Filetype PDF | Posted on 30 Jan 2023 | 2 years ago
Authentication
digital resources
130x Filetype PDF File size 0.08 MB Source: cas.tudelft.nl
Delft University of Technology
Faculty of Electrical Engineering, Mathematics, and Computer Science
Circuits and Systems Group
EE 4C03 STATISTICAL DIGITAL SIGNAL
PROCESSING AND MODELING
10 November 2016, 13:30–16:30
Open book exam: copies of the book by Hayes and the course slides allowed. No other
materials allowed.
This exam has four questions (40 points).
Question 1 (9 points)
Consider the complex random process
x(n) = Aej(ω0n+φ) +w(n)
where w(n) is a zero mean white Gaussian noise random process with variance σ2. For each
w
of the following cases,
– find the mean and the autocorrelation sequence of x(n);
– if the process is wide sense stationary (WSS), find the power spectrum.
(a) A is a Gaussian random variable with zero mean and variance σ2, and both ω0 and φ
A
are constants.
(b) φ is uniformly distributed over the interval [−π,π] and both A and ω0 are constants.
(c) ω0 is a random variable that is uniformly distributed over some interval [Ω0−∆,Ω0+∆],
and both A and φ are constants.
(d) ω0 is a random variable that is uniformly distributed over some interval [Ω0−∆,Ω0+∆],
φ is uniformly distributed over the interval [−π,π], and A is a constant.
Hint: you may need this DTFT pair:
x(n) = sin(∆n) ↔ X(ω)= 1, |ω| < ∆
πn 0, elsewhere
Solution
(a) When ω0 and φ are constants, then
mx(n) = E{Aej(ω0n+φ)+w(n)}
= E{A}ej(ω0n+φ)
= 0
and
r (k,l) = E{x(k)x∗(l)}
x
= E{Aej(ω0k+φ)Ae−j(ω0l+φ)}+σ2δ(k −l)
w
= E{A2}ejω0(k−l) +σ2δ(k −l)
w
= σ2ejω0(k−l) +σ2δ(k −l)
A w
This depends only on the difference k − l, and x(n) is a WSS process. The power
spectrum is
P (ejω) = 2πσ2u (ω −ω )+σ2
x A 0 0 w
(b) Because φ varies uniformly over [−π,π], again mx(n) = 0, and
rx(k,l) = E{A2ej(ω0(k−l)}+σ2δ(k −l)
w
= A2ejω0(k−l) +σ2δ(k −l)
w
The process is WSS, and
P (ejω) = 2πA2u (ω −ω )+σ2
x 0 0 w
(c) Now
m (n) = E{Aej(ω0n+φ)} = AejφE{ejω0n}
x
The expected value can be evaluated as
E{ejω0n} = 1 Z Ω0+∆ejω0ndω0 = ejΩ0nsin(∆n)
2∆ Ω0−∆ ∆n
The expected value is not independent of n, and x(n) is not WSS.
The autocorrelation is
rx(k,l) = A2E{ejω0(k−l)}+σ2δ(k−l)
w
= A2ejΩ0(k−l)sin(∆(k−l)) +σ2δ(k −l)
∆(k−l) w
Since it is not WSS, the power spectrum is not defined.
(d) If now φ is uniformly varying, we obtain again mx(n) = 0, while the autocorrelation is
as before only dependent on the difference k −l, hence
r (k) = A2ejΩ0ksin(∆k) +σ2δ(k)
x ∆k w
The process is WSS. Using the given DTFT pair, the power spectrum is
A2π +σ2, Ω0−∆<ω0<Ω0+∆
Px(ω) = ∆ w
0, elsewhere
Question 2 (10 points)
Suppose that x(n) is a wide-sense stationary process with autocorrelation sequence rx(n)
and our goal is to model x(n) using an optimal AR (all-pole) model with a system transfer
function of the form
b(0)
H(z) = 1+Pp ap(k)z−k ,
k=1
wherepisthemodelorder. WeconsidertheLevinson-Durbinrecursionandthecorresponding
FIR lattice filter.
(a) Write down the Yule-Walker equations for this scenario.
2
(b) What is the computational complexity (order of magnitude) of the Levinson method as
a function of the filter order p? And what is the complexity of solving the Yule-Walker
equations directly?
(c) Assume that the autocorrelation function of x(n) is given by
rx(−3) = 0, rx(−2) = 1, rx(−1) = 2, rx(0) = 3, rx(1) = 2, rx(2) = 1, rx(3) = 0.
Determine the reflection coefficients Γj, the model parameters aj(k) and the modeling
errors ǫj for j = 1,2.
(d) Conversely, given the reflection coefficients, it should be possible to recover the auto-
correlation sequence.
Recall that the Schur recursion resulted in a FIR lattice filter that showed how the
reflection coefficients are obtained from the autocorrelation sequence:
[0,rx(1),rx(2),···] [0,0,0,γ3,···]
Γ1 Γ2
Γ1 Γ2
[rx(0),rx(1),rx(2),···] z−1 z−1 [0,0,ǫ3,···]
Starting from this, derive a lattice filter implementation that, given the reflection coef-
ficients and rx(0), generates the rest of autocorrelation sequence at one of its outputs.
(Specify also the input of the filter.)
Nowsuppose that x(n) is a wide-sense stationary process and our goal is to predict x(n) one
step ahead. In other words, we want to estimate x(n+1) from x(n), x(n−1), ..., x(n−p)
using a linear filter w(n) of order p, i.e., xˆ(n+1) = w(0)x(n)+w(1)x(n−1)+···+w(p)x(n−p).
(e) Write down the Wiener-Hopf equations for this scenario (without solving them).
(f) How is the solution for the filter coefficients w(n) related to the optimal all-pole model
for the random process x(n)? Write down the system transfer function for this all-pole
model using the filter coefficients w(n).
Solution
(a)
rx(0) rx(−1) · · · rx(−p) 2
1 |b(0)|
.
. a (1) 0
rx(1) rx(0) . rx(−p+1) p
. = .
. . . . .
. . . . .
. . . ap(p) 0
rx(p) rx(p −1) ··· rx(0)
or (focusing only on the a-coefficients)
rx(0) rx(−1) · · · rx(−p+1) rx(1)
a (1)
. p
. r (2)
rx(1) rx(0) . rx(−p+2) . x
. =− .
.
. . . .
. . . .
. . . ap(p) rx(p)
rx(p−1) rx(p−2) ··· rx(0)
3
(b) The computational complexity of the Levinson method is O(p2), while directly solving
the Yule-walker equations costs O(p3).
(c) The solution can be obtained using the following steps:
Step 1:
a0(0) = 1, ǫ0 = rx(0) = 3
Step 2:
γ0 = rx(1) = 2
Γ1 = −γ0/ǫ0 = −2/3
a1(1) = Γ1 = −2/3
ǫ =ǫ (1−Γ2)=3(1−4/9)=5/3
1 0 1
Step 3:
γ1 = rx(2) +a1(1)rx(1) = 1 −2/3·2 = −1/3
Γ2 = −γ1/ǫ1 = 1/5
a2(1) = a1(1) +Γ2a1(1) = −2/3−1/5·2/3 = −4/5
a2(2) = Γ2 = 1/5
ǫ =ǫ (1−Γ2)=5/3(1−1/25) =8/5
2 1 2
(d) This is obtained by ’reversing’ the direction of the top arrows. The copying of the
autocorrelation coefficients is replaced by an addition. This gives
[0,rx(1),rx(2),···] [0,0,···]
−Γ1 −Γ2
Γ1 Γ2
[r (0),0,0,···] −1 −1
x z z
At the right side, the input is set to zero and the output is not used. This gives the
’maximum entropy’ solution. Instead, the output can be fed back to that input via
more lattice sections and arbitrary chosen reflection coefficients |Γj| < 1, j = 3,4,···,
showing that the recovered autocorrelation sequence is not unique for rx(j), j = 3,4,···.
(e) The (real-valued) Wiener-Hopf equations are of the form:
rx(0) rx(1) · · · rx(p) w(0) rx(1)
rx(1) rx(0) · · · rx(p −1)w(1) rx(2)
= .
. . . . . .
. . . . . .
. . . . . .
rx(p) rx(p−1) ··· rx(0) w(p) rx(p+1)
(f) This resembles the Yule-Walker equations for all-pole modeling if we replace a(n) by
−w(n−1). Hence, the all-pole system function related to the solution for w(n) will be
of order p + 1 and it is given by
H(z) = b(0) .
1−w(0)z−1−w(1)z−2···−w(p)z−p−1
4
no reviews yet
Please Login to review.
