Digital Signal Processing

Unit II:
Digital Filters: Design and Structures

Dr. Jigarkumar H. Shah

Asst. Prof., ICT department,
PDPU, Gandhinagar
1
 Topics to be covered:
 Ideal Digital Filters
 Practical Filters: Stability and Causality,
 FIR and IIR Filters
 Linear Phase and Implications
 Linear Phase FIR Filters
 Linear Phase FIR Filter Design: Methods
 Linear Phase FIR Filter Design: Steps
 Window Method
 IIR Filter Design:
 Pole-Zero Placement Method: Low-pass, High-pass,
Band-pass, Notch and All-pass Filters
 Overview of Laplace Transform and Analog Filter Design
 Analog Filter Standard Frequency Responses and Design
Equations
 IIR Filter Coefficients from Analog Filter: Bilinear
Transformation
 Basic Structures for FIR and IIR Filter
implementation:
 Direct, Transposed, Cascade and Parallel form Structures
 Effects of Co-efficient Quantization
Computer Aided Designs are essential to design
2
digital filters.
 References:

1. Proakis and Manolakis, “Digital Signal

Processing: Principles, Algorithm &
Application”, Pearson.
2. Oppenheim and Safer: “Discrete Time
Signal Processing”, PHI
3. Emmanuel Ifeacher, and Barrie W. Jervis,
“Digital Signal processing-A Practical
Approach”, Pearson Education

3
 Module I:
DT LTI Systems as Frequency
Selective Circuits:
Digital Filters

4
Ideal Digital Filters:
 An Ideal Digital Filter is a LTI system; which has frequency response
function that provides H ( )  1 within certain band of frequencies
(pass band) and H ( )  0 at other frequencies (stop band). Ideally
phase response is zero.
Frequency Response of Ideal Low-pass Filter:
Plot for one period of 2π: [- π, π )

Repeats periodically with

period 2π rad/sample outside
the plotted interval

5
 Frequency Response of Ideal High-pass
Filter and Ideal Band-pass Filter

𝐻ℎ𝑝 (𝜔) = 1 − 𝐻𝑙𝑝 (𝜔) for same cut off frequency

𝐻𝑏𝑝 𝜔 = 𝐻ℎ𝑝 𝜔 + 𝐻𝑙𝑝 𝜔 − 1
with LPF giving higher cut off (𝜔H)and HPF giving lower cut off (𝜔L) 6
Frequency Response of Ideal Band-
stop Filter

𝐻𝑏𝑠 (𝜔) = 1 − 𝐻𝑏𝑝 (𝜔) for same cut off frequencies

7
 Concept of Filtering:

 Let we have ideal LPF with frequency

response function:
1; −𝜔𝑐 < 𝜔 < 𝜔𝑐
𝐻𝑙𝑝 𝜔 =
0; 𝑒𝑙𝑠𝑒 𝑖𝑛 [−𝜋, 𝜋]
 Let input to the filter is: 𝑥 𝑛 =
𝐴1 cos 𝜔1 𝑛 + 𝜑1 + 𝐴2 cos 𝜔2 𝑛 + 𝜑2
 Given that 𝜔1 < 𝜔𝑐 and 𝜔2 > 𝜔𝑐
 The output is 𝑦 𝑛 = 𝐴1 cos 𝜔1 𝑛 + 𝜑1 .

8
 Solution:

1; −𝜔𝑐 < 𝜔 < 𝜔𝑐

 𝐻𝑙𝑝 𝜔 =
0; 𝑒𝑙𝑠𝑒
 ∠𝐻𝑙𝑝 𝜔 = 0
 𝑦 𝑛 =
𝐴1 𝐻𝑙𝑝 𝜔1 cos 𝜔1 𝑛 + 𝜑1 + ∠𝐻𝑙𝑝 𝜔1 +
𝐴2 𝐻𝑙𝑝 𝜔2 cos 𝜔2 𝑛 + 𝜑2 + ∠𝐻𝑙𝑝 𝜔2
 𝐻𝑙𝑝 𝜔1 =1; 𝐻𝑙𝑝 𝜔2 =0
 ∠𝐻𝑙𝑝 𝜔1 =∠𝐻𝑙𝑝 𝜔2 =0
 𝑦 𝑛 = 𝐴1 cos 𝜔1 𝑛 + 𝜑1 .
9
 Why Ideal Filters are non-
realizable?
 The frequency response of the ideal lowpass filter in [-π, π ] frequency
range is

 The unit impulse response can be written as

 Note: sinc(x) = sin(πx)/(πx); sinc(0)=1

 hlp(n)≠0 for n<0  Not causal
 Magnitude response is square integrable (i.e. having finite energy) 
Stable
 Note: hlp(n): Not absolute summable , but square summable (relaxed
DTFT existence condition)
10
Unit impulse response plot of ideal LPF:
Shown for n=-20 to 20 only

11
 Summary of ideal impulse response of
standard frequency selective filters

Filter type Ideal unit impulse response h(n)

Low Pass
High Pass

Band Pass

Band Stop

Note: fc, fL and fH are cut-off frequencies: fH > fL

 Ideal Unit Impulse Response Plots:
 Shown for n=-20 to 20 only: Non-causal but
stable; h(n) symmetrical w.r.t. n=0.
Causal LPF: Unit impulse response and
Frequency Response:

 Ideally we can write:

𝜔𝑐 𝜔𝑐 𝑛
 ℎ𝑙𝑝 𝑛 = sinc ; −∞ ≤ 𝑛 ≤ ∞
𝜋 𝜋
𝜔 𝜔𝑐 𝑛 −𝑗𝜔𝑛
 𝐻𝑙𝑝 𝜔 = 𝑐 ∞
𝑛=−∞ sinc ( )𝑒
𝜋 𝜋
 One approach is to provide ∞ delay to hlp(n) and
multiply it with unit step to make it zero for n < 0. But
Practically we can provide only finite delay.
𝜔𝑐 𝜔 𝑛−𝑀
 ℎ𝑙𝑝 𝑛 = ( sinc 𝑐 ) u(n); 𝑀 → ∞
𝜋 𝜋
𝜔𝑐 ∞ 𝜔𝑐 (𝑛−𝑀) −𝑗𝜔𝑛
 𝐻𝑙𝑝 𝜔 = 𝑛=0 sinc ( )𝑒 ;𝑀 → ∞
𝜋 𝜋
 This will deviate from ideal frequency response 𝐻𝑙𝑝 𝜔 .
 Write MATLAB script to check it with different M. (Take
finite length of filter unit impulse response)
14
 MATLAB Script:
 clc;
 close all;
 clear all;
 M=input('Enter the delay value ');
 wc=input('Enter the cut-off frequency ');
 n=0:2*M;
 k=1;
 for w=-pi:(pi/255):pi
 H(k)=(wc/pi)*sum(sinc((n-M)*wc/pi).*exp(-j*w*n));
 k=k+1;
 end
 magnitude=abs(H);
 phase=angle(H);
 w=-pi:(pi/255):pi
 figure;
 subplot(2,1,1);
 plot(w, magnitude);
 title('magnitude plot');
 xlabel('w in radians/sample');
 ylabel('|H(w)|');
 subplot(2,1,2);
 plot(w,phase);
 title('phase plot');
 xlabel('w in radians/sample');
 ylabel('<H(w)');

15
MATLAB Prog Results: Magnitude Responses
for several M are shown in figure.
Designed for cut-off frequency: pi/4
rad/sample.
Length of unit impulse response: 2M+1
Phase response is non-zero (not shown).

16
 Important implications in the design of
practical digital filters:
 The frequency response cannot be zero over any finite frequency
band, except at a finite set of points in frequency.
 The magnitude response cannot be constant in any finite range of
frequencies.
 The transition from pass band to stop band cannot be infinitely
sharp  There is always a transition band between pass band and
stop band.
 Oscillatory behaviour (Ripples) in Practical frequency responses in
pass band and stop band (around cut off) is called Gibbs
Phenomenon.
 The phase response is not zero. The magnitude and phase response
can’t be independently specified.
 Practically we can design a filter with finite length of unit impulse
response: Finite Impulse Response (FIR) filter: Length of unit
impulse response is finite. Always have Ripples.
 Practically it is also possible to design Infinite Impulse Response
(IIR) filter using feedback (recursion) to avoid ripples; but in that
case the unit impulse response will not follow the infinitely delayed
sinc function but some other practical function.

17
 Paley Wiener Theorem:

 Specifies necessary and sufficient condition for

magnitude response of a filter |H(𝜔)|to become
causal:
 For causal unit impulse response h(n), the
magnitude spectrum function |H(𝜔)| must satisfy
following inequality:
𝜋

−𝜋
|ln(
𝐻(𝜔) )|𝑑𝜔 < ∞
 Also, if |H( 𝜔)|
𝜋
is square integrable (having finite
energy) i.e. −𝜋 𝐻(𝜔) 2 𝑑𝜔 < ∞ then the system is
stable.
 So, if magnitude frequency response satisfies
Paley-Wiener Theorem and square integrable, it
gives causal as well as stable filter.
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 Module II:
FIR and IIR
Digital Filters: Examples

19
 FIR Filters:

 The unit impulse response of an FIR filter has finite duration and
corresponds to having no denominator in the rational function
H(z):
 There is no inherent feedback in the difference Equation
representation. This results in the reduced form:

M
y (n)   bk x(n  k )
k 0
 Unit impulse response of causal FIR filter is given by:

h(n)  bn ,0  n  M    M  1 taps
h(n)  0, otherwise

 The FIR systems are always stable. S  h  n  
n 
2
0
 Example of FIR filter: M-tap Moving
Average Filter
 Causal M-tap Moving Average filter is given by difference
equation: 1
y ( n)  ( x(n)  x(n  1)  .....  x(n  ( M  1)))
M
1 M 1
 
M k 0
x(n  k )

1
h( n)  ;0  n  M  1
M
1
H ( z)  (1  z 1  z  2  ......  z ( M 1) )
M
 H(z) contains only numerator polynomial of –ve power of z.
 Have zeros located anywhere in z-plane and poles only at
origin only, no non-zero poles Always stable.
 It can be implemented recursively or non-recursively.
 Recursive implementation reduces hardware/ computation
required.
21
  Non-Recursive implementation:
Const. Multiplier
M-1 -adders 1/
x(n) M
y(n)

Z-1 Z-1 Z-1

M-1 - Unit delay elements

 Recursive implementation:
1
y (n)  y (n  1)  ( x(n)  x(n  M ))
M
2 -adders y(n)
1/
x(n) M
Const. Multiplier
-1 Z-1
Z-M
1 – M Unit delay element 22
 Frequency Response of the Moving-
Average System

𝑀−1 −𝑗𝜔𝑛1
 𝐻 𝜔 = 𝑒
𝑀 𝑛=0
 MATLAB Plot for M=2, M=3 and M=20:
 Basically a LPF; Magnitude response is not flat and contain sidelobes
that increase with M, Linear Phase response in pass band.
 As M increases: BW reduces; more precise in detecting average.

magnitude plot
1 magnitude plot
1
|H(w)|

|H(w)|

0.5
0.5

0
-4 -3 -2 -1 0 1 2 3 4 0
w in radians/sample -4 -3 -2 -1 0 1 2 3 4
phase plot w in radians/sample
2 phase plot
4
<H(w)

<H(w)

1 2
0 0
-1 -2

-2 -4
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
w in radians/sample w in radians/sample

23
 Infinite-duration impulse response (IIR)
Filter:
N M
y (n)   ak y (n  k )   bk x(n  k )
k 1 k 0
M

Y ( z)  k
b z k

H ( z)   k 0
N
1   ak z  k
X ( z)
k 1

 The unit impulse response of the system is infinite in duration.

 There is inherent feedback and it is implemented by using recursive
structures only.
 IIR filters include the denominator term in H(z).
 So it has non-origin poles in addition to zeros and poles at origin.
 Stability needs to be investigated as they are not always stable.
 However, IIR filters are implemented using recursive structures only
and resulted in less hardware/computation compared to FIR filters
with same specifications.
 Also, with an IIR filter the magnitude frequency response can be
made more close to ideal.
 The phase response is non linear.
 The highest negative power of z in denominator polynomial (N)
specifies order of the filter and also number of non
24 – origin poles.
 Example of IIR Filter:

x(n) y(n)

3/4 Z-1

-1/8
Z-1
 Obtain governing difference equation, system transfer function, pole-zeros, stability, unit
impulse response and frequency response functions.
3 1
 𝑦 𝑛 =𝑥 𝑛 + 𝑦 𝑛−1 − 𝑦(𝑛 − 2)
4 8
1 𝑧2 𝑧2
 𝐻 𝑧 = 3 1 = 3 1 = 1 1
1− 𝑧 −1 + 𝑧 −2 𝑧2− 𝑧 + (𝑧− )(𝑧− )
2 4
4 8 4 8
 Stable as poles inside unit circle in z-plane.
1 1
 ℎ 𝑛 = (2( )𝑛 − ( )𝑛 )𝑢(𝑛) IIR
2 4
1
 𝐻 𝜔 = 3 1
1− 𝑒 −𝑗𝜔 + 𝑒 −2𝑗𝜔
4 8

25
 MATLAB Plot of unit impulse response
and frequency response:

26
 Comparison of FIR and IIR filters:
Parameter FIR Filter IIR Filter
Length of h(n) Finite Infinite
Stability Always Stable Stable or Unstable,
Stability needs to be
investigated
Pole-zero Contains zeros located Contains zeros and poles
anywhere in z-plane, poles located anywhere in z-
only at origin plane
Implementation Recursive or Non-recursive Recursive only
Feedback Not inherent Inherent
Magnitude Contains ripples in passband Possible to have ripple
Response and stopband less passband and
stopband
Phase response Possible to have linear phase Non linear phase
response response only
Order of the filter More Less
and components
reqd. for same
27
specifications
 Module III
Basics of Linear Phase FIR Filters

28
 Measure of Phase Linearity:
Phase Delay (𝝉p)and Group Delay(𝝉g)
 Phase Delay:
Practical filters have non-zero phase response.
A signal consists of several frequency components.
Phase response is a function of frequency.
So practically the group of frequency components
when passed through practical digital filter
experience group time delay.
The phase delay is the amount of time delay each
individual frequency components of the signal
suffer while transmitted through a system (digital
filter here).
Phase delay contd.
 Group Delay (Envelope Delay)

It is the average time delay of the

frequency components of the composite
signal.
Mathematically it is defined as:
𝝉g = -dh()/d
“the –ve derivative of phase wrt frequency”.
For no phase distortion, the group delay should be a constant; i.e.
h() must be linear.
The phase delay = group delay if h/ is a constant.
Non linear phase characteristics of a system results in phase distortion
at the output due to alteration in the phase relationship of frequency
components of the signal during processing.
Non linear phase delay is undesirable in hi-fi systems such as video,
bio-medical, data- transmission etc.
 Linear Phase Response:

 Linear Phase Response:

h    ; where  and  are
constants.
 Group delay 𝝉g = -dh()/d= 

 Example: Prove that M-tap Moving

Average Filter is linear phase having

group delay of (M-1)/2.
1 M 1

y ( n) 
M k 0
x(n  k )

1
h(n)  ;0  n  M  1
M
M
M 1  j
( M 1) sin( )
1 1
H ( ) 
M
e
n 0
 j n

M
e 2 2

sin( )
2
 Linear Phase FIR Filters:
Linear Phase Response: h    ; where  and 
are constants.
A sufficient condition for h(n) to obtain linear phase
FIR filter of length M:
The unit impulse response h(n) of the FIR filter
should be symmetrical or anti-symmetrical w.r.t.
center line:
h(n)= h(M-1-n) or h(n) = -h(M-1-n)
n=0,1,…,(M-1)
• If  = 0 or   Symmetrical h(n).
• If  = /2 or 3/2  Anti-Symmetrical h(n).
• For FIR filters the filter-length M, can be even or, odd.
• It returns four types of linear phase FIR filters.
 Four types of linear phase FIR filter:
Example of h(n):
Type III: Odd Anti-Symmetric
Type I: Odd Symmetric (M-1)/2 Middle sample is always 0.
Middle sample can take any value
h(n)
h(n) 0
M-1

0 M-1
(M-1)/2

(M-1)/2 (M-1)/2

Type IV: Even Anti-Symmetric

Type II: Even Symmetric (M)/2
h(n) h(n)
0
M-1
0 M-1

(M)/2
(M)/2 (M)/2
 Four types of linear phase FIR
filters:
Type Filter Symmetry/ Group
Length M Anti Symmetry Delay 𝝉g
I Odd h(n)= h(M-1-n) (M-1)/2
n=0,1,…,(M-3)/2
h((M-1)/2): Any value
II Even h(n)= h(M-1-n) (M-1)/2
n=0,1,…,(M-2)/2

III Odd h(n)= -h(M-1-n) (M-1)/2

n=0,1,…,(M-3)/2
h((M-1)/2)=0
IV Even h(n)= -h(M-1-n) (M-1)/2
n=0,1,…,(M-2)/2
 Transfer Function of Linear Phase
FIR Filters:
M 1
H ( z )   h(n) z  n  h(0)  h(1) z 1  .....  h( M  1) z ( M 1)
n 0

h( n)   h( M  1  n)
H ( z )   z  ( M 1) H ( z 1 )
 This result implies that the zeros of H(z) are identical to the
zeros of H(z –1).
 That means zeros of H(z) must occur in reciprocal way.
 In other words, if z1 is the zero of H(z) , then also has zero
at 1/z1.
 Figure shows the symmetry that exists in the location of the
zeros of the linear phase FIR filter.
 Single zero at zero at z=1 and a zero at z=-1 may occur.
 Zero Locations of Linear-Phase FIR
Transfer Functions
 Typical zero locations shown below
 Why IIR Filters can’t be Linear Phase?

 IIR filters have non-origin poles.

 To satisfy symmetry/anti symmetry
condition, the poles must be located in
reciprocal way.
 If pole p1 is inside unit circle, pole 1/p1 will
be outside unit circle.
 So, causal and stable linear phase IIR filter is
not possible.
 Module II
Frequency Response of Linear
Phase FIR Filters

39
 Type I: M odd, h(n)=h(M-1-n)





 Conclusion…

For Type I : Suitable for any kind of filter: Most

versatile FIR filter; Use Type II, III, IV for specific
purpose.
 MATLAB Frequency Response Plots:
 Type-I FIR filter  Type-II FIR filter

 Type-III FIR filter  Type-IV FIR filter

 Summary:
Linear Phase FIR filters
Type: I II III IV

LP  

HP  

BP    

BS 
 Module III
Linear Phase FIR Filter Design:
Co-efficient Calculation Methods

47
 Methods of Calculation of Linear
Phase FIR Coefficients
1. The Window Method: Truncate and shape unit
impulse response of ideal filters using some
window function
2. Frequency Sampling Method: Use DFT and
IDFT
3. Optimal ripple or, Min-max design method: Use
Optimization Criteria
If the mathematical model of h(n) is symmetrical or
anti-symmetrical, each method can lead to design of
a linear phase FIR Filter.
 Summery of ideal impulse response of
standard frequency selective filters
Filter type Ideal unit impulse response hd(n)

Low Pass
High Pass

Band Pass

Band Stop

Note: fc, fL and fH are cut-off frequencies: fH > fL

 Ideal Unit Impulse Response Plots:
 Shown for n=-20 to 20 only: Non-causal but symmetrical
w.r.t. n=0: Suitable to obtain practical Type I or II Linear
Phase FIR filters
 The Window method: Simple Design
Steps
• Specify or obtain cut off frequency/ies from given specifications.
• Begin with desired (ideal) magnitude frequency response Hd(ω) of a
filter.
• Obtain desired (ideal) unit impulse response hd(n).
• Obtain required filter length (M).
• Delay hd(n) by appropriate value α (=(M-1)/2 or M/2) to obtain
symmetrical (Type I & II) response in range n=0 to M-1:
hf(n)=hd(n-α).
• Then it is multiplied with suitable window function w(n) of length M
to obtain practical unit impulse response h(n).
• Window function is selected based on specifications of practical
frequency response required.
• Thus: h(n) = hf(n) w(n).
• Plot the response H(ω) and verify the compliance of specifications.
• Find H(z) and implement h(n) with proper structure.
Simplest window : Rectangular window


 Illustration Example 1:




 M=11
h(0)  -0.0417  h10 
h(1)  0.0000  h(9)
h(2)  0.0696  h(8)
h(3)  0.1476  h(7)
h(4)  0.2087  h(6)
h(5)  0.2318

 Higher Order: M=51 and 101

Why rectangular window method
provides smooth frequency response?
 Rectangular Window method …
Conclusions:
The spectrum of ideal low pass filter has a constant pass
band and a constant zero stop band and a jump discontinuity
(sharp cut off) at f=fc.
But the windowed spectrum (practical low pass FIR filter)
shows over-shoot in pass band, ripples in pass band and stop
band and a smooth transition from pass band to stop band but
no abrupt jump.
The side-lobs in the Window spectrum produces the ripple in
pass band and stop band.
The ripple in pass band and over-shoot is attributed to
Gibb’s phenomena.
Number of minima and maxima in the pass band and stop
band are equal to M.
The ripples have different heights, maximum near band
edges, decaying thereafter.
Ripples in stop band follow the attenuation at a rate of
 Rectangular Window method …
Conclusions:
The pass band to stop band transition is not sharp
but there is a finite width transition band between
pass band and stop band.
The normalized magnitude at f = fc is 0.5. It
corresponds to an attenuation of -6 dB.
The transition width is almost equal to width of
main lobe of window spectrum.
Narrower the main lobe (high M); narrower is the
transition band and cut-off is sharp.
 Rectangular Window: Disadvantages
and Improvements
• Rectangular window has abrupt discontinuity in time
domain (equivalently side-lobes in frequency domain)
and hence produce undesirable ripples in pass band and
stop band in FIR filter frequency response.
• It can be improved by using smoother window functions
which have no abrupt discontinuities in time domain.
• The highest magnitude ripple’s peak in the stop band
determines the stop band attenuation. The peak ripple
in stop band are approximately 9% or -21dB.
• By tapering the windows smoothly at each end, the
height of side-lobes can be diminished; however this can
be achieved at the expense of a wider main-lobe and
hence transition width.
 Some Commonly Used Windows:
Window Shape Expression w(n);
n= 0,1,…,M-1
M: Odd

Rectangular Rectangle 1

Bartlett Triangle 1–2{|n-(M-1)/2|/(M-1)}

Blackman Bell(zero touching) 0.42-0.50cos{2n/(M-1)}

+0.08cos{4n/(M-1)}

Hamming Bell(no zero touching) 0.54-0.46cos{2n/(M-1)}

Hanning Bell(zero touching) 0.50-0.50cos{2n/(M-1)}

MATLAB: Time Domain
Representation of window functions:
M=31.
Frequency Domain Representation of
window functions: MATLAB Plots
M=31
 Performance Evaluation of
commonly used windows:
Transition Width  Stopband attenuation
Name of window
Approximate Exact values AS(dB) (minimum)
4 1.8
Rectangular – 21
M M
8 6.1
Bartlett – 25
M M
12 11
Blackman – 74
M M
8 6.6
Hamming – 53
M M
8 6.2
Hanning – 44
M M

 From the characteristics, the hamming

window appears to be best choice for many
applications like speech processing.
 Illustration Example 2(a) to (d):

 Recalculate example 1 using:

 Bartlett window
 Blackman window
 Hamming window
 Hanning window
 Solution: Example 2 (a): Bartlett Window

For M = 7,
For M = 11,

h(0)  0  h(6) h(0)  0  h(10)

h(1)  0.0531  h(5) h(1)  0.000  h(9)
h(2)  0.1501  h(4) h(2)  0.03  h(8)
h(3)  0.25 h(3)  0.0955  h(7)
h(4)  0.1801  h(6)
M 1

 h(n)  0.6562
n 0
h(5)  0.25
M 1

 h(n)  0.8611
n 0
That means every sample of h(n) we
have to multiply with 1.5239 That means we have to multiply with 1.1613
h(0)  0  h(10)
h(0)  0  h(6) h(1)  0.000  h(9)
h(1)  0.0808  h(5) h(2)  0.0349  h(8)
h(2)  0.2287  h(4) h(3)  0.1109  h(7)
h(3)  0.381 h(4)  0.2091  h(6)
h(5)  0.2903
Solution: Example 2 (b): Blackman Window
 2 n   4 n 
w(n)  0.42  0.5 cos   0.08 cos 
 M 1  M 1
  2 n   4 n  M 1
h(n)  2 f c sinc2 f c (n   ) 0.42  0.5 cos   0.08 cos  , where  
  M 1  M  1  2
and 0  n  M  1
For M = 7 For M = 11,
h(0)  0  h(6)
h(1)  0.0207  h(5) h(0)  0  h(10)
h(2)  0.1418  h(4) h(1)  0.000  h(9)
h(3)  0.25 h(2)  0.0151  h(8)
h(3)  0.0811  h(7)
h(4)  0.1911  h(6)
M 1 h(5)  0.25
 h(n)  0.5750
n 0

thus multiply with 1.7392 M 1

h(0)  0  h(6)
 h(n)  0.8247
n 0

h(1)  0.036  h(5) thus multiply with 1.2126

h(2)  0.2466  h(4)
h(3)  0.4348 h(0)  0  h(10)
h(1)  0.000  h(9)
h(2)  0.0183  h(8)
h(3)  0.0984  h(7)
h(4)  0.2318  h(6)
h(5)  0.3031
 Solution: Example 2 (c) Hamming
Window
 2 n 
w(n)  0.54  0.46 cos 
 M 1
  2 n  M 1
h(n)  2 f c sinc2 f c (n   ) 0.54  0.46 cos  , where   and 0  n  M  1
  M  1  2
For M = 7
For M = 11,

h(0)  0.006  h(6)

h(0)  0.0036  h(10)
h(1)  0.0493  h(5)
h(1)  0.000  h(9)
h(2)  0.1733  h(4)
h(2)  0.0298  h(8)
h(3)  0.25
h(3)  0.1086  h(7)
h(4)  0.2053  h(6)
h(5)  0.25
M 1

 h(n)  0.7073
n 0

M 1
, thus multiply with 1.4138
 h(n)  0.9302
n 0

, thus multiply with 1.075

h(0)  0.0085  h(6)
h(1)  0.0698  h(5)
h(0)  0.0039  h(10)
h(2)  0.2450  h(4)
h(1)  0.000  h(9)
h(3)  0.3535
h(2)  0.0321  h(8)
h(3)  0.1167  h(7)
h(4)  0.2207  h(6)
h(5)  0.2687
 Solution: Example 2 (d) Hanning
Window
 2 n 
) w(n)  0.5  0.5 cos 
 M 1
  2 n  M 1
h(n)  2 f c sinc2 f c (n   ) 0.5  0.5 cos  , where   and 0  n  M  1
  M  1  2
For M = 7

h(0)  0  h(6) For M = 11,

h(1)  0.0398  h(5)
h(2)  0.1688  h(4) h(0)  0  h(10)
h(3)  0.25 h(1)  0.000  h(9)
h(2)  0.0259  h(8)
h(3)  0.1042  h(7)
M 1 h(4)  0.2036  h(6)
 h(n)  0.6672
n 0
h(5)  0.25
M 1
, thus multiply with 1.4988
 h(n)  0.9174
n 0
, thus multiply with 1.0901
h(0)  0  h(6)
h(0)  0.  h(10)
h(1)  0.0596  h(5)
h(1)  0.000  h(9)
h(2)  0.2530  h(4)
h(2)  0.0283  h(8)
h(3)  0.3747
h(3)  0.1136  h(7)
h(4)  0.2219  h(6)
h(5)  0.2725
 Module IV
Window Method: How to Choose
Window Function From Given
Specifications

71
 Generalized Frequency Response of
Practical Filters:

 Specs. of Practical Filter:

 Stopband Edge Frequency (rad/sample): ωs
 Passband Edge Frequency (rad/sample): ωp
 Transition width (rad/sample): ∆ω = ωs - ωp
 Minimum Stopband Attenuation (dB): As = 20 log10 (δs)
 Peak Passband Deviation (dB): Ap = 20 log10 (1+δp)
 Selecting window from given design
specifications:
 From the given specifications (ωs, ωp, As ,Ap), identify the desired
frequency response Hd(ω).
 Determine cutoff frequency ωc = (ωs+ ωp)/2
 Obtain the unit impulse response, hd(n) , for desired filter using inverse
DTFT.
 Select the appropriate window function w(n) that satisfies the stopband
attenuation As given in table.
 Determine the number of filter coefficients using appropriate relationship
between the filter length M and the transition width ∆ω, given in table.
Setect odd value of M. Also determine α=(M-1)/2.
 Obtain FIR filter coefficients h(n) = hd(n- α)w(n).
 Obtain H (ω).
 Verify for peak passband ripple Ap within the given specification.
 If not redesign by increasing M.
 Find H(z) and implement with suitable structure.
 Design Example 3:

Design a digital FIR highpass filter using

appropriate window that meets following
specifications:

4. As = –50dB. Both Hamming and Blackman can provide attenuation less than –
Solution: 50dB. We can choose the Hamming window, which provides the smaller transition
band and hence has the smaller length. Although we do not use the passband ripple
value of
Ap  0.25 dB

in the design, we will have to check the actual ripple from the design and
verify that it is indeed within the given tolerance with the use of MATLAB.

 =  p  s
6.6
=
M
6.6
M   66
0.1
, we are taking the odd value of M = 67.   33
hn   hd n    wn 

=  n  33  0.25 sinc0.25 n  33 0.54  0.46 cos 2 n , 0  n  66

  66 
 MATLAB Results:

 Passband ripple (dB) 0.1893

 Stopband attenuation (dB) -53
 Design Example 4:

Design digital filter with the use of appropriate

window, that meets following specifications:
H    0.01, 0    0.25
0.95  H    1.05, 0.3    0.6
,
H    0.01, 0.65    
Solution:
Filter is a bandpass L S  0.25 , L P  0.3 , H P  0.6 , H S  0.65
LP  LS
L   0.275 f L  0.1375
2
H P  H S
H   0.625 f H  0.3125
2

 s  0.01, As ( stopband attenuation in dB)  20 log10  s   40dB

1   p  1.05 Ap ( passband deviationin dB)  20 log10 1   p   0.4238dB

 0, 0    0.275

H d    1, 0.275    0.625
 0, 0.625    


hd (n)  2 f H sinc2 f H n   2 f L sinc2 f L n   0.625 sinc0.625 n   0.275 sinc0.275 n ,   n  

 As= – 40dB. Thus Hamming, Hanning and Blackman can provide attenuation less than – 40dB. We can choose the
Hanning window, which provides the smaller transition band and hence has the smaller length. Although we do not use
the passband ripple value of A  0.0864dB
p

in the design, we will have to check the actual ripple from the design and verify that it is indeed within the given tolerance
with the use of MATLAB

6.2
   L P   L S 
M
6.2
M   124
0.05
, we are taking the odd value of M = 125.

  62

hn   hd n    wn 
 0.625 sinc0.625 (n  62) 0.275 sinc0.275 (n  62) 
  2 n 
0.5  0.5 cos 124 , 0  n  124
  

 Passband ripple (dB) 0.1584

 Stopband attenuation (dB) -43
 Module V
Window Method: Design with
Kaiser Window

78
 Kaiser Window:
All The window functions considered so far are fixed shape, simple to
apply and simple to understand. But have following disadvantages:
Designing procedure includes only the stopband attenuation As. Designing
does not having control over passband ripples measure by Ap.
For a given window, stopband attenuation is fixed, even if we increase the
filter length M  No relationship between As and M or transition width.

In order to achieve desired performance we need a window function with

an appropriate filter length M to achieve prescribed transition band that
satisfies prescribed minimum stopband attenuation and passband ripple.
Optimization problem: Minimize M with given 𝛿𝑝 , 𝛿𝑠 , ∆𝜔  Window
shape needs to adjust as per specifications.
One such optimum window  Kaiser window.
Disadvantage of Kaiser Window: Computation is more and design
procedure is slightly complicated.
Other adjustable windows proposed by researchers: Lanczos, Tukey,
Dolph-Chebyshev etc.
.,

Kaiser Window: Design Equations

,

  2n  
2

I 0   1  1   

  M  1   0  n  M 1
wn  
I 0  

 where Io(.) is the modified zero-order Bessel function of first kind. Io(x)
is normally evaluated using the following power series expansion:
I x   1 
L 
 x / 2 
k 2

0  
 k! 
where typically L < 25. k 1

 The parameter 𝛽 is defined as  0.1102 A  8.7 , A  50dB


  0.5842 A  210.4  0.07886 A  21, 21dB  A  50dB
 A  21dB
 0,

A  7.95
A  20 log10     min  p ,  s  M 
2.2855 

 parameter 𝛽 controls the way the window function tapers at the edges in
the time domain
Kaiser Window: Time and Frequency
Domain Representation, M=51
 Kaiser Window: Design Steps
 From the given specifications (𝛿𝑝 , 𝛿𝑠 , 𝜔𝑠 , 𝜔𝑝 ) identify the desired
frequency response ( 𝐻𝑑 (𝜔)) .
 Determine cutoff frequency 𝜔𝑐 from specifications.
 Obtain unit impulse response, hd(n), for desired filter.
 Determine transition width ∆𝜔, A and δ from given specifications.
 Find the filter length M (odd) and α.
 Determine β parameter.
 Obtain h(n)= hd(n-α)w(n).
 Design Example 5: Kaiser Window

 Design a lowpass filter with a passband edge

of 1500Hz, a stopband edge of 2000Hz,
passband ripple of 0.01, stopband ripple of
0.1, and a sampling frequency of 8000Hz
using Kaiser window.     min  p ,  s  0.01
Solution:
2000 A  20 log10    40dB   s   p  0.125
 s  2  0.5
8000
A  7.95
1500 M   36
 p  2  0.375 2.2855 for odd value of M = 37 and
8000
  18
 p  s
c   0.4375
2
21dB  A  50dB
,
hd (n)  2 f c sinc2 f c n ,   n  
  0.5842 A  210.4  0.07886 A  21  3.3953
 0.4375 sinc0.4375n   2n  
2

I 0 3.3953 1  1   
  36  
hn   hd n    wn   0.4375sinc0.4375n  18)   , 0  n  36
I 0 3.3953
MATLAB Plot:
 Other Linear Phase FIR Filter Design
Methods:
 Frequency Sampling Method: Uses DFT
and IDFT  Consider later on
 Optimum Equi-ripple Filter Design:
Based on Optimum solution to Chebyshev
Approximation criterion  Implemented with
Remez Exchange Algorithm  A program
written by Parks-McClellan is available for
optimum equi-ripple linear phase FIR filter
design  No. of software packages for
designing equi-ripple linear-phase FIR filters
are now available.
 Module VI
FIR Filter: Basic Structures

86
 FIR Filter Structures:

• Direct form structure

• Direct form structure for linear phase FIR
filters
• Cascade form structure (second order
sections – SOS)
• Forth Order Sections for linear phase FIR filters
• Lattice structure: Useful for Speech
Processing and Linear Prediction Filtering
Applications
Direct Form FIR Filter Structure:
Direct form structure for linear
phase FIR filters
Cascade Form Structure: Modular
Approach
 Cascade Form Structure: Linear
Phase FIR Filter
 Zeros occur in reciprocal symmetry
 Module VII
IIR Filter Design:
Pole-Zero Placement Method

92
 POLE-ZERO PLACEMENT METHOD: IDEA
IIR Filter System Transfrer Function :
M

 z  z 
k
 Non Trivial zeros
H z   K z N M k 1
; z  e j ; z k , p k complex numbers
N
 Non Trivial poles
 z  p 
k 1
k

 
Gain Trivial zeros and poles

 When a zero is placed at a given point on the z-plane, the

frequency response will be zero at the corresponding point.
 A pole on the other hand produces a peak at the corresponding
frequency point.
 Poles that are close to the unit circle give rise to large peaks
 Whereas zeros close to or on the unit circle produces minima or
zero.
 Thus, by placing poles and zeros on the z-plane, we can obtain
frequency selective filter.
 It’s a rough and approximate method: Not using any
specification in design.
 POLE-ZERO PLACEMENT METHOD:
PRINCIPLE
 Locate poles near points of the unit circle
corresponding to frequencies to be emphasized,
and place zeros near the frequencies to be
deemphasized.
 Constrains:
 All poles should be placed inside the unit circle in
order for the filter to be stable. However, zeros can
be placed anywhere in the z-plane.
 All complex zeros and poles must occur in complex-
conjugate pairs in order for the filter coefficients to
be real.
 POLE-ZERO PLACEMENT METHOD:
Lowpass filter
 Lowpass poles should be placed near the unit
circle at point corresponding to low frequencies
(near ω=0) and zero should be placed near or on
the unit circle at points corresponding to high
frequencies (near ω=π).
1 a
Simple first order LPF:   K
 H z   , | a | 1
1 1
1 a z 1 a z
1

 The gain was selected as 1 – a, so that the filter

has unit gain at ω=0;i.e. H 1    1 at   0
 The addition of zero at z=-1 further attenuates the
response of the filter at high frequencies.
1  z 1 1 a
H 2 z   K , | a | 1 we get K
1  a z 1 2
 a=0.7
 r=0.7
 For higher order filter, for example second
order filter, having poles at p1, 2  re  j as shown
in figure, the system function   K

H z
1  re z 1  re z 
3 j 1  j 1

with 0 < r < 1

 angle  is very small value such that both
conjugate pair poles are near ω = 0, and 0 <
r < 1. Since, for lowpass filter H    1 for   0
 Then K  1  re j 1  re  j 

z   1  re 1  re 
j  j
 1  2r cos   r 2
H 3 z  
H3
1  re z 1  re z 
j 1  j 1
1  2r cos  z 1  r 2 z  2
 POLE-ZERO PLACEMENT METHOD:
Highpass filter
 The opposite holds true for HPF.
 Obtain simple HPF by reflecting (folding) the
pole-zero plot of LPF about the imaginary
axis in the z-plane.
 Thus we obtain the system function for
figure with |a\<1
H z  
K
1  az  1 1

 where, K =1-a; value is such that H 1    1 for  

H 2 z  
 
1  a 1  z 1
H 3 z  
K
H 3 z  
1  2r cos 1  r 2

2 1  az 1  1  2r cos 1 z 1  r 2 z  2 1  2r cos 1 z 1  r 2 z  2
 A simple lowpass-to-highpass filter
transformation
H hp    H lp    

b z k
k

H lp  z   k 0
N
1   a k z k
k 1

b e  jk
M M

 bk e  j   k   1 b e
k
 jk
H lp   
k
k 0
k
H hp    H hp   
N k 0 k 0
1   a k e  jk N N
1   a k e  j   k 1    1 a k e  jk
k
k 1
k 1 k 1

  1 b z k
k
k
H hp  z   k 0
N
1    1 ak z  k
k

k 1

H lp  z  

0.1 1  z 1  
0.1 1  z 1 
1  0.8 z 1 H hp  z  
1  0.8 z 1
 Examples:

1. A second order lowpass filter has the system function

H z  
K
1  a z  1
1 2

Determine the filter coefficient K and pole a1 such that the

 
filter frequency response satisfy the conditions H 0  1 H  2   0.25
Ans. a  2.2153 or a  0.451
K  0.301
1 1

select a1  0.451( stability )

2. Consider the filter having difference equation

y n   0.9 y n  1  Kx n 
(i) Is this filter is lowpass, highpass or bandpass?
(ii) Determine K so that H    1
(iii) Determine the 3-dB cutoff frequency.
Ans. Highpass, K  0.1  c  3.036
 Examples:
3. A highpass digital filter has one pole and one zero. The pole at the
distance 0.9 from the origin of the z-plane. DC signal do not pass through
the filter.
 Plot the pole-zero pattern of the filter and determines its system function
 Normalize the frequency response, so that | H   | 1.
 Determine the constant coefficient difference equation of the filter in
time domain.
 Determine the 3-dB cutoff frequency of the filter
 
 Compute the output of the system if the input is xn  2 cos 6 n  4 ,    n  

 Ans.
1  z 1 y n   0.9 y n  1  0.05xn   xn  1
H z   K K  0.05
 c  3.036
1  0.9 z 1

 
y (n)  0.0282 cos n  134.2  
6 
 Pole-zero Placement Method:
Bandpass filter
 Can’t be realized with first order transfer fun.
Second order: H z   1  2r cosK 1  zz  r z 
2
 1 1 2 2
0

  0 specify the center frequency of BPF

 Use |H1(ω0)|=1 to find K.
 Similarly we can specify for third, forth order
 Example:

 Design a second order IIR BPF that has the

centre of its passband at    2 , zero in its
frequency response characteristic at   0 and
  , and its 3-dB bandwidth is  9 . Also write its
constant coefficient difference equation.
 Ans.
H z  

0.15 1  z 2 
 
1  0 .7 z  2

y n   0.7 y n  2   0.15 xn   0.15 xn  2 

 Special Filters:

 Resonators: Narrowband pass (Tuned) Filters

 Notch Filter: A special band stop filter to
reject single frequency
 Comb Filter: Tuned at multiple frequencies or
reject multiple frequencies
 All Pass Filter: Magnitude response is flat
(unity) but phase response depends on
frequency.
 Digital Resonators: Narrowband pass
(tuned) filters
 j 0
 A special two pole BPF: p1, 2  re ,0 < r < 1, r  1
 Angle determines resonant (tuned)
frequency:  r   0
 Poles near to unit circle: High selectivity
H z  
K
1  re  j 0
z 1
1  re j 0
z 1 

H z  
K
1  2r cos  0 z 1  r 2 z  2

H   
K
1  re  j 0
e  j 0
1  re j 0
e  j 0 
H    1 at    0

K  1  r  1  r 2  2r cos  0
 Pole-zero placement method: Notch
Filter
 Special Band Stop Filter
 Notch filters are used in many applications where specific
frequency component is eliminated.
 For example, biomedical instrumentation and recording
systems signals are interfered by power line frequency 50-
Hz (that is called power-line Hum), needs to be eliminated
by notch filter.
 Contains one or more nulls in its frequency response
characteristic.
 For complete rejection at frequency  0 , we have to put
zeros at z  e 1, 2
 j 0

 For the perfect shape of notch filter or to reduce the width

of notch filter the poles are placed at p  re 0  r  1, r  1 . 1, 2
 j 0

1  e  j 0

z 1 1  e  j0 z 1  1  2 cos  z 1
 z 2  K
1  r  2r cos  0
2

H z   K H z   K K: H    1;    21  cos  0 
  
0

1  re  j0 z 1 1  re  j0 z 1 
1  2r cos  0 z 1  r 2 z  2 
 Pole-zero Placement Method: Comb
Filter
 Comb filter can be viewed as any filter which repeats pass
and stop bands periodically across the overall frequency
band [-π, π].
 The applications of comb filter in ionosphere measurement

and MTI (Moving Target Indicator) radars.

 For creating L-order comb filter, take a filter H(z) and
replace z by zL. Here, frequency response of L order comb
filter: HL(z)=H(zL)
HL(ω)=H(Lω)
LPF:
1
1 a 1 z
H z  
1
2 1 a z

Comb Filter:
L
1 a 1 z
L
H z  
2 1 a z
L

Let L = 3 and a = 0.7

 Notch Filter to Comb Filter:

H z  
1  r  2r cos  0
2
 
1  2 cos  0 z 1  z 2 
21  cos  0  
1  2r cos  0 z 1  r 2 z  2 

H z  
1  r  2r cos  0
2
 
1  2 cos  0 z  L  z 2 L 
21  cos  0  
1  2r cos  0 z  L  r 2 z  2 L 
 Pole-Zero Placement Method: All
Pass Filter
 All pass filter passes all frequency
components of its input with magnitude gain
unity, H    1,   
 The poles and zeros in APF are reciprocals of
one another.
B z  a  a z    a z z 1  N 1 N
H z    N 1

Nth -order APF:

N 1

Az  1 a z  a z 1 N
 1 N

Az  1
H z   z N

Az 

H z  
 a  z  , 1
1  a  1
 First-order APF: 1  az  1 1

 a cos  d h 1 a2
 h      2 tan 1    g     
 1  a cos  d 1  a 2  2a cos

 All poles inside unit circle but all zeros

outside unit circle for any APF.
 Applications of APF:
1. Delay Equalizer:
H(z)
Correct non-linear phase response of system Hd(z)
Hd(z) Hap(z)
such that overall H(z) has linear phase response
(constant group delay).
2. Magnitude Equalizer (Compensator) :
Design compensator Hc(z) such that overall system
G(z) is APF (Hap(z)) to inverse effect of system
Hd(z) on magnitude response. Hd(z) may not be
invertible system here (some or all zeros outside
unit circle). Select zeros of APF Hap(z) = zeros of
Hd(z) outside unit circle to get stable Hc(z).
 Minimum, Maximum and Mixed
Phase Systems: Invertible System: ?
 If system is having transfer function H(z); its
inverse has transfer function 1/H(z).
 If H(z) has all zeros (as well as poles) inside unit
circle  Minimum Phase  The system has stable
inverse system (invertible).
 If H(z) has all zeros outside unit circle (poles must
be inside unit circle) Maximum Phase  The
system does not have stable inverse system.
 If H(z) has some zeros inside and some zeros
outside unit circle (poles must be inside unit
circle) Mixed Phase  The system does not have
stable inverse system.
 Pole-Zero Placement Method:
Disadvantage and Solution
 Pole-Zero Placement: gives idea about effect of
placement of poles and zeros on frequency response
characteristics.
 This method is not good for designing digital filters with
well-specified passband and stopband specifications.
 However, all practical analog filters are IIR only and the field
of analog filter design is well investigated.
 To design IIR filter: Design a stable Analog Filter in Laplace
(s) domain and Map it to stable Digital filter using some s-
plane to z-plane transformation.
 Other methods for direct IIR digital filter design: Least-
square method, Pade Approximation Method, Fletcher-
Powell optimization method etc.
 Number of research papers are available for FIR and IIR
filter design for specific applications.
 Module VIII
IIR Filter Design:
A Brief of Laplace Transform and
Analog Filter Design

11
6
 Overview of Continuous Time Signals
and Systems in Laplace (s) domain
 The Laplace transform is a well established mathematical technique for
solving linear constant co-efficient differential equations (LCCDE).
 Continuous time LTI systems are described by using unit impulse response
function h(t) or using LCCDE which relates input and output.
 The continuous time LTI systems can be analysed and designed easily in s
domain rather than in time domain. Analog filters are continuous time LTI
systems.
 The Laplace transform changes a signal in the time domain into a signal in
the s-domain, also called complex s – plane; where s = σ + j Ω.
 A time domain signal, x(t), is transformed into an s-domain signal, X(s).

ℒ 𝑑𝑥 𝑡 ℒ
 𝑥(𝑡) 𝑋(𝑠) 𝑠𝑋(𝑠)
𝑑𝑡
 Laplace transform for CT signals and systems is analogous to Z-transform of
DT signals and systems with z = esTs ;where, Ts is sampling interval.
 Overview of Continuous Time Signals
and Systems in Laplace (s) domain

 For CT LTI system time domain relationship is: y(t) = x(t)*h(t) =

h(t)*x(t); where * denotes convolution integral operation.
 In Laplace (s) domain: Y(s) = X(s)H(s).
 H(s) = Y(s)/X(s) is called system transfer function of CT LTI system.
 For most practical CT LTI systems H(s) is in rational form given by:

 In a factored form:
 The roots of the numerator, z1, z2 , z3 ,.. are the zeros while the roots of
the denominator, p1, p2, p3,.. , are the poles of system transfer
function.
 Overview of Continuous Time Signals
and Systems in Laplace (s) domain

 A causal CT LTI system with system function H(s) is stable if all its poles
lie in the left half of s-plane; i.e. real part of pole is negative.
 For a stable CT LTI system the Laplace transform and Fourier transform
are related as: H(Ω) = H(s)|s= jΩ
 Where H(Ω) is frequency response function and it is Fourier transform of
h(t).
 H(Ω) is a complex function of frequency variable Ω with -∞< Ω< ∞. |H(Ω)|
is magnitude response and ᶿh(Ω) is phase response.
 The design of analog filters in the s-domain involves two steps:
 1. Specifying the number and location of the poles and zeros and hence specify H(s).
This is a pure mathematical problem, with the goal of obtaining the best
approximation to desired frequency response.
 2. An electronic circuit using OPAMP, R-L-C is derived that synthesize this s-domain
representation.
 IIR Digital Filter from Analog Filter

 Two approaches:
 For any kind of frequency response analog LPF is
designed first prototype filter
 Prototype Analog LPF Design:
 Determine filter transfer function H(s) to satisfy given specifications.
 A causal Analog LTI system with system function H(s) is stable if all its
poles lie in the left half of s-plane.
 This approach has no control over the phase response of the IIR filter.
 Therefore, IIR filter design will be treated as magnitude design only.
 Here, we specify the magnitude characteristics only and accept the phase
response that is obtained from the design.
 Specifications of Normalized Magnitude frequency response:
 Ωp  passband edge frequency
 Ωs  stopband edge frequency
 δs  Peak stopband ripple
 δp  Peak passband ripple
 N Order of filter depends on all other specs.
 Ωc  cutoff frequency  Depends on all.
 More sophisticated techniques,
which simultaneously approximate both the
magnitude and phase responses, require advanced optimization tools.
 Standard Magnitude Frequency
Response Shapes:

1. Buttorworth  monotonic in both passband and stopband

2. Chebyshev-I  equiripple in passband and monotonic in stopband
3. Chebyshev-II  equiripple in stopband and monotonic in passband
4. Elliptic  equiripple in both passband and stopband
5. Bessel  monotonic in both passband and stopband but higher
transition width, approx. linear phase in passband
Note: It is not possible to exactly satisfy all given specifications. We
take liberty with stopband specifications rather than with passband
specifications.
Buttorworth LPF: Normalized
Magnitude Frequency Response:
p
H   
2 1 1
 ; c 
1    c  1   2   p   1/ N
2N 2N

At    c ; H    c  
2 1
2
H    c  
1
 0.707  3dB
2
 Butterworth LPF: Transfer function,
poles and order
 The T.F. thatNsatisfy Butterworth frequency response:
c j  2 k  N 1
 
H s  N 1 sk   c e 2N
, k  0, 1, 2,, N  1
 s  s k 
k 0
N

log 1  s2   1 
The resulting denominator polynomial is Butterworth polynomial. 2 log s  c 
The angular spacing between the poles on the circle is  N
No zeros in H(s) 
 1  s2  1
log 


  2

N
2 log s  p 
or

1
2  1
(1   p ) 2
 Example: Analog Butterworth LPF
Design
 Design an analog  p  1000 rad / sec
lowpass Butterworth  s  4000 rad / sec
filter with a passband  p  0 .1   2 
1
 1  0.2346
edge of 500Hz, a 1    p
2

stopband edge of  s  0 .1

2kHz, passband ripple 

 1  s2  1
log 


 2 
of 0.1 and stopband N
2 log s  p 
 2.1802  N  3
ripple of 0.1.   
log 1  s2  1 
N  3  s  2.1509   c  5842.6rad / sec
 Solution: 2 log s  c  c
j  2 k  4 
s k  5842.6e 6
, k  0, 1, 2
s 0   2921.3 + j 5059.8
s1   5842.6
s 2   2921.3  j 5059.8
 cN 1.9944  1011
H s   
N 1
s  2921.3  j 5059.8s  2921.3  j 5059.8s  5842.6
 s  s 
k 0
k
 Chebyshev Type-I Low Pass
Normalized Magnitude Response:
H   
2 1
1   2 C N2   p 
CN (x) is the Nth order Chebyshev polynomial
cosh 
1
 s
 1  2 1

C N x   
 
cos N cos 1 x , x 1
C N 1  x   2 xC N  x   C N 1  x 
  
 1
cosh N cosh x ,  x 1
N
 
cosh 1  s 
 p 

N=6 N=7
 Chebyshev Type-I Low Pass
Normalized Magnitude Response:
H(s) has poles only.
The poles of Chebyshev-I filter lie on the ellipse in the s-plane with major axis.
1/ N
 2 1  1   2  1
 1
and minor axis
2
r1   p r2   p   
2 2   

The position of poles for Chebeshev-I filter lies on ellipse at the coordinates x k , y k 

x k  r2 cos  k , k 
2k  N  1
y k  r1 sin  k , k  1, 2,  , N  1 2N
 Chebyshev Type-II Low Pass
Normalized Magnitude Response:
 Contains zeros as well as poles.
H   
2 1
 C N2  s  p 
1   2
2

 N s
C    

For same specifications Chebyshev filter will result in lower order compared to
Buttorworth filter but have ripples in either of the band.
In other words for the same order of the filter, Chebyshev filter has smaller transition
width compared to Buttorworth filter
 Elliptic Low Pass Normalized
Magnitude Response
 Contains both zeros and poles
H   
1
UN (x) is a Jacobian elliptical function of order N
2

1  U N   P 
2

N even N odd

Providing smallest order filter for a given set of specifications but have ripples in
both the bands.
We can say that for a given order, an elliptical filter has the smallest transition width.
 Bessel Low Pass Normalized
Magnitude Response
b0
H s   B N (s) is the N - order Bessel polynomial
B N s 
B N s   s N  bN 1 s N 1  bN  2 s N  2    b1 s  b0

bk 
2 N  k ! k  0, 1,  , N  1
2 N  k k! N  k !
B N s   2 N  1B N 1 ( s )  s 2 B N  2 ( s )
B 0 (s) = 1 and B1 (s) = s + 1

• Bessel lowpass filters contain poles only.

• Bessel filters have a monotonically decreasing magnitude response, as do
Buttorworth filters.
• The Bessel filter has the larger transition width, but its phase is more linear
within the passband compared to Buttorworth filter of same order.
• Compared to the Butterworth, Chebyshev, and elliptic filters, the Bessel filter has
the slowest rolloff (highest transition width) for the same filter order and
requires the highest order to meet the same specifications.
 Comparison chart: Analog Filters; Std. Freq. Resp. Plots
Std. Freq. Buttorwor Chebyshev Chebyshev Elliptical Bessel
Resp. th -I -II
/Chara. ꜜ
Mag. Resp. Monotonic Equiripple Monotonic Equiripple Monotonic
Passband (No ripple) (No ripple) (No ripple)
Mag. Resp. Monotonic Monotonic Equiripple Equiripple Monotonic
Stopband (No ripple) (No ripple) (No ripple)
Phase Resp. Nonlinear Nonlinear Nonlinear Nonlinear Approx.
Linear
Pole-zeros Poles only Poles only Poles and Poles and Poles only
for LPF (circular (Elliptical Zeros Zeros
design pattern) pattern)
Polynomial Buttorworth Chebyshev Chebyshev Jacobian Bessel
for pole- Elliptical
zeros
Transition Higher High High Lowest Highest
width for
same order
Order for Higher High High Lowest Highest
same specs.
 Analog Filter Transformations: s  s
 Prototype analog LPF has pass-band edge
frequency Ωp (or cutoff frequency Ωc for
Buttorworth design only)

Note: The order of the BPF/BSF filter will be doubled after filter
transformation. In BPF/BSF filter design specification, the "order" refers to the
order of the prototype LPF.
For Buttorworth design we can also consider cut off frequencies in place of
band edge frequencies.
 Digital Filter Transformations: zz

 Prototype digital LPF has cutoff frequency ωc

 Module IX
IIR Filter Design:
sz Mapping

13
4
 Required Properties of sz mapping
(Transformation) for IIR Filter Design
 The left half of s-plane should be map into the inside the
unit circle of z-plane. Thus, stable and causal analog filter
will be converted into stable and causal digital filter.
 The jΩ axis in the s-plane should map into the unit circle in
the z-plane. Thus, there will be direct relationship between
the two frequency variables in two domains without
aliasing one to one mapping.
 The right half of s-plane will map into outside the unit circle
in z-plane.
 One such transformation which satisfy all these properties is
Bilinear transformation given by equation:
2  1  z 1 
s   
T  1  z 1 
T  Sampling time ( step size) for conversion of filter
from s  domain to z  domain; T is arbitrary and can
𝐻 𝑧 =𝐻 𝑠 | 2 1−𝑧 −1
𝑠=𝑇 be set to suitable value.
1+𝑧 −1
 Bilinear Transformation: Mapping
Characteristics
 The bilinear transformation is derived by
applying the trapezoidal numerical integration
approach to the differential equation
representation of H(s) that leads to the
difference equation representation of H(z).
s    j 2 r 2 1 
  
T  1  r  2r cos  
2

z  re j
2 2r sin  
  
2  re j  1  T  1  r  2r cos  
2
  j   j 
T  re  1 
If r < 1 then σ<0 and if r>1 then σ>0.
2  r cos   1  j sin  
   Therefore, the left hand s-plane maps inside the
T  r cos   1  j sin  
unit circle of z-plane and right hand s-plane maps
2
 
r 2 1
 j
2r sin  

outside the unit circle.
T  1  r 2  2r cos  1  r 2  2r cos   Again r=1 implies σ=0, the imaginary axis jΩ of s-plane
maps into unit circle of z-plane.
 Bilinear Transformation: Frequency
Warping Effect
2  sin   2   T 
 For r = 1:     tan
  2 tan
T  1  cos   T 2
1

 2


 The point Ω=∞ maps the point ω=π
and Ω=-∞ maps the point ω=-π.
 Thus the entire range of is mapped
only once into the range  No
aliasing and one to one mapping.
 However, the mapping is highly
nonlinear.
 The mapping is compressed at high
frequency end. This effect is called
frequency warping.
 Warping effect needs to be taken
into consideration while designing
IIR digital filter.
Illustration of Frequency Warping
Effect in BPF design from Analog
Frequency Response
 Example 1:
 Convert following first order analog Butterworth prototype
LPF to a digital filter with cut off frequency 0.2π using
Bilinear transformation.

 Solution: RC LPF: H(s)= Ωc /(s+ Ωc); Ωc = 1/RC; |𝐻 Ω |2 = 1

Ω 2
1+( )
Ω𝑐

 Take Ωc =1 rad/sec for prototype filter: H(s)= 1 /(s+1)

 Given specs.: ωc = 0.2π rad/sample
 Required: Ωc’=0.65/T; T can be assigned any arbitrary value: Let T=1.
 Equivalent analog filter: H’(s)=(0.65)/(s+0.65)
 Designed digital filter: H(z)=0.245(1+z-1)/(1-0.51z-1)
 Examples 2,3,4:

2. Using prototype Buttorworth LPF of previous

example 1 design a digital IIR HPF with cut off
frequency of 2π/3.
3. Using prototype Buttorworth LPF of previous
example 1 design a digital IIR BPF with cut off
frequencies of π/4 and 2π/3.
4. Using prototype Buttorworth LPF of previous
example 1 design a digital IIR BSF with cut off
frequencies of π/4 and 2π/3.
 Example 5:

 Design a second order digital IIR Buttorworth HPF with cut

off frequency π/2 rad/sample using Bilinear transformation.
 Solution:
 Given: N=2
 Prototype Analog Buttorworth LPF with Ωc=1 rad/sec: H(s)
= 1/(s2+√2s+1)
 Cut-off freq. of Analog HPF: Ωc’ =2/T rad/sec
 Take T=2 for simplicity
 Eq. Analog HPF :H’(s) = s2/(s2+√2s+1)
 Designed Digital HPF:
 H(z) = (1-z-1)2/[(2+√2)(1+z-2)]
 Example 6:

 Determine order of a digital Buttorworth IIR LPF using

Bilinear transformation which is processing a signal sampled
at 8000Hz having a passband edge of 1500Hz, a stopband
edge of 2000Hz. Filter should have max. passband ripple of
0.01 and max. stopband ripple of 0.1.
 Solution:
 Specs. of given digital IIR LPF:
 ωp=3π/8 rad/sample, ωs=π/2 rad/sample, δp=0.01, δs=0.1
 Equivalent Buttorworth Analog LPF specs.:
 Ωp=1.336/T rad/sec, Ωs=2/T rad/sec, δp=0.01, δs=0.1
 N=11
 Module X
IIR Filter Structures

14
4
 Basic IIR Filter Structures:
 Direct-Forms:
 Direct Form I
 Direct From II
 Transposed Form
 Cascade Form (Cascade of Second order sections)
 Parallel Form
 Lattice – Ladder Form
 Factors that influence our choice:
 Computational complexity in terms of number of arithmetic
operations (multiplications and additions)
 Memory requirements: number of locations required to store the co-
efficients, past inputs and outputs (delay elements) and any
intermediate computed values.
 Finite-word-length effects (Finite precision effects): Quantization
error effects due to finite precision of hardware/software
 Capability of Parallel or Pipelined processing
 Direct Form I:

 Generalized IIR Filter Transfer Function:

𝑌(𝑧) 𝑊(𝑧) 𝑌(𝑧)

 𝐿𝑒𝑡 𝐻 𝑧 = 𝐻1 𝑧 𝐻2 𝑧 ∴ =
𝑋(𝑧) 𝑋(𝑧) 𝑊(𝑧)

 H1(z): All zero (FIR) system, H2(z): All pole system:

Cascading H1(z) with H2(z): Direct Form I
 Requires M + N + 1 multiplications, M + N additions, and M
+ N delay elements.
 Direct Form II:
 Here H2(z): All pole structure preceded.
 Requires M + N + 1 multiplications, M + N additions, and the
maximum of {M,N } delay elements.
 Save the number of delay elements: If M=N reduced by 2
compared to Direct Form I.
 Disadv. of Direct Form Structures:
 Very Sensitive to co-efficient quantization errors due to
feedback (error is fed back and accumulated in the structure
which becomes severe when N (order) is high).
 When N (order) is large, a small change in a filter
coefficient due to quantization, results in a large change in
the location of the poles and zeros of the system:
Sometimes make the system unstable.
 Both structures are extremely sensitive to parameter
quantization, in general, and are not recommended in
practical applications.
 Transposed Form Structure
 A signal flow graph (SFG) provides an alternative, but equivalent,
graphical representation to a block diagram structure.
 In SFG all junctions and summing points are represented by
nodes. Each branch in SFG is having transmittance.
 Consider the two-pole and two-zero IIR system: M=N=2 with
Direct Form – II structure.
𝑏0 + 𝑏1 𝑧 −1 + 𝑏2 𝑧 −2
𝐻 𝑧 =
1 + 𝑎1 𝑧 −1 + 𝑎2 𝑧 −2
 Transposed Form Structure
 Transposition or flow-graph reversal theorem: if we
reverse the directions of all branch transmittances and
interchange the input (source node) and output (sink node)
in the flow graph, the system transfer function remains
unchanged.
 The resulting structure is called transposed direct form II
structure.
Transposed Direct Form II General
Structure
 The transposed direct form II
structure requires the same number of
multiplications, additions, and memory
locations as the original direct form II
structure.
 Both direct form II and its transposed
form are called canonical form:
Require number of delay elements =
order of system (N).
 Direct form I and its transpose are
non canonical form.
 Example:

 Obtain Direct form II, Transposed direct form

II structures:
i. 𝑦 𝑛 = −0.1𝑦 𝑛 − 1 + 0.72𝑦 𝑛 − 2 +
0.7𝑥 𝑛 − 0.252𝑥(𝑛 − 2)
Steps:
i. To obtain Direct form II: Obtain H(z)
ii. Express it as cascade of all pole and all zero system:
H(z)=H2(z)H1(z).
𝑌(𝑧) 𝑊(𝑧) 𝑌(𝑧)
iii. Express:𝐻 𝑧 = = = H2(z)H1(z)
𝑋(𝑧) 𝑋(𝑧) 𝑊(𝑧)
iv. Write time domain equation for H2(z) and H1(z) and draw
the block diagram.
v. Use SFG and transposition theorem to find transposed form.
 Cascade Form:
 The system can be factored into a cascade of second-order
sections (SOS) or Biquads.

 The coefficients {aki} and {bki} in SOS are real.

 We use the direct form II structure for each of the SOS.

 If N > M some of the SOS have numerator coefficients that

are zero.
 If N is odd, one of the section is of first order.
 Parallel Form:
 A parallel-form realization of an IIR system
can be obtained by performing a partial-
fraction expansion of H(z).
 where {pk} are the poles, {Ak} are the
coefficients (residues) in the partial-fraction
expansion, Constant C = bN/aN if M=N and
C=0 if M<N
 In general, some of the poles of H(z) may be complex
valued.
 In such a case the corresponding residues Ak are also
complex valued.
 To avoid multiplications by complex numbers we can
combine pairs of complex-conjugate poles to form two-pole
subsystems.

 where the coefficients {bki} and {aki} are real-valued

system parameters.
 When N is odd, one of the Hk(z) is really a single-pole
system.
 Disadv./Adv. of cascade and parallel
forms:
 Memory requirement is same as direct form-II.
 But the computational complexity is slightly more
than direct form II (assuming all sections are
implemented using direct form II).
 Less sensitivity to co-efficient quantization errors
(Highest feedback path is of 2nd order only: Less
error accumulation): More robust structures.
 Modularity (Cascade form).
 Parallel processing (Parallel form).
 Cascade of SOS is mostly used in practice.
 Example:

 Obtain Cascade and Parallel form structure:

(1 + 2𝑧 −1 + 𝑧 −2 )
𝐻 𝑧 =
(1 − 𝑧 −1 )(1 + 𝑧 −1 + 𝑧 −2 )
 Obtain Cascade and Parallel form structures:

2(1 − 𝑧 −1 )(1 + 2𝑧 −1 + 𝑧 −2 )
𝐻 𝑧 =
(1 + 0.5𝑧 −1 )(1 − 0.9𝑧 −1 + 0.81𝑧 −2 )
Solution: Residues for parallel form:
C=-4.938, K1=2.157, K2=4.78, K3=-1.595
 Module XI
Finite Precision Effects

15
9
 Finite Precision Effects: Finite Word
Length Effects
 In any digital hardware the word length (bit size) of register, memory
location and arithmetic unit (adder, multiplier) is finite can represent
finite precision number only.
 Finite Precision Effects in hardware implementation occur due to:
1. Input (Signal) Quantization  representation of signal sample 
problem of ADC  SQNR = α + 6n; where α is constant depending on
nature of signal and n is number of bits used to represent signal
amplitude.
2. Co-efficient (System) Quantization representation of filter co-
efficients using finite word length register. This effect is more serious in
IIR systems than in FIR  Change in pole locations due to finite
precision representation of system co-efficients change in stability.
3. Arithmetic (Process) Quantization  Overflow and Round off noise
in addition and multiplication, Limit cycles (undesirable oscillations in
output)  Results of addition and multiplication are represented by
using finite precision  Influenced by data format used by processor 
Fixed point Vs. Floating point.
 General model accounting for finite-
precision effects:

 Here, we focus on the effect of coefficient

quantization only.
 Effects of coefficient quantization in
M-tap FIR system:
x(n) y(n)
 𝐻 𝑧 = 𝑀−1𝑘=0 ℎ 𝑘 𝑧
−𝑘 = 𝑀−1 𝑏 𝑧 −𝑘
𝑘=0 𝑘 H(z)
 Due to co-efficient quantization:
𝑏𝑘 = 𝑏𝑘 + ∆𝑏𝑘 ∆ H(z)

 𝐻 𝑧 = 𝑀−1 𝑘=0 𝑏𝑘 𝑧
−𝑘 = 𝑀−1 𝑏 𝑧 −𝑘 + 𝑀−1 ∆𝑏 𝑧 −𝑘 = 𝐻 𝑧 +
𝑘=0 𝑘 𝑘=0 𝑘
∆𝐻 𝑧
 Effectively it is two FIR systems connected in parallel.
 System function of the quantized system ∆𝐻 𝑧 is linearly
related to the quantization errors in the filter coefficients.
 Frequency Response: 𝐻 ω = 𝐻 𝜔 + ∆𝐻 𝜔
 Designer: Needs to take care of some error specifications
(allowable tolerance) in 𝐻 𝜔 so that actual filter 𝐻 ω
meets the specifications within allowable tolerance.
 Error bound is used as guidelines in determining suitable
word length for a given FIR filter.
 Error Bound:
 Let we want to allow co-efficient quantization error ∆𝑏𝑘 in
range [-∆/2, ∆/2]: ∆𝑏𝑘 𝑚𝑎𝑥 ≤ ∆/2
 ∆𝐻 𝜔 = 𝑀−1 𝑘=0 ∆𝑏𝑘 𝑒
−𝑗𝜔𝑘

𝑀−1 −𝑗𝜔𝑘
 ∆𝐻 𝜔 = 𝑘=0 ∆𝑏𝑘 𝑒
𝑀−1 𝑀−1

≤ ∆𝑏𝑘 𝑒 −𝑗𝜔𝑘 ≤ ∆𝑏𝑘 𝑒 −𝑗𝜔𝑘

𝑘=0 𝑘=0
𝑀−1
 ∆𝐻 𝜔 ≤ 𝑘=0 ∆𝑏𝑘
 ∆𝐻 𝜔 𝑚𝑎𝑥 ≤ 𝑀∆/2
 Let us assume all co-efficient values are normalized between
-1 and +1 and we use 1 bit exclusively to represent sign of
co-efficient and b bits to represent fractional part.
 In this case Quantization step size: ∆ = 2-b.
− (𝑏 + 1) :
 ∆𝐻 𝜔 𝑚𝑎𝑥 ≤ 𝑀 2 This is useful in finding word
length reqd.
 Suitable structure for FIR Filter:
 Symmetry/Anti-symmetry properties of FIR filter co-efficients are
not affected by co-efficient quantization. Hence, direct form linear
phase filters still have linear phase after co-efficient quantization.

 However, magnitude response is affected but as they are always

stable; stability is not of concern here due to co-efficient
quantization.

 By analysis it is found that if the zeros of H(z) are tightly clustered,

then their locations will be highly sensitive to quantization errors in
the filter coefficients; but for most linear phase FIR filters, the
zeros are more or less uniformly spread in the z-lane.

 Direct form realizations of linear phase FIR filters are commonly

used since they tend to provide acceptable performance in most
cases.
 Co-efficient Quantization Effects in
IIR filters:
𝑀 −𝑘 𝑀 −1
𝑁(𝑧) 𝑘=0 𝑏𝑘 𝑧 𝑖=1(1−𝑧𝑖 𝑧 )
 𝐻 𝑧 = = =
𝐷(𝑧) 1+ 𝑁
𝑘=1 𝑎𝑘 𝑧
−𝑘 𝑁 −1
𝑖=1(1−𝑝𝑖 𝑧 )

 Due to co-efficient quantization:

𝑏𝑘 = 𝑏𝑘 + ∆𝑏𝑘 ; 𝑎𝑘 = 𝑎𝑘 + ∆𝑎𝑘 ; 𝑧𝑖 = 𝑧𝑖 + ∆𝑧𝑖 ; 𝑝𝑖 = 𝑝𝑖 + ∆𝑝𝑖
𝑀 −𝑘 𝑀 −1
𝑁 (𝑧) 𝑘=0 𝑏𝑘 𝑧 𝑖=1(1−𝑧𝑖 𝑧 )
 𝐻 𝑧 = = =
𝐷(𝑧) 1+ 𝑁
𝑘=1 𝑎𝑘 𝑧
−𝑘 𝑁 (1−𝑝 𝑧 −1 )
𝑖=1 𝑖

 Let us first do pole- sensitivity analysis:

𝐷 𝑧 =1+ 𝑁 −𝑘 = 𝑁 (1 − 𝑝 𝑧 −1 )
 𝑘=1 𝑎𝑘 𝑧 𝑖=1 𝑖
 Change in ith pole location pi w.r.t. changes in kth co-efficient 𝑎𝑘 =
𝜕𝑝𝑖
= Sensitivity of the ith pole to change in kth denominator co-
𝜕𝑎𝑘
efficient.
 Total error in location of pi is approximately given by (pole
displacement) :
𝑁 𝜕𝑝𝑖
 ∆𝑝𝑖 = 𝑘=1 𝜕𝑎 ∆𝑎𝑘
 𝜕𝐷 𝑧 𝜕𝐷 𝑧 𝜕𝑝𝑖
 = ( )
𝜕𝑎𝑘 𝑧=𝑝𝑖 𝜕𝑧 𝑧=𝑝𝑖 𝜕𝑎𝑘

𝜕𝐷 𝑧
𝜕𝑎𝑘
𝜕𝑝𝑖 𝑧=𝑝𝑖 −𝑝𝑖 𝑁−𝑘
 = 𝜕𝐷 𝑧 = 𝑁 (By analysis)
𝜕𝑎𝑘 𝑗=1,𝑗≠𝑖 (𝑝𝑖 −𝑝𝑗 )
𝜕𝑧 𝑧=𝑝𝑖

 Sensitivity is inversely proportional to the product of vector

distances (pi-pj) of all other poles to the concern pole pi.
 The same analysis applies to zeros.
 If the poles/zeros are tightly clustered, it is possible that
small quantization error in co-efficients can cause large
shifts in the poles/zeros.
 Suitable structure for IIR Filters:

 To minimize pole/zero displacement we need to maximize

vector distances.
 Achieved by implementing higher order structure by
series/parallel of single pole-zero (First order) or at the most
double pole-zero (second order) structures.
 Cascade form has direct control over both pole-zero
combination and order of connection while parallel form has
only direct control over poles. So, cascade form is most
robust from the co-efficient quantization effect point of view.
THE
END
168