Unit-3 FILTERS AND TUNED AMPLIFIERS

Chapter Outline
13.1 Filter Transmission, Types and Specification
13.2 The Filter Transfer Function
13.3 Butterworth and Chebyshev Filters
13.4 First-Order and Second-Order Filter Functions
13.5 The Second-Order LCR Resonator
13.6 Second-Order Active Filters Based on Inductor Replacement
13.7 Second-Order Active Filters Based on the Two-Integrator-Loop Topology
13.8 Single-Amplifier Biquadratic Active Filters
13.9 Sensitivity
13.10 Transconductance-C Filters
13.11 Switched-Capacitor Filters
 13.1 Filter Transmission, Types and Specifications

Filter Transfer Function

 A filter is a linear two-port network represented by the ratio of the output to input voltage
 Transfer function T(s)  Vo(s)/Vi(s)
 Transmission : evaluating T(s) for physical frequency s = j → T(j ) = |T(j )|e j()
 Gain: 20 log|T(j)| (dB)
 Attenuation: -20 log|T(j)| (dB)
Output frequency spectrum : |Vo(s)| = |T(s)| |Vi(s)|
Types of Filters
Filter Specification
 Passband edge : p
 Maximum allowed variation in passband transmission : Amax
 Stopband edge : s
 Minimum required stopband attenuation : Amin

Low-Pass Filter Band-Pass Filter

 The first step of filter design is to determine the filter specifications

 Then find a transfer function T(s) whose magnitude |T(j)| meets the specifications
 The process of obtaining a transfer function that meets given specifications is called filter
approximation
 13.2 Filter Transfer Function

Transfer Function
 The filter transfer function is written as the ratio of two polynomials:
aM s M  aM 1s M 1  ...  a0
T ( s) 
s N  bN 1s N 1  ...  b0
 The degree of the denominator → filter order
 To ensure the stability of the filter → N  M
 The coefficients ai and bj are real numbers
 The transfer function can be factored and expressed as:
aM ( s  z1 )(s  z2 )...(s  z M )
T ( s) 
( s  p1 )(s  p2 )...(s  p N )
 Zeros: z1 , z2 , … , zM and (NM) zeros at infinity
 Poles: p1 , p2 , … , pN
 Zeros and poles can be either a real or a complex number
 Complex zeros and poles must occur in conjugate pairs
 The poles have to be on the LHP of s-plane
 13.3 Butterworth and Chebyshev Filters

The Butterworth Filter

 Butterworth filters exhibit monotonically decreasing transmission with all zeros at  = 
 Maximally flat response → degree of passband flatness increases as the order N is increased
 Higher order filter has a sharp cutoff in the transition band
 The magnitude function of the Butterworth filter is:
1
| T ( j ) |
1   2 ( /  p ) 2 N
Required transfer functions can be defined based on filter specifications (Amax, Amin, p , s)
|T|
1

1
1  2


p
Natural Modes of the Butterworth Filter
 The natural modes (poles) locate on a circle
 The radius of the circle is 0   p (1 /  )1/ N

 Equal angle space
N

K
T (s) 
 s  s   s 
1  1  ...1  
 p1  p 2   p N 

K0N
 where 0   p (1 /  )1/ N
( s  p1 )(s  p2 )...(s  p N )
Design Procedure of the Butterworth Filters

Design Specifications
Amax, Amin, p, s

Design Procedure
1. Determine  (from Amax)
1
| T ( j p ) |  Amax [dB]  20 log 1   2    10 Amax /10  1
1  2
2. Determine the required filter order N (from p , s , Amin)
Attenuation A( s )[dB]  20 log[1 / 1   2 ( s /  p ) 2 N ]  10 log[1   2 ( s /  p ) 2 N ]  Amin
3. Determine the N natural modes (poles) with 0   p (1 /  )1/ N
p1 , p2 , .... p N

4. Determine T(s)
K0N
T (s)  where 0   p (1 /  )1/ N
( s  p1 )( s  p2 )...( s  p N )
 13.4 First-Order and Second-Order Filter Functions

Cascade Filter Design

 First-order and second-order filters can be cascaded to realize high-order filters
 Cascade design is one of the most popular methods for the design of active filters
 Cascading does not change the transfer functions of individual blocks if the output resistance is low
First-Order Filters
 Bilinear transfer function T ( s)  a1s  a0  a1s  a0
s  b0 s  0
First-Order Filters (Cont’d)
Second-Order Filters
 Biquadratic transfer function

a2 s 2  a1s  a0
T ( s) 
s 2  (0 / Q) s  02

 Pole frequency: 0
 Pole quality factor: Q
 Poles:
0 1
p1 , p2    j0 1 
2Q 4Q 2
 Bandwidth:
0
BW  2  1 
Q
Second-Order Filters (Cont’d)
Second-Order Filters (Cont’d)
 13.5 The Second-Order LCR Resonator

The Resonator Natural Modes

Parallel Resonator Current Excitation Voltage Excitation

Vo 1 1 s/C
Current Excitation    2 0  1 / LC
I i Y 1 / sL  sC  1 / R s  (1 / RC ) s  1 / LC

Vo ( R || 1 / sC ) 1 / LC
Voltage Excitation   2 Q  0 RC
Vi ( R || 1 / sC )  sL s  (1 / RC ) s  1 / LC

 The LCR resonator can be excited by either current or voltage source

 The excitation should be applied without change the natural structure of the circuit
 The natural modes of the circuits are identical (will not be changed by the excitation methods)
 The similar characteristics also applies to series LCR resonator
Realization of Transmission Zeros
 Values of s at which Z2(s) = 0 and Z1(s)  0
→ Z2 behaves as a short
 Values of s at which Z1(s) =  and Z2(s)  
→ Z1 behaves as an open

Realization of Filter Functions

Low-Pass Filter High-Pass Filter Bandpass Filter

Vo 1 / LC Vo s2 Vo (1 / RC ) s
T (s)   2 T (s)   2 T (s)   2
Vi s  (1 / RC ) s  1 / LC Vi s  (1 / RC ) s  1 / LC Vi s  (1 / RC ) s  1 / LC
 Notch Filter

Low-Pass Notch Filter High-Pass Notch Filter

 13.6 Second-Order Active Filters (Inductor Replacement)

Second-Order Active Filters by Op Amp-RC Circuits

 Inductors are not suitable for IC implementation
 Use op amp-RC circuits to replace the inductors
 Second-order filter functions based on RLC resonator
The Antoniou Inductance-Simulation Circuit
 Inductors are realized by op amp-RC circuits with negative feedbacks
 The equivalent inductance is given by
V1
Z in   sC4 R1 R3 R5 / R2  sLeq
I1
Leq  C4 R1 R3 R5 / R2
The Op Amp-RC Resonator
 The inductor is replaced by the Antoniou circuit
 The pole frequency and the quality factor are given by
C6 R2
0  1 / C4C6 R1 R3 R5 / R2 Q  0C6 R6  R6
C4 R1 R3 R5
 A simplified case where R1 = R2 = R3 = R5 = R and C4 = C6 = C
0  1 / RC Q  R6 / R
Filter Realization

Low-Pass Filter High-Pass Filter

Bandpass Filter Notch Filter

LPN Filter HPN Filter

All-Pass Filter
 13.10 Transconductance-C Filters

Limitations of Op Amp-RC Circuits

 Suitable for audio-frequency filters using discrete op amps, resistors and capacitors
 High-frequency applications limited by the relatively low bandwidth of general-purpose op amps
 Impractical for IC implementations due to:
 The need for large capacitors and resistors increases the IC cost
 The need for very precise values of RC time constant requires expensive trimming/tuning
 The need for op amps that can drive resistive and large capacitive loads
Methods for IC Filter Implementations
 Transconductance-C filters:
 Utilize transconductance amplifiers or transconductors together with capacitors for filters
 High-quality and high-frequency transconductors can be easily realized in CMOS technology
 Has been widely used for medium/high-frequency applications (up to hundreds of MHz)
 MOSFET-C filters:
 Replace resistors with MOSFETs in linear region
 Techniques have been evolved to obtain linear operation with large input signals
 Switched-capacitor filters:
 Replace the required resistor by switching a capacitor at a relatively high frequency
 The resulting filters are discrete-time circuits as opposed to the continuous-time ones
 Is ideally suited for implementation low-frequency filters in IC form using CMOS technology
Transconductors
 An ideal transconductor has infinite input resistance and infinite output resistance
 The output can be positive or negative depending the current direction
 Transconductor can be single-ended or fully differential
Basic Building Blocks
 Negative transconductor used to realize a resistance
 Transconductor loaded with a capacitor as an integrator

First-Order Gm-C Low-Pass Filter

Vo Gm1 G /G
   m1 m 2
Vi sC  Gm 2 1  sC / Gm 2
Second-Order Gm-C Low-Pass Filter

Gm 2
V2  V1
sC2
V1 sGm 4 / C1 G Gm1Gm 2
 2 BP center - freq gain  m 4 0 
Vi s  sGm3 / C1  Gm1Gm 2 / C1C2 Gm3 C1C2
V2 Gm 2Gm 4 / C1C2 Gm 4 Gm1Gm 2 C1
 2 LP dc gain   Q
Vi s  sGm3 / C1  Gm1Gm 2 / C1C2 Gm1 Gm3 C2
Simplified Circuit
 Gm1 = Gm2 = Gm
 C1 = C2 = C

Gm
0 
C
G
Q m
Gm 3

Fully Differential Circuit

 13.11 Switched-Capacitor Filters

Basic Principle
 A capacitor switched between two nodes at a sufficiently high rate is equivalent to a resistor
 The resistor in the active-RC integrator can be replaced by the capacitor and the switches
 Equivalent resistor:
C1vi v T
iav   Req  i  c
Tc iav C1

 Equivalent time constant for the integrator = Tc(C2/C1)

Practical Circuits
 Can realize both inverting and non-inverting integrator
 Insensitive to stray capacitances
 Noninverting switched-capacitor (SC) integrator

 Inverting switched-capacitor (SC) integrator

Filter Implementation
 Circuit parameters for the two integrators with the same time constant
C2 C
Tc  Tc 1  C1  C2  C  C3  C4  KC
C3 C4
1 1 C3 C4 K R5 C4 C
0    Q   C5  0Tc
C1C2 R3 R4 Tc C2 C1 Tc R4 C5 Q
UNIT - 3Butterworth and Chebyshev Filter

Dr. Sanjeev Mani Yadav

Assistant Professor
Dept. of ECE
SRM UNIVERSITY AP
 Butterworth and Chebyshev filters

Dr. Manjunatha. P (JNNCE) UNIT - 5: Analog Filter Design November 8, 2017 3 / 69

 IIR Filter Design Introduction

Magnitude Characteristic of filters

Figure 1: Magnitude response of a LPF,HPF,BPF,BSF,

Magnitude Characteristic of lowpass filter
The magnitude response can be expressed as

1 − δp ≤ |H(jΩ)| ≤ 1 for 0 ≤ Ω ≤ Ωp
Magnitude =
0 ≤ |H(jΩ) ≤ δs for |Ω| ≥ Ωs

Figure 2: Magnitude response of a LPF

 H(Ω) cannot have an infinitely sharp cutoff from passband to stopband, that is H(Ω)
cannot drop from unity to zero abruptly.
It is not necessary to insist that the magnitude be constant in the entire passband of the
filter. A small amount of ripple in the passband is usually tolerable.
The filter response may not be zero in the stopband, it may have small nonzero value or
ripple.
The transition of the frequency response from passband to stopband defines transition
band.
The passband is usually called bandwidth of the filter.
The width of transition band is Ωs − Ωp where Ωp defines passband edge frequency and
Ωs defines stopband edge frequency.
The magnitude of passband ripple is varies between the limits 1 ± δp where δp is the ripple
in the passband
The ripple in the stopand of the filter is denoted as δp
Ωp = Passband edge frequency in rad/second Ωs = Stopband edge frequency in rad/second
ωp = Passband edge frequency in rad/sample ωs = Stopband edge frequency in rad/sample
Ap = Gain at passband edge frequency As = Gain at stopband edge frequency
ωp ωs
Ωp = and Ωs =
T T
1
where T = fs
= Sampling frequency
 Butterworth Filter Design

The magnitude frequency re-

sponse of Butterworth filter is

1
|H(jΩ)|2 =   2N 
1 + ΩΩ
c

Figure 3: Frequency response of

Butterworth low pass filter

Properties of butterworth filter

|HN (jΩ)|2 |Ω=0 = 1 for all N
|HN (jΩ)|2 |Ω=Ωc = 0.5 for all finite N
|HN (jΩ)||Ω=Ωc = √1 = 0.707 20log |H(jΩ)||Ω = Ωc = −3.01 dB
2
|HN (jΩ)|2 is a monotonically decreasing function of for Ω
|HN (jΩ)|2 approaches to ideal response as the value of N increases
The filter is said to be normalized when cut-off frequency Ωc = 1 rad/sec.
From normalized transfer function LPF, HPF, BPF BSF can be obtained by suitable
transformation to the normalized LPF specification.
|HN (jΩ)|2 = HN (jΩ)HN (−jΩ) =   1  
2N
1+ ΩΩ
c
For normalized Butterworth lowpass filter Ωc = 1
1
HN (jΩ)HN (−jΩ) = h i
1 + (Ω)2N
s
Let s = jΩ ∴ Ω = j
1
HN (s)HN (−s) =  2N
s
1+ j

The poles of are determined by equating the denominator to zero

 2N
s
1+ =0
j
1
s = (−1) 2N j
-1 can be written as e jπ(2k+1) where k = 0, 1 . . . and j = e jπ/2
(2k+1)
sk = e jπ 2N e jπ/2 k = 0, 1 . . . 2N − 1
The poles are placed on a unit circle with radius unity and are placed at angles
kπ
sk = 1 N k = 0, 1 . . . 2N − 1 when N is odd
π kπ
= 1 2N
+ N k = 0, 1 . . . 2N − 1 when N is even
 Butterworth Filter Design
N=1 ∴ k=0,1 j
S-plane
kπ Unit circle
Sk = 1
s1
N
s0

S0 = 1 0, S1 = 1 π

The poles lying on left half of s plane is

LHP of H1(s) RHP of H1(-s)
1 1 1
HN (s) = Q = =
(s − sk ) (s − (s1 )) (s + 1) Figure 4: Poles of H1 (s)H1 (−s)
LHP
———————————————————————————————————————
N=2 ∴ k=0,1,2,3 N is Even

π kπ
Sk = 1 2N
+ N jΩ
S-plane
s1
π 3π Unit circle s0
S0 = 1 4, S1 = 1 4
5π 7π
S2 = 1 σ
4 , S3 = 1 4

The poles lying on left half of s plane is s2 s3

LHP of H2(s) RHP of H2(-s)
1 1
HN (s) = =
Figure 5: Poles of H2 (s)H2 (−s)
Q
[s − sk ] [s − (s1 )][s − (s2 )]
LHP
1
=
[s − (−0.707 + j0.707)][s − (−0.707 − j0.707)]
 1
H2 (s) =
[s + 0.707 − j0.707][s + 0.707 + j0.707]
1
=
[s + 0.707]2 − [j0.707]2
1
=
s 2 + 20.707s + (0.707)2 + (0.707)2
1
=
s 2 + 1.414s + 1
———————————————————————————————————–
Determine the poles of lowpass Butterworth filter for N=3. Sketch the location of poles on s
plane and hence determine the normalized transfer function of lowpass filter.
Solution:
jΩ
s2 S-plane
Unit circle s1
N=3 ∴ k=0,1,2,3,4,5
N is Odd s3 s0
Sk = 1 kπ σ
N
π 2π
S0 = 1 0, S1 = 1 3, S2 = 1 3
s5
s4
4π 5π
S3 = 1 π, S4 = 1 3 , S5 = 1 3
Left half poles of H3(s) Right half poles of H3(-s)

Figure 6: Poles of H3 (s)H3 (−s)

The poles lying on left half of s plane is

1 1
HN (s) = Q =
[s − sk ] [s − (s2 )][s − (s3 )][s − (s4 )]
LHP
1
H3 (s) =
[s − (−0.5 + j0.866)][s − (−1)][s − (−0.5 − j0.866)]
1
=
[s + 1][s + 0.5 − j0.866][s + 0.5 + j0.866)]
1 1
= =
[s + 1][(s + 0.5)2 − (j0.866)2 ]] (s + 1)(s 2 + s + 1)
The poles are distributed on unit circle in the s plane
They are distributed half on the left half plane and half on the right half plane.
1 1
HN (s) = Q =
(s − sk ) BN (s)
LHP

Table 1: Normalized Butterworth Polynomial

Order N Butterworth Polynomial
1 s+1 √
2 s 2 + 2s + 1
3 (s 2 + s + 1)(s + 1)
4 (s 2 + 0.76536s + 1)(s 2 + 1.84776s + 1)
5 (s + 1)(s 2 + 0.6180s + 1)(s 2 + 1.6180s + 1)
Design of Lowpass Butterworth Filter
The transfer function of normalized Butterworth lowpass filter is given by
1 1
HN (s) = Q =
(s − sk ) BN (s)
LHP

where BN (s) is nth order normalized Butterworth polynomial

The lowpass Butterworth filter has to meet the following frequency domain specifications

Kp ≤ 20log |H(jΩ)| ≤ 0 for all Ω ≤ Ωp

20log |H(jΩ)| ≤ Ks for all Ω ≥ Ωs
Kp =Attenuation at passband frequency Ωp in dB
Ks =Attenuation at stopband frequency Ωs in dB
Gain in dB

20 log H a ( jω )

0
KP

Ks
Ω
ΩP ΩS

Figure 7: LPF specifications

The magnitude frequency response is

1
|H(jΩ)| = 
 2N  12
Ω
1+ Ωc  2N
ΩP −Kp
= 10 10 −1 (1)
Ωc
Taking 20 log on both sides
  Ω = Ωs and K = Ks
 1  "  2N #
20 log |H(jΩ)| = 20 log  
  ΩS
 2N  12 KS = −10 log 1 +

Ωc
 
1 + ΩΩ
c

"  2N # 12 
ΩS
2N
−KS
Ω = 10 10 −1 (2)
= −20 log 1 +
Ωc Ωc
"  2N #
Ω Dividing Equation 1 by Equation
= −10 log 1 +
Ωc 2

 2N −Kp
Ω = Ωp and K = Kp Ωp 10 10 −1
= −KS
" # ΩS 10 10 −1
 2N
ΩP
Kp = −10 log 1 +
Ωc
 −Kp
 
 
Ωp 10 10 − 1 
2N log = log 
−KS
ΩS 10 10 − 1
" −K #
p
10 10 −1
log −KS
10 10 −1
N= h
Ωp
i
2log ΩS

where N is the order of the filter

The cutoff frequency ΩC is
Ωp
ΩC =   1
−Kp 2N
10 10 −1

OR
ΩS
ΩC =   1
−KS 2N
10 10 − 1
Design steps for Butterworth Lowpass Filter
From the given specifications
1 Determine the order of the Filter using

−Kp
" #
10 10 −1
log −KS
10 10 −1
N= h i
Ω
2log Ωp
S

2 Determine the cutoff frequency ΩC using

Ωp Ωs
ΩC =   1 OR ΩC =   1
−Kp 2N −Ks 2N
10 10 − 1 10 10 − 1

3 Determine the transfer function of normalized Butterworth filter by

1 1
HN (s) = Q =
(s − sk ) BN (s)
LHP

4 From analog lowpass to lowpass frequency transformation, find the desired transfer
function by substituting the following

Ha (s) = HN (s)|s→ s
ΩC
 Butterworth Filter Design

Design an analog Butterworth low pass filter to meet the following specifications T=1 second

0.707 = ≤ |H(e jω )| ≤ 1; for 0 ≤ ω ≤ 0.3π

= |H(e jω )| ≤ 0.2; for .75π ≤ ω ≤ π

Solution:
Passband edge frequency ωp = 0.3π rad/sample
Stopband edge frequency ωs = 0.75π rad/sample
ω
Passband edge analog frequency Ωp = 1p = 0.3π
1
= 0.3π rad/second
Stopband edge analog frequency Ωs = ω1s = 0.75π
1
= 0.75π rad/second
Kp =20log(0.707)=-3.01 dB, Ks =20log(0.2)=-13.97 dB,
The order of the filter is

Gain in dB
20 log H a ( jω )
−Kp
" #
10 10 −1
log −KS
10 10 −1 0
N = h i
Ω K P = −3.01dB
2log Ωp
S
3.01
 
10 10 −1 K s = −13.97dB
log 13.97
 1

10 10 −1 log 24 Ω
=  0.3π  = ΩP = 0.9425 rad/sec ΩS = 2.356 rad/sec
2log 0.75π
2 × (−0.398)
−1.38 Figure 8: LPF specifications
= = 1.7336 ' 2
−0.796
 OR
N=2 ∴ k=0,1,2,3 N is Even
For even N
π + kπ
Sk = 1 2N N
N
2
Y 1
H(sn ) = S0 = 1 π
4 , S1 = 1 3π
4
k=1
s 2 + bk s + 1
5π
S2 = 1 4 , S3 = 1 7π
4

The poles lying on left half of s plane

h i
(2k−1)π
where bk = 2sin 2N
N=2 1 1
HN (s) = =
k = N2 = 22 = 1 Q
[s − sk ] [s − (s1 )][s − (s2 )]
LHP
k=1 h i
(2−1)π 1
bk = 2sin 2×2 =1.4142 =
[s − (−0.707 + j0.707)][s − (−0.707 − j0.707)]
1 1
H(sn ) = =
s 2 + 1.4142s + 1 s 2 + 1.4142s + 1

Ωs jΩ
Ωc = −ks 1 S-plane
(10 10 − 1) 2N Unit circle s1
s0
2.3562
= 13.97 1 σ
(10 10 − 1) 4
= 1.0664 rad/sec s2 s3
LHP of H2(s) RHP of H2(-s)

Figure 9: Poles of H2 (s)H2 (−s)

Unnomalized transfer function, H(s)

Ha (s) = H2 (s)|s→ s
Ωc

= H2 (s)|s→ s
1.0644
1
=
s2
Ωc 2
+ 1.4142 Ωs + 1
c
1
=
s 2 +1.4142Ωc s+Ωc 2
Ω2c

Ω2c
=
s2 + 1.4142Ωc s + Ωc 2
1.06442
=
s 2 + 1.4142 × 1.0644s + 1.06442
1.133
=
s 2 + 1.5047s + 1.133
 Butterworth Filter Design

Design an analog Butterworth low pass filter to meet the following specifications T=1 second

0.9 = ≤ |H(e jω )| ≤ 1; for 0 ≤ ω ≤ 0.35π

= |H(e jω )| ≤ 0.275; for .7π ≤ ω ≤ π

Solution:
Passband edge frequency ωp = 0.35π rad/sample
Stopband edge frequency ωs = 0.7π rad/sample
ω
Passband edge analog frequency Ωp = 1p = 0.35π
1
= 0.35π rad/second
Stopband edge analog frequency Ωs = ω1s = 0.7π
1
= 0.7π rad/second
Kp =20log(0.9)=-0.9151 dB, Ks =20log(0.2)=-11.2133 dB,
The order of the filter is

Gain in dB
" −Kp
# 20 log H a ( jω )
10 10 −1
log −KS
10 10 −1 0
N = h i
Ω K P = −0.915dB
2log Ωp
S
0.9151
 
10 10 −1 K s = −11.213dB
log 11.213
 0.234 
10 10 −1 log 12.21 Ω
=  0.35π  = ΩP = 1.0996 rad/sec ΩS = 2.1991 rad/sec
2log 0.7π
2 × (−0.301)
−1.717 Figure 10: LPF specifications
= = 2.852 ' 3
−0.602
For odd N=3

N−1 OR
2 N=3 ∴ k=0,1,2,3,4,5 N is Even
1 Y 1
H(sn ) =
(s + 1) k=1 s 2 + bk s + 1 π + kπ
Sk = 1 2N N

S0 = 1 π S1 = 1 3π
h i
(2k−1)π 4 ,
where bk = 2sin 2N
4

N=3 S2 = 1 5π
4 , S3 = 1 7π
4
k = N−1
2
= 3−1
2
=1 The poles lying on left half of s plane
k=1 h i
(2−1)π
bk = b1 = 2sin 2×3 =1 1 1
HN (s) = Q =
[s − sk ] [s − (s1 )][s − (s2 )]
LHP

1 1
H(sn ) = =
[s − (−0.707 + j0.707)][s − (−0.707 − j0.707)]
(s + 1)(s 2
+ s + 1)
1
=
s 3 + 2s 2 + 2s + 1
jΩ
S-plane
Unit circle s1
s0

Ωs σ
Ωc = −ks 1
(10 10 − 1) 2N
s2 s3
2.2 2.2 LHP of H2(s)
= 11.21 1
= RHP of H2(-s)

(10 − 1)
10 6 1.515
= 1.45 rad/sec
Figure 11: Poles of H2 (s)H2 (−s)
Unnomalized transfer function, H(s) and Ωc = 1.45 rad/sec

1
Ha (s) = H3 (s)|s→ s = |s→ s
Ωc s 3 + 2s 2 + 2s + 1 Ωc

= H3 (s)|s→ s
Ωc

1
=
s3 2
Ωc 3
+ 2 Ωs 2 + 2 Ωs + 1
c c
1
=
s 3 +2Ωc s 2 +2Ω2c s+Ω3c
Ω3c

Ω3c
=
s 3 + 2Ωc s 2 + 2Ω2c s + Ω3c
1.453
=
s 3 + 2 × 1.45s 2 + 2 × 1.452 s + 1.453
3.048
=
s 3 + 2.9s 2 + 4.205s + 3.048
Design an analog Butterworth low pass filter to meet the following specifications T=1 second

0.8 = ≤ |H(e jω )| ≤ 1; for 0 ≤ ω ≤ 0.2π

= |H(e jω )| ≤ 0.2; for .32π ≤ ω ≤ π

Solution:
Passband edge frequency ωp = 0.2π rad/sample
Stopband edge frequency ωs = 0.32π rad/sample
ω
Passband edge analog frequency Ωp = 1p = 0.35π
1
= 0.6283 rad/second
Stopband edge analog frequency Ωs = ω1s = 0.7π
1
= 1.0053 rad/second
Kp =20log(0.8)=-1.9 dB, Ks =20log(0.2)=-13.97 dB,
The order of the filter is

Gain in dB
20 log H a ( jω )
−Kp
" #
10 10 −1
log −KS
10 10 −1 0
N = h i
Ω K P = −1.9dB
2log Ωp
S
1.9
 
10 10 −1 K s = −13.97dB
log 13.97
 0.548 
10 10 −1 log 24 Ω
=  0.6283  = ΩP = 0.6283 rad/sec ΩS = 1.0053 rad/sec
2log 1.0053
2 × (−0.204)
−1.641 Figure 12: LPF specifications
= = 4.023 ' 4
−0.408
For Even N=4
N
2
Y 1
H(sn ) =
k=1
s 2 + bk s + 1
h i
(2k−1)π
where bk = 2sin 2N
N=4
k = N2 = 42 = 2
k=1 h i
(2−1)π
bk = b1 = 2sin 2×4 =0.7654
k=2 h i
(4−1)π
bk = b2 = 2sin 2×4 =1.8478

1
H(sn ) =
(s 2 + 0.764s + 1)(s 2 + 1.8478s + 1)
1
=
s 4 + 2.6118s 3 + 3.4117s 2 + 2.6118s + 1

Ωs
Ωc = −ks 1
(10 10 − 1) 2N
1.0053 1.0053
= 13.97 1
=
(10 10 − 1) 8 1.4873
= 0.676 rad/sec
Unnomalized transfer function, H(s) and Ωc = 0.676 rad/sec

1
Ha (s) = H4 (s)|s→ s = |s→ s
Ωc s 4 + 2.6118s 3 + 3.4117s 2 + 2.6118s + 1 Ωc

= H4 (s)|s→ s
Ωc

1
=
s4 3 2
Ωc 4
+ 2.6118 Ωs 3 + 3.4117 Ωs 2 + 2.6118 Ωs + 1
c c c
1
=
s 4 +2.6118Ωc s 3 +3.4117Ω2c s 2 +2.6118Ω3c s+Ω4c
Ω4c

Ω4c
=
s 4 + 2.6118Ωc s 3 + 3.4117Ω2c s 2 + 2.6118Ω3c s + Ω4c
0.6764
=
s 4 + 1.7655s 3 + 1.559s 2 + 0.8068s + 0.2088
0.2088
=
s 4 + 1.7655s 3 + 1.559s 2 + 0.8068s + 0.2088
Design an analog Butterworth low pass filter which has -2 dB attenuation at frequency 20
rad/sec and at least -10 dB attenuation at 30 rad/sec.
Solution:
Passband edge analog frequency Ωp = 20 rad/second
Stopband edge analog frequency Ωs = 30 rad/second
Kp =-2 dB, Ks =-10 dB,
The order of the filter is

Gain in dB
20 log H a ( jω )
−Kp
" #
10 10 −1
log −KS
10 10 −1 0
N = h i
Ω K P = −1.9dB
2log Ωp
S
2
 
10 10 −1 K s = −13.97dB
log 10
 0.584 
10 10 −1 log 9 Ω
= = ΩP = 0.6283 rad/sec ΩS = 1.0053 rad/sec
2log 20
 
30
2 × (−0.176)
−1.1878 Figure 13: LPF specifications
= = 3.374 ' 4
−0.352
For Even N=4
N/2
Y 1
H(sn ) =
k=1
sn2 + bk sn + 1
h i
(2k−1)π
where bk = 2sin 2N
N=4
k = N2 = 42 = 2
k=1 h i
(2−1)π
bk = b1 = 2sin 2×4 =0.7654
k=2 h i
(4−1)π
bk = b2 = 2sin 2×4 =1.8478

1
H(sn ) =
(s 2 + 0.764s + 1)(s 2 + 1.8478s + 1)
1
=
s 4 + 2.6118s 3 + 3.4117s 2 + 2.6118s + 1

Ωs
Ωc = −ks 1
(10 10 − 1) 2N
30 30
= 10 1
=
(10 10 − 1) 8 1.316
= 22.795 rad/sec
Unnomalized transfer function, H(s) and Ωc = 22.795 rad/sec

1
Ha (s) = H4 (s)|s→ s = |s→ s
Ωc s 4 + 2.6118s 3 + 3.4117s 2 + 2.6118s + 1 22.795

1
=
s4 3 2
Ωc 4
+ 2.6118 Ωs 3 + 3.4117 Ωs 2 + 2.6118 Ωs + 1
c c c
1
=
s 4 +2.6118Ωc s 3 +3.4117Ω2c s 2 +2.6118Ω3c s+Ω4c
Ω4c

Ω4c
=
s4 + 2.6118Ωc s3 + 3.4117Ω2c s 2 + 2.6118Ω3c s + Ω4c
22.7954
=
s4 + 59.535s 3
+ 1772.76s 2 + 30935.611s + 22.7954
22.7954
=
s + 59.535s + 1772.76s 2 + 30935.611s + 22.7954
4 3
 Problems Problems

A Butterworth low pass filter has to meet the following specifications

i) passband gain KP =1 dB at ΩP =4 rad/sec
ii) Stop band attenuation greater than or equal to 20 dB at Ωs =8 rad/sec
Determine the transfer function Ha (s) of the lowest order Butterworth filter to meet the above
the specifications
Solution:
Ωp =4 rad/sec, Ωs =8 rad/sec,

Gain in dB
Kp =-1 dB, Ks =-20 dB, 20 log H a ( jω )
The order of the filter is
0
−Kp
" #
10 10 −1 K P = −1dB
log −KS
10 10 −1
N = h i
Ω
2log Ωp K P = −20dB
S
Ω
1 ΩP = 4 rad/sec ΩS = 8 rad/sec
 
10 10 −1
log 20
=
10 10
4
−1
= 4.289 ' 5 Figure 14: LPF specifications
2log 8
 Problems Problems

Sk = 1∠θk k = 0, 1 . . . 2N − 1
For odd N θk is
πk
θk =
N
jΩ
Unit circle S-plane
S0 = 1 0=1 0= s3 s2
S1 = 1 π
36◦ = 0.809 + j0.588 s4
5 =1 s1
2π
S2 = 1 5 =1
72◦ = 0.309 + j0.951 s5 s0
3π σ
S3 = 1 5 =1
108◦ = −0.309 + j0.951 s6
4π s9
S4 = 1 5 =1
144◦ = −0.809 − j0.588
S5 = 1 π=1 180◦ = −1 s7 s8
6π Left half poles of H5(s) Right half poles of H5(-s)
S6 = 1 5 = 1 216◦ = −0.809 − j0.588
7π
S7 = 1 5 = 1 252◦ = −0.309 − j0.951
8π
S8 = 1 5 = 1 288◦ = 0.309 − j0.951
9π
S9 = 1 5 = 1 324◦ = −0.809 − j0.588
 Problems Problems

1
H5 (s) = Q
(s − sk )
LHPonly
1
=
(s − s3 )(s − s4 )(s − s5 )(s − s6 )(s − s7 )
1
=
(s − 0.309 + j0.951)(s + 0.809 + j0.588)(s + 1)
(s + 0.809 − j0.588)(s + 0.309 + j0.951)
1
=
[(s − 0.309)2 + (0.951)2 ][(s + 0.809)2 + (0.588)2 ](s + 1)
1
=
[(s 2 + 0.618s + 1)(s 2 + 1.618s + 1)(s + 1)
1
=
s 5 + 3.236s 4 + 5.236s 3 + 5.236s 2 + 3.236s + 1

Ωp
Ωc = −kp
= 4.5784 rad/sec
1
(10 10 − 1) 2N

Ha (s) = H5 (s)|s→ s
4.5787
 Problems Problems

Ha (s) = H5 (s)|s→ s
Ωc

= H5 (s)|s→ s
4.5787
1
= s s s
( 4.5787 )5 + 3.236( 4.5787 )4 + 5.236( 4.5787 )3 +
s 2 s
5.236( 4.5787 ) + 3.236( 4.5787 ) + 1
2012.4
=
s 5 + 14.82s 4 + 109.8s 3 + 502.6s 2 + 1422.36s + 2012.4

Verification

2012.4
Ha (jΩ) =
(jΩ)5 + 14.82(jΩ)4 + 109.8(jΩ)3 + 502.6(jΩ)2 + 1422.3(jΩ) + 2012.4
2012.4
=
(14.82Ω4 − 502.6Ω2 + 2012.4) + j(Ω5 − 109.8Ω3 + 1422.3Ω)
2012.4
|Ha (jΩ)| = p
(14.82Ω4 − 502.6Ω2 + 2012.4)2 + j(Ω5 − 109.8Ω3 + 1422.3Ω)2
20 log |Ha (jΩ)|4 = −1 dB
20 log |Ha (jΩ)|8 = −24 dB
 Analog Filter Design Chebyshev Filter Design

Chebyshev Filter Design

 Chebyshev Filter Design

The magnitude frequency response of Chebyshev filter is

1
|H(jΩ)|2 = h  i
Ω
1+ 2 Tn2 Ωp

Properties of Chebyshev filter

If Ω)p = 1 rad/sec then it is called as type-I normalized Chebyshev lowpass filter.
HN (jΩ)|2 |Ω=0 = 1 for all N
|H(j0)| = 1 for odd N and |H(j0)| = √ 1 2 for even N
1+
The filter has uniform ripples in the passband and is monotonic outside the passband.
The sum of the number of maxima and minima in the passband equals the order of the
filter.

H ( jω ) H ( jω )
Gain

Gain
1 1
1 δp 1 δp
1+ ε 2 1+ ε 2

δs δs
Ω Ω
ΩP ΩS ΩP ΩS

(a) N Odd (b) N Even

Figure 15: Magnitude frequency response of LPF for Chebyshev

Order of the Filter

Kp Gain or Magnitude at passband in normal value(without dB) for frequency Ωp

Ks Gain or Magnitude at passband in normal value(without dB) for frequency Ωs
" 1 #
(1/Ks2 )−1 2
cosh−1 (1/Kp2 )−1
N1 =  
Ωs
cosh−1 Ωp

Kp Gain or Magnitude at passband in dB for frequency Ωp

Ks Gain or Magnitude at passband in dB for frequency Ωs
h
0.1Ks
i1 
cosh−1 100.1Kp −1
2
10 −1
N1 =  
Ωs
cosh−1 Ω
p

Chose the order of the filter N > N1

 Chebyshev Filter Design

Normalized Chebyshev lowpass filter transfer function

When N is Even When N is odd

N N−1
2 2
Y Bk B0 Y Bk
H(sn ) = H(sn ) =
k=1
s 2 + bk s + ck s + c0 k=1 s 2 + bk s + ck
h i h i
(2k−1)π (2k−1)π
where bk = 2yN sin 2N
, ck = yN2 + cos 2 2N
c0 = yN
"
 1 # N1 " 1 #− N1 
1 1 2 1 1 2 1
yN = +1 + − +1 + 
2 2  2 

1
where  = (1/Kp2 ) − 1 2


When N is Even the values of parameter Bk are evaluated using

1
H(sn )|s=0 = 1
(1 + 2 ) 2
When N is odd the values of parameter Bk are evaluated using

H(sn )|s=0 = 1
 Chebyshev Filter Design

Design steps for Chebyshev filter:

From the given specifications

1 Determine the order of the Filter
2 Determine the Normalized Chebyshev lowpass filter transfer function
3 From analog lowpass to lowpass frequency transformation, find the desired transfer
function by substituting the following

Ha (s) = HN (s)|s→ s
ΩC

where ΩC = ΩP
 Chebyshev Filter Design

Jan 2013, June 2015: Design a Chebyshev IIR analog low pass filter that has -3.0 dB frequency
100 rad/sec and stopband attenuation - 25 dB or grater for all radian frequencies past 250
rad/sec
Solution:
Passband ripple Kp =-3.0 dB or in normal value is Kp = 10Kp /20 = 10−3/20 = 0.707
Stopband ripple Ks =25.0 dB or in normal value is Ks = 10Ks /20 = 10−25/20 = 0.056
Passband edge frequency =100 rad/sec Stopband edge frequency =250 rad/sec

" 1 #
(1/Ks2 )−1 2
cosh−1 (1/Kp2 )−1
H ( jω )

Gain
N1 =  
Ωs
cosh−1 Ωp 1
" # 1 δp
i1
(1/0.0562 )−1
h
2
cosh−1 1+ ε 2
(1/0.7072 )−1
= 250


cosh−1 100
 1 δs
−1 317
cosh 1
2
cosh−1 [17.8] Ω
= = ΩP ΩS
cosh−1 (2.5) cosh−1 [2.5]
3.57 Figure 16: LPF specifications
= = 2.278 ' 3
1.566

N=3
 Chebyshev Filter Design

When N is odd
B0 Bk
H(sn ) = × 2
s + c0 s + b1 s + c1
N−1
2
B0 Y Bk
H(sn ) = 1
s + c0 k=1 s 2 + bk s + ck

(1/Kp2 ) − 1 2  =
1
(1/0.7072 ) − 1 2 = 1

=
N=3 k = N−1
2
= 3−1
2
= 1
——————————————————————————————————–

"
 1 # N1 " 1 #− N1 
1 1 2 1 1 2 1
yN = +1 + − +1 + 
2 2  2 
"
 1 # 13 " 1 #− 13 
1 1 2 1 1 2 1
= +1 + − +1 + 
2 12 1 12 1

1 h 1
i1 h 1
i− 1 
3 3
= (2) 2 + 1 − (2) 2 + 1
2
1h 1 1
i 1
= [1.414 + 1] 3 − [1.414 + 1]− 3 = [1.341 − 0.745] ' 0.298
2 2
h i
(2k−1)π
C0 = yN = 0.298 k=1 bk = 2 × yN sin 2×N
 Chebyshev Filter Design

k=1 h i
(2−1)π
b1 = 2 × 0.298sin 2×3 = 0.298
k=2
B0 B1
H(sn ) = × 2
s + c0 s + b1 s + c1

h
(2k−1)π
i B0 B1
ck = yN2 + cos 2 2N
H(sn ) = ×
s + 0.298 s 2 + 0.298s + 0.838
k=1
When N is odd the values of parameter Bk

(2 − 1)π
 are evaluated using
c1 = 0.2982 + cos 2
2×3 H(sn )|s=0 = 1
2 π
h i
= 0.088 + cos
6
" # B0 B1
1 + cos( 2π
6
) H(sn ) = =1
= 0.088 + 0.298 × 0.838
2
= 0.088 + 0.75 = 0.838 B0 B1 = 0.25
B0 = B1
B02 = 0.25
√
B0 = 0.25 = 0.5
 Chebyshev Filter Design

B0 B1
H(sn ) = × 2
s + 0.298 s + 0.298s + 0.838
0.25
H(sn ) =
s 3 + 0.596s 2 + 0.926s + 0.25
Unnomalized transfer function, H(s) and Ωp = 100 rad/sec

0.25
Ha (s) = H3 (s)|s→ s = |s→ s
Ωp s 3 + 0.596s 2 + 0.926s + 0.25 Ωp

= H3 (s)|s→ s
Ωp

0.25
=
s3 2
Ωp 3
+ 0.596 Ωs 2 + 0.926 Ωs + 0.25
p p

0.25
=
s 3 +0.596Ωp s 2 +0.926Ω2c s+0.25Ω3p
Ω3p

0.25 × Ω3p
=
s3 + 0.596Ωp s 2 + 0.926Ω2c s + 0.25Ω3p
0.25 × 1003
=
s3 + 0.596 × 100s 2
+ 0.926 × 1002 s + 0.25 × 1003
0.25 × 1003
=
s + 59.6s + 926s + 0.25 × 1003
3 2
 Chebyshev Filter Design

Design a Chebyshev IIR low pass filter that has to meet the following specifications
i) passband ripple ≤0.9151 dB and passband edge frequency 0.25π rad/sec
ii) Stopband attenuation ≥12.395 dB and Stopband edge frequency 0.5π rad/sec
Solution:
Passband ripple Kp =0.9151 dB
or in normal value is Kp = 10Kp /20 = 10−9151/20 = 0.9
Stopband ripple Ks =12.395 dB
or in normal value is Ks = 10Ks /20 = 10−12.395/20 = 0.24
Passband edge frequency 0.25π=0.7854 rad/sec
Stopband edge frequency 0.5π =1.5708rad/sec

" 1 #
(1/Ks2 )−1 2
cosh−1 (1/Kp2 )−1
N1 =  
Ωs
cosh−1 Ωp
" #
i1
(1/0.242 )−1
h
2
cosh−1 (1/0.92 )−1
=
cosh−1 1.5708


0.7854
−1 16.3611
 1
cosh 0.2346
2
cosh−1 [8.35] 2.8118
= 1.5708
= = = 2.135 ' 3
cosh−1 [2]


cosh−1 0.7854
1.3169

N=3
 Chebyshev Filter Design

When N is odd
B0 Bk
H(sn ) = × 2
s + c0 s + bk s + ck
N−1
2
B0 Y Bk
H(sn ) =
s + c0 k=1 s 2 + bk s + ck
1
(1/Kp2 ) − 1 2

 =
N=3 k = N−1
2
= 3−1
2
=1 =
 1
(1/0.92 ) − 1 2 = 0.4843
——————————————————————————————————–

"
 1 # N1 " 1 #− N1 
1 1 2 1 1 2 1
yN = +1 + − +1 + 
2 2  2 
"
 1 # 13 " 1 #− 13 
1 1 2 1 1 2 1
= +1 + − +1 + 
2 0.48432 0.4843 0.48432 0.4843

1 h 1
i1 h 1
i− 1 
3 3
= (5.2635) 2 + 2.064 − (5.2635) 2 + 2.064
2
1h 1 1
i 1
= [2.294 + 2.064] 3 − [2.294 + 2.064]− 3 = [1.6334 − 0.6122] ' 0.5107
2 2
h i
(2−1)π
C0 = yN = 0.5107 k=1 bk = 2 × 0.5107sin 2×3
= 0.5107
 Chebyshev Filter Design

h i
(2k−1)π
ck = yN2 + cos 2 2N
B0 B1
H(sn ) = ×
k=1 s + 0.5107 s 2 + 0.5107s + 1.0108

 When N is odd the values of parameter Bk

2 (2 − 1)π
2 are evaluated using
ck = 0.5107 + cos
2×3
2 π H(sn )|s=0 = 1
h i
2
= 0.5107 + cos
6
" 2π
#
2
1 + cos( 6
)
= 0.5107 +
2 B0 B1
H(sn ) = = 1.9372B0 B1 = 1
= 0.2608 + 0.75 = 1.0108 0.5107 × 1.0108

B0 B1 1
H(sn ) = × 2 B0 B1 = = 0.5162
s + c0 s + bk s + ck 1.9372
B0 = B1
B02 = 0.5162
B0 B1 √
H(sn ) = × 2
s + 0.5107 s + b1 s + c1 B0 = 0.5162 = 0.7185
 Chebyshev Filter Design

B0 B1 0.7185 0.7185
H(sn ) = × 2 = × 2
s + 0.5107 s + 0.5107s + 1.0108 s + 0.5107 s + 0.5107s + 1.0108
0.5162
H(sn ) =
s 3 + 1.0214s 2 + 1.2716s + 0.5162
Unnomalized transfer function, H(s) and Ωp = 0.7854 rad/sec

0.5162
Ha (s) = H3 (s)|s→ s = |s→ s
Ωp s 3 + 1.0214s 2 + 1.2716s + 0.5162 Ωp

0.5162
=
s3 2
Ωp 3
+ 1.0214 Ωs 2 + 1.2716 Ωs + 0.5162
p p

0.5162
=
s 3 +1.0214Ωp s 2 +1.2716Ω2p s+Ω3p
Ω3p

Ω3p
=
s3 + 1.0214Ωp s2 + 1.2716Ω2p s + Ω3p
0.5162 × 0.78543
=
s 3 + 1.0214 × 0.7854s 2 + 1.2716 × 0.78542 s + 0.78543
0.250
=
s 3 + 0.80229s 2 + 0.7844s + 0.2501
 Chebyshev Filter Design

Design a Chebyshev IIR low pass filter that has to meet the following specifications
i) passband ripple ≤1.0 dB and passband edge frequency 1 rad/sec
ii) Stopband attenuation ≥15.0 dB and Stopband edge frequency 1.5 rad/sec

Solution:
Passband ripple Kp =1.0 dB or in normal value is Kp = 10Kp /20 = 10−1/20 = 0.891
Stopband ripple Ks =15.0 dB or in normal value is Ks = 10Ks /20 = 10−15/20 = 0.177
Passband edge frequency =1 rad/sec Stopband edge frequency =1.5rad/sec

" 1 #
(1/Ks2 )−1 2
cosh−1 (1/Kp2 )−1
H ( jω )

Gain
N1 =  
Ωs
cosh−1 Ωp 1
" # 1 δp
h 2 i1
(1/0.177 )−1 2
cosh−1 1+ ε 2
(1/0.8912 )−1
= 1.5


cosh−1 1.0
1 δs
−1 31.0

cosh 0.26
2
cosh−1 [11.0] Ω
= = ΩP ΩS
cosh−1 (1.5) cosh−1 [1.5]
3.08 Figure 17: LPF specifications
= = 3.2 ' 4
0.96

N=4
 Chebyshev Filter Design

When N is Even B1 B2
H(sn ) = × 2
N s 2 + b1 s + c1 s + b2 s + c2
2
Y Bk
H(sn ) =
k=1
s 2 + bk s + ck
1
(1/Kp2 ) − 1 2

 =
N 4
N=4 k = 2
= 2
=2 1
(1/0.8912 ) − 1 2 = 0.51

=
——————————————————————————————————–

"
 1 # N1 " 1 #− N1 
1 1 2 1 1 2 1
yN = +1 + − +1 + 
2 2  2 
"
 1 # 14 " 1 #− 14 
1 1 2 1 1 2 1
= +1 + − +1 + 
2 0.512 0.51 0.512 0.51

1 h 1
i1 h 1
i− 1 
4 4
= (4.84) 2 + 1.96 − (4.84) 2 + 1.96
2
1h 1 1
i 1
= [2.2 + 1.96] 4 − [2.2 + 1.96]− 4 = [1.428 − 0.7] ' 0.364
2 2
h i
(2k−1)π
C0 = yN = 0.364 k=1 bk = 2 × yN sin 2×N
 Chebyshev Filter Design

k=1 h i
(2−1)π
b1 = 2 × 0.364sin 2×4 = 0.278
k=2 h i
(2×2−1)π
b2 = 2 × 0.364sin 2×4
= 0.672
h i
2 2 (2k−1)π
ck = yN + cos 2N
k=2
k=1
 
(2 × 2 − 1)π
c2 = 0.3642 + cos 2
 
(2 − 1)π
c1 = 0.3642 + cos 2 2×4
2×4  
3π
2 π = 0.132 + cos 2
h i
= 0.132 + cos 8
8 " #
1 + cos( 6π )
" #
1 + cos( 2π
8
) = 0.132 + 8
= 0.132 + 2
2
= 0.132 + 0.853 = 0.132 + 0.146 = 0.278
= 0.132 + 0.853 = 0.985
 Chebyshev Filter Design

B1 B2
H(sn ) = × 2
s 2 + b1 s + c1 s + b2 s + c2

B1 B2
H(sn ) = ×
s 2 + 0.278s + 0.985 s 2 + 0.672s + 0.278
When N is odd the values of parameter Bk
are evaluated using

1 1
H(sn )|s=0 = = = 0.89
(1 + 2 )1/2 (1 + 0.512 )1/2

B1 B2
H(sn ) = = 0.89
0.985 × 0.278

B1 B2 = 0.244
B1 = B2
B12 = 0.244
√
B1 = 0.264 = 0.493
 Chebyshev Filter Design

B1 B2 0.493 0.493
H(sn ) = × = 2 ×
s 2 + 0.278s + 0.985 s 2 + 0.672s + 0.278 s + 0.278s + 0.985 s 2 + 0.672s + 0.278

0.243
H(sn ) =
s 4 + 0.95s 3 + 1.45s 2 + 1.434s + 0.2738
Unnomalized transfer function, H(s) and Ωp = 1.0 rad/sec

0.243
Ha (s) = H4 (s)|s→ s = |s→ s
Ωp s 4 + 0.95s 3 + 1.45s 2 + 1.434s + 0.2738 1

0.263
=
s 4 + 0.95s 3 + 1.45s 2 + 1.434s + 0.2738
 Chebyshev Filter Design

July 2014, Dec 2014 Design a Chebyshev IIR low pass filter that has to meet the following
specifications
i) passband ripple ≤2 dB and passband edge frequency 1 rad/sec
ii) Stopband attenuation ≥20 dB and Stopband edge frequency 1.3 rad/sec
Solution:
Passband ripple Kp =2 dB
or in normal value is Kp = 10Kp /20 = 10−2/20 = 0.7943
Stopband ripple Ks =12.395 dB
or in normal value is Ks = 10Ks /20 = 10−20/20 = 0.1
Passband edge frequency 1 rad/sec
Stopband edge frequency 1.3 rad/sec

" 1 #
(1/Ks2 )−1 2
cosh−1 (1/Kp2 )−1
N1 =  
Ωs
cosh−1 Ωp
" #
i1
(1/0.12 )−1
h
2
cosh−1 (1/0.79432 )−1
=
cosh−1 1.3


1.0
−1 99.0
 1
cosh 0.585
2
cosh−1 [13.00] 3.256
= 1.3
= = = 4.3 ' 5
cosh−1 [1.3]


cosh−1 1.0
0.756

N=5
Dr. Manjunatha. P (JNNCE) UNIT - 5: Analog Filter Design November 8, 2017 51 / 69
 Chebyshev Filter Design

When N is odd
B0 B1 B2
H(sn ) = × ×
s + c0 s 2 + b1 s + c1 s 2 + b2 s + c2
N−1
2
B0 Y Bk
H(sn ) =
s + c0 k=1 s 2 + bk s + ck
1
(1/Kp2 ) − 1

 = 2

N−1 5−1 1
N=5 k = = =2 2 = (1/0.79432 ) − 1 2 = 0.7648
2

——————————————————————————————————–

"
 1 # N1 " 1 #− N1 
1 1 2 1 1 2 1
yN = +1 + − +1 + 
2 2  2 
"
 1 # 15 " 1 #− 15 
1 1 2 1 1 2 1
= +1 + − +1 + 
2 0.76482 0.7648 0.76482 0.7648

1 h 1
i1 h 1
i− 1 
5 5
= (2.71) 2 + 1.307 − (2.71) 2 + 1.307
2
1h 1 1
i 1
= [1.646 + 1.307] 5 − [1.646 + 1.307]− 5 = [1.241 − 0.805] ' 0.218
2 2
h i h i
(2k−1)π (2−1)π
C0 = yN = 0.218 bk = 2 × yN sin 2×N
k=1 b1 = 2 × 0.218sin 2×5
= 0.134
 Chebyshev Filter Design

k=2 h i
(4−1)π
b2 = 2 × 0.218sin 2×5 = 0.352
h
(2k−1)π
i k=2
ck = yN2 + cos 2 2N  
(4 − 1)π
k=1 c2 = 0.2182 + cos 2
2×5
 

(2 − 1)π
 3π
c1 = 0.2182 + cos 2 = 0.047 + cos 2
2×5 10
" #
2 π 1 + cos( 2×3π )
h i
= 0.047 + cos = 0.047 + 10
10 2
" #
1 + cos( 2π
10
) = 0.047 + 0.345 = 0.392
= 0.047 +
2
= 0.047 + 0.904 = 0.951
————————————————————–

B0 B1 B2
H(sn ) = × 2 × 2
s + c0 s + b1 s + c1 s + b2 s + c2
B0 B1 B2
H(sn ) = × 2 × 2
s + 0.218 s + 0.134s + 0.951 s + 0.352s + 0.392
When N is odd the values of parameter Bk are evaluated using

H(sn )|s=0 = 1
 Chebyshev Filter Design

When N is odd the values of parameter Bk are evaluated using

H(sn )|s=0 = 1

B0 B1 B2
H(sn ) = = 12.3B0 B1 B2 = 1
0.218 × 0.951 × 0.392

1
B0 B1 B2 = = 0.081
12.3
√
3 1
B0 = B1 = B2 Then B03 = 0.081 B0 = 0.081 = 0.081 3 = 0.432
0.432 0.432 0.432
H(sn ) = × 2 × 2
s + 0.218 s + 0.134s + 0.951 s + 0.352s + 0.392
0.081
H(sn ) =
s 5 + 0.7048s 4 + 1.496s 3 + 0.689s 2 + 0.456s + 0.081
Unnomalized transfer function, H(s) and Ωp = 1 rad/sec
Hence
0.081
H(sn ) = 5
s + 0.7048s 4 + 1.496s 3 + 0.689s 2 + 0.456s + 0.081
 Chebyshev Filter Design

Dec 2014: Design A Chebyshev I low pass filter that has to meet the following specifications
i) passband ripple ≤2 dB and passband edge frequency 1 rad/sec
ii) Stopband attenuation ≥20 dB and Stopband edge frequency 1.3 rad/sec
Solution:
Ωp =1 rad/sec, Ωs =1.3 rad/sec,
Kp =–2 dB, Ks =–20 dB,
H ( jω )

Gain
 
1
Kp = 20 log √ = −2 1
1 + 2 1 δp
=0.76478 1+ ε 2

1
δp = 1 − √ = 0.20567
1 + 2 δs
Ω
ΩP ΩS
Ks = 20 log δs = −20
δs =0.1 Figure 18: LPF specifications
s
(1 − δp )−2 − 1 The order of the filter is
d= = 0.077
δs−2 − 1
1
cosh−1


d
N=
= 4.3 ' 5
Ωp 1 cosh−1 1
K = = = 0.769 K
Ωs 1.3
 √ !1
N
√ !− 1
N
1 1+ 1 + 2 1 1+ 1 + 2
a= − = 0.21830398
2  2 

√ !1
N
√ !− 1
N
1 1+ 1 + 2 1 1+ 1 + 2
b= + = 1.0235520
2  2 
h π i h πi
Ωk = b cos (2k − 1) = b cos (2k − 1)
2N 10
h π i h πi
σk = −a sin (2k − 1) = −a sin (2k − 1)
2N 10
where k = 1, 2, . . . 2N i.e., k = 1, 2, . . . 10
The poles those are lie on left half of the s plane is

k σk Ωk
1 -0.0674610 0.9734557
2 -0.1766151 0.6016287
3 -0.2183083 0
4 -0.1766151 -0.6016287
5 -0.0674610 -0.9734557
 Chebyshev Filter Design

KN KN
H5 (s) = Q =
(s − sk ) (s − s1 )(s − s2 )(s − s3 )(s − s4 )
LHPonly
KN
=
(s + 0.067461 − j0.9734557)(s + 0.067461 + j0.9734557)
(s + 0.1766151 − j0.6016287)(s + 0.1766151 + j0.6016287)(s + 0.2180383)
KN
=
(s 2 + 0.134922s + 0.95215)(s 2 + 0.35323s + 0.393115)(s + 0.2183083)
KN
=
s 5 + 0.70646s 4 + 1.4995s 3 + 0.6934s 2 + 0.459349s + 0.08172
N is odd KN = bo =0.08172
0.08172
H5 (s) =
s 5 + 0.70646s 4 + 1.4995s 3 + 0.6934s 2 + 0.459349s + 0.08172
Verification
0.08172
Ha (jΩ) =
(jΩ)5 + 0.70646(jΩ)4 − 1.49(jΩ)3 − 0.693(jΩ)2 + 0.4593(jΩ) + 0.08172
0.08172
|Ha (jΩ)| = p
(.7064Ω4 − .693Ω2 + .0817)2 + j(Ω5 − 1.499Ω3 + .4593Ω)2
20 log |Ha (jΩ)|1 = −2 dB
20 log |Ha (jΩ)|1 .3 = −24.5 dB
References

J. G. Proakis and D. G. Monalakis, Digital signal processing Principles Algorithms &

Applications, 4th ed. Pearson education, 2007.
Oppenheim and Schaffer, Discrete Time Signal Processing. Pearson education, Prentice
Hall, 2003.
S. K. Mitra, Digital Signal Processing. Tata Mc-Graw Hill, 2004.

L. Tan, Digital Signal Processing. Elsivier publications, 2007.