Digital Signal Processing
IIR Filter
IIR Filter Design by Approximation of
Derivatives
Analogue filters having rational transfer function H(s) can be
described by the linear constant coefficient differential
equation. One of the simplest methods for converting an
analog filter into a digital filter is to approximate the
differential equation by an equivalent difference equation.
This approach is often used to solve a linear constant
coefficient differential equation numerically on a digital
computer.
For the derivative dy(t)/dt at a time t = nT we substitute the
backward difference [y[n]-y[n-1)]/T. Thus
dy ( t )
dt
t  nT
y (n )  y (n  1)

T
(1)
Where T represents the sampling interval and y(n) = y(nT).
The analogue differentiator with output dy(t)/dt
has the system function H(s) = s, while the
digital system that produces the output [y(n)y(n-1)]/T has the transfer function H(s) = (1 –
z-1)/T. Consequently, the frequency domain
equivalent is:
1  z 1
s
T
(2)
The second derivative d2y(t)/dt2 is replaced by the
second difference which is given by
d 2 y(t ) y(n)  2y(n  1)  y(n  2)

2
dt
T2
(3)
In the frequency domain, (3) is equivalent to
1
1  2z  z
s 
T2
2
2
1 z
 
 T
1



2
(4)
It is easy to generalize equation (2) and (4) as
sk
 1  z 1


T





k
(5)
Implications: From (2) we have
z 
1
1  sT
(6)
If we substitute s = jw in (6), we have
1
1
wT
z

j
2 2
1  jwT 1  w T
1  w 2T2
(7)
As w varies from - to + , the corresponding
locus of points in the z-plane is a circle of radius
½ and with centre at z =1/2, as shown in the figure
given on the next slide.
jw
s-plane
1/2

The above figure shows that the mapping in (7) takes points in
the LHP of s into corresponding points inside this circle (i.e.
circle with radius ½ and the points in RHP plane in s are
mapped into points outside this circle. Consequently, this
mapping has the desirable property that a stable analog filter is
Transformed into a stable digital filter.
However, the possible location of the poles of the digital filter
are confined to relatively small frequencies and, thus, the
mapping is restricted to low-pass and band-pass filters having
relatively small resonant frequencies.
Example1: Convert the analogue bandpass filter with system function given below
into a digital IIR filter by use of the
backward difference for the derivative.
H(s) 
1
s  0.12  9
Solution: substituting s = (1-z-1)/T yields
H( z ) 
1
1 z

 T
1
2

 0.1   9

Bilinear
Method)
Transformation
(Tustin’s
This is the most widely used transformation
which is suitable for the design of low-pass,
high-pass, band-pass and band-stop filters.
In the Bilinear Transformation (BLT) method, the
basic operation required to convert an
analogue filter is to replace s as follows:
sk
z 1
,
z 1
k = 1 or 2/T
(8)
To
investigate the characteristics
transformation, let
z = rejwT
s =  + jw’
Now
rejwT 1
z 1
z 1
rejwT 1
sk
of
the
bilinear
k


r2  1
2r si nwT

 k 

j
2
2
1  r  2r cos wT 
 1  r  2r cos wT
Consequently,
k
r 2 1
1 r 2  2 r cos wT
wT
w'  k 1r22rsin
2r cos wT
(9)
(10)
Note that if r < 1,  <0, and if r > 1,  > 0. Consequently, the
LHP in s maps into the inside of the unit circle in the z-plane
and the RHP in s maps into the outside of the unit circle.
When r = 1, then  = 0 and
w'  k 1sincoswTwT  k tan wT2 
(11)
2
1 w '
or w  tan  k 
T
(12)
The relationship between the frequency variables in the two
domains is illustrated in the following figure:
20
18
16
14
w’ 12
10
8
6
4
2
00
0.5
1
1.5
2
2.5
3
w
3.5
From the figure we observe that the mapping between w
and w’ is almost linear for small values of w, but
becomes non-linear for larger values of w, leading to a
distortion (or warping) of the digital frequency
response. This effect is normally compensated for by
pre-warping the the analog filter before applying the
bilinear transformation.
To compensate for the effect, we pre-warp one or more
critical frequencies before applying the BLT. For
example, for a low-pass filter, we often pre-warp the
cutoff or bandedge frequency as follows:
 w pT 

w  k tan 
 2 
'
p
where wp = specified cutoff frequency
w’p = prewarped cutoff frequency
k = 1 or 2/T
T = sampling period.
(13)
Summary of the BLT method of coefficient
calculation:
For standard, frequency selective IIR filters, the
steps for using the BLT method may be
summarized as follows:
1.
Use the digital filter specifications to find a
suitable normalized, prototype, analogue
lowpass filter, H(s).
2.
Determine and prewarp the bandedge or
critical frequencies of the desired filter.
3.
Denormalize the analog prototype filter by
replacing s in the transfer function H(s),
using one of the following transformations,
depending on the type of filter required:
s
s
s
w 'p
lowpass to lowpass
(14)
lowpass to highpass
(15)
lowpass to bandpass
(16)
lowpass to bandstop
(17)
w 'p
s
s 2  w 02
s
Ws
Ws
s 2
s  w 02
where
W  w'p2  w'p1
and
w02  w'p2w'p1
4. Apply the BLT to obtain the desired digital filter transfer
function.
Example 1:Design a digital low-pass filter to approximate
the following transfer function:
1
H (s )  2
s  2s  1
Using the BLT method obtain the transfer function, H(z),
of the digital filter, assuming a 3 dB cutoff frequency of
150 Hz and a sampling frequency of 1.28kHz.
Solution: wp = 2150 rad/sec, T = 1/Fs = 1/1280, giving a
prewarped critical frequency of
w’p = tan(wpT/2) = 0.3857.
The frequency scaled analog filter is given by
H' s  
1
s / w  
' 2
p
2sw  1
'
p

w 'p
s 2  2sw'p  1
0.1488
 2
s  0.5455 s  0.1488
Applying the BLT gives
H( z )  H' (s) s  z1

z 1
0.0878z 2  0.1756z  0.0878

z 2  1.0048z  0.3561

0.08781  2z 1  z 2

1  1.0048z 1  0.3561z  2
Example 2: The normalized transfer function of a simple,
analog lowpass, filter given by
H (s ) 
1
s1
Starting from the s-plane, determine, using the BLT method,
the transfer function of an equivalent discrete time high-pass
filter. Assume a sampling frequency of 150 Hz and a cutoff
frequency of 30 Hz.
Solution: The cut-off frequency of the digital
filter is wp = 230 rad/sec. The cut-off
frequency, after prewarping is w’p =
tan(wpT/2). With T = 1/150, w’p = tan(/5) =
0.7265.
Using the low to high-pass transformation of
equation (15), the denormalized analog transfer
function is obtained as
H( z )  H' (s)
w'p
s s
s

s  0.7265
The z-transfer function is obtained by applying
the BLT:
Hz   H' (s) s  z1
z 1

z  1 / z  1
0.13671  z 1 


z  1 / z  1  0.7265 1  0.1584z 1
Example 3: A discrete time bandpass filter with
Butterworth characteristics meeting the specifications
given below is required. Obtain the coefficients of the
filter using the BLT method.
passband
200 – 300 Hz
sampling frequency
2 kHz
filter order
2
Solution: A first-order normalized analog low-pass filter is
required (since the frequency band transformation for
for bandpass filters – equation (16) – will double the
filter order. Thus
1
H (s ) 
s1
The prewarped critical frequencies are:
 w p 1T 
 2  200
  tan
w 'p1  tan
  0.3249
 2  2000
 2 
 w p 2T 
 2  300
'


w p 2  tan
  tan 2  2000  0.5095
2




w 02  w 'p1w 'p 2  0.1655
W  w 'p 2  w 'p1  0.1846
Using the lowpass-to-bandpass transformation, equation
(16), we have
H' (s)  H(s)
1
s2  w02
s  Ws
Ws
 s2  w 2
 2
2
0
s

Ws

w
0
Ws  1
Applying the BLT gives
H( z )  H' s  s  z1
z 1
1  z 2
 0.1367
1  1.2362z 1  0.7265z  2