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ADSP-15-MSP-QMF-EC623 by harshi446

VIEWS: 8 PAGES: 35

This deals about the digital processing programs

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									                                                                   www.jntuworld.com
   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
                                                  www.jntuworld.com
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
                              www.jntuworld.com
  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
                                          www.jntuworld.com
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
                                  www.jntuworld.com
  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
                                                                                      www.jntuworld.com
           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.
                                                                   www.jntuworld.com
• 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)
                                                                             www.jntuworld.com
                        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.
                                                                               www.jntuworld.com
• In   fact, |H1(e(jω))| is a mirror image of |H0(ejω )| with respect to the
                             Π
 quadrature frequency        2   , justifying the name quadrature mirror
 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
                                                                          www.jntuworld.com
 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
                                                               www.jntuworld.com
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)
                                                                                               www.jntuworld.com
                     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
                                                   www.jntuworld.com
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)
                                                                           www.jntuworld.com
• 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.
                                                                                                                    www.jntuworld.com
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
                                                             www.jntuworld.com
                     (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
                                                                                      www.jntuworld.com
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)
                               www.jntuworld.com
LHS:

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


       Y3(z) = (↑ L)V1(z)

              = V1(z L )


       Y3(z) = X(z L )G(z L)
                                                                                   www.jntuworld.com
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-
                                                                      www.jntuworld.com
    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)
                                                                                                        www.jntuworld.com
• 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
                                                                                                      www.jntuworld.com
• 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
                                                                                                www.jntuworld.com
                              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.
                                                                                          www.jntuworld.com
                    −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)
                                                                           www.jntuworld.com
                                     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
                                                                               www.jntuworld.com
                                   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)
                                                                  www.jntuworld.com
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.
                                                                                         www.jntuworld.com
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
                                                                            www.jntuworld.com
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
                                                    www.jntuworld.com
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        )
                                                                             www.jntuworld.com
• 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
                                                                                         www.jntuworld.com
  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
                                                                                www.jntuworld.com
                                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.
                                                                                    www.jntuworld.com
 • 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
www.jntuworld.com

								
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