Simple Digital Filters
Simple FIR 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
• 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
1
2
Copyright © 2001, S. K. Mitra
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
H 0 ( z) = 12 (1 + z −1) =
2z
• The above transfer function has a zero at
z = −1and a pole at z = 0
• Note that here the pole vector has a unity
magnitude for all values of ω
Simple FIR Digital Filters
• On the other hand, as ω increases from0 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
fromω = 0 to ω = π
4
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
Simple FIR Digital Filters
• The magnitude response | H0 ( e jω )| = cos(ω / 2)
can be seen to be a monotonically
decreasing function of ω
• The maximum value of the magnitude
function is 1 at ω = 0, and the minimum
value is 0 at ω = π, i.e.,
| H 0 (e j0 )| = 1,
| H 0 (e jπ )| = 0
First-order FIR lowpass filter
1
• The frequency response of the above filter
is given by
Magnitude
0.8
H 0 (e j ω ) = e − jω / 2 cos(ω / 2)
5
0.4
0.2
6
Copyright © 2001, S. K. Mitra
0.6
0
0
0.2
0.4
0.6
ω/π
0.8
1
Copyright © 2001, S. K. Mitra
1
Simple FIR Digital Filters
7
• The frequency ω = ωc at which
1
H 0 (e j ωc ) =
H 0 (e j 0 )
2
is of practical interest since here the gainG( ωc )
in dB is given by
G( ωc ) = 20 log10 H (e jωc )
= 20 log10 H ( e j 0 ) − 20 log10 2 ≅ −3 dB
since the dc gain G ( 0) = 20 log H (e j0 ) = 0
10
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, ω c is called the 3-dB cutoff
frequency
• To determine the value ofωc we set
| H0 ( e j ωc )| 2 = cos2 (ωc / 2 ) = 12
which yields ωc = π / 2
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
Simple FIR Digital Filters
• A cascade of the simple FIR filter
H 0 ( z) = 12 (1 + z −1 )
results in an improved lowpass frequency
response as illustrated below for a cascade
of 3 sections
• 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 fromπ/2 to π
• Note: H 0 (z ) has a zero atz = −1 or ω = π,
which is in the stopband of the filter
First-order FIR lowpass filter cascade
1
Magnitude
0.8
0.6
0.4
0.2
9
10
Copyright © 2001, S. K. Mitra
0
0
0.2
0.4
0.6
ω/ π
0.8
1
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
Simple FIR Digital Filters
• The 3-dB cutoff frequency of a cascade of
M sections is given by
ω c = 2 cos−1(2 −1/ 2M )
• For M = 3, the above yields ωc = 0 .302 π
• 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
• 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
11
12
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
2
Simple FIR Digital Filters
Simple FIR Digital Filters
• Corresponding frequency response is given
by
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 (e j ω ) = j e − j ω / 2 sin( ω / 2)
whose magnitude response is plotted below
First-order FIR highpass filter
1
0.8
Magnitude
H1( z ) = 1 (1 − z −1 )
2
0.6
0.4
0.2
13
14
0
0
0.2
0.4
0.6
ω/π
Copyright © 2001, S. K. Mitra
0.8
1
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
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
• 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
−1
H1( z) = 1 ∑M
( −1)n z − n
n =0
15
M
is obtained by replacing z with − z in the
transfer function of a moving average filter
16
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
Simple FIR Digital Filters
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)
• The clutter can be removed by filtering the
radar return signal through a two-pulse
canceler, which is the first-order FIR
highpass filterH1( 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
17
18
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
3
Simple IIR Digital Filters
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 of1 − 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
Lowpass IIR Digital Filters
• A first-order causal lowpass IIR digital
filter has a transfer function given by
1 − α  1 + z −1 


H L P ( 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
20
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
Simple IIR Digital Filters
| H LP ( e j π )| = 0
• i.e., | H LP (e j 0 )| = 1,
jω
• Therefore, | H LP( e )| is a monotonically
decreasing function of ω fromω = 0 to ω = π
as indicated below
1
α = 0.8
α = 0.7
α = 0.5
0
-5
Gain, dB
Magnitude
0.8
0.6
-10
0.4
21
α = 0.8
α = 0.7
α = 0.5
-15
0.2
0
• The squared magnitude function is given by
(1 − α) 2 (1 + cosω)
| H LP (e jω )| 2 =
2 (1 + α 2 − 2 α cosω)
• The derivative of | H LP (e j ω )| 2 with respect
to ω is given by
d | H LP ( e j ω )| 2 − (1 − α)2 (1 + 2α + α 2 ) sin ω
=
dω
2(1 − 2α cosω + α 2 )2
0
0.2
0.4
0.6
ω/ π
0.8
1
-20
-2
10
-1
ω/π
10
0
10
22
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
Simple IIR Digital Filters
(1 − α) 2 (1 + cosωc )
d | H LP (e j ω )| 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
1
| H LP (e jω c )| 2 =
2
in the expression for the square magnitude
function resulting in
23
=
1
2
2(1 + α − 2α cosωc )
or
2
(1 − α) (1 + cosωc ) = 1 + α 2 − 2 α cosωc
which when solved yields
2
cos ω c =
2α
1 + α2
• The above quadratic equation can be solved
for α yielding two solutions
24
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
4
Simple IIR Digital Filters
Simple IIR Digital Filters
• The solution resulting in a stable transfer
function H LP (z) is given by
α=
1 − sin ωc
cosωc
• It follows from
| H LP (e jω )| 2 =
(1 − α) 2 (1 + cosω)
2 (1 + α 2 − 2 α cosω)
that H LP (z) is a BR function for |α| < 1
25
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
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
Simple IIR Digital Filters
• Its 3-dB cutoff frequency ωc is given by
α = (1 − sin ωc ) / cosωc
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
0.6
-5
Gain, dB
Magnitude
1
0.8
• 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
α = 0.8
α = 0.7
α = 0.5
-10
0.4
-15
0.2
27
0
0
-20
0.2
0.4
0.6
0.8
1
10 2-
10-1
100
ω/π
ω/ π
28
• Therefore
α = (1 − sin ωc ) / cosωc = − 0.5095245
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
Simple IIR Digital Filters
• Therefore,
Bandpass IIR Digital Filters
• A 2nd-order bandpass digital transfer
function is given by
1 + α  1 − z −1 
2 1 − α z −1


1 − z −1

= 0.245238 
−1 
1 + 0. 5095245 z 
H HP ( z ) =
H BP ( z ) =
• Its squared magnitude function is
H BP (e j ω )
=
29
30
Copyright © 2001, S. K. Mitra

1 − α 
1 − z −2


−
1
−
2
2 1 − β(1 + α ) z + α z 
2
(1 − α )2 (1 − cos 2ω)
2[1 + β 2 (1 + α) 2 + α 2 − 2β (1 + α )2 cos ω + 2α cos 2ω]
Copyright © 2001, S. K. Mitra
5
Simple IIR Digital Filters
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 frequencyof the bandpass
filter, where
ω o = cos−1(β)
• The frequencies ωc1 and ωc 2 where | HBP (e j ω )| 2
becomes 1/2 are called the 3-dB cutoff
frequencies
31
• The difference between the two cutoff
frequencies, assuming ωc 2 > ωc1 is called
the 3-dB bandwidth and is given by
2α 
Bw = ωc 2 − ωc1 = cos−1

 1 + α2 
• The transfer function H BP (z ) is a BR
function if |α| < 1 and |β| < 1
32
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
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(0.4 π) = 0.309017
and
2α
= cos(Bw ) = cos(0.1π) = 0.9510565
1 + α2
• The solution of the above equation yields:
α = 1.376382 and α = 0.72654253
• Plots of | HBP (e jω )| are shown below
α = 0.6
β = 0.34
1
α = 0.8
α = 0.5
α = 0.2
0.6
0.4
0.2
0
β = 0.8
β = 0.5
β = 0.2
1
Magnitude
Magnitude
0.8
0.8
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
33
34
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
Simple IIR Digital Filters
• The corresponding transfer functions are
1 − z− 2
H 'BP( z ) = −0.18819
1 − 0.7343424 z −1 + 1.37638 z −2
and
• Thus, the poles of H 'BP ( z ) are outside the
unit circle making the transfer function
unstable
• On the other hand, the poles of H "BP ( z ) are
at z = 0 .2667655 ± j 0.8095546 and have a
magnitude of 0.8523746
• Hence H "BP ( z ) is BIBO stable
• Later we outline a simpler stability test
1− z −2
1 − 0.533531z−1 + 0.72654253z −2
• The poles of H 'BP ( z ) are at z = 0.3671712 ±
H "BP( z ) = 0.13673
j1.11425636 and have a magnitude > 1
35
36
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
6
Simple IIR Digital Filters
Simple IIR Digital Filters
• Figures below show the plots of the
magnitude function and the group delay of
H "BP ( z )
Bandstop IIR Digital Filters
• A 2nd-order bandstop digital filter has a
transfer function given by
7
1
Group delay, samples
Magnitude
0.8
0.6
0.4
0.2
0
37
H BS ( z) =
6
5
4
• The transfer function H BS (z) is a BR
function if |α| < 1 and |β| < 1
3
2
1
0
0
0.2
0.4
0.6
0.8
-1
1
0
0.2
ω/π
0.4
0.6
0.8
1
ω/π
38
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
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
0.4
β = 0.8
β = 0.5
β = 0.2
0.2
1
• 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
ω o = cos−1(β)
• The digital transfer function H BS (z) is more
commonly called a notch filter
0.6
0
0
0.2
0.4
0.6
0.8
1
ω/π
39
40
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
Simple IIR Digital Filters
• The frequencies ωc1 andωc 2 where | HBS ( e jω )| 2
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
41

1 + α  1 − 2β z −1 + z −2

2 1 − β(1 + α) z −1 + α z −2 
2α 
Bw = ωc 2 − ωc1 = cos −1

 1+ α 2 
Copyright © 2001, S. K. Mitra
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 ofK first-order lowpass
sections characterized by the transfer
function
1 − α  1 + z−1 


H L P ( z) =
2 1 − α z−1
Copyright © 2001, S. K. Mitra
7
Simple IIR Digital Filters
Simple IIR Digital Filters
• The overall structure has a transfer function
given by
K
 1 − α 1 + z −1 

GLP ( z) = 
⋅
−1 
 2 1 −α z 
43
• The corresponding squared-magnitude
function is given by
K
 (1 − α )2 (1 + cosω) 
|GLP (e jω )| 2 = 

2
 2(1 + α − 2α cosω )
44
• To determine the relation between its 3-dB
cutoff frequency ωc and the parameter α,
we set
K
 (1 − α) 2 (1 + cosω c ) 
1
=


2
2
2 (1 + α − 2 α cosωc ) 
which when solved for α, yields for a stable
G LP (z):
1 + (1 − C ) cosω c − sin ω c 2C − C 2
α=
1 − C + cosωc
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
where
Simple IIR Digital Filters
C = 2( K −1) / K
• It should be noted that the expression for α
given earlier reduces to
α=
1 − sin ω c
cosωc
for K = 1
45
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 ω c 1 + sin(0.4 π)
α=
=
= 0. 1584
cosωc
cos(0 .4π)
Copyright © 2001, S. K. Mitra
Copyright © 2001, S. K. Mitra
Simple IIR Digital Filters
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
• 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
1 + (1 − C ) cosω c − sin ω c 2C − C 2
α=
1 − C + cosωc
=
47
0
K=4
-20 -2
10
0
K=1
-2
10
-1
10
ω/π
K=4
-2
-4
-15
48
Copyright © 2001, S. K. Mitra
Gain, dB
Gain, dB
-10
1 + (1 − 1. 6818) cos( 0. 4π ) − sin( 0.4 π) 2(1. 6818) − (1. 6818 )2
1 − 1. 6818 + cos( 0 .4π )
= − 0.251
Passband details
K=1
-5
-1
10
ω/π
0
10
0
10
Copyright © 2001, S. K. Mitra
8