(Fall EE236A 2007-08)
Lecture 6 FIR lter design
FIR lters linear phase lter design magnitude lter design equalizer design
61
FIR lters
nite impulse response (FIR) lter:
n1
y (t) =
=0
h u(t ),
tZ
u : Z R is input signal ; y : Z R is output signal hi R are called lter coecients ; n is lter order or length lter frequency response: H : R C H ( ) = h0 + h1ej + + hn1ej (n1)
n1 t=0 n1 t=0
=
ht cos t j
ht sin t
(j =
1)
periodic, conjugate symmetric, so only need to know/specify for 0 FIR lter design problem: choose h so H and h satisfy/optimize specs
FIR lter design 62
example: (lowpass) FIR lter, order n = 21 impulse response h:
0.2 0.1 0 0.1 0.2 0 2 4 6 8 10 12 14 16 18 20
h(t)
t
frequency response magnitude |H ( )| and phase H ( ):
10
1
3 2
|H ( )|
10
H ( )
0
1 0 1 2 3 0
10
1
10
2
10
3
0
0.5
1
1.5
2
2.5
3
0.5
1
1.5
2
2.5
3
FIR lter design
63
Linear phase lters
suppose n = 2N + 1 is odd and impulse response is symmetric about midpoint: ht = hn1t, t = 0, . . . , n 1 then H ( ) = h0 + h1ej + + hn1ej (n1) = ejN H ( )
= ejN (2h0 cos N + 2h1 cos(N 1) + + hN )
term ejN represents N -sample delay H ( ) is real |H ( )| = |H ( )| called linear phase lter ( H ( ) is linear except for jumps of )
FIR lter design 64
Lowpass lter specications
1 1/1
2 p s
specications: maximum passband ripple (20 log10 1 in dB): 1/1 |H ( )| 1, 0 p
minimum stopband attenuation (20 log10 2 in dB): |H ( )| 2,
FIR lter design
s
65
Linear phase lowpass lter design
sample frequency (k = k/K , k = 1, . . . , K ) 1/1 H (k ) 1 design for maximum stopband attenuation: minimize 2 subject to 1/1 H (k ) 1, 0 k p 2 H (k ) 2, s k passband ripple 1 is given an LP in variables h, 2 known (and used) since 1960s
can assume wlog H (0) > 0, so ripple spec is
can add other constraints, e.g., |hi|
FIR lter design 66
example linear phase lter, n = 31 passband [0, 0.12 ]; stopband [0.24, ] max ripple 1 = 1.059 (0.5dB) design for maximum stopband attenuation impulse response h and frequency response magnitude |H ( )|
0.2 0.15
10
1
10
0
|H ( )|
0.1
h(t)
10
1
0.05 0 0.05 0.1
10
2
10
3
0
5
10
15
20
25
30
10
4
0
0.5
1
1.5
2
2.5
3
t
FIR lter design
67
Some variations
H ( ) = 2h0 cos N + 2h1 cos(N 1) + + hN minimize passband ripple (given 2, s, p, N ) minimize 1 subject to 1/1 H (k ) 1, 0 k p 2 H (k ) 2, s k minimize transition bandwidth (given 1, 2, p, N ) minimize s subject to 1/1 H (k ) 1, 0 k p 2 H (k ) 2, s k
FIR lter design
68
minimize lter order (given 1, 2, s, p) minimize N subject to 1/1 H (k ) 1, 0 k p 2 H (k ) 2, s k can be solved using bisection
each iteration is an LP feasibility problem
FIR lter design
69
Filter magnitude specications
transfer function magnitude spec has form L( ) |H ( )| U ( ), where L, U : R R+ are given and
n1 n1
[0, ]
H ( ) =
t=0
ht cos t j
ht sin t
t=0
arises in many applications, e.g., audio, spectrum shaping not equivalent to a set of linear inequalities in h (lower bound is not even convex) can change variables and convert to set of linear inequalities
FIR lter design 610
Autocorrelation coecients
autocorrelation coecients associated impulse with response h = (h0, . . . , hn1) Rn are
n1t
rt =
=0
h h +t
(with hk = 0 for k < 0 or k n)
rt = rt and rt = 0 for |t| n; hence suces to specify r = (r0, . . . , rn1) Fourier transform of autocorrelation coecients is
n1
R( ) =
e
j
r = r0 +
t=1
2rt cos t = |H ( )|2
can express magnitude specication as L( )2 R( ) U ( )2, . . . linear inequalities in r
FIR lter design 611
[0, ]
Spectral factorization
question: when is r Rn the autocorrelation coecients of some h Rn? answer (spectral factorization theorem): if and only if R( ) 0 for all spectral factorization condition is convex in r (a linear inequality for each ) many algorithms for spectral factorization, i.e., nding an h such that R( ) = |H ( )|2 magnitude design via autocorrelation coecients: use r as variable (instead of h) add spectral factorization condition R( ) 0 for all optimize over r use spectral factorization to recover h
FIR lter design 612
Magnitude lowpass lter design
maximum stopband attenuation design with variables r becomes minimize 2 subject to 1/1 R( ) 1, [0, p] R( ) 2, [s, ] R( ) 0, [0, ]
2 (i corresponds to i in original problem)
now discretize frequency: minimize 2 subject to 1/1 R(k ) 1, 0 k p R(k ) 2, s k R(k ) 0, 0 k . . . an LP in r, 2
FIR lter design 613
Equalizer design
g (t) h(t)
(time-domain) equalization: given g (unequalized impulse response) gdes (desired impulse response) design (FIR equalizer) h so that g = h g gdes common choice: pure delay D: gdes(t) = as an LP: 1 t=D 0 t=D
minimize maxt=D |g (t)| subject to g (D) = 1
614
FIR lter design
example unequalized system G is 10th order FIR:
1 0.8 0.6
g (t)
0.4 0.2 0 0.2 0.4 0 1 2 3 4 5 6 7 8 9
t
10
1
3 2
| G( ) |
G( )
1 0 1 2
10
0
10
1
0
0.5
1
1.5
2
2.5
3
3 0
0.5
1
1.5
2
2.5
3
FIR lter design
615
design 30th order FIR equalizer with G( ) ej 10 minimize maxt=10 |g (t)| equalized system impulse response g
1 0.8
g (t)
0.6 0.4 0.2 0 0.2 0 5 10 15 20 25 30 35
t
equalized frequency response magnitude |G| and phase G
10
1
3 2
e | G( ) |
e G( )
1 0 1 2
10
0
10
1
0
0.5
1
1.5
2
2.5
3
3 0
0.5
1
1.5
2
2.5
3
FIR lter design
616
Magnitude equalizer design
H ( ) G( )
given system frequency response G : [0, ] C design FIR equalizer H so that |G( )H ( )| 1: minimize max[0,] |G( )H ( )|2 1 use autocorrelation coecients as variables: minimize subject to |G( )|2R( ) 1 , R( ) 0, [0, ] when discretized, an LP in r, , . . .
FIR lter design 617
[0, ]
Multi-system magnitude equalization
given M frequency responses Gk : [0, ] C design FIR equalizer H so that |Gk ( )H ( )| constant: minimize maxk=1,...,M max[0,] |Gk ( )H ( )|2 k subject to k 1, k = 1, . . . , M use autocorrelation coecients as variables: minimize subject to |Gk ( )|2R( ) k , R( ) 0, [0, ] k 1, k = 1, . . . , M . . . when discretized, an LP in k , r,
FIR lter design 618
[0, ],
k = 1, . . . , M
example. M = 2, n = 25, k 1
unequalized and equalized frequency responses
2.5 2.5
2
2
1.5
Gk ( ) H ( ) | 2
0.5 1 1.5 2 2.5 3
Gk ( ) | 2
1.5
1
1
0.5
0.5
0 0
0 0
0.5
1
1.5
2
2.5
3
FIR lter design
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FIR lter design
620