Simple Digital Filters
• Later in the course we shall review various
methods of designing frequency-selective
filters satisfying prescribed specifications
• We now describe several low-order FIR and
IIR digital filters with reasonable selective
frequency responses that often are
satisfactory in a number of applications
1
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
• FIR digital filters considered here have
integer-valued impulse response coefficients
• These filters are employed in a number of
practical applications, primarily because of
their simplicity, which makes them amenable
to inexpensive hardware implementations
2
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
3
Lowpass FIR Digital Filters
• The simplest lowpass FIR digital filter is the
2-point moving-average filter given by
z +1
−1
1
H 0 ( z ) = 2 (1 + z ) =
2z
• The above transfer function has a zero at
z = −1 and a pole at z = 0
• Note that here the pole vector has a unity
magnitude for all values of ω
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
• On the other hand, as ω increases from 0 to
π, the magnitude of the zero vector
decreases from a value of 2, the diameter of
the unit circle, to 0
• Hence, the magnitude response | H 0 (e jω )| is
a monotonically decreasing function of ω
from ω = 0 to ω = π
4
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
• The maximum value of the magnitude
function is 1 at ω = 0, and the minimum
value is 0 at ω = π, i.e.,
j0
jπ
| H 0 (e )| = 1, | H 0 (e )| = 0
• The frequency response of the above filter
is given by
jω
H 0 (e ) = e
− jω / 2
cos(ω / 2)
5
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
jω
• The magnitude response | H 0 (e )| = cos(ω / 2)
can be seen to be a monotonically
decreasing function of ω
First-order FIR lowpass filter
1
Magnitude
0.8
0.6
0.4
0.2
6
0
0
0.2
0.4
0.6
ω/π
0.8
1
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
• The frequency ω = ωc at which
1
jωc
H 0 (e ) =
H 0 (e j 0 )
2
is of practical interest since here the gain G (ωc )
in dB is given by
jωc
(
ω
)
G c = 20 log10 H (e )
= 20 log10 H (e ) − 20 log10 2 ≅ −3 dB
j0
since the dc gain G (0) = 20 log H (e ) = 0
j0
7
10
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
8
• Thus, the gain G(ω) at ω = ωc is
approximately 3 dB less than the gain at ω
=0
• As a result, ωc is called the 3-dB cutoff
frequency
• To determine the value of ωc we set
jωc 2
2
| H 0 (e )| = cos (ωc / 2) = 1
2
which yields ωc = π / 2
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
• The 3-dB cutoff frequency ωc can be
considered as the passband edge frequency
• As a result, for the filter H 0 (z ) the passband
width is approximately π/2
• The stopband is from π/2 to π
• Note: H 0 (z ) has a zero at z = −1 or ω = π,
which is in the stopband of the filter
9
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
• A cascade of the simple FIR filter
−1
1
H 0 ( z ) = 2 (1 + z )
results in an improved lowpass frequency
response as illustrated below for a cascade
of 3 sections
First-order FIR lowpass filter cascade
1
Magnitude
0.8
0.6
0.4
0.2
10
0
0
0.2
0.4
0.6
ω/π
0.8
1
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
• The 3-dB cutoff frequency of a cascade of
M sections is given by
ωc = 2 cos −1 (2−1 / 2 M )
• For M = 3, the above yields ωc = 0.302π
• Thus, the cascade of first-order sections
yields a sharper magnitude response but at
the expense of a decrease in the width of the
passband
11
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
12
• A better approximation to the ideal lowpass
filter is given by a higher-order movingaverage filter
• Signals with rapid fluctuations in sample
values are generally associated with highfrequency components
• These high-frequency components are
essentially removed by an moving-average
filter resulting in a smoother output
waveform
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
Highpass FIR Digital Filters
• The simplest highpass FIR filter is obtained
from the simplest lowpass FIR filter by
replacing z with − z
• This results in
H1( z ) = 1 (1 − z −1 )
2
13
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
• Corresponding frequency response is given
by
jω
− jω / 2
H1 (e ) = j e
sin(ω / 2)
whose magnitude response is plotted below
First-order FIR highpass filter
1
Magnitude
0.8
0.6
0.4
0.2
14
0
0
0.2
0.4
0.6
ω/π
0.8
1
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
• The monotonically increasing behavior of
the magnitude function can again be
demonstrated by examining the pole-zero
pattern of the transfer function H1(z )
• The highpass transfer function H1(z ) has a
zero at z = 1 or ω = 0 which is in the
stopband of the filter
15
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
• Improved highpass magnitude response can
again be obtained by cascading several
sections of the first-order highpass filter
• Alternately, a higher-order highpass filter of
the form
M −1
n −n
1
H1( z ) = ∑n =0 (−1) z
M
is obtained by replacing z with − z in the
transfer function of a moving average filter
16
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
• An application of the FIR highpass filters is
in moving-target-indicator (MTI) radars
• In these radars, interfering signals, called
clutters, are generated from fixed objects in
the path of the radar beam
• The clutter, generated mainly from ground
echoes and weather returns, has frequency
components near zero frequency (dc)
17
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
• The clutter can be removed by filtering the
radar return signal through a two-pulse
canceler, which is the first-order FIR
highpass filter H1( z ) = 1 (1 − z −1 )
2
• For a more effective removal it may be
necessary to use a three-pulse canceler
obtained by cascading two two-pulse
cancelers
18
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
Lowpass IIR Digital Filters
• A first-order causal lowpass IIR digital
filter has a transfer function given by
−1 

1− α 1+ z


H LP ( z ) =
2  1 − α z −1
where |α| < 1 for stability
• The above transfer function has a zero at z = −1
i.e., at ω = π which is in the stopband
19
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
• H LP (z ) has a real pole at z = α
• As ω increases from 0 to π, the magnitude
of the zero vector decreases from a value of
2 to 0, whereas, for a positive value of α,
the magnitude of the pole vector increases
from a value of 1 − α to 1 + α
• The maximum value of the magnitude
function is 1 at ω = 0, and the minimum
value is 0 at ω = π
20
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
j0
jπ
|
H
e
=
H
e
(
)
|
1
,
|
(
)| = 0
• i.e., LP
LP
jω
• Therefore, | H LP (e )| is a monotonically
decreasing function of ω from ω = 0 to ω = π
as indicated below
1
α = 0.8
α = 0.7
α = 0.5
-5
0.6
Gain, dB
Magnitude
0.8
0
0.4
-15
0.2
0
21
-10
α = 0.8
α = 0.7
α = 0.5
0
0.2
0.4
0.6
ω/π
0.8
1
-20
10
-2
10
-1
ω/π
Copyright © 2001, S. K. Mitra
10
0
Simple IIR Digital Filters
• The squared magnitude function is given by
2
(1 − α) (1 + cos ω)
jω 2
| H LP (e )| =
2
2(1 + α − 2α cos ω)
jω 2
• The derivative of | H LP (e )| with respect
to ω is given by
2
2
jω 2
d | H LP (e )|
− (1 − α) (1 + 2α + α ) sin ω
=
2 2
dω
2(1 − 2α cos ω + α )
22
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
d | H LP (e jω )| 2 / dω ≤ 0 in the range 0 ≤ ω ≤ π
verifying again the monotonically decreasing
behavior of the magnitude function
• To determine the 3-dB cutoff frequency
we set
jωc 2 1
| H LP (e )| =
2
in the expression for the square magnitude
function resulting in
23
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
(1 − α) 2 (1 + cos ωc ) 1
=
2(1 + α 2 − 2α cos ωc ) 2
or
(1 − α) 2 (1 + cos ωc ) = 1 + α 2 − 2α cos ωc
which when solved yields
2α
cos ωc =
2
1+ α
• The above quadratic equation can be solved
for α yielding two solutions
24
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
• The solution resulting in a stable transfer
function H LP (z ) is given by
1 − sin ωc
α=
cos ωc
• It follows from
2
(
1
−
α
)
(1 + cos ω)
jω 2
| H LP (e )| =
2(1 + α 2 − 2α cos ω)
that H LP ( z ) is a BR function for |α| < 1
25
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
26
Highpass IIR Digital Filters
• A first-order causal highpass IIR digital filter
has a transfer function given by
1 + α  1 − z −1 
H HP ( z ) =
2  1 − α z −1 
where |α| < 1 for stability
• The above transfer function has a zero at z = 1
i.e., at ω = 0 which is in the stopband
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
• Its 3-dB cutoff frequency ωc is given by
α = (1 − sin ωc ) / cos ωc
which is the same as that of H LP (z )
• Magnitude and gain responses of H HP (z )
are shown below
0
α = 0.8
α = 0.7
α = 0.5
0.8
0.6
-5
Gain, dB
Magnitude
1
-10
0.4
-15
0.2
0
27
α = 0.8
α = 0.7
α = 0.5
0
0.2
0.4
0.6
ω/π
0.8
1
-20
-2
-1
10
10
10
0
ω/π
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
• H HP (z ) is a BR function for |α| < 1
• Example - Design a first-order highpass
digital filter with a 3-dB cutoff frequency of
0.8π
• Now, sin(ωc ) = sin(0.8π) = 0.587785 and
cos(0.8π) = − 0.80902
28
• Therefore
α = (1 − sin ωc ) / cos ωc = − 0.5095245
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
• Therefore,
1 + α  1 − z −1 
H HP ( z ) =
2  1 − α z −1 
−1


1
z
−

= 0.245238 
−1 
 1 + 0.5095245 z 
29
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
Bandpass IIR Digital Filters
• A 2nd-order bandpass digital transfer
function is given by

1− α 
1 − z −2


H BP ( z ) =
2  1 − β(1 + α) z −1 + α z −2 
• Its squared magnitude function is
jω 2
H BP (e )
30
(1 − α) 2 (1 − cos 2ω)
=
2[1 + β2 (1 + α) 2 + α 2 − 2β(1 + α) 2 cos ω + 2α cos 2ω]
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
• | HBP (e jω )|2 goes to zero at ω = 0 and ω = π
• It assumes a maximum value of 1 at ω = ωo ,
called the center frequency of the bandpass
filter, where
ωo = cos −1 (β)
jω 2
ω
ω
• The frequencies c1 and c 2 where | HBP (e )|
becomes 1/2 are called the 3-dB cutoff
frequencies
31
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
• The difference between the two cutoff
frequencies, assuming ωc 2 > ωc1 is called
 2αand is given by
the 3-dB bandwidth

2
 1 + α  −1 2α 
Bw = ωc 2 − ωc1 = cos 
2
1 + α 
• The transfer function H BP ( z ) is a BR
function if |α| < 1 and |β| < 1
32
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
• Plots of | HBP (e jω )| are shown below
β = 0.34
1
α = 0.8
α = 0.5
α = 0.2
0.6
0.4
0.2
0
0
β = 0.8
β = 0.5
β = 0.2
1
Magnitude
0.8
Magnitude
α = 0.6
0.8
0.6
0.4
0.2
0.2
0.4
0.6
ω/π
0.8
1
0
0
0.2
0.4
0.6
0.8
ω/π
33
Copyright © 2001, S. K. Mitra
1
Simple IIR Digital Filters
• Example - Design a 2nd order bandpass
digital filter with center frequency at 0.4π
and a 3-dB bandwidth of 0.1π
• Here β = cos(ωo ) = cos(0.4π) = 0.309017
and
2α
= cos( Bw ) = cos(0.1π) = 0.9510565
2
1+ α
• The solution of the above equation yields:
α = 1.376382 and α = 0.72654253
34
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
• The corresponding transfer functions are
−2
1− z
'
H BP ( z ) = −0.18819
−1
−2
1 − 0.7343424 z + 1.37638 z
and
−2
1
−
z
" ( z ) = 0.13673
H BP
1 − 0.533531z −1 + 0.72654253z − 2
' ( z ) are at z = 0.3671712 ±
• The poles of H BP
j1.11425636 and have a magnitude > 1
35
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
' ( z ) are outside the
• Thus, the poles of H BP
unit circle making the transfer function
unstable
" ( z ) are
• On the other hand, the poles of H BP
at z = 0.2667655 ± j 0.8095546 and have a
magnitude of 0.8523746
" ( z ) is BIBO stable
• Hence H BP
• Later we outline a simpler stability test
36
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
• Figures below show the plots of the
magnitude function and the group delay of
" ( z)
H BP
7
1
Group delay, samples
6
Magnitude
0.8
0.6
0.4
0.2
37
0
0
5
4
3
2
1
0
0.2
0.4
0.6
ω/π
0.8
1
-1
0
0.2
0.4
0.6
0.8
ω/π
Copyright © 2001, S. K. Mitra
1
Simple IIR Digital Filters
Bandstop IIR Digital Filters
• A 2nd-order bandstop digital filter has a
transfer function given by

1 + α  1 − 2β z −1 + z −2


H BS ( z ) =
2  1 − β(1 + α) z −1 + α z −2 
• The transfer function H BS (z ) is a BR
function if |α| < 1 and |β| < 1
38
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
1
1
0.8
0.8
Magnitude
Magnitude
• Its magnitude response is plotted below
0.6
0.4
α = 0.8
α = 0.5
α = 0.2
0.2
0
0
0.2
0.4
0.6
ω/π
0.8
0.6
0.4
β = 0.8
β = 0.5
β = 0.2
0.2
1
0
0
0.2
0.4
0.6
0.8
ω/π
39
Copyright © 2001, S. K. Mitra
1
Simple IIR Digital Filters
• Here, the magnitude function takes the
maximum value of 1 at ω = 0 and ω = π
• It goes to 0 at ω = ωo , where ωo , called the
notch frequency, is given by
ωo = cos −1 (β)
• The digital transfer function H BS (z ) is more
commonly called a notch filter
40
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
jω 2
• The frequencies ωc1 and ωc 2 where | HBS (e )|
becomes 1/2 are called the 3-dB cutoff
frequencies
• The difference between the two cutoff
frequencies, assuming ωc 2 > ωc1 is called
the 3-dB notch bandwidth and is given by
2α 
Bw = ωc 2 − ωc1 = cos 
2
1 + α 
−1 
41
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
42
Higher-Order IIR Digital Filters
• By cascading the simple digital filters
discussed so far, we can implement digital
filters with sharper magnitude responses
• Consider a cascade of K first-order lowpass
sections characterized by the transfer
function
1 − α  1 + z −1 


H LP ( z ) =
2  1 − α z −1
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
• The overall structure has a transfer function
given by
K
−
1
1 − α 1 + z 

GLP ( z ) = 
⋅
−1 
2
1− α z 

43
• The corresponding squared-magnitude
function is given by
K
2
jω 2  (1 − α ) (1 + cos ω) 
|GLP (e )| = 

2
 2(1 + α − 2α cos ω) 
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
44
• To determine the relation between its 3-dB
cutoff frequency ωc and the parameter α,
we set
K
2
 (1 − α) (1 + cos ωc ) 
1

 =
2
2
 2(1 + α − 2α cos ωc ) 
which when solved for α, yields for a stable
GLP ( z ):
2
1 + (1 − C ) cos ωc − sin ωc 2C − C
α=
1 − C + cos ωc
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
where
C = 2( K −1) / K
• It should be noted that the expression for α
given earlier reduces to
1 − sin ωc
α=
cos ωc
for K = 1
45
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
46
• Example - Design a lowpass filter with a 3dB cutoff frequency at ωc = 0.4π using a
single first-order section and a cascade of 4
first-order sections, and compare their gain
responses
• For the single first-order lowpass filter we
have
1 + sin ωc 1 + sin(0.4π)
α=
=
= 0.1584
cos ωc
cos(0.4π)
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
• For the cascade of 4 first-order sections, we
substitute K = 4 and get
C = 2( K −1) / K = 2( 4−1) / 4 = 1.6818
• Next we compute
2
1 + (1 − C ) cos ωc − sin ωc 2C − C
α=
1 − C + cos ωc
1 + (1 − 1.6818) cos(0.4π) − sin(0.4π) 2(1.6818) − (1.6818) 2
=
1 − 1.6818 + cos(0.4π)
47
= − 0.251
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
• The gain responses of the two filters are
shown below
• As can be seen, cascading has resulted in a
sharper roll-off in the gain response
0
-5
0
Gain, dB
Gain, dB
Passband details
K=1
-10
K=4
K=1
-2
-4
-15
10
48
-20 -2
10
-1
10
ω/π
K=4
10
-2
-1
10
ω/π
10
0
Copyright © 2001, S. K. Mitra
0