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H 0 (z) x(n)x(n) o(n)o(n) M G 0 (z) M + vo(n)vo(n) yo(n)yo(n) H 1 (z) 1(n)1(n) M G 1 (z) M...
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Transcript of H 0 (z) x(n)x(n) o(n)o(n) M G 0 (z) M + vo(n)vo(n) yo(n)yo(n) H 1 (z) 1(n)1(n) M G 1 (z) M...
H0(z)
x(n)
o(n)
M G0(z)M
+
vo(n) yo(n)
H1(z)1(n)
M G1(z)M
v1(n) y1(n)
fo(n)
f1(n)
y(n)
Figure 31
HM-2(z)
M-2(n)
M GM-2(z)M
vM-2(n)
HM-1(z)
M-1(n)
M GM-1(z)M
vM-1(n)
fM-2(n)
fM-1(n)yM-1(n)
Without the decimator and interpolator, the input/output relationship is straightforward
Y z G z H z X zk kk
M
0
1
(32)
With the decimator and interpolator, we have to take it step-by-step
X z z V z Y zk k
k kz H z X z k M 0 1 2 1, , ,....., (33)
k kz H z X z k M 0 1 2 1, , ,....., (33)
V zM
z Wk kr
MM r
1
0
1 1
(34)
Decimated signal
Images
W er j r M 2 /
k kz H z X z k M 0 1 2 1, , ,....., (33)
V zM
z Wk kr
MM r
1
0
1 1
(34)W er j r M 2 /
F z V zk kM (35)
k kz H z X z k M 0 1 2 1, , ,....., (33)
V zM
z Wk kr
MM r
1
0
1 1
(34)W er j r M 2 /
F z V zk kM (35)
Y z G z F zk kk
M
0
1
(36)
k kz H z X z k M 0 1 2 1, , ,....., (33)
V zM
z Wk kr
MM r
1
0
1 1
(34)W er j r M 2 /
F z V zk kM (35)
Y z G z F zk kk
M
0
1
(36)
Y z G z V zk kM
k
M
0
1
k kz H z X z k M 0 1 2 1, , ,....., (33)
V zM
z Wk kr
MM r
1
0
1 1
(34)W er j r M 2 /
F z V zk kM (35)
Y z G z F zk kk
M
0
1
(36)
Y z G zM
zWk kr
r
M
k
M
1
0
1
0
1
(37)
k kz H z X z k M 0 1 2 1, , ,....., (33)
V zM
z Wk kr
MM r
1
0
1 1
(34)W er j r M 2 /
F z V zk kM (35)
Y z G z F zk kk
M
0
1
(36)
(37) Y z G zM
H zW X zWk kr
r
M
k
Mr
1
0
1
0
1
k kz H z X z k M 0 1 2 1, , ,....., (33)
V zM
z Wk kr
MM r
1
0
1 1
(34)W er j r M 2 /
F z V zk kM (35)
Y z G z F zk kk
M
0
1
(36)
(37) Y zM
X zW H zW G zrk
rk
r
M
k
M
1
0
1
0
1
Y z
MX z X zW
H z H z H z
H zW H zW H zW
H zW H zW H zW
G z
G z
G z
M
o M
o M
oM M
MM
M
1 1
1 1
1 1
11
11
1
0
1
1
,....
.....
....
....
.
zzzzzzzM
zY TTT xAAxgHx 1
(38)
(39)
Alias Component (AC) matrix
Note Y(z) is a 1x1 matrix!
zG
zG
zG
zWHzWHzWH
zWHzWHzWH
zHzHzH
zWXzXM
zzzY
MM
MMM
o
Mo
Mo
M
T
1
1
0
11
11
1
11
11
11.
....
....
.....
,....
Ax
Aliasing error
0
0
zT
zz gH Aliasing error free condition
(40)
Y z
MX z X zW
H z H z H z
H zW H zW H zW
H zW H zW H zW
G z
G z
G z
M
o M
o M
oM M
MM
M
1 1
1 1
1 1
11
11
1
0
1
1
,....
.....
....
....
.
Aliasing error
Aliasing error free condition
(41)
0
0
onz
zzz tgH
Condition for perfect reconstruction is simple in theory, as
zzzzzz tHgtgH 1
However,
zz
zAdjzzz t
Hdet
HtHg 1
Is complicated to solve and resulted in IIR synthesis filters.
An effective solution employing Polyphase decomposition
Mkl
M
l
lk zEzzH
1
0
Recalling,
1
1
111101
111110
100100
1
1
0 1
MMMM
MM
MM
MM
MM
MM
MM
M z
z
zEzEzE
zEzEzE
zEzEzE
zH
zH
zH
.
....
....
.....
,,,
,
,
Or simply
i.e.,
zzz M eEh
Similar treatment to the synthesis filter gives
Mlk
M
l
lMk zRzzG
1
0
1
.
....
....
.....
,,,
,
,
MMM
MM
MM
MM
MM
MM
MMT
M
MT
M zRzRzR
zRzRzR
zRzRzR
z
z
zG
zG
zG
111101
111110
100100
2
1
1
1
0
1
Or simply
i.e.,
MTMT zzzz Reg 1
Diagram illustration:
zH0
zHM 1
zH1
MzE
z-1
z-1
z-1
Diagram illustration:
zG0
zGM 1
zG1
MzR
z-1
z-1
z-1
Can be replaced with their polyphase components, as
zH0
zHM 1
zH1
zG0
zGM 1
zG1
M
M
M
M
M
M
Can be replaced with their polyphase components, as
nx ny
M
M
M
M
M
M
MzR
z-1
z-1
z-1
MzE
z-1
z-1
z-1
nx
ny
M
M
M
M
M
M
zR
z-1
z-1
z-1
zE
z-1
z-1
z-1
nx
ny
M
M
M
M
M
M
z-1
z-1
z-1
zP
z-1
z-1
z-1
nx
ny
IczzEzRzP m Results in perfect reconstruction
1
1
0
0
z
Iz M
Implementing FIR filters for M-channel filter banks, to begin with,
10
011
MIzz
MM Iz
z
Izzz 1
11
1
0
0
Noted that
01 RzzRzRzE JJ .....
Let
111
10
J
RzRzRzR .....
zzRzzRzzRRzzRzR
zEzRzP
JJJ
0111
11
0 ..........
Compute P(z)
zzRzzRzzzRzR J
011
11
0 ..........
I
zzRzzRRzRzR JJ
.......
.......... 0111
11
10
121
012
001
0R
It can be seen that both analysis and synthesis filters are FIR
100
210
121
01TRzR
123
012
0011
0R
111
111
111
01
2
24122
225
zzz
zzz
zzz
RzRzE
100
210
3211
1R
All the analysis filters can be generated from
111
111
111
01
2
24122
225
zzz
zzz
zzz
RzRzE
54310 225 zzzzzH
54311 2422 zzzzzH
5432 2 zzzzH
All the analysis filters can be generated from
111
111
111
11
13183
852
32
10
zzz
zzz
zzz
RzRzR
5430 23 zzzzF
5431 258 zzzzF
5432 38131 zzzzF
Spectral AnalysisSpectral Analysis
sk
N
sf k k s N
0
1
0 1*
Given f(n) = [f0, f1, ..... , fN-1] and an orthonormal basis
0 1 1n n nN, ,....,
rk
N
sk k r s otherwise
0
1
1* = for = and 0 i.e.,
The spectral (generalized Fourier) coefficients of f(n) are defined as
f k k k Nss
N
s
0
1
0 1
(66)
(67)
Eqn. 66 and 67 define the orthonormal transform and its inverse
Spectral AnalysisSpectral Analysis
If the members of are sinusoidal sequences, the transform is known as the Fourier Transform
The Parseval theorem - Conservation of Energy in orthonormal transform
f kk
N
nn
N
0
1 2
0
1 2
(68)
An Application - Spectral AnalysisAn Application - Spectral Analysis
0 1* N n N
N
N
1 1* N n
N N n 1 1*
f(n)
0
1
N-1
Orthonormal spectral analyser implemented with multirate filter bank
Figure 32
An Application - Spectral AnalysisAn Application - Spectral Analysis
Transform efficiency - measured by decorrelation and energy compactness
Correlation - Neighboring samples can be predicted from the current sample : an inefficient representation.
Energy Compactness - The importance of each sample in forming the entire signal. If every sample is equally important, everyone of them has to be included in the representation: again an inefficient representation.
An ideal transform: 1. Samples are totally unrelated to each other.2. Only a few samples are necessary to represent the
entire signal.
How to derive the optimal transform?How to derive the optimal transform?
E f n n
Given a signal f(n), define the mean and autocorrelation as
and E f n f n k R n k n ,
Assume f(n) is wide-sense stationary, i.e. its statistical properties are constant with changes in time
cons t and R n k n R ktan ,
Define and
(69)
(70)
1
1
1
1
1
0
Nf
f
f
f
How to derive the optimal transform?How to derive the optimal transform?
R
R R R N
R R R N
R N R N R
N
N
N N
0 1 1
1 0 2
1 2 0
1 1 1
1 1 2
1 2 1
....
....
.....
....
....
.....
= 2
where R k k and 2 0 1 (71)
Equation 69 can be rewritten as
C conv f E f fT
The covariance of f is given by
R E f f T (72)
(73) R T
How to derive the optimal transform?How to derive the optimal transform?
The signal is transform to its spectral coefficients with eqn 66
sk
N
sf k k s N
0
1
0 1*
Comparing the two sequences:
f n f f f andN N 0 1 1, ,..., , ,...., 0 1 -1
How to derive the optimal transform?How to derive the optimal transform?
The signal is transform to its spectral coefficients with eqn 66
sk
N
sf k k s N
0
1
0 1*
Comparing the two sequences:
f n f f f andN N 0 1 1, ,..., , ,...., 0 1 -1
a. Adjacent terms are relatedb. Every term is important
a. Adjacent terms are unrelatedb. Only the first few terms are
important
How to derive the optimal transform?How to derive the optimal transform?
The signal is transform to its spectral coefficients with eqn 66
sk
N
sf k k s N
0
1
0 1*
similar to f, we can define the mean, autocorrelation and covariance matrix for
R E T
How to derive the optimal transform?How to derive the optimal transform?
f n f f f andN N 0 1 1, ,..., , ,...., 0 1 -1
a. Adjacent terms are related a. Adjacent terms are unrelated
Adjacent terms are uncorrelated if every term is only correlated to itself, i.e., all off-diagonal terms in the autocorrelation function is zero.
Define a measurement on correlation between samples:
f fjj i
N
i
N
jj i
N
i
N
R i j and R i j
, , 1
1
1
1
1
1
1
1
(74)
How to derive the optimal transform?How to derive the optimal transform?
We assume that the mean of the signal is zero. This can be achieved simply by subtracting the mean from f if it is non-zero.
The covariance and autocorrelation matrices are the same after the mean is removed.
How to derive the optimal transform?How to derive the optimal transform?
f n f f f andN N 0 1 1, ,..., , ,...., 0 1 -1
b. Every term is important b. Only the first few terms are important
0
1
1
0
1
1N
r
r
r
r N
Note:
If only the first L-1 terms are used to reconstruct the signal, we have
f L r rr
L
0
1
(75)
How to derive the optimal transform?How to derive the optimal transform?
If only the first L-1 terms are used to reconstruct the signal, the error is
The energy lost is given by e eLT
L rr L
N
21
e f fL L r rr L
N
1
r rk
NT
rf k k f
*
0
1
but,
hence r rT T
rf f2
(76)
(77)
(78)
How to derive the optimal transform?How to derive the optimal transform?
Eqn. 78 is valid for describing the approximation error of a single sequence of signal data f. A more generic description for covering a collection of signal sequences is given by:
J E e e E
E f f R
L LT
L rr L
N
rT T
rr L
N
rT
f rr L
N
'
21
1 1
(79)
An optimal transform mininize the error term in eqn. 79. However, the solution space is enormous and constraint is required. Noted that the basis functions are orthonormal, hence the following objective function is adopted.
How to derive the optimal transform?How to derive the optimal transform?
J RrT
f r r rT
rr L
N
11
(80)
The term r is known as the Lagrangian multiplier
The optimal solution can be found by setting the gradient of J to 0 for each value of r, i.e.,
rr
JJ
0
Eqn 81 is based on the orthonormal property of the basis functions.
(81)
How to derive the optimal transform?How to derive the optimal transform?
R f r r r
The solution for each basis function is given by
(82)
ris an eigenvector of Rf and r is an eigenvalue
Grouping the N basis functions gives an overall equation
R fT T
N where 0 1,......., (83)
R = Rf= which is a diagonal matrix.The decorrelation criteria is satisfied
(84)