Lecture 06: Design of Recursive Digital Filters
John Chiverton
School of Information Technology
Mae Fah Luang University
1st Semester 2009/ 2552
Lecture Contents
Introduction
IIR Filter Design
Pole-Zero Placement Method
IIR Filter Design from Analogue Filters
Summary
Outline
Introduction
IIR Filter Design
Pole-Zero Placement Method
IIR Filter Design from Analogue Filters
Summary
What is a Digital Filter?
( Ω)
|H ( Ω) |
Often used to remove some frequencies from a signal X(Ω) and to
allow other frequencies to pass through to the output Y (Ω).
π
Ω
π
Ω
H(Ω
Inv ), Line
aria
a
nt S r Tim
e
yst
em
|H ( Ω) |
Low Frequencies
Ou
tpu
t, Y
Low Pass
Band Pass
X(Ω
)
|H ( Ω) |
Middle Frequencies
π
High Frequencies
Inp
ut,
High Pass
Ω
Recursive digital filters
Input, x[n]
Linear Time
Invariant System
Output, y[n]
What is a Recursive digital filter?
I
“Recursive” comes from the word “to recur”
Meaning: to repeat
A recursive filter uses past output values (y[n − i]) for the
calculation of the current output y[n]:
I
Recursive Filter Example
y[n] = 0.5y[n − 1] + 0.5x[n].
A non-recursive filter only uses input values x[n − i]:
I
Non-recursive Filter Example
y[n] = 0.5x[n − 1] + 0.5x[n].
Generalised Difference Equation
Recall the generalised difference equation for causal LTI systems
(see Lecture 02):
N
X
a[k]y[n − k] =
k=0
M
X
b[k]x[n − k]
k=0
If a[0] = 1, this can then be changed to:
y[n] =
M
X
b[k]x[n − k] −
k=0
N
X
a[k]y[n − k].
k=1
The Frequency Response (see Lecture 03) of such a system can be
described by:
M
P
H(Ω) =
M
P
b[k] exp(−jkΩ)
k=0
exp(0) +
N
P
k=1
=
a[k] exp(−jkΩ)
b[k] exp(−jkΩ)
k=0
N
P
1+
k=1
.
a[k] exp(−jkΩ)
z-transform Representation
Fourier based frequency representation:
M
P
H(Ω) =
b[k] exp(−jkΩ)
k=0
N
P
1+
.
a[k] exp(−jkΩ)
k=1
Can also be represented in the z-domain (z-transform):
M
P
H(z) =
b[k]z −k
k=0
N
P
1+
.
a[k]z −k
k=1
Both describe a type of frequency response of the system.
Comparison of IIR and FIR System Structures
The system structure of an IIR filter demonstrates the feedback of
the output into the input again.
IIR Filter
FIR Filter
x[n]
x[n−1]
T
x1/4
x[n−2]
T
x[n−3]
x[n]
x1/4
x1/4
x1/4
x1/4
x[n−1]
T
T
y[n]
y[n] =
1
(x[n] + x[n − 1] + x[n − 2] + x[n − 3])
4
y[n−1]
x[n−3]
T
x1/4
x1/4
x(−1/10)
+
x[n−2]
T
x1/4
+
T
y[n]
y[n] =
(x[n] + x[n − 1] + x[n − 2] + x[n − 3])
−
1
y[n − 1]
10
Unit Delay in the z-plane
A single z−1 is the same as a unit delay, “T” in a system diagram.
x[n]
x[n−1]
T
x1/4
x[n−2]
T
x1/4
x1/4
x(−1/10)
y[n−1]
x[n−3]
T
x1/4
+
T
y[n]
Unit Delay in the z-plane
A single z−1 is the same as a unit delay, “T” in a system diagram.
x[n]
z−1
x1/4
x[n−1]
x[n−2]
z−1
x1/4
x1/4
x(−1/10)
y[n−1]
z−1
x1/4
+
z−1
x[n−3]
y[n]
Recursive Digital Filters
Recursive digital filters are often known as
I
Infinite Impulse Response (IIR) Filters
as the impulse response of an IIR filter often has an infinite
number of coefficients.
IIR Filters
I
Require fewer calculations than FIR filters.
I
∴ Faster response to the input signal,
I
and ∴ shorter frequency response transition width.
However!
I
IIR filters can become unstable.
I
∴ Need to think carefully about stability when designing IIR
Filters.
Outline
Introduction
IIR Filter Design
Pole-Zero Placement Method
IIR Filter Design from Analogue Filters
Summary
IIR Filter Design Overview
1. Filter specification
2. Coefficient calculations
3. Convert transfer function to suitable filter structure
4. Error analysis
5. Implementation
Filter Specification
Frequency domain parameters - for filter description.
1+ δ p
passband ripple
1
1− δ p
I Stop band - high
attenuation
I Pass band - low
attenuation (gain≈1)
Ω s1
Ω p1
Pass Band
Tr
a
W nsit
idt ion
h
Stop Band
Tr
an
W sitio
idt
h n
δs
Ω p2
Ω s2
Stop Band
Filter Specification
Frequency domain parameters - for filter description.
1+ δ p
passband ripple
1
I δp passband ripple
1− δ p
I δs stopband ripple
I Ωs1 lower stop band
edge frequency
I Ωp1 lower pass band
edge frequency
I Ωp2 upper pass band
edge frequency
δs
Ω s1
Ω p1
Pass Band
Tr
a
W nsit
idt ion
h
Stop Band
Tr
an
W sitio
idt
h n
I Ωs2 upper stop band
Ω p2
Ω s2
Stop Band
edge frequency
Outline
Introduction
IIR Filter Design
Pole-Zero Placement Method
IIR Filter Design from Analogue Filters
Summary
Pole-Zero Placement Method
A filter can be described in the z-plane with Poles and Zeros
(Lecture 04):
H(z) =
K(z − z1 )(z − z2 )(z − z3 )...
zeros
Y (z)
=
=
X(z)
(z − p1 )(z − p2 )(z − p3 )...
poles
I
Poles located at:
z1 , z2 , z3 , ...
I
Zeros located at:
p1 , p2 , p3 , ...
Pole-Zero Placement Method
A filter can be described in the z-plane with Poles and Zeros
(Lecture 04):
H(z) =
I
Poles (X) close to unit circle
I
I
K(z − z1 )(z − z2 )(z − z3 )...
zeros
Y (z)
=
=
X(z)
(z − p1 )(z − p2 )(z − p3 )...
poles
make large peaks
Zeros (O) close to unit circle
I
make troughs or minima
Pole-Zero Placement Method
Angle of poles and zeros on z-plane correspond to frequencies that can be used
for filter specification.
I A bandpass filter, with centre frequency Ω0 radians can have two poles at
±Ω0 radians in the z-plane1 .
I Complete attenuation at two frequencies, Ωr1 = 0 radians and Ωr2 = π
radians can have two zeros at 0 and π radians.
calculated with:
r∼
= 1 − Ωcf
|H ( Ω) |
I The radius of the poles can be
or
Ωbw
r∼
=1−
2
where Ωbw = 2Ωcf is the -3dB
bandwidth of the filter.
1
Band Pass
Ω
0
(DC)
Ω0−Ωcf
Ω0+Ωcf
Ω0
Complex conjugate pair to make real filter coefficients, when Ω0 6=0 or
Ω0 6=π radians (on the real line).
π
Pole-Zero Placement Method: Example
Q. Design a bandpass filter using the Pole-zero placement method with:
I centre frequency at Ω0 = π/2;
I a bandwidth of Ωbw = π/8;
I complete attenuation at Ωr1 = 0 and Ωr2 = π;
I and peak unity pass band gain.
A. Bandpass filter has x2 poles at ±Ω0 = ±π/2 radians.
∴ H(z) = K
zeros
(z − r exp(jπ/2))(z − r exp(−jπ/2))
The radii of the poles are given by:
Ωbw
π/8
r∼
=1−
= 0.80365;
=1−
2
2
and the zeros are at Ωr1 = 0 and Ωr2 = π, so that
H(z) = K
(z − exp(jΩr1 ))(z − exp(jΩr2 ))
.
(z − 0.80365 exp(jπ/2))(z − 0.80365 exp(−jπ/2))
Pole-Zero Placement Method: Example cont’d.
As
I exp(Ωr1 ) = exp(j0) = cos(0) + j sin(0) = 1 − j0 = 1
I exp(Ωr2 ) = exp(jπ) = cos(π) + j sin(π) = −1 + j0 = −1,
then the transfer function becomes:
H(z) = K
(z − 1)(z + 1)
.
(z − 0.80365 exp(jπ/2))(z − 0.80365 exp(−jπ/2))
Using Euler’s identity,
I exp(jπ/2) =
cos(π/2) + j sin(π/2) = +j
I and exp(jπ/2) =
cos(π/2) − j sin(π/2) = −j,
so that
H(z) = K
(z − 1)(z + 1)
.
(z − 0.80365j)(z + 0.80365j)
the pole zero diagram can then be
plotted.
Pole-Zero Placement Method: Example cont’d.
Recall that H(z) =
H(z) =
Y (z)
,
X(z)
Y (z)
(z − 1)(z + 1)
z2 − 1
=K
=K 2
.
X(z)
(z − 0.80365j)(z + 0.80365j)
z + 0.64585
Then
Y (z)(z 2 + 0.64585) = X(z)K(z 2 − 1).
Remembering that each z −1 is a unit delay, so that each z is a unit advance, then
the difference equation is:
y[n + 2] + 0.64585y[n] = K(x[n + 2] − x[n])
which can be made causal by making n = n − 2 so that
y[n] + 0.64585y[n − 2] = K(x[n] − x[n − 2]).
K is not known, but can be used to make the peak pass band gain to be unity.
Pole-Zero Placement Method: Example cont’d.
The frequency response of the filter can be determined from the difference equation:
y[n] + 0.64585y[n − 2] = K(x[n] − x[n − 2]),
in combination with:
M
P
H(Ω) =
Magnitude Frequency Response
b[k] exp(−jkΩ)
k=0
N
P
1+
.
a[k] exp(−jkΩ)
k=1
So that (using Euler’s identity):
H(Ω) =
K(1 − cos(2Ω) + j sin(2Ω))
1 + 0.64585(cos(2Ω) − j sin(2Ω))
which has magnitude frequency
response:
Mag(H(Ω))2 =
K((1 − cos(2Ω))2 + sin2 (2Ω))
.
(1 + 0.64585 cos(2Ω))2 + sin2 (2Ω)
where K = 0.17708.
Pole-Zero Placement Method: Example cont’d.
Relating the digital frequencies for previous example to
actual frequencies...
If the sampling frequency is fs = 500Hz, the sampling frequency
corresponds to Ω = 2π, therefore the filter parameters become:
I
centre frequency at Ω0 = π/2, so actual centre frequency
f0 = π/2
2π fs = 125Hz;
I
a bandwidth of Ωbw = π/8, so actual bandwidth
fbw = 31.25Hz;
I
complete attenuation at Ωr1 = 0 and Ωr2 = π, with actual
π
frequencies fr1 = 0Hz and fr2 = 2π
500 = 250Hz.
Pole-Zero Placement Method,
Example 2: Band Stop Filter
|H ( Ω) |
Q. Design digital bandstop filter using pole-zero placement
method with the following parameters:
I Centre frequency, Ω0 = π/10 radians (complete attenuation)
I Band stop width, Ωw = 2Ωcf = π/20 radians
tion
No
ua
tten
Band Stop
A
n
atio
No
nu
Atte
Attenuation
0
(DC)
Ω0−Ωcf
Ω0+Ωcf
Ω0
Ω
π
Pole-Zero Placement Method,
Example 2: Band Stop Filter
I Complete attenuation at Ω0 = π/10, ∴ x2 zeros (complex-conjugate pair)
at ±Ω0 = ±π/10:
H(z) = K
(z − exp(jπ/10))(z − exp(−jπ/10))
poles
I Centre frequency at Ω0 = π/10 radians, ∴ x2 poles (complex-conjugate
pair) at ±Ω0 = ±π/10,
H(z) = K
(z − exp(jπ/10))(z − exp(−jπ/10))
(z − r exp(jπ/10))(z − r exp(−jπ/10))
I The poles are scaled with radius r to control the width of the band
stop,
π/20
Ωw
r∼
=1−
= 0.92146
=1−
2
2
I resulting in:
H(z) = K
(z − exp(jπ/10))(z − exp(−jπ/10))
(z − 0.92146 exp(jπ/10))(z − 0.92146 exp(−jπ/10))
Pole-Zero Placement Method,
Example 2: Band Stop Filter cont’d.
I Transfer function is then (using Euler’s identity like before):
H(z) = K
z 2 − 1.9021z + 1
z 2 − 1.7527z + 0.84909
I As before, each z is a unit advance, so
y[n+2]−1.7527y[n+1]+0.84909y[n] = K(x[n+2]−1.9021x[n+1]+x[n])
letting n = n − 2, making it causal:
y[n]−1.7527y[n−1]+0.84909y[n−2] = K(x[n]−1.9021x[n−1]+x[n−2]).
I With frequency response:
M
P
H(Ω) =
b[k] exp(−jkΩ)
k=0
N
P
1+
k=1
=
a[k] exp(−jkΩ)
K(1 − 1.9021 exp(−jΩ) + exp(−j2Ω))
.
1 − 1.7527 exp(−jΩ) + 0.84909 exp(−j2Ω)
Pole-Zero Placement Method,
Example 2: Band Stop Filter cont’d.
Using Euler’s identity:
H(z) =
K(1 − 1.9021(cos Ω − j sin Ω) + cos 2Ω − j sin 2Ω)
.
1 − 1.7527(cos Ω − j sin Ω) + 0.84909(cos 2Ω − j sin 2Ω)
Magnitude Frequency response is then:
Mag(H(z))2 =
K((1 − 1.9021 cos Ω + cos 2Ω)2 + (1.9021 sin Ω − sin 2Ω)2 )
.
(1 − 1.7527 cos Ω + 0.84909 cos 2Ω)2 + (1.7527 sin Ω − 0.84909 sin 2Ω)2 )
Magnitude frequency
response of the notch or
bandstop filter:
Outline
Introduction
IIR Filter Design
Pole-Zero Placement Method
IIR Filter Design from Analogue Filters
Summary
Converting Analogue Filters to Digital Filters
Most common approach for IIR filter design.
I
Use well-established analogue filter specifications to design
digital IIR filters
Two common approaches:
I
Impulse invariant method
I
Bilinear transformation
Laplace Transform
I
Analogue filter transfer function h(t) can be specified in the
s-plane with the Laplace transform L(h(t)) = H(s)
I
Hence, the s-plane is for analogue instead of the z-plane (for
digital).
Can be used to analyse stability of analogue filters,
I
I
Similar to the z-transform for digital filters.
⇐⇒
Discrete (z-plane)
Imaginary Axis
ON
GI
RE
BL
E
EG
I
ON
UN
ST
A
jω
ER
I
EG
ON
UN
ER
BL
Real Axis
A
E
BL
A
ST
ST
Analogue (s-plane)
UN
ST
AB
L
ON
ON
GI
RE
GI
ST
UN
LE
AB
RE
ON
GI
RE
E
L
σ
AB
L
AB
ST
N
IO
EG
ER
T
S
UN
Laplace Transform
I
Analogue filter transfer function h(t) can be specified in the
s-plane with the Laplace transform L(h(t)) = H(s)
I
Hence, the s-plane is for analogue instead of the z-plane (for
digital).
Can be used to analyse stability of analogue filters,
I
I
Similar to the z-transform for digital filters.
Discrete
z-transform Z(h[n])
z-plane H[z]
Difference equation, h[n]
⇐⇒
⇐⇒
⇐⇒
⇐⇒
Analogue
Laplace transform L(h(t))
s-plane H(s)
Differential equation h(t)
How to Convert Analogue Frequency to Digital?
Problem!
I
Analogue frequency, ω = 0...∞.
I
But digital frequency, Ω = 0...2π.
So how to convert analogue frequency to digital?
Need to swap analogue frequencies with digital frequencies...
I
If Ω → 2π then Very high analogue frequencies (ω → ∞)
I
If Ω → 0 then Very low analogue frequencies (ω → 0).
Bilinear Transformation IIR Filter Design
Bilinear Transformation method replaces analog frequency s or
jω with digital frequency Ω using:
Ω
2
.
s = jω = j tan
Ts
2
where ω is analogue frequency, Ω is digital frequency and
Ts = 1/fs is the sampling period.
OR
to find the z-transform H(z):
s = jω = 2fs
1 − z −1
1 + z −1
.
Bilinear Transformation IIR Filter Design
Example:
Given an analog filter with:
H(ω) = H(s)|s=jω =
K(jω − z1 )(jω − z2 )...
(jω − p1 )(jω − p2 )...
Then bilinear transformation gives
H(Ω) = H(s)|s=j2fs tan( Ω )
2
Ω
K(j2fs tan 2 − z1 )(j2fs tan Ω2 − z2 )...
=
(j2fs tan Ω2 − p1 )(j2fs tan Ω2 − p2 )...
or...
H(z) = H(s)|s=2f “ 1−z−1 ”
s
1+z −1
−1 −1 1−z
K(2fs 1−z
−
z
)(2f
1
s 1+z −1 − z2 )...
1+z −1
=
−1
1−z −1
(2fs 1−z
−
p
)(2f
− p2 )...
1
s
−1
−1
1+z
1+z
Bilinear Transformation Example
Q. Convert the single pole low pass analog filter:
H(s) =
ωcf
s + ωcf
into a digital filter (z-plane form) with digital cut-off frequency Ωcf = 0.2π
using the bilinear transformation.
A.
1. Calculate analogue cut-off frequency ωcf from digital cut-off frequency
Ωcf = 0.2π:
ωcf = 2fs tan(Ωcf /2) = 2fs tan(0.1π) = 2fs A
2. Therefore analogue transfer function:
H(s) =
2fs A
s + 2fs A
−1
3. Apply bilinear transformation: s = 2fs 1−z
:
1+z −1
H(z) =
2fs A
−1
2fs 1−z
1+z −1
+ 2fs A
„
=
2fs
2fs
«
A(1 + z −1 )
(1 −
+ A(1 + z −1 )
z −1 )
Bilinear Transformation Example cont’d.
The z-transform transfer function of the filter is then:
H(z) =
Y (z)
A + Az −1
=
X(z)
1 + A + (A − 1)z −1
Stability Analysis
Rearranging to determine the poles for stability analysis gives:
H(z) =
I So there is 1 pole at z +
A−1
1+A
z+1
A
.
1 + A z + A−1
1+A
= 0 or z = − A−1
.
1+A
I Remember A = tan(0.1π), so the pole is: z = −0.50953,
I the magnitude is less than 1, so the filter is stable.
(1)
Bilinear Transformation Example cont’d.
Difference Equation
The difference equation can now be found.
Multiplying both sides by both denominators of equation (1) results in
˘
¯
˘
¯
Y (z) 1 + A + (A − 1)z −1 = X(z) A + Az −1
Remembering that each z −1 is a unit delay, so that
(1 + A)y[n] + (A − 1)y[n − 1] = Ax[n] + Ax[n − 1]
Dividing through by (1 + A) and rearranging gives
y[n] =
A−1
A
(x[n] + x[n − 1]) −
y[n − 1],
1+A
1+A
where A = tan(0.1π).
This is now a difference equation we can use to filter a signal.
Bilinear Transformation Example cont’d.
Frequency Response
The frequency response can be found directly using the bilinear transformation
or from the z-transform transfer function. We will compare both approaches.
Bilinear Transformation
The analogue transfer function from step 2 in earlier slide was:
H(s) =
2fs A
s + 2fs A
The s-plane variable s can be replaced by the Fourier complex frequency
variable jω,
˛
2fs A
˛
=
H(ω) = H(s)˛
.
jω + 2fs A
s=jω
The Fourier `frequency
can then be converted to the digitial frequency Ω using
´
ω = 2fs tan Ω2 (see earlier slide):
˛
˛
H(Ω) = H(ω)˛
ω=2fs tan(
Ω
2
)
=
2fs A
A
`Ω´
`Ω´
=
j2fs tan 2 + 2fs A
j tan 2 + A
Bilinear Transformation Example cont’d.
So the magnitude frequency response calculated directly from the Bilinear
transformation is:
s
s
(tan(0.1π))2
A2
= `
|H(Ω)| = `
` Ω ´´2
` Ω ´´2
2
+A
+ (tan(0.1π))2
tan 2
tan 2
The designed cut-off frequency Ωcf = 0.2π is confirmed by this plot.
Bilinear Transformation Example cont’d.
Frequency Response from z-Transform Transfer Function
Remember the z-transform transfer function calculated earlier (equation (1)):
H(z) =
Y (z)
A + Az −1
.
=
X(z)
1 + A + (A − 1)z −1
This can be converted to the frequency response using (see earlier slides and
lecture 04)
M
P
˛
˛
H(Ω) = H(z)˛
=
z=exp(−jkΩ)
k=0
N
P
b[k] exp(−jkΩ)
a[k] exp(−jkΩ)
k=0
So:
I b[0] = b[1] = A,
I a[0] = 1 + A,
I a[1] = A − 1.
So the frequency response is given by (using Euler’s identity):
H(Ω) =
A + A(cos(Ω) − j sin(Ω))
(1 + A) + (A − 1)(cos(Ω) − j sin(Ω)).
Bilinear Transformation Example cont’d.
So the magnitude is given by:
s
|H(Ω)| =
(A + A cos(Ω))2 + (A sin(Ω))2
((1 + A) + (A − 1) cos(Ω))2 + ((A − 1) sin(Ω)))2 .
which is the same as the frequency response calculated directly from the
bilinear transformation. The Bilinear transformation is more quicker here.
Famous Analogue Filters
Butterworth
I
Magnitude frequency response: |H(ω)| =

1
”2n ff1/2
“
ω
1+ ω
cf
Chebyshev
Magnitude frequency response: |H(ω)| =
|H ( ω) |
magnitude response
I
n
2
1+2 Cn
Ideal
1.0 −1/2
2
(1+ ε )
Chebyshev
0.707
Butterworth
ωcf
frequency ω
1
“
ω
ωcf
”o1/2
Butterworth and Chebyshev Analog Filters
Bilinear transformation: jω = j2fs tan(Ω/2), therefore:
Magnitude Frequency Response
(Analogue)
(Digital)
|H(ω)|
|H(Ω)|
Butterworth
»
1
“
”2n –1/2
1+ ωω
cf
Chebyshev
h
2
1+2 Cn
1
“
1
» “
”2n –1/2
tan(Ω/2)
1+ tan(Ω /2)
cf
1
ω
ωcf
”i1/2
h
2
1+2 Cn
“
tan(Ω/2)
tan(Ωcf /2)
”i1/2
Comparison of IIR and FIR filters
Characteristic
Multiplications
Coefficient quantification sensitivity
Overflow errors
Stability
Linear phase
Simulate analog filter
Coefficient memory
Design complexity
IIR
least
can be high
FIR
most
very low
can be high
by design
no
yes
very low
always
always
no
least
moderate
most
simple
adapted from “Understanding digital signal processing” by R. G. Lyons
Outline
Introduction
IIR Filter Design
Pole-Zero Placement Method
IIR Filter Design from Analogue Filters
Summary
Lecture Summary
This lecture has covered:
I
Introduction to IIR filters.
I
Frequency domain parameters.
I
Pole-zero placement method for IIR filter design
I
Band stop filter design
I
IIR filter design from analog filters using bilinear
transformation method