Introduction to Probability and Stochastic Processes I 6/ARMA Model … · One exception is the...
Transcript of Introduction to Probability and Stochastic Processes I 6/ARMA Model … · One exception is the...
Introduction to Probability andStochastic Processes I
Lecture 5
Henrik Vie Christensen
Department of Control Engineering
Institute of Electronic Systems
Aalborg University
Denmark
Slides originally by: Line Ørtoft Endelt
Introduction to Probability and Stochastic Processes I – p. 1/43
LLTIVC Systems
1. Lumped: A dynamic system is called lumped if it can bemodeled by a set of ordinary differential or differenceequations.
2. Linear:
f(a1x1(t) + a2x2(t)) = a1f(x1(t)) + a2f(x2(t))
3. Time Invariant: If y(t) = f(x(t)), then
y(t − t0) = f [x(t − t0)]
4. Causal:y(t0) = f [x(t); t ≤ t0]
Introduction to Probability and Stochastic Processes I – p. 2/43
Deterministic LTIVC I
The input-output relation can be written as
N∑
m=0
amy[(m + n)Ts] =N
∑
m=0
bmx[(m + n)Ts]
It is assumed that x, y, ai’s and bi’s are real-valued.
And the input-output relation can also be written
y(n) = x(n) ∗ h(n) =∞
∑
m=−∞
h(n − m)x(m) =∞
∑
m=−∞
h(m)x(n − m)
where h is the impulse response, since the system iscausal, then h(k) = 0 for k < 0.
Introduction to Probability and Stochastic Processes I – p. 3/43
Deterministic LTIVC II
For a stable system
∞∑
k=0
|h(k)| < ∞
The transfer function is found as
H(f) = F{h(n)} =∞
∑
m=−∞
h(n) exp(−j2πnf), |f | <1
2
The impulse response is recovered by
h(n) = F−1{H(f)} =
∫1
2
−1
2
H(f) exp(j2πnf) df
Introduction to Probability and Stochastic Processes I – p. 4/43
Deterministic LTIVC III
Assume that F (x(n)) and F (y(n)) exists, then
YF (f) =∞
∑
n=−∞
[
∞∑
m=−∞
h(m)x(n − m)]
exp(−j2πnf)
If the system is stable, this can be rewritten to
YF (f) = XF (f)H(f)
The output y(n) is recovered as
y(n) = F−1{YF (f)} =
∫1
2
−1
2
XF (f)H(f) exp(j2πnf)df
Introduction to Probability and Stochastic Processes I – p. 5/43
Deterministic LTIVC IV
The Z-transform of a discrete sequence is defined as
XZ(z) =
∞∑
n=0
x(n)z−n
HZ(z) =
∞∑
n=0
h(n)z−n
The Z-transform is only applicable when the sequence isdefined on the nonnegative integers.
For a stable system, the fourier transform and theZ-transform are equal if z = exp(j2πf).
The Z-transform of a stable system will exists if |z| > 1.
Introduction to Probability and Stochastic Processes I – p. 6/43
Random Input I
If the input to the system is a random sequence X(n), theinput-output relation may be written as
Y (n) =
∞∑
m=−∞
h(m)X(n − m)
The mean of the output is found as
E{Y (n)} = µY (n) =∞
∑
m=−∞
h(m)E{X(n − m)}
Introduction to Probability and Stochastic Processes I – p. 7/43
Random Input II
The autocorrelation of the output is found as
RY Y (n1, n2) = E{Y (n1)Y (n2)}
=∞
∑
m1=−∞
∞∑
m2=−∞
h(m1)h(m2)RXX(n1 − m1, n2 − m2)
TheDistribution function of the output sequence are ingeneral very difficult to obtain, and cannot in general beexpressed in closed functional form.
One exception is the Gaussian process, since Y (n) isobtained as a linear combination of Gaussians, and henceGaussian. The mean-value and the Variance can be foundfrom the two equations above.
Introduction to Probability and Stochastic Processes I – p. 8/43
WSS input
If X(n) is WSS, then
µY = E{Y (n)} =∞
∑
m=−∞
h(m)µX = µX
∞∑
m=−∞
h(m) = µXH(0)
and
RY Y (n1, n2) =∞
∑
m1=−∞
∞∑
m2=−∞
h(m1)h(m2)
× RXX [(n2 − n1) − (m2 − m1)]
Hence the output of a LTIVC system is WSS when the inputis WSS. (It can be shown that SSS → SSS).
Introduction to Probability and Stochastic Processes I – p. 9/43
Correlation and psd of Output I
Assume that X(n) is a real WSS input to a LTIVC system,the cross-correlation function between the input and output
RY X(k) = E{Y (n)X(n + k)}
= E{[
∞∑
m=−∞
h(m)X(n − m)]
X(n + k)}
=∞
∑
m=−∞
h(m)E{X(n − m)X(n + k)}
=∞
∑
m=−∞
h(m)RXX(k + m)
=∞
∑
n=−∞
h(−n)RXX(k − n) = h(−k) ∗ RXX(k)
Introduction to Probability and Stochastic Processes I – p. 10/43
Correlation and psd of Output II
Since RY X(τ) = RXY (−τ) and RXX is an even function
RXY (k) = h(k) ∗ RXX(k)
It can also be shown that
RY Y (k) = RY X(k) ∗ h(k)
and hence
RY Y (k) = RXX(k) ∗ h(k) ∗ h(−k)
Introduction to Probability and Stochastic Processes I – p. 11/43
Correlation and psd of Output III
The psd of Y (n) is the fourier transform of theautocorrelation function of Y (n)
SY Y (f) =∞
∑
n=−∞
RY Y (n) exp(−j2πnf), |f | <1
2
also
SY Y (f) = F{RY Y (k)}= F{RXX(k) ∗ h(k) ∗ h(−k)}= F{RXX(k)}F{h(k)}F{h(−k)}= SXX(f)H(f)H(−f)
= SXX(f)H(f)H∗(f)
= SXX(f)|H(f)|2Introduction to Probability and Stochastic Processes I – p. 12/43
Example
X(n) is a zero-mean white noise random sequence, withvariance σ2. Let X(n) be the input sequence to a LTIVCsystem, then
SXX(f) = F (RXX(n)) = σ2
since RXX(0) = σ2 and RXX(n) = 0 for n 6= 0, so
SY Y (f) = SXX(f)|H(f)|2 = σ2|H(f)|2
Also
RXY (k) = h(k) ∗ RXX(k) = h(k)σ2
So if the input and the output is known, the impulseresponse can be found.
Introduction to Probability and Stochastic Processes I – p. 13/43
Correlation and psd of Output IV
From the definition of the Z-transform
HZ [exp(j2πf)] = H(f)
Defining
S#XX(z) =
∞∑
n=0
z−nRXX(n)
S#XX [exp(j2πf)] = SXX(f)
It can be shown that
S#Y Y (z) = S#
XX(z)|H(z)|2
= S#XX(z)H(z)H(z−1)
Introduction to Probability and Stochastic Processes I – p. 14/43
Example a I
X(n): Stationary random sequence, is input to a LLTIVCsystem, with µX = 0, RXX(0) = 1 and RXX(k) = 0 for k 6= 0.
h(k) =
{
1 for k = 0, 1
0 for k > 1
The mean of the output Y (n):
µY (n) = µXH(0) = 0
The Transfer Function
H(f) =∞
∑
k=0
h(k) exp(−j2πkf)
= 1 + exp(−j2πf)
Introduction to Probability and Stochastic Processes I – p. 15/43
Example a II
The psd for X
SXX(f) = F{RXX(k)}
=∞
∑
k=0
RXX(k) exp(−j2πkf)
= 1, |f | <1
2
The psd of Y is found as
SY Y (f) = SXX(f)|H(f)|2
= 1 · |1 + exp(−j2πf)|2
= 2 + 2 cos 2πf, |f | <1
2Introduction to Probability and Stochastic Processes I – p. 16/43
Example a III
Using the Inverse Fourier Transform:
RY Y (k) = F−1{SY Y (f)}
=
∫1
2
−1
2
(2 + 2 cos 2πf) exp(j2πkf) df
=
∫1
2
−1
2
(2 + exp(−j2πf) + exp(j2πf)) exp(j2πkf) df
soRY Y (0) = 2
RY Y (±1) = 1
RY Y (k) = 0 |k| > 1
Introduction to Probability and Stochastic Processes I – p. 17/43
Example b I
X(n) is a zero-mean white noise random sequence, withvariance σ2. X(n) is input to a filter with transfer function
HZ(z) =a0 + a1z
−1 + a2z−2
1 + b1z−1
µY (n) = µXH(0) = 0
The psd of X(n) is
S#XX(z) =
∞∑
n=0
z−nRXX(n) = σ2
Introduction to Probability and Stochastic Processes I – p. 18/43
Example b II
The psd of Y (n) is
S#Y Y (z) = σ2
(a0 + a1z−1 + a2z
−2
1 + b1z−1
)(a0 + a1z + a2z2
1 + b1z
)
= σ2a20 + a2
1 + a22 + a1(a0 + a2)(z + z−1) + a0a2(z
2 + z−2)
1 + b21 + b1(z + z−1)
Switching to the Fourier domain, subst. z = exp(j2πf):
SY Y (f) = σ2a20 + a2
1 + a22 + 2a1(a0 + a2) cos 2πf + 2a0a2 cos 4πf
1 + b21 + 2b1 cos 2πf
|f | <1
2
Introduction to Probability and Stochastic Processes I – p. 19/43
Autoregressive Process I
An Autoregressive process is one represented by adifference equation of the form
X(n) =
p∑
i=1
φp,iX(n − i) + e(n)
Where X(n) is a real random sequence, φp,i areparameters, with φp,p 6= 0, and e(n) is a sequence of iidzero-mean white noise, i.e.
E{e(n)} = 0
E{(e(n))2} = σ2N
fe(n)(λ) =1√
2πσN
exp{
− λ2
2σ2N
}
Introduction to Probability and Stochastic Processes I – p. 20/43
Autoregressive Process II
Autoregressive models are also called state models,recursive digital filters, and all-pole models.The AR-model can be transformed to a state model
X(n) = ΦX(n − 1) + E(n)
Often the AR-model is written as
X(n) =
p∑
i=1
hiX(n − i) + e(n)
The transfer function of the AR-model is
H(f) =1
1 −∑p
i=1 φp,i exp(−j2πfi), |f | <
1
2
Introduction to Probability and Stochastic Processes I – p. 21/43
First-order Autoregressive Model I
First-order Autoregressive model:
X(n) = φ1,1X(n − 1) + e(n)
The mean of X(n) is found as
µX = E{X(n)} = E{φ1,1X(n − 1) + e(n)}= φ1,1µX + 0
hence for φ1,1 6= 1
µX = 0
Introduction to Probability and Stochastic Processes I – p. 22/43
First-order Autoregressive Model II
0 20 40 60 80 100−3
−2
−1
0
1
2
3
Sample function of a first-order AR-process:X(n) = 0.48X(n − 1) + e(n).
Introduction to Probability and Stochastic Processes I – p. 23/43
First-order Autoregressive Model III
0 20 40 60 80 100−10
−5
0
5
Sample function of a first-order AR-process:X(n) = 0.97X(n − 1) + e(n).
Introduction to Probability and Stochastic Processes I – p. 24/43
First-order Autoregressive Model IV
The variance of the AR-model is
σ2X = E{X(n)2}
= E{φ21,1X(n − 1)2 + e(n)2 + 2φ1,1X(n − 1)e(n)}
= φ21,1σ
2X + σ2
N
hence
σ2X =
σ2N
1 − φ21,1
Since σ2X is finite and nonnegative, then
−1 < φ1,1 < 1
Introduction to Probability and Stochastic Processes I – p. 25/43
First-order Autoregressive Model V
The autocorrelation function is given by
RXX(m) = E{X(n)X(n − m)}, m ≥ 1
= E{[φ1,1X(n − 1) + e(n)][X(n − m)]
= φ1,1RXX(m − 1)
Hence
RXX(m) = φm1,1RXX(0) = φm
1,1σ2X
and the autocorrelation coefficient of the process is
rXX(m) =RXX(m)
RXX(0)= φm
1,1
Introduction to Probability and Stochastic Processes I – p. 26/43
First-order Autoregressive Model VI
0 2 4 6 8 10 120
0.2
0.4
0.6
0.8
1
Corrolation coefficient of the first-order AR-process.Introduction to Probability and Stochastic Processes I – p. 27/43
First-order Autoregressive Model VII
The psd can be found using
SXX(f) = |H(f)|2See(f)
Since
See(f) = σ2N |f | <
1
2
H(f) =1
1 − φ1,1 exp(−j2πf)|f | <
1
2
|H(f)|2 =1
1 − 2φ1,1 cos(2πf) + φ21,1
|f | <1
2
Introduction to Probability and Stochastic Processes I – p. 28/43
First-order Autoregressive Model VIII
Therefore
SXX(f) =σ2
N
1 − 2φ1,1 cos(2πf) + φ21,1
|f | <1
2
SXX(f) =σ2
X(1 − φ21,1)
1 − 2φ1,1 cos(2πf) + φ21,1
|f | <1
2
If z−1 is defined to be the backshift operator, which can bewritten as:
z−1[X(n)] = X(n − 1), z−1[e(n)] = e(n − 1)
z−k[X(n)] = X(n − k), z−k[e(n)] = e(n − k)
Introduction to Probability and Stochastic Processes I – p. 29/43
First-order Autoregressive Model IX
Then the first-order autoregressive model can be written as
X(n) = φ1,1z−1[X(n)] + e(n)
Hence
X(n) =e(n)
1 − φ1,1z−1
and, then
X(n) =[
∞∑
i=0
φi1,1z
−i]
e(n) =∞
∑
i=0
φi1,1e(n − i)
A weighted infinite sum of white noise.
Introduction to Probability and Stochastic Processes I – p. 30/43
Second-order Autoregressive Model IThe second-order autoregressive process is given by
X(n) = φ2,1X(n − 1) + φ2,2X(n − 2) + e(n)
0 20 40 60 80 100−3
−2
−1
0
1
2
3
Introduction to Probability and Stochastic Processes I – p. 31/43
Second-order Autoregressive Model II
The mean value of the process is
µX = φ2,1µX + φ2,2µX
hence if φ2,1 + φ2,2 6= 1, then
µX = 0
Introduction to Probability and Stochastic Processes I – p. 32/43
Second-order Autoregressive Model III
The variance
σ2X = E{X(n)X(n)}
= E{φ2,1X(n)X(n − 1) + φ2,2X(n)X(n − 2) + X(n)e(n)}= φ2,1RXX(1) + φ2,2RXX(2) + σ2
N
Since RXX(k) = σ2XrXX(k), then
σ2X =
σ2N
1 − φ2,1rXX(1) − φ2,2rXX(2)
Since σ2X is finite and positive, then
φ2,1rXX(1) + φ2,2rXX(2) < 1
Introduction to Probability and Stochastic Processes I – p. 33/43
Second-order Autoregressive Model IVRXX(m) for m ≥ 1:
RXX(m) = E{X(n − m)X(n)}= E{φ2,1X(n − m)X(n − 1) + φ2,2X(n − m)X(n − 2)
+ X(n − m)e(n)}= φ2,1RXX(m − 1) + φ2,2RXX(m − 2)
This is a second-order linear homogeneous differenceequation, having the solution
RXX(m) = A1λm1 + A2λ
m2 if λ1 6= λ2
= B1λm + B2mλm if λ1 = λ2
λ1 and λ2 are roots of the characteristic equation obtainedby assuming RXX(m) = λm for m ≥ 1, which gives . . .
Introduction to Probability and Stochastic Processes I – p. 34/43
Second-order Autoregressive Model V
λ2 = φ2,1λ + φ2,2
hence
λ =φ2,1 ±
√
φ22,1 + 4φ2,2
2
If the model must satisfy the initial condition, RXX(0) = σ2X ,
and we require stationarity, then
RXX(1) = φ2,1RXX(0) + φ2,2RXX(−1)
RXX(1) =φ2,1
1 − φ2,2σ2
X
Introduction to Probability and Stochastic Processes I – p. 35/43
Second-order Autoregressive Model VI
If λ1 6= λ2, then
RXX(0) = σ2X = A1 + A2
RXX(1) = A1λ1 + A2λ2 =φ2,1
1 − φ2,2σ2
X
Now A1 and A2 can be found, hence RXX is known.
With ai = Ai/σ2X for i = 1, 2
rXX(m) =RXX(m)
σ2X
= a1λm1 + a2λ
m2
Introduction to Probability and Stochastic Processes I – p. 36/43
Second-order Autoregressive Model VII
So
rXX(1) =φ2,1
1 − φ2,2
rXX(2) =φ2
2,1
1 − φ2,2+ φ2,2
Substituting this in the expression for the variance gives
σ2X =
σ2N (1 − φ2,2)
(1 + φ2,2)(1 − φ2,1 − φ2,2)(1 + φ2,1 − φ2,2)
This will be finite if
φ2,2 6= −1, φ2,1 + φ2,2 6= 1, φ2,2 − φ2,1 6= 1
Introduction to Probability and Stochastic Processes I – p. 37/43
Second-order Autoregressive Model VIII
And positive if
−1 < φ2,2 < 1, φ2,1 + φ2,2 < 1, −φ2,1 + φ2,2 < 1
The psd is given by
SXX(f) = |H(f)|2σ2N , |f | <
1
2
Where
H(f) =1
1 − φ2,1 exp(−j2πf) − φ2,2 exp(−j4πf), |f | <
1
2
Introduction to Probability and Stochastic Processes I – p. 38/43
Second-order Autoregressive Model IX
Giving
SXX(f) =σ2
N
|1 − φ2,1 exp(−j2πf) − φ2,2 exp(−j4πf)|2 , |f | <1
2
This could also be found as the Fourier transform ofRXX(m).
Introduction to Probability and Stochastic Processes I – p. 39/43
General Autoregressive Model I
The general autoregressive model:
X(n) =
p∑
i=1
φp,iX(n − i) + e(n)
or using the backshift operator
X(n) =
p∑
i=1
φp,iz−iX(n) + e(n)
By using the same technique as before, the mean
µX = 0
Introduction to Probability and Stochastic Processes I – p. 40/43
General Autoregressive Model II
The variance
σ2X = E{X(n)X(n)} = E
{
X(n)
p∑
i=1
φp,iX(n − i) + X(n)e(n)}
=
p∑
i=1
φp,iRXX(i) + σ2N
The autocorrelation coefficient is given by
rXX(k) =RXX(k)
σ2X
=E{X(n − k)X(n)}
σ2X
=
p∑
i=1
φp,irXX(k − i), for k ≥ 1
Introduction to Probability and Stochastic Processes I – p. 41/43
General Autoregressive Model III
rXX(k) =
p∑
i=1
φp,irXX(k − i), for k ≥ 1
This is a pth order difference equation, which can beexpressed in matrix form
rXX = RΦ
where R is the correlation coefficient matrix, rXX is thecorrelation coefficient vector, and Φ is the autoregressivecoefficient vector.The matrix equation is called the Yule-Walker equation.Since R is invertible
Φ = R−1
rXX
Introduction to Probability and Stochastic Processes I – p. 42/43
General Autoregressive Model IV
The psd can be shown to be
SXX(f) = See(f)|H(f)|2
=σ2
N∣
∣1 −∑p
i=1 φp,i exp(−j2πfi)∣
∣
2 , |f | <1
2
Introduction to Probability and Stochastic Processes I – p. 43/43