Environmental Data Analysis with MatLab
Lecture 14:
Applications of Filters
SYLLABUS
Lecture 01
Lecture 02
Lecture 03
Lecture 04
Lecture 05
Lecture 06
Lecture 07
Lecture 08
Lecture 09
Lecture 10
Lecture 11
Lecture 12
Lecture 13
Lecture 14
Lecture 15
Lecture 16
Lecture 17
Lecture 18
Lecture 19
Lecture 20
Lecture 21
Lecture 22
Lecture 23
Lecture 24
Using MatLab
Looking At Data
Probability and Measurement Error
Multivariate Distributions
Linear Models
The Principle of Least Squares
Prior Information
Solving Generalized Least Squares Problems
Fourier Series
Complex Fourier Series
Lessons Learned from the Fourier Transform
Power Spectral Density
Filter Theory
Applications of Filters
Factor Analysis
Orthogonal functions
Covariance and Autocorrelation
Cross-correlation
Smoothing, Correlation and Spectra
Coherence; Tapering and Spectral Analysis
Interpolation
Hypothesis testing
Hypothesis Testing continued; F-Tests
Confidence Limits of Spectra, Bootstraps
purpose of the lecture
further develop the idea of the
Linear Filter
and its applications
from last lecture
present output ∝ past and present values of input
output
input
filter
“convolution”, not
multiplication
Part 1: Predicting the Present
or
Part 1: Predicting the Present
or
very close to a
convolution
output
input
“prediction error”
filter
strategy for predicting the future
1. take all the data, d, that you have up to today
2. use it to estimate the prediction error filter, p
(use generalized least-squares to solve p*d=0)
3. use the filter, p, and all the data, d, to predict
dtomorrow
application to the Neuse River hydrograph
4
discharge, cfs
x 10
2
1
fs)2 per cycle/day
0
0
500
9
x 10
8
6
4
1000
1500
2000
2500
time, days
3000
3500
4000
pef(t) (std)
error filter, p(t)
prediction
here’s the best fit filter, p
1.5
1
0.5
0
-0.5
0
10
20
30
40
50
60
time t, days
time t, days
1.5
g)
1
70
80
90
pef(t) (std)
error filter, p(t)
prediction
here’s the best fit filter, p
in this case, only the
first few coefficients
are large
1.5
1
what’s that?
0.5
0
-0.5
0
10
20
30
40
50
60
time t, days
time t, days
1.5
g)
1
70
80
90
importance of the prediction error
since one is using least squares, the equation
0=p*d
is not solved exactly
the prediction error, e=p*d
tells you what aspects of the data cannot be predicted
on the basis of past behavior
d(t)
prediction
e(t) error, e(t) discharge,
d(t)
A)
15000
10000
5000
0
-5000
0
50
100
250
300
350
150
200
time t, days
250
300
350
time t, days
B)
15000
150
200
time t, days
10000
5000
0
-5000
0
50
100
time t, days
d(t)
prediction
e(t) error, e(t) discharge,
d(t)
A)
15000
10000
5000
0
-5000
0
50
100
the error is
small
10000
250
300
350
time t, days
B)
15000
150
200
time t, days
the error is
spiky
5000
many spikes are at
times when discharge
increases
0
-5000
0
50
100
150
200
time t, days
time t, days
250
300
350
Part 2: Inverse Filters
Can a convolution be undone?
if θ = g * h
is there another filter ginv for which
h = ginv * θ
?
convolution c=a*b by hand: step 1
for simplicity, suppose a and b are of length 3
write a backward in time
write b forward in time
overlap the ends by one, and
multiply. That gives c1
convolution c=a*b by hand: step 2
slide a right one place
multiply and add. That gives c2
convolution c=a*b by hand: step 3
slide a right another place
multiply and add. That gives c3
convolution c=a*b by hand: keep going
Multiply. That gives c5
an important observation
this is the same pattern that we obtain when we
multiply polynomials
z-transform
turn a filter into a polynomial
g = [g1, g2, g3, …
g(z) = g1 + g2 z + g3
2
z
T
gN]
+ … gN
N-1
z
inverse z-transform
turn a polynomial into a filer
g(z) = g1 + g2 z + g3 z2 + … gN zN-1
g = [g1, g2, g3, …
T
gN]
why would we want to do this?
because we know a lot about
polynomials
the fundamental theorem of algebra
a polynomial of n-th order has exactly n roots
and thus can be factored into the product of
n factors
the fundamental theorem of algebra
a polynomial of n-th order has exactly n-roots
largest power, zn
solutions to g(z)=0
and can be factored into the product of n factors
g(z) ∝ (z-r1) (z-r2) … (z-rn)
in the case of a polynomial constructed
from a length-N filter, g
where r1, r2, … rN-1 are the roots
so, the filter g
is equivalent a “cascade” of
N-1 length-2 filters
now let’s try to find the inverse of a
length-2 filter
the filter that undoes convolution by
[-ri, 1]T is … ?
z-transform
the function that undoes multiplication
by z-ri is1 /(z-ri)
problem:
1/(z-ri) is not a polynomial
solution: compute its Taylor series
Taylor series
Taylor series
contains all powers of z
so the filter that undoes convolution by
[-ri, 1]T is …
an indefinitely long filter
this filter will only be useful if its
coefficients fall off
must decrease
this happens when |ri|-1 > 1
or
|ri| < 1
this filter will only be useful if its
coefficients fall off
must decrease
this happens when |ri|-1 < 1
or
|ri| > 1
the root, ri, must lie
outside the “unit circle”
the inverse filter for a length-N filter g
step 1: find roots of g(z)
step 2: check that roots are inside
the unit circle
step 3: construct inverse filter
associated with each root
step 4: convolve them all together
example
construct inverse filter of:
6
g(j)
gj
4
2
0
5
10
15
20
25
30
element j
ginv(j)
element j
0.1
0
-0.1
35
40
45
50
only hard part of the process is finding
the roots of a polynomial
% find roots of g
r = roots(flipud(g));
fortunately, MatLab does this easily
6
g(j)
gj
4
2
0
5
10
15
20
25
30
element j
35
40
45
50
25
30
elementjj
element
element j
ginv(j)
0
-0.1
[g*ginv](j)
gjinv
0.1
[ginv*g]j
0
5
10
15
20
35
40
45
50
0
5
10
15
20
35
40
45
50
1
0
-1
25
30
elementjj
element
short time series
6
g(j)
gj
4
2
0
5
10
15
20
25
30
element j
35
40
45
50
20
35
40
45
50
20
35
40
45
50
element j
long timeseries
ginv(j)
gjinv
0.1
0
-0.1
[g*ginv](j)
0
[ginv*g]j
5
10
15
25
30
elementjj
element
spike
1
0
-1
0
5
10
15
25
30
elementjj
element
Part 3: Recursive Filters
a way to make approximate a long filter
with
two short ones
in the standard filtering formula
we compute the output θ1, θ2, θ3, … in sequence
but without using our knowledge of
θ1 when we compute θ2
or θ2 when we compute θ3
etc
that’s wasted information
suppose we tried to put the
information to work, as follows
here we’ve introduced two new
filters, u and v
convention
al filter, g
new but
conventional filter v that acts on
filter, u
already computed
values of θ
now define v1=1, so
if we can find short filters u and v such that
vinv * u ≈ g
then we can speed up the convolution process
an example
g*h is the weighted average of recent
values of h
if g is truncated to N≈10 elements, then each time step
takes 10N multiplications and 10N additions
try
this works, since the inverse of a length-2
filter is
the convolution then becomes
which requires only one addition and one
multiplication per time step
a savings of a factor of about ten
A)
h(t)
and q(t)
h1(t) and q1(t)
1
0
-1
0
h(t)
and q(t)
h2(t) and q2(t)
5
10
20
30
40
20
30
40
50
time,
time t, t
60
70
80
90
100
60
70
80
90
100
B)
0
-5
0
10
50
time,, tt
time