Digital Signal Processing
Moslem Amiri, Václav Přenosil
Masaryk University
Understanding Digital Signal Processing, Third Edition, Richard Lyons
(0-13-261480-4) © Pearson Education, 2011.
Infinite Impulse Response Filters
2
Introduction
 Infinite impulse response (IIR) digital filters
 Are fundamentally different from FIR filters
 FIR filter output samples depend only on past input
samples
 Each IIR filter output sample depends on previous input
samples and previous filter output samples
 IIR filters have memory of past outputs (require feedback)
 As in all feedback systems, perturbations at IIR
filter input could cause filter output to become
unstable and oscillate indefinitely
 Infinite impulse response
 IIR filters have more complicated structures (block
diagrams), are harder to design and analyze, and
do not have linear phase responses
3
Introduction
 Why use an IIR filter?
 Because they are very efficient
 IIR filters require far fewer multiplications per filter
output sample to achieve a given frequency magnitude
response
 From a hardware standpoint, IIR filters can be very
fast, allowing us to build real-time IIR filters that
operate over much higher sample rates than FIR filters
4
Introduction
where the 19-tap FIR filter requires
19 multiplications per filter output
sample, the 4th-order IIR filter
requires only 9 multiplications for
each filter output sample
reduced passband ripple
and a sharper filter rolloff,
with less than half the
multiplication workload
5
Introduction
 IIR vs. FIR
 An FIR filter’s frequency response with very steep
transition regions requires a very long impulse
response
 The maximum number of FIR filter taps we can
have (length of impulse response) depends on
how fast our hardware can perform the required
number of multiplications and additions to get a
filter output before the next input sample arrives
 IIR filters can be designed to have impulse
responses longer than their number of taps
 Thus, IIR filters can give much better filtering for a
given number of multiplications per output sample
6
An Introduction to IIR Filters
 If IIR filter’s input suddenly becomes all
zeros, its output could remain nonzero
forever
 This is because of feedback structure of their
delay units, multipliers, and adders
7
An Introduction to IIR Filters
)3()3()2()2(
)1()1()()0()(
−+−+
−+=
nxhnxh
nxhnxhny
difference equation
8
An Introduction to IIR Filters
)3()3()2()2()1()1(
)3()3()2()2()1()1()()0()(
−+−+−+
−+−+−+=
nyanyanya
nxbnxbnxbnxbny
difference equation
describing this IIR
filter
9
An Introduction to IIR Filters
 How determine a(k) and b(k) IIR filter coefficients
 Window method of lowpass FIR filter design
 Define frequency response of desired FIR filter  take
inverse Fourier transform  shift that transform result
 we get filter’s time-domain impulse response
 Due to the nature of transversal FIR filters, the desired
h(k) filter coefficients turn out to be exactly equal to the
impulse response sequence
 Following that same procedure with IIR filters
 Desired frequency response of IIR filter  inverse
Fourier transform  time-domain impulse response
 But there’s no direct method for computing IIR filter’s
a(k) and b(k) coefficients from impulse response
 FIR filter design techniques cannot be used here
10
An Introduction to IIR Filters
 Standard IIR filter design techniques
 Fall into three basic classes: impulse invariance,
bilinear transform, and optimization methods
 These design methods use a discrete sequence,
mathematical transformation process known as
the z-transform whose origin is Laplace transform
used in analyzing of continuous systems
11
The Laplace Transform
 Laplace transform
 A mathematical method of solving linear
differential equations
 Process of using Laplace transform
 A time-domain differential equation is written that
describes input/output relationship of a physical
system
 The differential equation is Laplace transformed,
converting it to an algebraic equation
 Standard algebraic techniques are used to determine
desired output function’s equation in Laplace domain
 The desired Laplace output equation is, then, inverse
Laplace transformed to yield the desired time-domain
output function’s equation
12
The Laplace Transform
 Laplace transform of a continuous timedomain
function f(t)
 Variable s is the complex number s = σ + jω
 f(t) is defined only for positive time (t > 0)
 Systems where system conditions for negative time (t
< 0) are not needed (one-sided ) are referred to as
causal systems
 Causal systems may have initial conditions at t = 0
that must be taken into account, but we don’t need
to know what the system was doing prior to t = 0
∫
∞
−
=
0
)()( dtetfsF st
13
The Laplace Transform
 s = σ + jω
 σ is a real number
 ω is frequency in radians/second
 e−st is dimensionless  s has dimension of
1/time, or frequency
 Laplace variable s is often called a complex frequency
14
The Laplace Transform
 Laplace transform
 e−jωt is a unity-magnitude phasor rotating clockwise
around origin of a complex plane at a frequency of ω
radians/second
 eσt is a real number whose value is one at t = 0
 As t increases, eσt gets larger (when σ is positive), and
complex e−st phasor’s magnitude gets smaller as phasor
rotates on complex plane
 The tip of that phasor traces out a curve spiraling in
toward origin of complex plane
ttt
tj
tjttjst
st
e
t
j
e
t
e
e
eeee
dtetfsF
σσσ
ω
ωσωσ ωω )sin()cos(
)()(
)(
0
−====
=
−
−−+−−
∞
−
∫
15
The Laplace Transform
real part of F(s), for
a particular value of
s, is correlation of
f(t) with a cosine
wave of frequency
ω and a damping
factor of σ, and
imaginary part of
F(s) is correlation
of f(t) with a
sinewave of
frequency ω and a
damping factor of σ
16
The Laplace Transform
 s-plane
 If we associate each of different values of
complex s variable with a point on a complex
plane, called s-plane, we could plot real part of
F(s) correlation as a surface above (or below)
that s-plane and generate a second plot of
imaginary part of F(s) correlation as a surface
above (or below) s-plane
 We can also graph magnitude |F(s)| as a function
of s
17
The Laplace Transform
)(
)(
)(
)()(
01012
2
2 txb
dt
tdx
btya
dt
tdy
a
dt
tyd
a +=++
We’ll use Laplace transform toward
our goal of figuring out what the y(t)
output will be for any given x(t) input
18
The Laplace Transform
)()()()(
)(
)(
)(
)(
)(functiontransfer
)()()()(or
)()()()()(
)(
)(
)(
)()(
)(
,...,
)(
,
)(
,
)(
01
2
2
01
01
2
2
01
0101
2
2
0101
2
2
and,offunctionsbeandletweIf
01012
2
2
3
3
3
2
2
2
sHsX
asasa
bsb
sXsY
asasa
bsb
ex
ey
sX
sY
sH
exbsbeyasasa
exbesxbeyaesyaeysa
txb
dt
tdx
btya
dt
tdy
a
dt
tyd
a
es
dt
ed
es
dt
ed
es
dt
ed
se
dt
ed
ts
ts
tsts
tststststs
)y(e)x(eey(t)x(t)
tsn
n
tsn
ts
ts
ts
ts
ts
ts
ststst
=
++
+
=
++
+
===
+=++
+=++
 →
+=++
====
19
The Laplace Transform
)()()()(
)(
)(
)(
)(
)(functiontransfer
01
2
2
01
01
2
2
01
sHsX
asasa
bsb
sXsY
asasa
bsb
ex
ey
sX
sY
sH ts
ts
=
++
+
=
++
+
===
20
The Laplace Transform
 Laplace analysis technique is based on
system’s x(t) input being some function of est,
or x(est)
 All practical x(t) input functions can be
represented with complex exponentials, e.g.,
 A constant, c = ce0t
 Sinusoids, sin(ωt) = (ejωt − e−jωt)/2j or cos(ωt) = (ejωt +
e−jωt)/2
 A monotonic exponential, eat
 An exponentially varying sinusoid, e−at cos(ωt)
21
The Laplace Transform
 System’s transfer function H(s)
 If we know H(s), we can take Laplace transform
of any x(t) input to determine X(s), multiply that
X(s) by H(s) to get Y(s), and then inverse Laplace
transform Y(s) to yield time-domain expression
for the output y(t)
 Not needed because it’s H(s) in which we’re interested
 Being able to express H(s) mathematically or
graph the surface |H(s)| as a function of s will tell
us two important properties we need to know
about system:
 Is system stable
 And if so, what is its frequency response
22
The Laplace Transform
 Poles and zeros on s-plane and stability
 A system is stable if, given any bounded input,
the output will always be bounded
 Instability would result in a filter output that’s not
at all representative of filter input—a situation
we’d like to avoid if we can
 Example
 If s = −a0/a1, denominator equals zero and H1(s) would
have an infinite magnitude
 This s = −a0/a1 point on s-plane is called a pole
10
10
01
0
1
/
/
)(
)(
)(functiontransferLaplacesSystem'
aas
ab
asa
b
sX
sY
sH
+
=
+
===
23
The Laplace Transform
the pole is located
exactly on negative
portion of real σ axis
H1(s) is stable
because its y(t)
output approaches
zero as time passes
distance of pole from
σ = 0 axis, a0/a1 for
our H1(s), gives the
decay rate of y(t)
impulse response
If the system described by H1 were at rest and we disturbed it with an
impulse like x(t) input at time t = 0, its continuous time-domain y(t)
output would be the damped exponential curve shown in (b)
24
The Laplace Transform
Laplace transform is
a more general case
of Fourier transform
because if σ = 0,
then s = jω
intersection of
vertical σ = 0 plane
(jω axis plane) and
|H1(s)| surface,
gives us frequency
magnitude response
|H1(ω)| of system
25
The Laplace Transform
 Example: 2nd-order transfer function H2(s)
 Order of transfer function is the largest exponential
order of s in either numerator or denominator
 If s is equal to −p or −p*, one of polynomial roots in the
denominator will equal zero, and H2(s) will have an
infinite magnitude  two complex poles







 −
−+⋅







 −
++=++=
−=+==
++
=
++
=
++
==
∗
∗
2
2
2
2
2
imagrealimagreal21
2
2021
2
21
01
2
2
1
2
4
4
24
4
2
)(
,,/where
))((
)(
/)/(
)/(
)(
)(
)(
a
acb
a
b
s
a
acb
a
b
scbsassf
jpppjpppabA
psps
As
sH
aasaas
sab
asasa
sb
sX
sY
sH
26
The Laplace Transform
the two complex
poles are located off
the negative portion
of real σ axis
If H2 system were at rest and we disturbed it with an impulse-like x(t)
input at t = 0, its continuous time-domain y(t) output would be the
damped sinusoidal curve shown in (b)
H2(s) is stable
because its oscillating
y(t) output approaches
zero as time increases
distance of poles from σ = 0 axis
(−preal) gives decay rate of sinusoidal
y(t) impulse response. Likewise,
distance of poles from jω = 0 axis
(±pimag) gives frequency of sinusoidal
y(t) impulse response
27
The Laplace Transform
28
The Laplace Transform
stable systems: when
disturbed from their
at-rest condition, they
respond and, at some
later time, return to
that initial condition
1/s transfer function
conditional stability:
if an H(s) transfer
function has
conjugate poles
located exactly on
jω axis (σ = 0), the
system will go into
oscillation when
disturbed from its
initial condition
instability: the poles
lie to the right of jω
axis. When disturbed
from their initial atrest
condition by an
impulse input, their
outputs grow without
bound
29
The Laplace Transform
for a system to be
stable, all of its
transfer-function
poles must lie on the
left half of s-plane
30
The z-Transform
 z-transform
 Discrete-time cousin of continuous Laplace
transform
 While Laplace transform is used to simplify analysis of
continuous differential equations, z-transform facilitates
analysis of discrete difference equations
 z-transform is performed on a discrete h(n)
sequence, converting that sequence into a
continuous function H(z) of the continuous
complex variable z
∑
∞
−∞=
−
=
n
n
znhzH )()(
31
The z-Transform
 z-transform
 This Equation can be interpreted as Fourier transform
of product of original sequence h(n) and exponential
sequence r−n
 When r = 1, it simplifies to Fourier transform
 Thus on z-plane, the contour of H(z) surface for those
values where |z| = 1 is Fourier transform of h(n)
 If h(n) represents a filter impulse response sequence,
evaluating H(z) transfer function for |z| = 1 yields
frequency response of filter
∑∑
∑
∞
−∞=
−−
∞
−∞=
−=
∞
−∞=
−
=== →
=
n
njn
n
njjrez
n
n
ernhrenhreHzH
znhzH
j
)()())(()()(
)()(
ωωωω
32
The z-Transform
33
The z-Transform
jω frequency axis on continuous Laplace s-plane is linear and ranges from
− ∞ to + ∞ radians/second. The ω frequency axis on complex z-plane,
however, spans only the range from −π to +π radians  z-plane frequency
axis is equivalent to coiling s-plane’s jω axis about the unit circle on z-plane
34
The z-Transform
 Poles, zeros, and digital filter stability
 Region of filter stability is mapped to the inside of
unit circle on z-plane
 Given H(z) transfer function of a digital filter, we
can examine that function’s pole locations to
determine filter stability
 If all poles are located inside unit circle, filter will
be stable
35
The z-Transform
 Example
 If a causal filter’s H(z) transfer function has a
single pole at location p on z-plane, its transfer
function can be represented by
 Filter’s time-domain impulse response sequence
 u(n) represents a unit step (all ones) sequence
beginning at time n = 0
 When |p| < 1, h(n) impulse response sequence is
unconditionally bounded in amplitude
 |p| < 1 means that pole must lie inside z-plane’s unit circle
1
1
1
)( −
−
=
pz
zH
)()( nupnh n
⋅=
36
The z-Transform
point z = −1
corresponds to π
radians (or πfs
radians/second =
fs/2 Hz)  ωo = π/4
means that fo = fs/8
and our y(n) will
have eight samples
per cycle of fo
37
Using z-Transform to Analyze IIR Filters
 Representing delay operation
 Assume we have a sequence x(n) whose ztransform
is X(z) and a sequence y(n) = x(n−1)
whose z-transform is Y(z)
 Thus, effect of a single unit of time delay is to multiply
z-transform of undelayed sequence by z−1
)]([)(
)()()(
)1()()(
1)(1
1)1(1letweif
zXzzkxz
zzkxzkxzY
znxznyzY
k
k
k
k
k
knk
n n
nn
−
∞
−∞=
−−
∞
−∞=
−−
∞
−∞=
+−−=
∞
−∞=
∞
−∞=
−−
==
== →
−==
∑
∑∑
∑ ∑
38
Using z-Transform to Analyze IIR Filters
X(z)z−k is z-transform of x(n)
delayed by k samples. So a
transfer function of z−k is
equivalent to a delay of kts
seconds from the instant
when t = 0, where ts = 1/fs
Because a delay of one
sample is equivalent to
factor z−1, the unit time
delay symbol is usually
indicated by z−1 operator
39
Using z-Transform to Analyze IIR Filters
40
Using z-Transform to Analyze IIR Filters
 Fig. 6-17
 Is a general Mth-order IIR filter
 This IIR filter structure is called Direct Form I
∑
∑
∑∑
∑∑
=
−
=
−
=
−
=
−
=
−
=
−
−−−
−−−
−
== →=








−
+=
++++
++++=
−++−+−+
−++−+−+=
M
k
k
N
k
k
N
k
k
M
k
k
M
k
k
N
k
k
M
N
zka
zkb
zX
zY
zHzkbzXzkazY
zkazYzkbzXzY
zzYMazzYazzYa
zzXNbzzXbzzXbzXbzY
MnyManyanya
NnxNbnxbnxbnxbny
1
0function
transfer
01
10
21
21
)(1
)(
)(
)(
)()()()(1)(
)()()()()(
)()(...)()2()()1(
)()(...)()2()()1()()0()(
)()(...)2()2()1()1(
)()(...)2()2()1()1()()0()(
order of H(z) and order of filter =
the largest exponential order of z in
either numerator or denominator
41
Using z-Transform to Analyze IIR Filters
 Two things to know about an IIR filter: its
frequency response and stability
 We can evaluate denominator of H(z) to
determine positions of filter’s poles on z-plane
indicating filter’s stability
 From H(z) we develop an expression for IIR
filter’s frequency response
 H(z) is a complex-valued surface above, or below, the
z-plane
 Intersection of H(z) surface and perimeter of a cylinder
representing z = ejω unit circle is the filter’s complex
frequency response
42
Using z-Transform to Analyze IIR Filters
∑∑
∑∑
∑
∑
∑
∑
==
==−=
=
−
=
−
=
−
=
−
+−
−
= →
−
=
=
=→
−
==
−
M
k
M
k
N
k
N
kje
M
k
jk
N
k
jk
jM
k
k
N
k
k
kkajkka
kkbjkkb
H
eka
ekb
ez
zHH
zka
zkb
zX
zY
zH
j
11
00)sin()cos(
1
0
1
0
)sin()()cos()(1
)sin()()cos()(
)(
)(1
)(
)()(
)(1
)(
)(
)(
)(
ωω
ωω
ω
ω
ωω
ω
ω
ω
ω
43
Using z-Transform to Analyze IIR Filters
44
Using z-Transform to Analyze IIR Filters
 Fig. 6-18
 (a) is a 2nd-order lowpass IIR filter whose
positive cutoff frequency is ω = π/5 (fs/10 Hz)
)]2sin(436.0)1sin(194.1[)2cos(436.0)1cos(194.11
)]2sin(0605.0)1sin(121.0[)2cos(0605.0)1cos(121.00605.0
)(
436.0194.11
0605.0121.00605.0
)(
436.0194.11
0605.0121.00605.0
)(
)(
)(
)(436.0)(194.1
)(0605.0)(121.0)(0605.0)(
)2(436.0)1(194.1
)2(0605.0)1(121.0)(0605.0)(
21
210
21
210
21
21
ωωωω
ωωωω
ω
ω ωω
ωωω
⋅−⋅+⋅+⋅−
⋅+⋅−⋅+⋅+
=
⋅+⋅−
⋅+⋅+⋅
=
⋅+⋅−
⋅+⋅+⋅
==
⋅−⋅+
⋅+⋅+⋅=
−⋅−−⋅+
−⋅+−⋅+⋅=
−−
−−−
−−
−−
−−
−−
j
j
H
ee
eee
H
zz
zzz
zX
zY
zH
zzYzzY
zzXzzXzXzY
nyny
nxnxnxny
jj
jjj
45
Using z-Transform to Analyze IIR Filters
although both filters
require the same
computational workload,
five multiplications per
filter output sample,
lowpass IIR filter has
the superior frequency
magnitude response
phase nonlinearity is
inherent in IIR filters
knowing that the
filter’s phase
response is nonlinear,
we should expect the
impulse response to
be asymmetrical
infinite impulse response:
if we used infiniteprecision
arithmetic in our
filter implementation, h(k)
impulse response would
be infinite in duration
46
Using z-Transform to Analyze IIR Filters
 To determine our IIR filter’s stability
 So when z = p0 = 0.597 − j0.282, or when z = p1
= 0.597 + j0.282, filter’s H(z) transfer function’s
denominator is zero and |H(z)| is infinite
 Because those pole locations are inside the unit circle
(their magnitudes are less than one), our example IIR
filter is unconditionally stable
)282.0597.0)(282.0597.0(
)1)(1(
))((
))((
)(
436.0194.1
0605.0121.00605.0
)(
436.0194.11
0605.0121.00605.0
)(
)(
)(
10
10
2
2
bymultiply
21
210
22
jzjz
zz
pzpz
zzzz
zH
zz
zz
zH
zz
zzz
zX
zY
zH
/zzH(z)
−−+−
++
=
−−
−−
=
+⋅−
+⋅+⋅
= →
⋅+⋅−
⋅+⋅+⋅
== −−
−−
47
Using z-Transform to Analyze IIR Filters
if we were to unwrap the
bold |H(ω)| curve and lay
it on a flat surface, we
would have the |H(ω)|
curve in Fig. 6-19(a)
48
Using Poles and Zeros to Analyze IIR Filters
 IIR filter transfer function algebra
 Several ways to write H(z) = Y(z)/X(z) z-domain
transfer function
))()()((
))()()((
)(
)4()3()2()1(
)4()3()2()1()0(
)(
)4()3()2()1(1
)4()3()2()1()0(
)(
3210
3210formfactored
234
234
)formpolynomial(
/bygmultiplyin
4321
4321
44
pzpzpzpz
zzzzzzzz
zH
azazazaz
bzbzbzbzb
zH
zazazaza
zbzbzbzbb
zH
zz
−−−−
−−−−
= →
++++
++++
= →
++++
++++
= −−−−
−−−−
useful because we can
replace z with ejω to
obtain an expression for
frequency response of
filter
necessary so we can factor (find
roots of) polynomials to obtain
values (locations) of numerator
zeros and denominator poles
49
Using Poles and Zeros to Analyze IIR Filters
 Using poles/zeros to obtain transfer functions
 We can analyze an IIR filter’s frequency-domain
performance based solely on poles and zeros
 Given zk zeros and pk poles, we can write the
factored form of filter’s transfer function as
 G = G1/G2 is an arbitrary gain constant
 Filter zeros are associated with decreased frequency
magnitude response, and filter poles are associated with
increased frequency magnitude response
)...)()()()((
)...)()()()((
)...)()()()((
)...)()()()((
)(
43210
43210
432102
432101
pzpzpzpzpz
zzzzzzzzzzG
pzpzpzpzpzG
zzzzzzzzzzG
zH
−−−−−
−−−−−
=
−−−−−
−−−−−
=
50
Using Poles and Zeros to Analyze IIR Filters
if a filter has no z-plane
zeros, and one pole at p0
= 0.8, we can write its
transfer function as
H1(z)=G/(z-0.8)
|H1(ω)| is normalized.
P0 is closest to ω = 0
radians/sample (z = 1)
on unit circle 
lowpass filter.
|p0| < 1  filter is
unconditionally stable
if a filter has a zero at z0
= 1, and a pole at p0 =
−0.8, we write its
transfer function as
8.0)8.0(
)1(
)(2
+
−
=
−−
−
=
z
GGz
z
zG
zH
pole is closest to ω = π
radians/sample (z = −1)
on unit circle 
highpass filter.
the zero located at z =
1 causes filter to have
infinite attenuation at ω
= 0 radians/sample
(zero Hz)
51
Using Poles and Zeros to Analyze IIR Filters
 Fig. 6-21(c)
 Consider a filter having two complex conjugate
zeros at −0.707 ± j0.707, as well as two complex
conjugate poles at 0.283 ± j0.283
 The two poles on the right side of z-plane make this a
lowpass filter having a wider passband than H1(z)
 Two zeros are on unit circle at angles of ω = ±3π/4
radians, causing filter to have infinite attenuation at
frequencies ω = ±3π/4 radians/sample (±3fs/8 Hz)
)283.0283.0()283.0283.0(
)707.0707.0()707.0707.0(
)]283.0283.0([)]283.0283.0([
)]707.0707.0([)]707.0707.0([
)(3
jzjz
jzjzG
jzjz
jzjzG
zH
+−⋅−−
++⋅−+
=
−−⋅+−
−−−⋅+−−
=
52
Using Poles and Zeros to Analyze IIR Filters
53
Using Poles and Zeros to Analyze IIR Filters
 Fig. 6-21(d)
 If we add a z-plane zero at z = 1 to H3(z)
 The zero at z = 1 yields infinite attenuation at ω =
0 radians/sample (zero Hz), creating a bandpass
filter
 Because p0 and p1 poles are oriented at angles of
θ = ±π/4 radians, filter’s passbands are centered
in the vicinity of frequencies ω = ±π/4
radians/sample (±fs/8 Hz)
)283.0283.0()283.0283.0(
)707.0707.0()707.0707.0()1(
)(4
jzjz
jzjzzG
zH
+−⋅−−
++⋅−+⋅−
=
54
Using Poles and Zeros to Analyze IIR Filters
 Fig. 6-21(e)
 If we increase magnitude of H4(z) filter’s poles,
making them equal to 0.636 ± j0.636, we position
the conjugate poles much closer to unit circle
 Poles near unit circle now have a much more
profound effect on filter’s magnitude response
 The poles’ infinite gains cause H5(z) passbands to be
very narrow (sharp)
 When a pole is close to unit circle, center
frequency of its associated passband can be
accurately estimated to be equal to pole’s angle
)636.0636.0()636.0636.0(
)707.0707.0()707.0707.0()1(
)(5
jzjz
jzjzzG
zH
+−⋅−−
++⋅−+⋅−
=
55
Using Poles and Zeros to Analyze IIR Filters
 Fig. 6-21(f)
 Consider an FIR filter—a digital filter whose H(z)
transfer function denominator is unity
 For an FIR filter to have linear phase, each z-plane
zero located at z = z0 = Mejα, where M ≠ 1, must be
accompanied by a zero having an angle of −α and a
magnitude of 1/M
 z0 is accompanied by z3
 If FIR filter’s transfer function polynomial has realvalued
bk coefficients, then a z0 zero not on the zplane’s
real axis will be accompanied by a complex
conjugate zero at z = z2
 Likewise, for FIR filter to have linear phase, z2 zero
must be accompanied by z1 zero
 z1 and z3 zeros are complex conjugates of each other
56
Using Poles and Zeros to Analyze IIR Filters
 z-plane pole/zero properties
 Filter poles are associated with increased
frequency magnitude response (gain)
 Filter zeros are associated with decreased
frequency magnitude response (attenuation)
 To be unconditionally stable, all filter poles must
reside inside the unit circle
 Filter zeros do not affect filter stability
 The closer a pole (zero) is to unit circle, the
stronger will be its effect on filter’s gain
(attenuation) at the frequency associated with the
pole’s (zero’s) angle
57
Using Poles and Zeros to Analyze IIR Filters
 z-plane pole/zero properties
 A pole (zero) located on unit circle produces
infinite filter gain (attenuation)
 If a pole is at the same z-plane location as a zero,
they cancel each other
 Poles or zeros located at origin of z-plane do not
affect frequency response of filter
 Filters whose transfer function denominator
(numerator) polynomial has real-valued
coefficients have poles (zeros) located on real zplane
axis, or pairs of poles (zeros) that are
complex conjugates of each other
58
Using Poles and Zeros to Analyze IIR Filters
 z-plane pole/zero properties
 For an FIR filter (transfer function denominator is
unity) to have linear phase, any zero on z-plane
located at z0 = Mejα, where z0 is not on unit circle
and α is not zero, must be accompanied by a
reciprocal zero whose location is 1/z0 = e−jα/M
 If an FIR filter has real-valued coefficients, is
linear phase, and has a z-plane zero not located
on real z-plane axis or on unit circle, that z-plane
zero is a member of a “gang of four” zeros
 If we know z-plane location of one of those four zeros,
then we know location of the other three zeros
59
Alternate IIR Filter Structures
 Direct Form I, Direct Form II, and transposed
structures
 Direct Form I structure of an IIR filter can be
converted to several alternate forms
60
Alternate IIR Filter Structures
thinking of
feedforward and
feedback portions as
two separate filter
stages, because both
stages are linear and
time invariant, we can
swap them with no
change in y(n) output
because sequence
g(n) is shifted down
along both delay lines
in (b), we can
eliminate one of delay
paths and arrive at
filter structure shown
in (c), where only half
the delay storage
registers are required
compared to Direct
Form I structure
61
Alternate IIR Filter Structures
 Transposition theorem
 There is a process in DSP that allows us to
change structure of an LTI digital network without
changing network’s transfer function (its
frequency response)
 That network conversion process follows transposition
theorem
 A transposed version of some digital network
might be easier to implement, or may exhibit
more accurate processing, than the original
network
62
Alternate IIR Filter Structures
 Steps to transpose a digital filter (starting with
Direct Form II)
 1. reverse direction of all signal-flow arrows
 2. convert all adders to signal nodes
 3. convert all signal nodes to adders
 4. swap x(n) input and y(n) output labels
63
Alternate IIR Filter Structures
By convention, we flip
the network in (b) from
left to right so that x(n)
input is on the left as
shown in (c)
64
Alternate IIR Filter Structures
 Fig. 6-23
 Transposed filter contains the same number of
delay elements, multipliers, and addition
operations as the original filter, and both filters
have the same transfer function given by
 When implemented using infinite-precision
arithmetic, Direct Form II and transposed Direct
Form II filters have identical frequency response
properties
21
21
)2()1(1
)2()1()0(
)(
)(
)( −−
−−
−−
++
==
zaza
zbzbb
zX
zY
zH
65
Alternate IIR Filter Structures
 Fig. 6-23
 Transposed Direct Form II structure is less
susceptible to errors that can occur when finiteprecision
binary arithmetic is used to represent
data values and filter coefficients within a filter
implementation
 Direct Form II filters implement (possibly high-gain)
feedback pole computations before feedforward zeros
computations  large intermediate data values which
must be truncated
 Transposed Direct Form II filters implement zeros
computations first followed by pole computations
 Direct Form I filter has the most resistance to
coefficient quantization and stability problems
66
Impulse Invariance IIR Filter Design Method
 Impulse invariance method
 Is based upon the notion that we can design a
discrete filter whose time-domain impulse
response is a sampled version of impulse
response of a continuous analog filter
 If that analog filter (called prototype filter) has
some desired frequency response, then our IIR
filter will yield a discrete approximation of that
desired response
67
Impulse Invariance IIR Filter Design Method
68
Impulse Invariance IIR Filter Design Method
 Impulse invariance method
 Our goal is to design a digital filter whose impulse
response is a sampled version of analog filter’s
continuous impulse response
 We can map each pole on s-plane for analog
filter’s Hc(s) transfer function to a pole on z-plane
for discrete IIR filter’s H(z) transfer function
 Impulse invariance method yields useful IIR filters
as long as sampling rate is high relative to
bandwidth of signal to be filtered
 IIR filters designed using impulse invariance method
are susceptible to aliasing problems because practical
analog filters cannot be perfectly band-limited
69
Impulse Invariance IIR Filter Design Method
we prefer to make fs as
large as possible to
minimize the overlap
between primary
frequency response
curve and its replicated
images spaced at
multiples of ±fs Hz
Due to aliasing
behavior of impulse
invariance design
method, this filter
design process should
never be used to
design highpass digital
filters
70
Impulse Invariance IIR Filter Design Method
 Two methods for designing IIR filters using
impulse invariance
 Method 1
 Requires that an inverse Laplace transform as well as
a z-transform be performed
 Method 2
 Uses a direct substitution process to avoid inverse
Laplace and z-transformations at the expense of
needing partial fraction expansion algebra necessary
to handle polynomials
71
Impulse Invariance IIR Filter Design Method
 Method 1
 Step 1: Design a prototype analog filter with
desired frequency response
 In a lowpass filter design, for example, filter type
(Chebyshev, Butterworth, elliptic), filter order (number
of poles), and cutoff frequency are parameters to be
defined in this step
 Result of this step is a continuous Laplace transfer
function Hc(s) expressed as ratio of two polynomials,
such as
 N < M, and a(k) and b(k) are constants
∑
∑
=
=
−
−
=
+++−+
+++−+
= M
k
k
N
k
k
MM
NN
c
ska
skb
asasMasMa
bsbsNbsNb
sH
0
0
1
1
)(
)(
)0()1(...)1()(
)0()1(...)1()(
)(
72
Impulse Invariance IIR Filter Design Method
 Method 1
 Step 2: Determine analog filter’s continuous timedomain
impulse response hc(t) from Hc(s)
Laplace transfer function
 Can be done using Laplace tables as opposed to
actually evaluating an inverse Laplace transform
equation
 Step 3: Determine digital filter’s fs, and calculate
ts = 1/fs
 fs is chosen based on absolute frequency, in Hz, of
prototype analog filter
 Because of aliasing problems associated with this
impulse invariance design method, fs should be made
as large as is practical
73
Impulse Invariance IIR Filter Design Method
 Method 1
 Step 4: Find z-transform of continuous hc(t) to
obtain IIR filter’s z-domain transfer function H(z)
in form of a ratio of polynomials in z
 Step 5: Substitute the value ts for the continuous
variable t in H(z) transfer function obtained in
Step 4
 In performing this step, we are ensuring that IIR filter’s
discrete h(n) impulse response is a sampled version of
continuous filter’s hc(t) impulse response so that h(n) =
hc(nts), for 0 ≤ n ≤ ∞
74
Impulse Invariance IIR Filter Design Method
 Method 1
 Step 6: H(z) from Step 5 will now be of the
general form
 Because process of sampling continuous impulse
response results in a digital filter frequency response
that’s scaled by a factor of 1/ts, we include ts factor in this
equation
 Incorporating ts makes IIR filter time-response scaling
independent of sampling rate, and discrete filter will have
the same gain as prototype analog filter
∑
∑
=
−
=
−
−−−−
−−−−
−
=
+++−+
+++−+
= M
k
k
N
k
k
MM
NN
zka
zkb
azazMazMa
bzbzNbzNb
zH
1
0
1)1(
1)1(
)(1
)(
)0()1(...)1()(
)0()1(...)1()(
)(
∑
∑
=
−
=
−
−
== M
k
k
N
k
k
s
zka
zkbt
zX
zY
zH
1
0
)(1
)(
)(
)(
)(
75
Impulse Invariance IIR Filter Design Method
 Method 1
 Step 7: By inspection, we can express filter’s
time-domain difference equation as
)()(...)2()2()1()1(
)]()(...)2()2()1()1()()0([)(
)(1
)(
)(
)()(...)2()2()1()1(
)()(...)2()2()1()1()()0()(
)(1
)(
)(
1
0
1
0
MnyManyanya
NnxNbnxbnxbnxbtny
zka
zkbt
zH
MnyManyanya
NnxNbnxbnxbnxbny
zka
zkb
zH
s
M
k
k
N
k
k
s
M
k
k
N
k
k
−++−+−+
−++−+−+⋅=
→
−
=
−++−+−+
−++−+−+=
→
−
=
∑
∑
∑
∑
=
−
=
−
=
−
=
−
these time-domain
expressions apply only
to the filter structure in
Fig. 6-18. The a(k) and
b(k), or ts · b(k),
coefficients, however,
can be applied to the
improved IIR structure
shown in Fig. 6-22 to
complete our design
76
Impulse Invariance IIR Filter Design Method
77
Impulse Invariance IIR Filter Design Method
 Method 2
 Step 1: Obtain Laplace transfer function Hc(s) for
prototype analog filter
 Same as Method 1, Step 1
 Step 2: Select an appropriate sampling frequency
fs and calculate sample period as ts = 1/fs
 Same as Method 1, Step 3
∑
∑
=
=
−
−
=
+++−+
+++−+
= M
k
k
N
k
k
MM
NN
c
ska
skb
asasMasMa
bsbsNbsNb
sH
0
0
1
1
)(
)(
)0()1(...)1()(
)0()1(...)1()(
)(
78
Impulse Invariance IIR Filter Design Method
 Method 2
 Step 3: Express analog filter’s Laplace transfer
function Hc(s) as sum of single-pole filters
 This requires us to use partial fraction expansion
methods
where M > N, individual Ak factors are constants, and
the kth pole is located at −pk on s-plane
 Hk(s) = kth single-pole analog filter
M
M
M
k k
k
MM
NN
c
ps
A
ps
A
ps
A
ps
A
asasMasMa
bsbsNbsNb
sH
+
++
+
+
+
=
+
=
+++−+
+++−+
=
∑=
−
−
...
)0()1(...)1()(
)0()1(...)1()(
)(
2
2
1
1
1
1
1
k
k
k
ps
A
sH
+
=)(
79
Impulse Invariance IIR Filter Design Method
 Method 2
 Step 4: Substitute 1−e−pktsz−1 for s+pk in Hc(s)
equation in Step 3
 This mapping of each Hk(s) pole, located at s = −pk on
s-plane, to z = e−pkts location on z-plane is how we
approximate the impulse response of each single-pole
analog filter by a single-pole digital filter
 So, the kth analog single-pole filter Hk(s) is
approximated by a single-pole digital filter whose zdomain
transfer function is
∑ ∑= =
−−
−−
−
== →
−
=
M
k
M
k
tp
k
k
tp
k
k
ze
A
zHzH
ze
A
zH
sk
sk
1 1
1
H(z)combinedfinal
1
1
)()(
1
)(
80
Impulse Invariance IIR Filter Design Method
 Method 2
 Step 5: Calculate z-domain transfer function of
the sum of M single-pole digital filters in the form
of a ratio of two polynomials in z
 Because H(z) in Step 4 will be a series of fractions,
we’ll have to combine those fractions over a common
denominator to get a single ratio of polynomials in form
of
∑
∑
=
−
=
−
−
== M
k
k
N
k
k
zka
zkb
zX
zY
zH
1
0
)(1
)(
)(
)(
)(
81
Impulse Invariance IIR Filter Design Method
 Method 2
 Step 6: As in Method 1, Step 6, by inspection, we
express filter’s time-domain equation in general
form of
)()(...)2()2()1()1(
)]()(...)2()2()1()1()()0([)(
)(1
)(
)(
)()(...)2()2()1()1(
)()(...)2()2()1()1()()0()(
)(1
)(
)(
1
0
1
0
MnyManyanya
NnxNbnxbnxbnxbtny
zka
zkbt
zH
MnyManyanya
NnxNbnxbnxbnxbny
zka
zkb
zH
s
M
k
k
N
k
k
s
M
k
k
N
k
k
−++−+−+
−++−+−+⋅=
→
−
=
−++−+−+
−++−+−+=
→
−
=
∑
∑
∑
∑
=
−
=
−
=
−
=
−
Finally, we implement
the improved IIR
structure shown in Fig.
6-22 using a(k) and
b(k) coefficients or a(k)
and ts·b(k) coefficients
from these time-domain
expressions
82
Impulse Invariance IIR Filter Design Method
 Impulse invariance design Method 1 example
 Design an IIR filter that approximates a 2nd-order
Chebyshev prototype analog lowpass filter whose
passband ripple is 1 dB
 fs = 100 Hz (ts = 0.01)
 Filter’s 1 dB cutoff frequency = 20 Hz
83
Impulse Invariance IIR Filter Design Method
 Method 1 example
 Given above filter requirements, assume that
analog prototype filter design effort results in
 It’s this transfer function that we intend to approximate
with our discrete IIR filter
145.1741094536.137
145.17410
)( 2
++
=
ss
sHc
22222
22
2)(
)sin(
)(
:)(:)(oftransformLaplace),(
ωαα
ω
ωα
ω
ω
ωα
ω α
+++
=
++
⋅↔
++
−
ss
A
s
A
tAe
s
A
txtxsX
t
84
Impulse Invariance IIR Filter Design Method
)485173.112sin(77724.154
)sin()(
)485173.112()972680.68(
)485173.112)(77724.154(
)(
77724.154
145.17410
485173.112145.17410145.17410
972680.68
2
94536.137
2145.1741094536.137
145.17410
)(
972680.68
filteranalogprototypeof
responseimpulse
domain-time
22
222
2222
te
tAeth
s
sH
A
ss
A
ss
sH
t
t
c
c
c
⋅=
⋅= →
++
=
==
=−=→=+
==
+++
=
++
=
−
−
ω
ω
αωωα
α
ωαα
ω
α
85
Impulse Invariance IIR Filter Design Method
21
1
268972680.02168972680.0
168972680.0
201.0972680.682101.0972680.68
101.0972680.68
variable
continuous
for01.0
substitute
2972680.6821972680.68
1972680.68
972680.68
221
1
25171605.043278805.01
059517.70
)(
)(
)]12485173.1cos([21
)12485173.1sin(77724.154
)]01.0485173.112cos([21
)01.0485173.112sin(77724.154
)(
)]485173.112cos([21
)485173.112sin(77724.154
)(
)485173.112sin(77724.154)(
)]cos([21
)sin(
)sin(
:)(oftransform-z),(:)(
−−
−
−×−−−
−−
−××−−×−
−×−
=
−×−−−
−−
−
−−−−
−−
−
+−
==
+⋅−
⋅
=
+×⋅−
×⋅
= →
+⋅−
⋅
=
→⋅=
+⋅−
⋅
↔⋅
zz
z
zX
zY
zeze
ze
zeze
ze
zH
zezte
zte
zH
teth
zezte
ztCe
tCe
txzXtx
t
t
tt
t
t
c
tt
t
t
s
αα
α
α
ω
ω
ω
86
Impulse Invariance IIR Filter Design Method
)2(25171605.0)1(43278805.0)1(70059517.0)(
)2(25171605.0)1(43278805.0)1(059517.7001.0)(
)(25171605.0)(43278805.0)(059517.70)(
)059517.70()()25171605.043278805.01()(
25171605.043278805.01
059517.70
)(
)(
)(
scalingfor010bytcoefficien)1(Multiplyfilter.IIRforexpressiondomain-Get time
211
121
21
1
−⋅−−⋅+−⋅=
−⋅−−⋅+−⋅⋅=
 →
⋅−⋅+⋅=
⋅=+−⋅
+−
==
−−−
−−−
−−
−
nynynxny
nynynxny
zzYzzYzzXzY
zzXzzzY
zz
z
zX
zY
zH
.=tn-x s
these coefficients are what we use
in implementing the improved IIR
structure shown in Fig. 6-22 to
approximate the original 2nd-order
Chebyshev analog lowpass filter
87
Impulse Invariance IIR Filter Design Method
 Impulse invariance design Method 2 example
 Given original prototype filter’s Hc(s) as
and ts = 0.01
145.1741094536.137
145.17410
)( 2
++
=
ss
sHc
)2/)(2/(
)(
4
4
24
4
2
)(
)(
|4/)4|(and1
22
2
14517410
94536137
2
jRbsjRbs
c
sH
cbb
s
cbb
s
c
sH
csbs
c
sH
c
cbRj
c
c
.c =
.b =
−+++
= →







 −
−+⋅







 −
++
=
++
= →
−=−=
If we substitute the values for b
and c, the quantity under radical
sign is negative  factors in
denominator are complex
88
Impulse Invariance IIR Filter Design Method
)2/(
2/
)2/(
2/
)(
222/2/
)2/(
)2/)(2/(
222/2/
)2/(
)2/)(2/(
)2/()2/()2/)(2/(
)(
2/
2
2/
1
21
jRbs
Rjc
jRbs
Rjc
sH
R
jc
jR
c
jRbjRb
c
jRbs
jRbsjRbs
c
K
R
jc
jR
c
jRbjRb
c
jRbs
jRbsjRbs
c
K
jRbs
K
jRbs
K
jRbsjRbs
c
sH
c
jRbs
jRbs
c
−+
−
+
++
=
−
==
+++−
=






−+×
−+++
=
=
−
=
−+−−
=






++×
−+++
=
−+
+
++
=
−+++
=
+−=
−−=
89
Impulse Invariance IIR Filter Design Method
212/
12/
21)2/()2/(
1)2/()2/(
21)2/(1)2/(
1)2/(1)2/(
1)2/(1)2/(
1)2/(1)2/(
1)2/(1)2/(
termsfor1substitute
:plane-ztoplane-sfrom2and2polesmap
)(1
)()2/(
)(
)(1
))(2/(
)(
1
)11)(2/(
)(
)1)(1(
)1)(2/()1)(2/(
)(
1
2/
1
2/
)(
)2/(
2/
)2/(
2/
)(
1
21
−−−−−
−−−
−−−+−−−
−−−+−
−−−+−−−−
−+−−−−
−−−−+−
−+−−−−
−−−−+−
+−
−−−
++−
−
=
++−
−
=
+−−
+−−
=
−−
−−−
=
−
−
+
−
=
 →
−+
−
+
++
=
−−
zezeee
zeeeRjc
zH
zezee
zeeRjc
zH
zezeze
zezeRjc
zH
zeze
zeRjczeRjc
zH
ze
Rjc
ze
Rjc
zH
jRbs
Rjc
jRbs
Rjc
sH
ssss
sss
sss
ss
sss
ss
ss
ss
ss
k
stkp
btjRtjRtbt
jRtjRtbt
bttjRbtjRb
tjRbtjRb
bttjRbtjRb
tjRbtjRb
tjRbtjRb
tjRbtjRb
tjRbtjRb
psze
+jRb/=pjRb/=p
c
90
Impulse Invariance IIR Filter Design Method
21
1
21
1
010and,48517112,94536137,14517410
212/
12/
212/
12/
2/)()(osand2/)()sin(:Euler
212/
12/
25171605.043278805.01
059517.70
25171605.0)86262058.0)(50171312.0(1
)902203655.0)(50171312.0)(77724.154(
)(
)]cos(2[1
)][sin()/(
)]cos(2[1
)]sin(2[)2/(
)(
)(1
)()2/(
)(
−−
−
−−
−
−−−−
−−
−−−−
−−
+=−=
−−−−−
−−−
+−
=
+−
=
 →
+−
=
+−
−
=
 →
++−
−
=
−−
zz
z
zz
z
zH
zezRte
zRteRc
zezRte
zRtjeRjc
zH
zezeee
zeeeRjc
zH
.=t.R=.b=.c=
bt
s
bt
s
bt
bt
s
bt
s
bt
eecjee
btjRtjRtbt
jRtjRtbt
s
ss
s
ss
s
jjjj
ssss
sss
φφφφ
φφ
91
Impulse Invariance IIR Filter Design Method
)2(25171605.0)1(43278805.0)1(70059517.0)(
)2(25171605.0)1(43278805.0)1(059517.7001.0)(
)2(25171605.0)1(43278805.0)1(059517.70)(
)(25171605.0)(43278805.0)(059517.70)(
)059517.70()()25171605.043278805.01()(
25171605.043278805.01
059517.70
)(
)(
)(
bytcoefficien)1(hemultiply t
211
121
21
1
−⋅−−⋅+−⋅=
−⋅−−⋅+−⋅⋅=
 →
−⋅−−⋅+−⋅=
⋅−⋅+⋅=
⋅=+−⋅
+−
==
−
−−−
−−−
−−
−
nynynxny
nynynxny
nynynxny
zzYzzYzzXzY
zzXzzzY
zz
z
zX
zY
zH
stnx
92
Impulse Invariance IIR Filter Design Method
)2/(
2/
)2/(
2/
)(
jRbs
Rjc
jRbs
Rjc
sHc
−+
−
+
++
=
93
Impulse Invariance IIR Filter Design Method
 IIR filter’s z-plane pole locations


64.450.5017:poleconjugate
64.45-0.5017
radians
)2/()2/(
)1(
)2/(
)1(
)2/(
)(
1
2/
1
2/
)(
poleplane-zupper
2/2/)2/(poleplane-zlower
)2/()2/(
1)2/(1)2/(
1)2/(1)2/(
+∠= →
∠
−∠=== →
−
−
+
−
=
−
−
+
−
=⋅
−
−
+
−
=
−−−+−
−−+−
−−−−+−
−−−−+−
s
btjRtbttjRb
tjRbtjRb
tjRbtjRb
tjRbtjRb
Rteeeez
ez
zRjc
ez
zRjc
zez
Rjcz
zez
Rjcz
z
z
zH
ze
Rjc
ze
Rjc
zH
ssss
ss
ss
ss
94
Impulse Invariance IIR Filter Design Method
b(0) coefficient is zero here equivalent
95
Impulse Invariance IIR Filter Design Method
 In general, Method 2 is more popular for two
reasons
 (1) inverse Laplace and z-transformations can be
very difficult for higher-order filters
 (2) unlike Method 1, Method 2 can be coded in a
software routine or a computer spreadsheet
96
Impulse Invariance IIR Filter Design Method
 Fig. 6-35(b)
 Roll-off is not particularly steep
 Attenuation slope is so gradual that it doesn’t appear to
be of much use as a lowpass filter
 Any frequency representation (be it a digital filter
magnitude response or a signal spectrum) that has
nonzero values at +fs/2, most probably has aliasing
problem
 We also see that passband ripple is greater than
the desired value of 1 dB in Fig. 6-34
 It’s not the low order of filter that contributes to its
poor performance, but the sampling rate used
97
Impulse Invariance IIR Filter Design Method
Increasing sampling
rate to 400 Hz results in
the much improved
frequency response
Sampling rate changes do not affect filter
order or implementation structure; it only
changes ts in our design equations, resulting
in a different set of filter coefficients
the smaller we make ts (larger fs), the
better the resulting filter because the
replicated spectral overlap indicated in
Fig. 6-32(b) is reduced due to the larger fs
impulse invariance IIR filter
design techniques are most
appropriate for narrowband
filters, that is, lowpass filters
whose cutoff frequencies are
much smaller than the
sampling rate
98
Bilinear Transform IIR Filter Design Method
 Bilinear transform method
 A popular analytical IIR filter design technique
 Like impulse invariance method, this design
technique approximates a prototype analog filter
defined by continuous Laplace transfer function
Hc(s) with a discrete filter whose transfer function
is H(z)
99
Bilinear Transform IIR Filter Design Method
 Bilinear transform method has great utility
 It allows to substitute a function of z for s in Hc(s)
to get H(z), eliminating the need for Laplace and
z-transformations as well as any necessity for
partial fraction expansion algebra
 It maps the entire s-plane to z-plane, enabling us
to completely avoid frequency-domain aliasing
problems we had with impulse invariance design
method
 It induces a nonlinear distortion of H(z)’s
frequency axis, relative to the original prototype
analog filter’s frequency axis, that sharpens the
final roll-off of digital lowpass filters
100
Bilinear Transform IIR Filter Design Method
 Bilinear transform method
 If transfer function of a prototype analog filter is
Hc(s), we obtain discrete IIR filter z-domain
transfer function H(z) by substituting the following
for s in Hc(s)
where ts is discrete filter’s sampling period (1/fs)








+
−
= −
−
1
1
1
12
z
z
t
s
s
101
Bilinear Transform IIR Filter Design Method
 Bilinear transform’s s- to z-plane mapping
 Any pole on the left side of s-plane will map to the
inside of unit circle in z-plane
 If σ < 0, |z| will be less than 1
 If σ > 0, |z| will be greater than 1
 This confirms that when using bilinear transform, any
pole located on the left side of s-plane (σ < 0) will map to
a z-plane location inside unit circle
22
22
rdenominato
numerator
1
1
)2/()2/1(
)2/()2/1(
Mag
Mag
2/)2/1(
2/)2/1(
2/2/1
2/2/1
)2/(1
)2/(1
1
12
sas
sas
sas
sas
sas
sasjs
s
s
s
tt
tt
z
tjt
tjt
tjt
tjt
z
st
st
z
z
z
t
s
a
ωσ
ωσ
ωσ
ωσ
ωσ
ωσωσ
+−
++
==
−−
++
=
−−
++
= →
−
+
=→








+
−
=
+=
−
−
analog
102
Bilinear Transform IIR Filter Design Method
 Bilinear transform’s s- to z-plane mapping
 jωa axis of s-plane maps to perimeter of unit
circle in z-plane
1
)2/()1(
)2/()1(
Mag
Mag
2/1
2/1
2/)2/1(
2/)2/1(
22
22
rdenominato
numerator
0
0
=
+
+
==
−
+
= →
−−
++
=
=
=
sa
sa
sa
sa
sas
sas
t
t
z
tj
tj
z
tjt
tjt
z
ω
ω
ω
ω
ωσ
ωσ
σ
σ
103
Bilinear Transform IIR Filter Design Method
 Bilinear transform’s s- to z-plane mapping
 Frequency mapping from jωa axis of s-plane to
perimeter of unit circle in z-plane is not linear
)2/tan(
2
)2/cos(
)2/sin(
2
22
)]2/cos(2[
)]2/sin(2[2
)(
)(2
1
12
1
12
2/
2/
2/
2/
2/)()cos(and
2/)()sin(:Euler
2/2/2/
2/2/2/
1circleuniton
1
1
d
sd
d
j
j
s
a
d
j
d
j
s
a
ee
jee
jjj
jjj
s
a
j
j
s
ez
s
t
jj
e
e
t
js
e
je
t
js
eee
eee
t
js
e
e
t
s
z
z
t
s
d
d
d
djj
jj
ddd
ddd
d
ddj
ω
ω
ω
ωσ
ω
ω
ωσ
ωσ
ω
ω
ω
ω
φ
φ
ωωω
ωωω
ω
ω
φφ
φφ
ω
=⋅⋅=+=
⋅=+= →
+
−
⋅=+=








+
−
= →








+
−
=
−
−
−
−
+=
−=
−−
−−
−
−
=
−
−
−
−
104
Bilinear Transform IIR Filter Design Method
 Bilinear transform’s s- to z-plane mapping
 Range of ωd is ±π, and dimensions of digital
frequency ωd are radians/sample
 Range of ωa is ±∞, and dimensions of analog
frequency ωa are radians/second
πωπ
ω
ω
ωωω
σ
ωωσ
≤≤−





=




∞≤≤∞−=
=
→=+=
−
d
sa
d
ad
s
a
d
s
a
t
t
t
j
js
2
tan2
)2/tan(
2
0
)2/tan(
2
1
105
Bilinear Transform IIR Filter Design Method
πωπ
ω
ω ≤≤−





= −
d
sa
d
t
2
tan2 1
frequency warping due
to bilinear transform
no matter how large s-plane’s analog ωa
becomes, z-plane’s ωd will never be
greater than π radians/sample (fs/2 Hz)
106
Bilinear Transform IIR Filter Design Method
bilinear transform maps s-plane’s
entire jωa axis onto the unit circle
in z-plane
107
Bilinear Transform IIR Filter Design Method
 Frequency warping
 ωd = π radians/sample corresponds to fd = fs/2 Hz
 ωd = 2π(fd / fs)
Hz
)/tan(
2
tan
2
Hz
)/(tan
2
tan2
2
)/(2
1
2
)/(2
1
π
πω
ω
π
πω
ω
πω
πω
πω
πω
sds
a
f
ff
d
s
a
sas
d
f
ff
sa
d
fff
f
t
fff
f
t
aa
sdd
aa
sdd
= →





=
= →





=
=
=
−
=
=
−
108
Bilinear Transform IIR Filter Design Method
Hz
)/(tan 1
π
π sas
d
fff
f
−
=
fd frequency warping
(compression)
becomes more severe
as fd approaches fs/2
if a bilinear-transformdesigned
digital
bandpass filter is desired
to have an upper cutoff
frequency of fd1 Hz, then
the original prototype
analog bandpass filter
must be designed
(prewarped) to have an
upper cutoff frequency of
fa1 Hz using:
Hz
)/tan(
π
π sds
a
fff
f =
no IIR filter response
aliasing can occur with
bilinear transform
design method
109
Bilinear Transform IIR Filter Design Method
 Steps to perform an IIR filter design using
bilinear transform method
 Step 1: Obtain Laplace transfer function Hc(s) for
the prototype analog filter
 Step 2: Determine digital filter’s equivalent fs and
establish ts = 1/fs
 Step 3: In Laplace Hc(s) transfer function,
substitute the expression
for the variable s to get IIR filter’s H(z) transfer
function








+
−
−
−
1
1
1
12
z
z
ts
110
Bilinear Transform IIR Filter Design Method
 Step 4: Multiply numerator and denominator of
H(z) by appropriate power of (1 + z−1) and collect
terms of like powers of z in the form
 Step 5: By inspection, express IIR filter’s timedomain
equation in the general form of
 Although this expression only applies to filter structure
in Fig. 6-18, we can apply a(k) and b(k) coefficients to
improved IIR structure shown in Fig. 6-22
∑
∑
=
−
=
−
−
= M
k
k
N
k
k
zka
zkb
zH
1
0
)(1
)(
)(
)()(...)2()2()1()1(
)()(...)2()2()1()1()()0()(
MnyManyanya
NnxNbnxbnxbnxbny
−++−+−+
−++−+−+=
111
Bilinear Transform IIR Filter Design Method
 Bilinear transform design example
 Design an IIR filter that approximates 2nd-order
Chebyshev prototype analog lowpass filter whose
passband ripple is 1 dB
 fs = 100 Hz (ts = 0.01)
 Filter’s 1 dB cutoff frequency = 20 Hz
 Original prototype filter’s Laplace transfer function is
given as
145.1741094536.137
145.17410
)( 2
++
=
ss
sHc
112
Bilinear Transform IIR Filter Design Method
22122
21
zofpowerslikecollecting
2111212
21
)1(byrdenominato
andnumeratormultiply
1
1
2
1
1
2
/2
:simplifyto
1
1
2
1
1
1
12
2
14517410
94536137
2
)()22()(
)21(
)(
)1()1)(1()1(
)1(
)(
1
1
1
1
)(
1
12
1
12
)(
)(
145.1741094536.137
145.17410
)(
21
1
1
−−
−−
−−−−
−
+
−
−
−
−
=
−
−
−
−








+
−
=
−++−+++
++
= →
++−++−
+
= →
+







+
−
+







+
−
= →
+







+
−
+







+
−
⋅
= →
++
= →
++
=
−
−
−
zabcazaccaba
zzc
zH
zczzabza
zc
zH
c
z
z
ab
z
z
a
c
zH
c
z
z
t
b
z
z
t
c
zH
csbs
c
sH
ss
sH
z
ta
ss
z
z
t
s
c
.c =
.b =
c
s
s
113
Bilinear Transform IIR Filter Design Method
)2(35083938.0)1(53153089.0
)2(20482712.0)1(40965424.0)(20482712.0)(
35083938.053153089.01
20482712.040965424.020482712.0
)(
35083938.053153089.01
)21(20482712.0
)(
)(
)(
)(
)22(
1
)21(
)(
)(
filterIIRfor
expression
domain-time
21
21
21
21
14517410
94536137
2002
2
2
2
1
2
2
21
2
c)ab(abyrdenominatoandnumeratordivide
r,denominatoinoneofermconstant tagetto
2
−⋅−−⋅+
−⋅+−⋅+⋅= →
+−
++
=
+−
++
= →
++
−+
+
++
−
+
++
++
= →
−−
−−
−−
−−
−−
−−
++
nyny
nxnxnxny
zz
zz
zH
zz
zz
zH
z
caba
abca
z
caba
ac
zz
caba
c
zH
.c =
.b =
=/ta = s
114
Bilinear Transform IIR Filter Design Method
bilinear-transformdesigned
filter’s
magnitude response
approaches zero at
folding frequency of
fs/2 = 50 Hz
115
Bilinear Transform IIR Filter Design Method
bilinear transform design method gives a much sharper roll-off for our lowpass filter
for two reasons: 1) frequency warping of bilinear transform method compresses
(sharpens) roll-off portion of a lowpass filter; 2) the price we pay in terms of
additional complexity of implementation of our IIR filter: our new filter requires five
multiplications per filter output sample where impulse invariance design filter in Fig.
6-36(a) required only three multiplications
116
Bilinear Transform IIR Filter Design Method
 Prewarping
 If cutoff frequency is a large percentage of fs,
resultant |Hd(fd)| cutoff frequency will be below
the desired value
 To avoid this, we prewarp prototype analog filter’s
cutoff frequency requirement before the analog Hc(s)
transfer function is derived in Step 1
 In that way, they compensate for the bilinear transform’s
frequency warping before it happens
 To determine prewarped analog filter lowpass cutoff
frequency that we want mapped to the desired IIR
lowpass cutoff frequency, use






=
2
tan
2 d
s
a
t
ω
ω
we plug desired IIR cutoff frequency
ωd in here to calculate ωa cutoff
frequency used to derive prototype
analog filter’s Hc(s) transfer function
117
Optimized IIR Filter Design Method
 Optimization methods
 Developed for situation when the desired IIR filter
frequency response is not of standard lowpass,
bandpass, or highpass form
 Closed-form expressions for filter’s z-transform do not
exist  no explicit equations to work with
 Designer should describe, in some way, the desired
IIR filter frequency response
 The algorithm, then, assumes a filter transfer function
H(z) as a ratio of polynomials in z and starts to
calculate filter’s frequency response
 Based on some error criteria, the algorithm iteratively
adjusts filter’s coefficients to minimize the error
between desired and actual filter frequency response
118
Optimized IIR Filter Design Method
 Optimized IIR filter design routines
 Are used to design the simpler lowpass,
bandpass, or highpass forms even though
analytical techniques exist
 They only require the designer to specify a few
key amplitude and frequency values, and the
desired order of IIR filter (the number of feedback
taps), and the software computes the final
feedforward and feedback coefficients
119
Optimized IIR Filter Design Method
In specifying a lowpass IIR
filter, a software design routine
might require us to specify the
values for δp, δs, f1, and f2
120
Comparison of IIR and FIR Filters
Characteristic IIR FIR (nonrecursive)
Number of necessary multiplications Least Most
Sensitivity to filter coefficient quantization Can be high Very low
Probability of overflow errors Can be high Very low
Stability Must be designed in Guaranteed
Linear phase No Guaranteed
Can simulate prototype analog filters Yes No
Required coefficient memory Least Most
Hardware filter control complexity Moderate Simple
Availability of design software Good Very good
Ease of design, or complexity of design
software
Moderately complicated Simple
Difficulty of quantization noise analysis Most complicated Least complicated
Supports adaptive filtering With difficulty Yes
So long as FIR
coefficients are
symmetrical (or
antisymmetrical)