Multistage Implementation
Problem: There are cases in which the filter requirements call for a digital filter of high
complexity, in terms of number of stages.
Example: a signal has a bandwidth of 450Hz and it is sampled at 96kHz. We want to resample
it at 1kHz:
| X (F ) |
Fx  96kHz
Fy  1 kHz
H ( z)
96
x ( n)
y ( m)
F (kHz)
045
.
| H () |
 
045
.
050
.
0.9

96
96


 
01
.
96
48 F kHz


Solution: for an FIR filter designed by the window method, the order of the filter
is determined by the size of the transition region. With a Hamming Window the
order is determined by the equation
8
01
.
 

M
96
which yields
M  960  8  7,680
Disdvantage: a lot of computations at a high freq. rate
7,680  96,000  0.74  109 multiplies / sec
One Stage Implementation: we decimate in one shot.
Fx
Fy 
LPF
x ( n)
Fx
D
D
y ( m)
Multistage Implementation: we decimate the signal in several stages.
Fy 
Fx  F1
F2
x ( n)
H1
D1
FL
F3
H2
D2
 
HL

D  D1 D2  DL
Fx
D
DL
y ( m)
See the last stage first:
FL
x L (n)
y ( m)
HL
DL
FL Fx
Fy 

DL D
H L ( )
p
Fp
Passband:
0  Fp
 
DL
Fy / 2

F

Fy
FL
F
Stopband: Fy / 2  FL / 2 , since D  2 F  F  2D  2
L
L
L
This filter clears everything above Fy / 2 .
Problem: we can design the low pass filters
that the spectrum X ( ) is bandlimited.
Hi
in a clever way, by taking into consideration
xi [n]
Fi
Hi
Di
X i ( )
xi 1[m]
Fi
Fi 1 
Di
X i 1 ()
pass
stop
Fy
aliased
Fy Fi1
2 2

Di
2
Fi 1
Fy
Fi
2
F


2
Fi1
2

F

Problem: we can design the low pass filters
that the spectrum X ( ) is bandlimited.
Hi
in a clever way, by taking into consideration
xi 1 (m)
xi ( n)
Fi
Hi
Specs for
• pass:
• stop:
H i ( )
:
Fi
Fi 1 
Di
Di
Hi
0  Fp
Fi 1  Fy 2  Fi 2
Fp
Fi 1 
Fy
2
Fi
2
F
Example: same problem we saw before:
Pass Band
Stop Band
Sampling Freq.
Use Multistage.
F1  96kHz
F2  12kHz
H1
x ( n)
8
[0, 450] Hz
> 500 Hz
96 kHz
F3  2 kHz
H2
6
F4  1kHz
H3
2
y ( m)
115
.
045
.
0.9 23
96
96
15
.
045
.
0.9 3
12
12
045
.
050
.

0.9
2
2

 


 


 

22.1
96
2.1
12
0.1
2
35
46
160
3.4  10
6
0.5  106
F

kHz

rad
rad
order M
0.3  106 mult./sec
Efficient Multirate Implementation
Goal: we want to determine an efficient implementation of a multirate system. For example in
Decimation and Interpolation:
s(n)
x(n)
H (z )
y (m)
D
 


 


you have to compute
every D samples.
s(m D)
only, ie. one

x(n)
y (m)
H (z )
I
s (m)
  
  
most of the values of s(mD) are zero

Noble Identities:
x(n)
y (m)
G( z D )

D
x(n)
y (m)   g D (mD  k ) x(k )

k
y (m)
G (z )
D
y(m)   g (m  ) x(D)
 0 if k  D

y(m)   g D (mD D) x(D)

g (m)
g2 (n)


 
1
2
G( z )

 n
g 2 (2m)  g (m)



1
G (z )
 m
For example take an FIR Filter
x ( n)
h(0)
zD
h(1)

y ( m)
D

x ( n)
y ( m)
h(0)
D
z 1

zD
z 1
h(2)
since
z
D
D

D
z 1
h(1)
h(2)


Similarly:
x(n)
y (m)
G( z M )
M
y (m)   g M (m  kM ) x(k )

 

k
 0 if m  M

x(n)
y (m)
G (z )
M
 g (  k ) x(k ) if m  M

y(m)   k

otherwise
0
 g M (  k )M x(k ) if m  M

y(m)   

otherwise
0
g (m)
g2 (n)


 
1
G( z 2 )

 n
g 2 (2m)  g (m)



1
G (z )
 m
x ( n)
y ( m)
h(0)
M
z M
z M
h(1)


x ( n)
h(0)
z 1

h(1)


z 1
h(2)
h(2)
since
M
z M

z 1
M
y ( m)
M
Application: POLYPHASE Filters
x ( n)
y ( m)
H ( z)
Decimator: take, for example, D=2
2
H ( z)   h(n) z n   h(2m) z 2 m  z 1  h(2m  1) z 2 m
n
m
m
H ( z)  He ( z 2 )  z 1 Ho ( z 2 )
even
odd
x ( n)
Therefore this system becomes:
He ( z 2 )
y ( m)
z 1
Ho ( z 2 )

2
x ( n)
He ( z 2 )
2
Filtering at High
Sampling Rate
z 1
2
Ho ( z )
2

y ( m)
Filtering at Low
Sampling rate
x ( n)
2
He ( z)
z 1
2
Ho ( z)

y ( m)
Similarly:
x ( n)
y ( m)
x ( n)
2
H ( z)
2
Ho ( z 2 )
z 1
2
He ( z )
x ( n)
Ho ( z)
Filtering at High
Sampling Rate
2
z 1
He ( z)
Filtering at Low
Sampling Rate
2

y ( m)

y ( m)
Example: Consider the Filter/Decimator structure shown below, with
s(n)
x(n)
y (m)
H (z )
H ( z )  1.2  0.6 z 1  1.5 z 2  2.0 z 3  4 z 4 
2
 2.0 z 5  1.5 z 6  0.6 z 7  1.2 z 8
He ( z2 )
This can be written as

H ( z )  1.2  1.5 z 2  4 z 4  1.5 z 6  1.2 z 8 


 0.6

2.0
z 
2
.
0
z

0
.
6
z

 z 1
2
4
6
Ho ( z2 )
and implemented in Polyphase form:
x ( n)
2
1.2  1.5z 1  4 z 2  1.5z 3  1.2 z 4
z 1
2
1
2
0.6  2.0 z  2 z  0.6 z
3

y ( m)
General Polyphase Decomposition
Given any integer N:

H  z 
 h[n]z
n
n 
  z
H ( z)  H0 z
N
1



 Nk 
  z   h[kN  ]z 
0
 k 

N 1
  ... z
H1 z
N
H
z 
( N 1)
N
 
H N 1 z
N
Example: take N=3
H ( z)  2.1  .5z 1  .3z 2  4.2z 3  0.6z 4 1.4z 5  0.2z 6  0.7 z 7

H0 ( z 3 )  2.1  4.2z 3  0.2z 6
H1 ( z 3 )  .5  0.6z 3  0.7 z 6
H 2 ( z 3 )  .3 1.4 z 3
Apply to Downsampling…
H z 
x[n]
y[m]
N
POLYPHASE
H N 1  z N 
z 1
H2  zN 
z 1
H1  z N 
x[n]
z 1
H0  zN 
y[m]
N
… apply Noble Identity
N
H N 1  z 
N
H2  z 
N
H1  z 
z 1
z 1
x[n]
z 1
N
H0  z 
y[m]
Serial to Parallel (Buffer)
N
z 1

N
x1[m]  x[mN  1]
N
x0 [m]  x[mN ]
z 1
x[n]
xN 1[m]  x[(m  1) N  1]
Serial to Parallel (Buffer):
x[n]
1
2
3
4
5
S/P
6
N
n  N
n0
xN 1[m]
1
3
5
2
4
6
x0[m]
m0
Same for Upsampling…
N
x[n]
H z 
y[m]
POLYPHASE
x[n]
N
H0 z N 
 
y[m]
z 1
H1 z N

 
H N 1 z N
z 1
… apply Noble Identity
x[n]
H z 
N
y[m]
NOBLE IDENTITY
x[n]
H 0 z 
H1 z 
z 1
N

H N 1 z 
y[m]
N
z 1
N
Parallel to Serial (Unbuffer or Interlacer)
N
y0 [ n]
N
yN 1[n]
y[ m ]
z 1
This is a Parallel to Serial (an Unbuffer):
1
3
5
2
4
6
y0 [ n ]
yN 1[n]
n0
n 1
y[ m ]
P/S
N
1
m0
2
3
mN
4
5
6