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Pictorial View Point of Spectra at Various
Stages in 2-Channel QMF

x [n]       v [n]       y0[n]           x [n]
0           0
H (Z)           2           2           F (Z)    0
0                                       0

x[n]                                                             x[n]

H (Z)                                   F (Z)
1              2           2            1
x [n]       v [n]       y [n]           x [n]
1           1           1               1
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Upper Channel (Low Pass)
X(e jw )
1

0           2   w
2
H0(e   jw   )       1

0           2   w

X0 (ejw )
1
2

0           2   w

V0 (ejw )           1
2

0           2   w
2
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Y0 (ejw )
1
2

0       2   w
2
F 0 (ejw )   1

0       2   w

X0 (ejw )
1
2

0       2   w

aliasterm at
output
F (Z)       1
0
2

0       2   w
2
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Lower Channel (High Pass)
X (ejw )
1

0           2   w
2
H1(e jw )   1

0           2   w

X1 (ejw )
1
2

0           2   w

V1 (ejw )
1
2

0           2   w
2
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Y1 (ejw )

1
2

0           2   w
2
F 1 (ejw )   1

0           2   w

X1 (ejw )    1
2

0           2   w

alias term at
output
F (Z)      1
1
2

0           2   w
2
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Perfect Reconstruction QMF Bank
• If   a QMF bank is free from aliasing, amplitude and phase distor-
tions, then it is said to have the perfect reconstruction property.

Mathematically:

• QMF:

1                                  1
X(z) = 2 [H0(z)F0(z) + H1 (z)F1(z)]X(z) + 2 [H0(−z)F0(z) + H1(−z)F1(z)]X(−z)

• Let          X(z) = T (z)X(z) + A(z)X(−z)

where, T (z) = 1 [H0(z)F0(z) + H1(z)F1(z)] is the T.F representing ampli-
2

tude and phase distortion.

• A(z) = 1 [H0(−z)F0(z) + H1(−z)F1(z)]
2
is the T.F representing aliasing.
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• Alias   free QMF:

A(z) = 0

ˆ
X(z) = T (z)X(z)

• Alias   and Distortion free QMF:
(T (ejw ))
T (ejw ) = |T (ejw )|ej

|T (ejw )| = c   and (T (ejw )) = a + bw
T (ejw ) = cej(a+bw) = ceja ejbw

T (ejw ) = ce−jn0 w

T (z) = cz −n0        &            c=0

• Therefore,   for a perfect reconstruction QMF

ˆ
X(z) = cz −n0 X(z)
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x(n) = c.x[n − n0]
ˆ                           &         c=0

∀   possible inputs x[n].

• In    other words, x[n] is merely a scaled and delayed version of x[n].
ˆ

A Simple Alias free QMF system:

• In   the earliest known QMF banks the analysis ﬁlters were related
as:
H1(z) = H0 (−z)                   (1)

|H1(ejω | = |H0 (ej(Π−w))|            (2)

• This    ensures that H1(z) is a good high pass ﬁlter if H0(z) is a good
low pass ﬁlter.
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• In   fact, |H1(e(jω))| is a mirror image of |H0(ejω )| with respect to the
Π
ﬁlters.

• For   alias cancelation we have:

H0 (−z)F0(z) + H1(−z)F1(z) = 0

⇒ F0(z) = H1(−z)      &       F1(z) = −H0(−z)          (3)

(1)&(3) ⇒

F0(z) = H0(z)     &      F1(z) = −H1(z)             (4)

• Thus     all four ﬁlters are completely determined by a single ﬁlter
H0 (z).   The designer has to concentrate on the design of only this
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ﬁlter.

• The   system with four ﬁlters related as above was known as QMF
Bank.

• But   the term QMF has since been used to indicate generalized
versions for ex., M -channel system.

• From (4) =⇒ F0(z)    and F1(z) are low pass and high pass ﬁlters, re-
spectively.

• Therefore,   the distortion function is

T (z) = 1 [H0(z).F0(z) + H1(z).F1(z)]
2

1
= 2 [H0(z).H0(z) + H1(z).(−H1(z))]

= 1 [H0 (z) − H1 (z)] = 1 [H0 (z) − H0 (−z)]
2
2        2
2
2        2
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Polyphase Representation:
• Let       H0(z) = E0(z 2 ) + z −1 .E1(z 2 )

H1(z) = H0(−z)

H1(z) = E0(z 2) − z −1 .E1(z 2)

• That   is, in matrix-vector notation,

                            
2
 H0 (z)   1 1   E0(z ) 
        =                 
−1    2
H1 (z)     1 −1    z .E1(z )

• Similarly,

F0(z) = H0(z)

F0(z) = E0(z 2) + z −1 .E1(z 2)

F1(z) = −H1(z)
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F1(z) = −(E0(z 2 ) − z −1 E1(z 2))

ˆ      ˆ       ˆ
X(z) = X0(z) + X1(z)

ˆ
X(z) = Y0(z)F0(z) + Y1(z)F1(z)

ˆ
X(z) = Y0(z)(z −1 E1(z 2 ) + E0(z 2 )) + Y1(z)(z −1 E1(z 2) − E0(z 2))

ˆ
X(z) = z −1 E1(z 2)(Y0(z) + Y1(z)) + E0(z 2)(Y0(z) − Y1(z))

• That   is, in matrix-vector notation,

         
1 1 
F0(z) F1(z)   =    z −1 .E1(z 2) E0(z 2)       
1 −1
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x(n)
E (Z2)        2   2        E (Z2)
0                          1

−1                                           −1
Z                                            Z

E (Z2)        2   2        E (Z2)
1                          0
−1           −1            x(n)
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• By   using the noble identities,
G(z M )(↓ M ) ≡ (↓ M )G(z)

(↑ L)G(z L) ≡ G(z)(↑ L)

• We have        the modiﬁed structure as
x(n)
2    E (Z )                        E (Z )    2
0                             1

−1                                                            −1
Z                                                             Z

2    E (Z )                        E 0(Z )   2
1        −1             −1                 x(n)

• The        polyphase components are now operating at the lowest pos-
sible rate, so that the number of multiplications and additions per
unit time is minimized.
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Noble Identities:

1. ((↓ M )X[z])G(z) ≡ (X(z).G(z M ))(↓ M )

x[n]             v1[n]            y1[n]                x[n]                       v2[n]                    y2[n]
M
M              G(Z)                                      G(Z )                              M

RHS:

y2[n] = (↓ M )v2[n]

Y2(z) = (↓ M )V2(z)

= (↓ M )(X(z)G(z M ))
(M−1)             j2πl               j2πl
1               1                  1
=   M           X(z M e    M     )G((z M e    M     )M )
l=0
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(M−1)                   j2πl
1                   1
Y2(z) =   M
X(z M e      M     )G(z)
l=0

LHS:

v1[n] = (↓ M )x[n]

(M−1)                 j2πl
1                   1
V1(z) =      M            X(z M .e      M     ).
l=0

Y1(z) = V1(z)G(z)

(M−1)                   j2πl
1                  1
Y1(z) =   M
X(z M .e       M     ).G(z)
l=0
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2. (X(z)G(z))(↑ L) ≡ ((↑ L)X(z))G(z L)

x[n]            v1[n]            y3[n]           x[n]        x4[n]           y4[n]
L
G(Z)                 L                           L           G(Z )

RHS:

x4[n] = (↑ L).x[n]

X4(z) = (↑ L).X(z)

X4(z) = X(z L ) −→   Compression in the Frequency Domain

Y4(z) = X(z L )G(z L)
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LHS:

V1(z) = X(z)G(z)

Y3(z) = (↑ L)V1(z)

= V1(z L )

Y3(z) = X(z L )G(z L)
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M-Channel Filter Bank:
v (n)         y (n)
x(n)                    x (n)        0             0               x (n)
0      M            M            F (Z)     0
H (Z)                                       0
0

x (n)        v (n)       y (n)            x 1(n)
1            1           1       F (Z)
H (Z)           M            M              1
1

x (n)        v (n)        y (n)           x (n)
H (Z)    M−1          M−1          M−1            M−1
M−1            M            M            F (Z)
M−1             x(n)

• x[n]   is split into M subband signals xk [n] by the M analysis ﬁlters
Hk (z).

• Each      signal is then decimated by M to obtain vk [n].

• The    decimated signals are eventually passed through M -fold ex-
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jw
H (e )

0                     2                       (M−1)       w

M         M                        M

panders and recombined via the synthesis ﬁlters Fk (z) to produce
x[n].
ˆ

Expression for the Reconstructed Signal :

• Each   subband signal is given by

Xk (z) = Hk (z).X(z)               (5)
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• The   decimated signal vk [n] has Z-Transform
(M−1)                   −j2πl
1                    1
Vk (z) =     M
Xk (z M .e    M      )
l=0

(M−1)
1                      1     −j2πl              1   −j2πl
Vk (z) =              Hk (z M .e       M      ).X(z M .e      M      )        (6)
M
l=0

• The   outputs of the expander are therefore given by

(M−1)
1                            −j2πl             −j2πl
Yk (z) = Vk (z M ) =                   Hk (z.e     M      ).X(z.e    M       )   (7)
M
l=0

• Therefore,   the reconstructed signal is
(M−1)
ˆ
X(z) =           Fk (z).Yk (z)
k=0
(M−1)              (M−1)                 −j2πl           −j2πl
1
=         Fk (z). M               (Hk (z.e    M      ).X(z    M        ))
k=0                  l=0
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• Rearranging    the terms we have

M−1                           M−1
ˆ      1                     −j2πl                        −j2πl
X(z) =             X(z.e      M      ).         Hk (z.e    M      ).Fk (z)    (8)
M
l=0                          k=0

• Let,

M−1
1                   −j2πl
Al (z) =           Hk (z.e    M      ).Fk (z)            0≤l ≤M −1              (9)
M
k=0

• Therefore,

M−1
−j2πl
ˆ
X(z) =             X(z.e        M      ).Al (z)                   (10)
l=0

• The    spectrum of reconstructed signal is therefore given by
M−1                    2πl
ˆ
X(ejw ) =             X(ej(w− M )).Al (ejw )
l=0
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2πl
• Where, X(ej(w− M ) )                 for l = 0 represents a shifted version of X(ejw ).

• Therefore,                                     ˆ
the reconstructed spectrum X(ejw ) is a linear combina-
tion of X(ejw ) and its (M − 1) uniformly shifted versions.

Errors Created in M -Channel QMF Bank
Aliasing Errors:

• The     output of synthesis ﬁlter bank i.e. reconstructed signal is
given by
M−1
−j2πl
ˆ
X(z) =         X(z.e    M      ).Al (z)      (11)
l=0
−j2πl
• X(z.e    M      ), l > 0    is due to the decimation and interpolation opera-
tions.
−j2πl
• For l > 0, X(z.e        M        )   represents the lth aliasing term.
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−j2πl
• Each X(z.e        M      )   is associated with term Al (z).

• Aliasing    can be eliminated for every possible input x[n], iﬀ

Al (z) = 0          1≤l ≤M −1

Illustration:                   M =3⇒    Three parallel ﬁlters.

M−1
−j2πl
ˆ
X(z) =            X(z.e    M      ).Al (z)   (12)
l=0
−j2π
Let, W = e    M

M−1
ˆ
X(z) =          X(z.W l ).Al (z)
l=0
2
ˆ
X(z) =         X(z.W l ).Al (z)
l=0

ˆ
X(z) = X(z).A0(z) + X(zW ).A1(z) + X(zW 2).A2(z)              (13)
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M−1
1
Al (z) = .             Hk (z.W l ).Fk (z)   (14)
M
k=0

2
1
A0(z) = .             Hk (z).Fk (z)       (15)
3
k=0

2
1
A1(z) = .             Hk (z.W ).Fk (z)     (16)
3
k=0

2
1
A2(z) = .          Hk (z.W 2).Fk (z)        (17)
3
k=0

• A1 (z)   and A2(z) are the terms associated with aliasing. For the
aliasing free condition,

A1(z) = 0
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2
1
3.          Hk (z.W ).Fk (z) = 0
k=0

2
⇒            Hk (z.W ).Fk (z) = 0
k=0

H0 (z.W ).F0(z) + H1 (z.W ).F1(z) + H2 (z.W ).F2(z) = 0   (18)

Similarly,

A2(z) = 0
2
1
3
.         Hk (z.W 2).Fk (z) = 0
k=0

H0(z.W 2).F0(z) + H1(z.W 2).F1(z) + H2(z.W 2).F2(z) = 0    (19)
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Plot of     H0(z), H1(z)   and   H2(z):

H0(Z)             H1(Z)        H2(Z)

o              π                2π
3                 3           π   ω

→   Uniform ﬁlter bank.
→ H0 (z)   is a Low Pass Filter.
→ H1 (z)   and H2(z) are Band pass Filters.
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Plot of   H0(z), H0(zW )        and    H0(zW 2):

−j2π                 −j2π
1.   H0(zW ) = H0 (z.e    M     ) = H0 (z.e    3     )
−j4π
2.   H0(zW 2) = H0 (z.e    3     )

H0 (Z)

−π       − 2π
3             −π
3                  0        π
3
2π
3        π   ω
H0 (ZW )

0                           ω
H0(ZW 2)

0                           ω
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Plot of   H1(z), H1(zW )        and      H1(zW 2):

−j2π
1.   H1(zW ) = H1 (z.e    3     )
−j4π
2.   H1(zW 2) = H1 (z.e    3        )

H0(Z)

−π        − 2π
3                −π
3
0         π
3
2π
3   π   ω
H1(ZW )

0                      ω

H1(ZW 2 )

0                      ω
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Plot of   H2(z), H2(zW )        and     H2(zW 2):

−j2π
1.   H2(zW ) = H2 (z.e    3     )
−j4π
2.   H2(zW 2) = H2 (z.e    3        )
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• Signal    which enters F0(z) is coming through the channel having the
analysis ﬁlter H0(z).

• Due     to the decimation and interpolation the signal which enters
F0(z)    contains the terms H0(z)X(z), H0(zW )X(zW ) and H0(zW 2)X(zW 2).
→    The ﬁrst term is the required one, and other represents alias
terms.

• F0(z)   is used to eliminate the alias terms X(zW ) and X(zW 2). For
this response |F0(ejw )| should resemble H0(ejw ).

• Responses     of F1(z) and F2(z) also follow the responses of H1(z) and
H2 (z)   for the similar reasoning.

• Since    the ﬁlters Fk (z) and Hk (z) are not ideal in practice, they do
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not eliminate the shifted replicas Hk (zW ) and Hk (zW 2) completely.

• Responses   of H0(zW ).F0(z), H1(zW ).F1(z) and H2(zW ).F2(z) overlap.

• Objective   of alias cancelation is to chose the synthesis ﬁlter such
that these overlapping terms cancel out.

Amplitude and Phase Distortions:
Output of Synthesis Bank:
M−1                   M−1
ˆ
X(z) =           ˆ
Xk (z) =              Vk (z).Fk (z)
k=0                   k=0

M−1         M−1
ˆ
X(z) =           1
(M         X(zW l ).Hk (zW l )).Fk (z)
k=0        l=0

M−1                          M−1
l       1
=          X(zW          ).( M         Fk (z).Hk (zW l ))
l=0                         k=0
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M−1
ˆ
X(z) =           X(zW l ).Al (z)            0≤l ≤M −1
l=0

For Alias Free M -Channel QMF System:
Al (z) = 0         1 ≤ l ≤ M − 1.

• Therefore,               ˆ
X(z) = X(z).A0 (z)

ˆ
X(z) = X(z).T (z)

• Where, T (z) = A0(z)        is the distortion function.
M−1
.              1
T (Z) = A0(z) =      M
Hk (z).Fk (z).
k=0

• Substituting Z = ejw ,        we have

T (ejw )
T (ejw ) = |T (ejw )|.ej

• If |T (ejw )| = C = 0,        ˆ
then X(z) will not have any amplitude distortion.
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• Similarly,     T (ejw ) = a + bω ,                               ˆ
a linear phase factor, then X(z) will not
have any phase distortion.

Perfect Reconstruction System:
If Hk (z) and Fk (z) are such that aliasing is completely canceled and
T (z)   is a pure delay (i.e. T (z) = C.z −n0 , C = 0), then the system is free
from aliasing, amplitude and phase distortions. Such system will
have x[n] = C.x[n − n0] and is called a prefect reconstruction system.
ˆ
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