Digital Signal Processing - eng.staff.alexu.edu.eg
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Transcript of Digital Signal Processing - eng.staff.alexu.edu.eg
• Reading:
• Mitra
– Sections: 10.10.3&10.10.4
– Sections: 10.12 (10.12.1&10.12.2)
11/29/2018 14
perfect reconstruction 𝑴− band filter bank
• The DFT filter bank has the disadvantage that the synthesis filters have a much higher order than the analysis filters, in case of perfect reconstruction.
• In general, for perfect reconstruction filter bank, the filters 𝑯𝒌(𝒛) are NOT a modulated version of 𝑯𝟎(𝒛)
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Polyphase representation of the 𝑴−band filter bank
• polyphase representation of
the analysis filter bank 𝐻𝑘(𝑧)
• polyphase representation of
the synthesis filter bank 𝐺𝑘(𝑧)
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= 𝒛−(𝑴−𝟏) 𝒛−(𝑴−𝟐) ⋯ 𝟏 𝑹(𝒛𝑴)
=
=
−−−−
−
−
−−−−
−
−
)()()(
)()()(
)()()(
)(
,
)()()(
)()()(
)()()(
)(
1,11,10,1
1,11110
1,00100
1,11,10,1
1,11110
1,00100
zRzRzR
zRzRzR
zRzRzR
z
zEzEzE
zEzEzE
zEzEzE
z
MMMM
M
M
MMMM
M
M
R
E
Polyphase representation of the 𝑴−band filter bank
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Perfect reconstruction 𝑴−band filter bank
• perfect reconstruction (PR) condition is– 𝐸 𝑧 is the polyphase matrix of 𝐻𝑘 𝑧
– 𝑅(𝑧) is the polyphase matrix of 𝐺𝑘(𝑧)
• 𝑐 is a constant
• In general, the PR filter-bank is equivalent to a pure delay ∆
∆= 𝑴− 𝟏 + 𝒇𝒊𝒍𝒕𝒆𝒓 𝒅𝒆𝒍𝒂𝒚 > 𝟎
11/29/2018 18
IRE−= czzz )()(
)((z)
)()(
1 zc
czz
−=
=
ER
IRE
• example:
determine the analysis and
synthesis filters given 𝑬(𝒛𝟑)?
• The analysis filters 𝑯𝒊 𝒛 :
• For perfect reconstruction, we have
𝑬 𝒛 𝑹 𝒛 = 𝑰
• Synthesis filters 𝑮𝒊 𝒛 :
• = 𝒛−(𝑴−𝟏) 𝒛−(𝑴−𝟐) ⋯ 𝟏 𝑹(𝒛𝑴)
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• Analysis filters:
• PR condition:
• Synthesis filters:
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22)( ,21)( ,21)(
211
022
211
1
)(
)(
)(
211
022
211
4)(
)(4)( 4)()(
1)( ,1)( ,1)(
1
101
111
111
)(
)(
)(
101
111
111
)(
2
2
21
1
21
0
12
2
1
0
3
1
2
2
21
1
21
0
2
1
2
1
0
3
−−−−−
−−
−
−−−−−
−
−
+−=+−=++=
−
−=
−
−=
==
−=+−=++=
−
−=
−
−=
zzGzzzGzzzG
zz
zG
zG
zG
z
zzzz
zzHzzzHzzzH
z
z
zH
zH
zH
z
R
ERIRE
E
• example: determine the synthesis filters of a three-band prefect reconstruction filter bank if the analysis filters are:
• Constructing the polyphase matrix
• For perfect reconstruction we must have:
11/29/2018 21
1)(
2)(
64)(
2
1
1
21
0
=
+=
++=
−
−−
zH
zzH
zzzH
=
=
=
−
−
−
−
001
012
164
)(
1
001
012
1641
)(
)(
)(
)(
3
2
1
2
13
2
1
0
z
z
z
z
zz
zH
zH
zH
E
E
)()(
)()(
1 zz
zz
−=
=
ER
IRE
• The synthesis filters in polyphase form
• Synthesis filters transfer functions:
11/29/2018 22
−
−=
==
−
−
861
210
100
001
012
164
)()(
1
1 zz ER
−
−=
=
−−
−−
821
610
100
1
)(
)(
)(
)( 1
)(
)(
)(
12
2
1
0
312
2
1
0
zz
zG
zG
zG
zzz
zG
zG
zG
R
82)(
6)(
1)(
12
2
1
1
0
+−=
−=
=
−−
−
zzzG
zzG
zG
• example
• Polyphase components
• Perfect reconstruction condition
• Polyphase representation
11/29/2018 23
211
211)( ,
11
2121)(
−=
−= zz RE
=
=
−
−=
10
01
211
211
11
2121)()( zz RE
• Analysis filters
• Synthesis filters
Not ideal filters!!11/29/2018 24
211
211)( ,
11
2121)(
−=
−= zz RE
)1(2
1)(
)()()(
1)(
)()()(
1
1
2
11
2
01
1
1
1
0
2
10
2
00
1
0
−
−
−
−
−−=
+=
+=
+=
zzG
zRzRzzG
zzG
zRzRzzG
𝐻0(𝜔) 𝐻1(𝜔)
1
1
2
11
12
101
1
0
2
01
12
000
1)(
)()()(
)1(2
1)(
)()()(
−
−
−
−
−=
+=
+=
+=
zzH
zEzzEzH
zzH
zEzzEzH
• example: find the analysis and synthesis filters?
• Perfect reconstruction condition:
• This filter bank is a perfect reconstruction
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1 2 1
1
1
1 1 2
1 3 1 1 1
8 4 8 4 4( ) ,
1 11
2 2
1 11
4 4( )
1 1 1 3 1
2 2 8 4 8
z z z
z
z
z
z
z z z
− − −
−
−
− − −
− − + + +
=
+ −
+
=
+ − +
E
R
=
= −1
1-
-1
z0
0z)()( zzz RE
𝐻0(𝜔)
𝐻1(𝜔)
• Analysis filters:
• Synthesis filters:
• Not ideal filters !!
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21
1
2
11
12
101
4321
0
2
01
12
000
2
1
2
1)(
)()()(
8
1
4
1
4
3
4
1
8
1)(
)()()(
−−
−
−−−−
−
+−=
+=
−+++−
=
+=
zzzH
zEzzEzH
zzzzzH
zEzzEzH
4321
1
2
11
2
01
1
1
21
0
2
10
2
00
1
0
8
1
4
1
4
3
4
1
8
1)(
)()()(
2
1
2
1)(
)()()(
−−−−
−
−−
−
++−+=
+=
++=
+=
zzzzzG
zRzRzzG
zzzG
zRzRzzG
The delay of this filter bank is 3 samples11/29/2018 27
𝑥[𝑛]
𝑢0[𝑛]
𝑦 𝑛
𝑢1[𝑛]
Two-channel Quadrature-Mirror filterbank (QMF)
• Maximally “critically” sampled filterbank
• Linear Time Varying (LTV) filterbank
• 𝐻0 𝑧 & 𝐺0(𝑧) are lowpass filters
• 𝐻1 𝑧 & 𝐺1(𝑧) are highpass filters
Analysis of the QMF filterbank
The filterbank output
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)(ˆ)()(ˆ)()(
10 )()()()(2
1
)()( 2
1 )(ˆ
10 )()(ˆ
)()( 2
1)(
)()()(
1100
2
2121
zVzGzVzGzY
kzXzHzXzH
zVzVzV
kzUzV
zVzVzU
zXzHzV
kk
kkk
kk
kkk
kk
+=
−−+=
−+=
=
−+=
=
The reconstructed signal at the filterbank output
The distortion transfer function
The aliasing component
11/29/2018 29
)()()()(2
1)(
)()()()(2
1)(
)()()()()(
)()()()()(2
1
)()()()()(2
1)(
)(ˆ)()(ˆ)()(
1100
1100
1100
1100
1100
zGzHzGzHzA
zGzHzGzHzT
zXzAzXzTzY
zXzGzHzGzH
zXzGzHzGzHzY
zVzGzVzGzY
−+−=
+=
−+=
−−+−+
+=
+=
• alias-cancellation condition
“LPF”
“HPF”
• “ the filterbank becomes LTI”
• perfect reconstruction (PR) condition
11/29/2018 30
l
l
l
zzHzHzHzH
zzGzHzGzH
zzTlnxny
zXzTzY
zHzHzHzH
zHzG
zHzG
zGzHzGzH
zGzHzGzHzA
zGzHzGzHzT
zXzAzXzTzY
−
−
−
=−−−
=+
=−=
=
=−−−−−
−−=
−=
=−+−
−+−=
+=
−+=
2)()()()(
2)()()()(
)(][][
)()()(
0)()()()(
)()(
)()(let
0)()()()(
)()()()(2
1)(
)()()()(2
1)(
)()()()()(
0110
1100
0110
01
10
1100
1100
1100
spectrum components in the 2-channels QMF filterbank
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alias term
alias term
P. P. Vaidyanathan, "Multirate digital filters, filter banks, polyphase networks, and applications: a tutorial," in Proceedings of the
IEEE, vol. 78, no. 1, pp. 56-93, Jan. 1990
0}{2
1)(
)}()()()({2
1)(
}{2
1)(
)}()()()({2
1)(
1)( )(
)( 1)(
11
1100
111
1100
1
1
1
1
00
=+=
−+−=
=+=
+=
==
==
−−
−−−
−
−
zzzA
zGzHzGzHzA
zzzzT
zGzHzGzHzT
zGzzH
zzGzH
• example: trivial two-channel PR QMF filterbank
• distortion transfer function
• aliasing term
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0})1()1()1()1{(4
1)(
)}()()()({2
1)(
})1()1()1()1{(4
1)(
)}()()()({2
1)(
)()1(2
1)( )1(
2
1)(
)()1(2
1)( )1(
2
1)(
1111
1100
11111
1100
0
1
1
1
1
1
1
0
1
0
=+−+++−=
−+−=
=+−−+++=
+=
−−=+−=−=
−=+=+=
−−−−
−−−−−
−−
−−
zzzzzA
zGzHzGzHzA
zzzzzzT
zGzHzGzHzT
zHzzGzzH
zHzzGzzH
• example
• distortion transfer function
• alias term
11/29/2018 33
analysis synthesis
Polyphase implementation of the two-channels QMF
• The polyphase decomposition of the analysis filters 𝐻0 𝑧 &𝐻1(𝑧) and noting that 𝐻1 𝑧 = 𝐻0(−𝑧)
• In matrix form
• distortion transfer function
• The polyphase decomposition of the synthesis filters 𝐺0 𝑧 & 𝐺1(𝑧) and noting that 𝐺0 𝑧 = 𝐻0 𝑧 , and 𝐺1 𝑧 = −𝐻1 𝑧 = −𝐻0(−𝑧)
• In matrix form
11/29/2018 34
−=
−=
+=
−
−
−
)(
)(
11
11
)(
)(
)()()(
)()()(
2
1
1
2
0
1
0
2
1
12
01
2
1
12
00
zEz
zE
zH
zH
zEzzEzH
zEzzEzH
−=
+−=
+=
−
−
−
11
11 )()()()(
)()()(
)()()(
2
0
2
1
1
10
2
1
12
01
2
1
12
00
zEzEzzGzG
zEzzEzG
zEzzEzG
)()(2)(
)}()()()({2
1)(
2
1
2
0
1
1100
zEzEzzT
zGzHzGzHzT
−=
+=
• example:
11/29/2018 35
−=
−=
=
=
+=
−
−
−
11
111)()(
1
11
11
)(
)(
1)(
1)(
1)(
1
10
1
1
0
2
1
2
0
1
0
zzGzG
zzH
zH
zE
zE
zzH
