Tut3_solutions

Copyright 2000 by A. Bouzerdoum
N–1 n=0
p ( a, ϕ ) = pA ( a ) p φ ( ϕ )
The autocorrelation function of x ( n ) is given by
r xx ( l ) = -r xx ( l + N ) = -1 N
N–1 n=0
∑
x ( n ) x ( n – l ) = -1 N
n = –M
∑
x ( n ) x ( n ) = !x
The average power:
(v) From part (i), the autocorrelation of a periodic signal is periodic with the same period. Therefore, we can determine the period of a signal by finding the period of its autocorrelation sequence. Page 3 Copyright 2000 by A. Bouzerdoum Page 4
x(n – l)
The crosscorrelation sequence can be written as follows.
r xy ( l ) = 2 N + 1 – l – n 0 ,
n0 + N – l
Exercise 3.
x(m ) x ( n ) = A cos ( ω 0 n + φ ) A and φ are random variables with pdfs, respectively, pA ( a ) = 1 ⁄ 3,
∑
∞
x ( k ) x ( l + k ) = r xx ( l )
r xy ( l ) =
∑
n = –∞Biblioteka x(n)y(n – l) =
∑
n = n0 – N
y(n – l) =
m = n0 – N – l
∑
y(m)
(i) For n 0 + N – l < – N ⇒
( l > 2 N + n 0 ) , r xy ( l ) = 0. ⇒ ( n0 ≤ l ≤ 2N + n0 ),
2
∑
x ( n + l + N ) x ( n ) = --
∑
x ( n + l ) x ( n ) = r xx ( l )
cos ( α ) cos ( β ) = - [ cos ( α + β ) + cos ( α – β ) ]
we can write 1 cos ( ω 0 n + φ ) cos ( ω 0 ( n – l ) + φ ) = - [ cos ( 2 ω 0 n – ω 0 l + 2 φ ) + cos ( ω 0 l ) ] 2 Therefore, the autocorrelation is given by
r xy ( l ) =
m = –N
∑
1 = 2N + 1 – l + n0 ⇒ ( n0–2N ≤ l < n0 ) , 1 = 2N + 1 + l – n0
(iii) For n 0 + N – l > N and n 0 – N – l ≤ N
N
The autocorrelation of the sequence x ( n ) is given by
1 2π 0
2π
= 0
sion (in terms of a ) must be nonpositive.
∆′ = b r xx ( l ) – b x ≤ 0
(iv) Energy signal
∞
2 2
2
2
⇒
r xx ( l ) ≤ x = r xx ( 0 )
2
2
2
µ x = E [ x ] = E [ A ] E [ cos ( ω 0 n + φ ) ] = 0
y = ∑ y ( n ) y ( n ) = ∑ [ ax ( n ) + bx ( n – l ) ] 2
n = –∞ n = –∞
∞
r xx ( l ) = - cos ( ω 0 l )
The mean: Since
3 2
y = x a 2 + 2 br xx ( l ) a + x b 2
r xx ( l ) = 2 N + 1 – l ,
The cross-correlation r xy ( l ) is given by
∞
Let k = 3 – n , then r yy ( l ) can be written as
n0 + N
n0 + N – l
r yy ( l ) =
k = –∞
n0 + N – l
Solutions of Tutorial No. 3
Solutions of Tutorial No. 3
Exercise 1.
The autocorrelation sequences of the signals x ( n ) = { 1, 2, 1, 1 } and y ( n ) = { 1, 1, 2, 1 } .
r xy ( l ) =
(iv) For n 0 – N – l > N
m = n0 – N – l n0 + N
∑
r xx ( l ) =
Let m = n – l , then we have
n = –∞
∑
∞
x(n)x(n – l) =
n = n0 – N
∑
⇒ ( l < n 0 – 2 N ) , r xx ( l ) = 0. for n 0 –2 N ≤ l ≤ n0 + 2 N
ENS2342: Digital Signal Processing
Solutions of Tutorial No. 3
ENS2342: Digital Signal Processing (ii) For n 0 + N – l ≥ n 0 – N and n 0 – N – l ≤ n 0 – N
⇒ ( l < – 2 N ) , r xx ( l ) = 0. for – 2N ≤ l ≤ 2 N
r yy ( l ) =
n = –∞
∑
∞
y( n )y (n – l ) =
n = –∞
∑ x(3 – n)x(3 – (n – l))
∞
Note that the correlation is an even function, and hence it can be written as follows:
n0 + N – l
Exercise 2.
1; x(n) = 0; 1; y(n) = 0; n0 – N ≤ x ≤ n0 + N otherwise –N ≤ x ≤ N otherwise
(ii) For n 0 + N – l ≥ – N and n 0 – N – l ≤ – N
Therefore, the autocorrelation is periodic with period N . (ii) The autocorrelation sequence is an even function.
r xx ( l ) =
n = –∞ ∞
∑
∞
x( n )x( n – l ) =
n = –∞
∑
x( n + l )x( n )
∫0 ( a
3
2
⁄ 3 ) da = a ⁄ 9
3
3 0
= 3
∞
1 2π - cos ( 2 ω 0 n – ω 0 l + 2 ϕ ) dϕ = 0 E [ cos ( 2 ω 0 n – ω 0 l + 2 φ ) ] = ---2 π ∫0
(iii) Consider an energy signal y ( n ) = ax ( n ) + bx ( n – l ) . The energy of y ( n ) is given by
The autocovariance: 3 C xx ( l ) = r xx ( l ) – µ x = - cos ( ω 0 l ) 2 The variance:
2 σx
r xx ( l ) ≤ r xx ( 0 )
r xx ( 0 ) =
n = –∞
∑
x ( n ) x ( n ) = x
⇒ (0 ≤ l ≤ 2N),
r xx ( l ) = r xx ( l ) = { 1, 3, 5, 7, 5, 3, 1 } r yy ( l ) = { 1, 3, 5, 7, 5, 3, 1 }
m = n0 – N
∑
1 = 2N + 1 – l ⇒ ( –2 N ≤ l < 0 ) ,
(iii) For n 0 + N – l > n 0 + N and n 0 – N – l ≤ n 0 + N
Since the energy of a signal is always positive, the discriminant ∆ of the previous quadratic expres1 2π