1
Nyquist Filters
Lth-Band Filters
• Under certain conditions, a lowpass filter
can be designed to have a number of zerovalued coefficients
• When used as interpolation filters these
filters preserve the nonzero samples of the
up-sampler output at the interpolator output
• Moreover, due to the presence of these
zero-valued coefficients, these filters are
computationally more efficient than other
lowpass filters of same order
• These filters, called the Nyquist filters or
Lth-band filters, are often used in single-rate
and multi-rate signal processing
• Consider the factor-of-L interpolator shown
below
Copyright © 2010, S. K. Mitra
x[n ]
L
xu [n]
H (z )
y[n ]
• The input-output relation of the interpolator
in the z-domain is given by
Y ( z) = H ( z) X ( z L )
2
Copyright © 2010, S. K. Mitra
Lth-Band Filters
Lth-Band Filters
• Then we can express Y(z) as
• If H(z) is realized in the L-band polyphase
form, then we have
H ( z ) = ∑iL=−01z −i Ei ( z L )
• Assume that the k-th polyphase component
of H(z) is a constant, i.e., Ek (z ) = α:
Y ( z ) = α z −k X ( z L ) +
• As a result,
+ αz − k + z −( k +1) Ek +1 ( z L ) + ... + z −( L −1) EL −1( z L )
Copyright © 2010, S. K. Mitra
5
Copyright © 2010, S. K. Mitra
y[ Ln + k ] = α x[n]
4
Copyright © 2010, S. K. Mitra
Lth-Band Filters
• A filter with the above property is called a
Nyquist filter or an Lth-band filter
• Its impulse response has many zero-valued
samples, making it computationally
attractive
• For example, the impulse response of an
Lth-band filter for k = 0 satisfies the
following condition
α,
n=0
h[Ln] = ⎧⎨
⎩ 0, otherwise
∑ z −l El ( z L ) X ( z L )
l =0
l≠k
• Thus, the input samples appear at the output
without any distortion for all values of n,
whereas, in-between ( L − 1) output samples
are determined by interpolation
H ( z ) = E0 ( z L ) + z −1E1 ( z L ) + ... + z −( k −1) Ek −1 ( z L )
3
L −1
Lth-Band Filters
• Figure below shows a typical impulse
response of a third-band filter (L = 3)
h[n]
_6
_3
n
0
3
6
• Lth-band filters can be either FIR or IIR
filters
6
Copyright © 2010, S. K. Mitra
1
Lth-Band Filters
Half-Band Filters
• If the 0-th polyphase component of H(z) is a
constant, i.e., E0 ( z ) = α then it can be shown
that
L −1
∑k =0 H ( zWLk ) = Lα = 1 (assuming α = 1/L)
• Since the frequency response of H ( zWLk ) is
the shifted version H (e j (ω− 2 πk / L ) ) of H (e jω) ,
the sum of all of these L uniformly shifted
versions of H (e jω) add up to a constant
7
with its impulse response satisfying
α,
n=0
h[2n] = ⎧⎨
0
,
otherwise
⎩
_
H(zWL L 1) H(z)
H(zWL) H(zWL2)
H(z)
• An Lth-band filter for L = 2 is called a halfband filter
• The transfer function of a half-band filter is
thus given by
H ( z ) = α + z −1E1 ( z 2 )
2π
0
ω
Copyright © 2010, S. K. Mitra
8
Copyright © 2010, S. K. Mitra
Half-Band Filters
Half-Band Filters
H (e j ( π / 2−θ) ) and H (e j ( π / 2+θ) ) add up
to 1 for all θ
• Or, in other words, H (e jω) exhibits a
symmetry with respect to the half-band
frequency π/2, hence the name “half-band
filter”
• The condition
•
H ( z ) = α + z −1E1 ( z 2 )
reduces to
H ( z ) + H (− z ) = 1 (assuming α = 0.5)
• If H(z) has real coefficients, then
H (−e jω ) = H (e j ( π−ω) )
• Hence
9
H ( e j ω ) + H ( e j ( π − ω) ) = 1
Copyright © 2010, S. K. Mitra
10
Half-Band Filters
Half-Band Filters
• Figure below illustrates this symmetry for a
half-band lowpass filter for which passband
and stopband ripples are equal, i.e., δ p = δs
and passband and stopband edges are
symmetric with respect to π/2, i.e., ω p + ωs = π
• Attractive property: About 50% of the
coefficients of h[n] are zero
• This reduces the number of multiplications
required in its implementation significantly
• For example, if N = 101, an arbitrary Type 1
FIR transfer function requires about 50
multipliers, whereas, a Type 1 half-band
filter requires only about 25 multipliers
H(e jω )
1+ δ
1_ δ
δ
11
ωp
π/2
ωs
π
Copyright © 2010, S. K. Mitra
ω
Copyright © 2010, S. K. Mitra
12
Copyright © 2010, S. K. Mitra
2
Half-Band Filters
Half-Band Filters
• An FIR half-band filter can be designed
with linear phase
• However, there is a constraint on its length
• Consider a zero-phase half-band FIR filter
for which h[ n] = α * h[ − n] , with | α | = 1
• Let the highest nonzero coefficient be h[R]
13
15
17
Copyright © 2010, S. K. Mitra
• Then R is odd as a result of the condition
α,
n=0
h[2n] = ⎧⎨
⎩ 0, otherwise
• Therefore R = 2K+1 for some integer K
• Thus the length of h[n] is restricted to be of
the form 2R+1 = 4K+3 [unless H(z) is a
constant]
14
Copyright © 2010, S. K. Mitra
Design of Linear-Phase
Lth-Band Filters
Design of Linear-Phase
Lth-Band Filters
• A lowpass linear-phase Lth-band FIR filter
can be readily designed via the windowed
Fourier series approach
• In this approach, the impulse response
coefficients of the lowpass filter are chosen
as h[n] = hLP [n] ⋅ w[ n] where hLP [n] is the
impulse response of an ideal lowpass filter
with a cutoff at π/L and w[n] is a suitable
window function
• Now, the impulse response of an ideal Lthband lowpass filter with a cutoff at ωc = π / L
is given by
sin( πn / L)
, −∞ ≤ n ≤∞
hLP [n] =
πn
Copyright © 2010, S. K. Mitra
• It can be seen from the above that
hLP [n] = 0 for n = ± L, ± 2 L, ...
16
Copyright © 2010, S. K. Mitra
Design of Linear-Phase
Lth-Band Filters
Design of Linear-Phase
Lth-Band Filters
• Hence, the coefficient condition of the Lthband filter
α,
n=0
h[Ln] = ⎧⎨
⎩ 0, otherwise
• There are many other candidates for Lthband FIR filters
• Program 13_8 can be used to design an Lthband FIR filter using the windowed Fourier
series approach
• The program employs the Hamming
window
• However, other windows can also be used
is indeed satisfied
• Hence, an Lth-band FIR filter can be
designed by applying a suitable window
w[n] to hLP [n]
Copyright © 2010, S. K. Mitra
18
Copyright © 2010, S. K. Mitra
3
Design of Linear-Phase
Lth-Band Filters
Design of Linear-Phase
Lth-Band Filters
• Figure below shows the gain response of a
half-band filter of length-23 designed using
Program 13_8
• The filter coefficients are given by
h[ −11] = h[11] = − 0.002315; h[ −10] = h[10] = 0;
h[ −9] = h[9] = 0.005412; h[ −8] = h[8] = 0;
h[ −7 ] = h[ 7 ] = − 0.001586; h[ −6] = h[ 6] = 0;
0
h[ −5] = h[5] = 0.003584; h[ −4] = h[ 4] = 0;
Gain, dB
-20
h[ −3] = h[3] = − 0.089258; h[ −2] = h[ 2] = 0;
h[ −1] = h[1] = 0.3122379; h[ 0] = 0.5;
-40
-60
-80
0
0.2
0.4
0.6
0.8
1
ω/π
19
Copyright © 2010, S. K. Mitra
20
Design of Linear-Phase
Lth-Band Filters
• The problem of designing a real-coefficient
half-band FIR filter can be transformed into
the design of a single passband FIR filter
with no stopband which can be easily
designed using the Parks-McClellan
algorithm
• An inverse transformation of the wideband
filter then yields the half-band FIR filter
Gain response of 4th-band FIR filter
Gain, dB
0
R = 0.1→ ← R = 0.4
-40
-60
-80
0
0.2
0.4
0.6
0.8
Copyright © 2010, S. K. Mitra
Design of Linear-Phase
Half-Band Filters
• We show below the gain response of a
length 23 4th-band lowpass filter designed
using the M-file firnyquist with 2
different roll-off factors
-20
• As expected, h[n] = 0 for
n = ± 2, ± 4, ± 6, ± 8, ±10
1
ω/π
21
23
Copyright © 2010, S. K. Mitra
22
Copyright © 2010, S. K. Mitra
Design of Linear-Phase
Half-Band Filters
Design of Linear-Phase
Half-Band Filters
• Let the specifications of the real-coefficient
half-band filter G(z) of order N be as
follows:
• Passband edge at ωp , stopband edge at ωs ,
passband ripple of δp , and stopband ripple
of δ s
• Now for a half-band filter δ p = δ s = δ ,
ω p + ωs = π and the order N is even with
N/2 odd
• Now, consider the design of a wide-band
linear-phase FIR filter F(z) of degree N/2
with a passband from 0 to 2ωp , a transition
band from 2ωp to π, and a passband ripple
of 2δ
• Since N/2 is odd, F(z) has a zero at z = −1
• Let f[n] denote the impulse response of F(z)
Copyright © 2010, S. K. Mitra
24
Copyright © 2010, S. K. Mitra
4
Design of Linear-Phase
Half-Band Filters
Design of Linear-Phase
Half-Band Filters
• Define
25
• The plots below the magnitude response of an
wide-band lowpass filter of degree 13 with a
passband from 0 to 0.85π and a transition band
from 0.9π to π and the magnitude response of the
derived half-band filter
Wide-Band Lowpass Filter
26
29
0.8
0.6
0.4
0.6
0.4
0.2
0
0.2
0.4
0.6
ω/π
0.8
1
0
0
0.2
0.4
0.6
0.8
1
ω/π
Copyright © 2010, S. K. Mitra
Design of Half-Band IIR Filters
G ( z )G ( z −1 ) + G (− z )G (− z −1 ) = 1
can be decomposed in the form
Copyright © 2010, S. K. Mitra
0.8
0
• Recall that an odd-order bounded real (BR)
lowpass IIR transfer function G(z) =
P(z)/D(z) with a symmetric numerator and
satisfying the power-symmetry condition
27
1
0.2
Design of Half-Band IIR Filters
G ( z ) = 1 [A0 ( z 2 ) + z −1A1 ( z 2 )]
2
where A0 (z ) and A1 (z ) are stable allpass
functions
Half-Band Lowpass Filter
1
Magnitude
Magnitude
G ( z ) = 1 [ z − N / 2 + F ( z 2 )]
2
• G(z) can be seen to be the transfer function
of a causal half-band lowpass FIR filter
with an impulse response
⎧1 f [n / 2],
n even
⎪⎪2
g[n] = ⎨ 0,
n odd , n ≠ N
2
⎪ 1
N
,
n
=
⎪⎩ 2
2 Copyright © 2010, S. K. Mitra
28
• It follows from the power-symmetry
condition that G(z) is a half-band lowpass
transfer function
• A Butterworth half-band lowpass IIR filter
G(z) can be designed by first designing an
odd-order analog Butterworth lowpass filter
with a 3-dB cutoff frequency at Ωc = 1 and
then applying a bilinear transformation
• We next consider the design of an elliptic
IIR half-band filter
Copyright © 2010, S. K. Mitra
Design of Half-Band IIR Filters
Design of Half-Band IIR Filters
• It can be shown that any odd-order elliptic
lowpass half-band filter G(z) with a
frequency response specification given by
1 − 2δ p ≤ G (e jω ) ≤ 1,
for 0 ≤ ω ≤ ω p
• It can be shown that the poles of the elliptic
lowpass half-band filter lie on the imaginary
axis
• Using the pole-interlacing property, we can
readily identify the expressions for A0 ( z )
and A1 ( z )
G ( e jω ) ≤ δ s ,
for ωs ≤ ω ≤ π
and satisfying the conditions
ω p + ωs = π, δ2s = 4δ p (1 − δ p )
is a power-symmetric transfer function
Copyright © 2010, S. K. Mitra
30
Copyright © 2010, S. K. Mitra
5
31
Design of Half-Band IIR Filters
Design of Half-Band IIR Filters
• Design Steps:
• Since ω p + ωs = π, δ2s = 4δ p (1 − δ p ) , only
one of the bandedges and one of the ripples
can be specified
• Let the specified stopband edge and
stopband ripple be ωs and δs , respectively
• Then ωp and δ p are determined using the
equations at the top of the slide
•
Copyright © 2010, S. K. Mitra
32
35
r=
tan(ω p / 2)
tan(ωs / 2)
r' = 1 − r2
(1 − r ')
q0 =
2(1 + r ')
and compute
q = q0 + 2 q05 + 15q09 + 150 q13
0
2
⎛ 1 − δ2s ⎞
D = ⎜⎜ 2 ⎟⎟
⎝ δs ⎠
Copyright © 2010, S. K. Mitra
Design of Half-Band IIR Filters
Design of Half-Band IIR Filters
• Next, the estimate of the order of G(z) is
determined by choosing the smallest odd
integer satisfying
log (16 D)
N ≥ 10
log10 (1 / q )
• As a result, the corresponding value of δs
will be smaller than the original specified
value
• To determine the actual value of δs , the
actual value of the parameter D is first
computed from
• Now the integer value of N is almost always
higher than the quantity on the RHS of the
above equation
33
Define
Copyright © 2010, S. K. Mitra
D=
10 N log10 (1 / q )
16
34
Copyright © 2010, S. K. Mitra
Design of Half-Band IIR Filters
Design of Half-Band IIR Filters
• From the actual value of D, the actual value
of δs is computed by solving
2
⎛ 1 − δ2 ⎞
D = ⎜⎜ 2 s ⎟⎟
⎝ δs ⎠
• From the new value of δs , the actual value
of δ p is obtained from
δ2s = 4δ p (1 − δ p )
• Next, the poles of the two allpass filters are
computed as follows:
2 q1 / 4 ∑i∞=0 (−1)i qi (i +1) sin((2i + 1)kπ / N )
λk =
2
1 + 2 ∑i∞=0 (−1)i qi cos(2πki / N )
λ2
bk = (1 − rλ2k )⎛⎜1 − k ⎞⎟
r ⎠
⎝
2
ck = 2bk /(1 + λ k )
Copyright © 2010, S. K. Mitra
36
α k −1 = (2 − ck ) /(2 + ck )
Copyright © 2010, S. K. Mitra
6
Design of Half-Band IIR Filters
Design of Half-Band IIR Filters
• In general, the two infinite sums in the
expression for
converge after the
addition of 5 or 6 terms
• The poles of the two allpass filters are on
the imaginary axis at z = ± j α k and are
inside the unit circle, as the parameters α k
are distinct with magnitudes less than 1
• Using the pole-interlacing property, then
poles of A0 (z ) and A1 (z ) are selected
• Their corresponding zeros are at the mirrorimage locations
• Example
• We consider the design of an elliptic halfband lowpass filter
37
Copyright © 2010, S. K. Mitra
Copyright © 2010, S. K. Mitra
Design of Half-Band IIR Filters
Design of Half-Band IIR Filters
• The specifications are:
ωs = 0.6 π
δ s = 0.016
• The transfer functions of the two allpass
sections of the half-band filter designed
using Program 13_9 are given by
• The pole-zero plot and the magnitude
response of the designed elliptic lowpass
half-band filter are shown below:
1 + 0.23647 z −1
, A1 ( z ) =
0.71454 + z
−1
1 + 0.71454 z −1
Copyright © 2010, S. K. Mitra
Half-Band IIR Lowpass Filter
1
0.5
Magnitude
0.23647 + z
−1
The pole-zero plot
1
Imaginary Part
A0 ( z ) =
39
38
0
-0.5
0.8
0.6
0.4
0.2
-1
-1.5
40
-1
-0.5
0
0.5
Real Part
1
1.5
0
0
0.2
0.4
0.6
0.8
1
ω/ π
Copyright © 2010, S. K. Mitra
7