
5
10
15
20
25
30
35
40
45
50
-10
-15
s[n], n  0,...,49
-20
-50
-40
-30
-20
-10
0
10
20
30
rss [n], n  49,...,49
40
50
Example: Chirp
30
1
25
0.8
0.6
20
0.4
15
0.2
0
10
-0.2
5
-0.4
-0.6
0
-0.8
-1
-5
0
5
10
15
20
25
30
35
40
45
50
-10
-50
s[n], n  0,...,49
s=chirp(0:49,0,49,0.1)
-40
-30
-20
-10
0
10
20
30
rss [n], n  49,...,49
40
50
Example: Pseudo Noise
50
2.5
40
2
1.5
30
1
0.5
20
0
-0.5
10
-1
-1.5
0
-2
-2.5
0
5
10
15
20
25
30
35
40
45
50
-10
-20
-50
-40
-30
-20
-10
0
10
20
s[n], n  0,...,49
s=randn(1,50)
rss [n], n  49,...,49
30
40
50
Compare them
chirp
cos
pseudonoise
1
s[n]
2.5
1
0.8
2
0.8
0.6
1.5
0.6
0.4
1
0.4
0.2
0.2
0.5
0
0
0
-0.2
-0.2
-0.4
-0.4
-0.6
-0.6
-0.5
-1
-1.5
-0.8
-1
-0.8
-2
0
5
10
15
20
25
30
35
40
45
-1
50
0
5
10
15
20
25
30
35
40
45
50
-2.5
30
rss [n]
0
5
10
15
20
25
30
35
0
10
20
40
45
50
50
25
25
20
15
20
10
15
40
30
20
5
10
0
5
10
0
0
-5
-10
-5
-15
-20
-50
-10
-40
-30
-20
-10
0
10
20
30
40
50
-10
-50
-40
-30
-20
-10
0
10
20
30
40
50
-20
-50
-40
Two best!
-30
-20
-10
30
40
50
Detection with Noise
Now see with added noise
y[n]  As[n  n0 ]  w[n]
h[n]
rˆ[n]  rys[n  n0  N  1]  ryw[n]
h[n]  s*[ N 1  n], n  0,...,N 1
White Noise
A first approximation of a disturbance is by “White Noise”.
White noise is such that any two different samples are
uncorrelated with each other:
3
w[n]
2
1
0
-1
-2
-3
-4
0
100
200
300
400
500
600
700
800
900
1000
White Noise
The autocorrelation of a white noise signal tends to be a
“delta” function, ie it is always zero, apart from when n=0.
rss [n]
n
White Noise and Filters
The output of a Filter
N 1
w[n]   h[]w[n  ]
w[n]
 0
h[n]
1
M
M 1
 N 1 N 1

   h[ 1 ]h[ 2 ]w[n   1 ]w[n   2 ] 



n 0  1 0  2 0

N 1 N 1
 1 M 1

   h[ 1 ]h[ 2 ]  w[n   1 ]w[n   2 ] 
1 0  2 0
 M n 0

1
w[ n ] 

M
n 0
2
M 1
 N 1
2  1
   h[] 
  0
 M

w[n] 

n 0

M 1
2
White Noise
The output of a Filter
N
w[n]   h[]w[n  ]
w[n]
 0
h[n]
In other words the Power of the Noise at the ouput is
related to the Power of the Noise at the input as
 N 1
2
PW    h[n]  Pw
 n 0

Back to the Match Filter
rˆ[n]  Arss [n  n0  N  1]  w[n]
y[n]  As[n  n0 ]  w[n]
h[n]
h[n]  s*[ N 1  n], n  0,...,N 1
At the peak:
rˆ[n0  N 1]  Arss [0]  w[n0  N 1]
Match Filter and SNR
At the peak:
rˆ[n0  N 1]  Arss [0]  rsw [n0  N 1]
N 1
 N 1

2 
Arss [0]   | As[n] |  | s[n] |2 
 n 0
 n0

2
 N 1
2
PW   | s[n] |  PW
 n 0

 N 1
2
N  PS    s[n] 
 n 0
  N  SNR
SNRpeak 
 N 1
2
  s[n]   PW
 n 0

Example
Transmit a Chirp of length N=50 samples, with SNR=0dB
2
30
1.5
25
20
1
15
0.5
10
0
5
-0.5
0
-1
-5
-1.5
-10
-2
-15
0
50
100
150
200
Transmitted
250
300
0
200
400
600
800
1000
Detected with
Matched Filter
1200
Example
Transmit a Chirp of length N=100 samples, with SNR=0dB
50
2
40
1.5
30
1
0.5
20
0
10
-0.5
0
-1
-1.5
-2
-10
0
50
100
150
200
250
Transmitted
300
-20
0
200
400
600
800
1000
Detected with
Matched Filter
1200
Example
Transmit a Chirp of length N=300 samples, with SNR=0dB
160
2
140
1.5
120
1
100
0.5
80
0
60
-0.5
40
-1
20
-1.5
0
-2
0
50
100
150
200
250
300
-20
-40
Transmitted
0
200
400
600
800
1000
Detected with
Matched Filter
1200
1400