Ministry of Higher Education and Scientific research
Al-Iraqia University
Engineering College
Network Engineering Department
(Analog Filter Design Using Butterworth Analysis)
A Project Submitted to the Department of Networks Engineering in Partial Fulfilment
for the Requirements of the Success Degree in DSP course for the Third year
By
1. Ismaeel Laith Fahmi
2. Abdullah Khalid Abdalwahab
3. Ahmed Hashim Khaleal
4. Haytham Ali Ibraheem
5. Mohammed Sattar Jaber
6. Jaffar Mohammed Jaber
Supervised By
Asst. Prof. Dr. Mohammed Nasser Hussain Al-Turfi
2019-2020
Contents
Contents
Subject Name
Student Name
Page
No.
Ismaeel Laith Fahmi
1
LPF design &examples
Abdullah Khalid Abdalwahab
6
HPF design &examples
Ahmed Hashim Khaleal
11
BPF design &examples
Haytham Ali Ibraheem
16
BSF design &examples
Mohammed Sattar Jaber
20
Electronic design for analog filters
Jafaar Mohammed Jaber
25
Way of design and system pools
evaluation
Name: Ismaeel Laith Fahmi
Class: 3rd – Morning – A
Introduction to Filter Approximation:
Ideal filters have brick wall type of frequency response. So, they have flat
passband and monotonic stopband. And the transition from passband to stopband
very abrupt (vertical transition). But in actual filter design, it is almost impossible
to achieve frequency response like an ideal filter. So, in filter design (particularly
for higher order filter design), five different types of filter approximations are quite
commonly used which are (Butterworth, Chebyshev, Inverse Chebyshev, Elliptic
and Bessel) Filter Approximations [1].
Butterworth Filter Approximation:
This filter approximation was first designed by British engineer Stephen
Butterworth. It is also known as the maximally flat response approximation as it
provides the flat passband response. It does not have any ripple in the stopband
(also flat) and the roll-off rate is 20n dB /decade (Wide transition band). As the
order of the filter ‘n’ increases, the response of the filter is closer to the ideal
response (narrow or near closed transition band). Its
maximum gain occurs at 𝑤 = 0 and it is 𝐻(𝑤) = 1. It is
most often used as tracking high frequency Low Pass
Filters, smoothing filters and Antialiasing filters [2].
System way of design and poles evolution:
Figure.1 Effect of N on the
characteristics
We will design a Normalized Low Pass Filter [3] (i.e. 𝒘𝒄 = 𝟏) and then Scale
Filter to desired Cutoff frequency. After that we can easily transform it into a highpass filter by replacing all terms of the form
𝒘
𝒘𝒄
to
−𝒘𝒄
𝒘
The Normalized Amplitude Response |𝐻𝑛 (𝑗𝑤)| =
Page | 1
and from them we get other.
1
√1+𝑤 2𝑛
Note: Where the maximum
gain in the passband equal 1
Name: Ismaeel Laith Fahmi
Class: 3rd – Morning – A
Note That |𝐻𝑛 (𝑗𝑤)|2 = 𝐻𝑛 (𝑗𝑤)𝐻𝑛 (−𝑗𝑤) → 𝑯𝒏 (𝒋𝒘)𝑯𝒏 (−𝒋𝒘) =
𝟏
𝟏+𝒘𝟐𝒏
The Transfer Function of a filter is denoted 𝐻(𝑠). ∵ 𝑠 = 𝜎 + 𝑗𝑤.
The Frequency Response 𝐻(𝑗𝑤) can be obtained from 𝐻(𝑠) by evaluating it at
𝑠 = 𝑗𝑤 (i.e. 𝜎 = 0).
The Transfer Function 𝐻(𝑠) can be obtained from 𝐻(𝑗𝑤) by evaluating it at
𝑠
𝑤= .
𝑗
So, we have |𝐻𝑛 (𝑗𝑤)|2 = 𝐻𝑛 (𝑗𝑤)𝐻𝑛 (−𝑗𝑤) → |𝐻𝑛 (𝑠)|2 = 𝐻𝑛 (𝑠)𝐻𝑛 (−𝑠)
→ |𝑯𝒏 (𝒔)|𝟐 =
Since 𝐻𝑛 (𝑠)𝐻𝑛 (−𝑠) =
Note: To find the poles, the denominator of the
𝟏
𝟏+(𝒔/𝒋)𝟐𝒏
1
1+(𝑠/𝑗)2𝑛
transfer function should equal to zero.
, The 2n Poles Of 𝐻𝑛 (𝑠)𝐻𝑛 (−𝑠) occur when
(𝑠/𝑗)2𝑛 = −1
Isolating s Yields 𝑠 2𝑛 = −(𝑗)2𝑛
Since −1 = 𝑒 𝑗𝜋(2𝑘−1) for integer k, and 𝑗 = 𝑒 2𝜋/2 , we have 𝑠 2𝑛 = 𝑒 𝑗𝜋(2𝑘−1+𝑛)
Taking the 1/2𝑛 root of each side of the equation yields 𝑠𝑘 = 𝑒
→ 𝒔𝒌 = 𝐜𝐨𝐬 (
𝝅
𝟐𝒏
(𝟐𝒌 + 𝒏 − 𝟏)) + 𝒋 𝐬𝐢𝐧 (
H(s)H(-s) Poles for n=4
Page | 2
𝝅
𝟐𝒏
𝑗𝜋
(2𝑘+𝑛−1)
2𝑛
(𝟐𝒌 + 𝒏 − 𝟏)), for 𝑘 = 1,2, … ,2𝑛.
H(s)H(-s) Poles for n=7
Name: Ismaeel Laith Fahmi
Class: 3rd – Morning – A
n Poles in Left Hand Plane correspond to 𝐻𝑛 (𝑠) which represent the stability of
the system, when correspond to 𝑘 = 1,2, … , 𝑛 [3].
n Poles in Right Hand Plane correspond to 𝐻𝑛 (−𝑠).
Finally, we’ve developed The Transfer Function of The Normalized Butterworth
Low Pass Filter As
𝑯𝒏 (𝒔) = (𝒔−𝒔
𝟏
𝟏 )(𝒔−𝒔𝟐 )…(𝒔−𝒔𝒏 )
→ 𝑯𝒏 (𝒔) =
𝟏
𝒔𝒏 +𝒂
𝒏−𝟏 𝒔𝒏−𝟏 +⋯+𝒂𝟏 𝒔+𝟏
→ 𝑯𝒏 (𝒔) =
𝟏
𝑩𝒏 (𝒔)
Where 𝐵𝑛 (𝑠) is the nth Order Butterworth Polynomial [3].
We can compute 𝐻𝑛 (𝑠) “By Hand” or Lookup 𝐵𝑛 (𝑠) in a Table
So, we know how to design an nth Order Normalized Butterworth Low Pass Filter
What if we want a Cutoff Frequency other than wc = 1 [4]?
We Replace 𝒘 with 𝒘/𝒘𝒄 in 𝐻𝑛 (𝑤) and Replace 𝒔 with 𝒔/𝒘𝒄 in 𝐻𝑛 (𝑠).
So, |𝐻𝑛 (𝑗𝑤)| =
1
√1+( 𝑤 )
→ |𝐻𝑛 (𝑗𝑤)|2 = 𝐻𝑛 (𝑗𝑤)𝐻𝑛 (−𝑗𝑤) =
2𝑛
𝑤𝑐
→ 𝐻𝑛 (𝑠)𝐻𝑛 (−𝑠) =
→ 𝑠𝑘 =
1
𝑛
𝑠2
1+(− 2 )
𝑤𝑐
𝑗𝜋
±𝑗𝑤𝑐 𝑒 2𝑛(2𝑘−1)
𝒔𝒌 = 𝒘𝒄 [− 𝐬𝐢𝐧
𝝅
𝟐𝒏
→ 𝑝𝑜𝑙𝑒𝑠: (−
𝑠2
𝑛
1
1+(
𝑤 2𝑛
)
𝑤𝑐
𝑗𝜋
) = −1 → 𝑠 2 = −𝑤𝑐2 𝑒 𝑛 (2𝑘−1)
𝑤2
𝑐
Note: represent the left-hand side poles
so we take it and leave the minus.
(𝟐𝒌 − 𝟏) + 𝒋 𝐜𝐨𝐬
𝝅
𝟐𝒏
(𝟐𝒌 − 𝟏)], Where 𝑘 = 1 → 𝑛, [5].
If we want to specify a Low Pass Filter, typically we specify 4 different numbers:
 Passband Gain 𝑮𝒑 at Passband Frequency 𝒘𝒑
 Stopband Gain 𝑮𝒔 at Stopband Frequency 𝒘𝒔
Usually we specify Gain in dB [4].
Page | 3
Name: Ismaeel Laith Fahmi
Class: 3rd – Morning – A
The Gain at any Frequency 𝑤𝑥 is [4]:
𝐺𝑥 = 20 log10 |𝐻(𝑗𝑤𝑥 )| → 𝐺𝑥 = 20 log10 (1/√1 + (
𝑤𝑥 2𝑛
) )
𝑤𝑐
𝑤𝑥 2𝑛
𝑤𝑥 2𝑛
→ 𝐺𝑥 = 0 − 20 log10 (√1 + ( ) ) → 𝐺𝑥 = −10 log10 [1 + ( ) ]
𝑤𝑐
𝑤𝑐
So, Gains at Frequency 𝑤𝑝 𝑎𝑛𝑑 𝑤𝑠 are:
𝑤𝑝 2𝑛
𝑤𝑝 2𝑛
= 10−𝐺𝑝,𝑑𝐵/10 − 1 …. equation (1)
𝐺𝑝,𝑑𝐵 = −10 log10 [1 + ( ) ] → ( )
𝑤
𝑤
𝑐
𝑤
𝑐
2𝑛
𝑤
2𝑛
𝐺𝑠,𝑑𝐵 = −10 log10 [1 + ( 𝑠 ) ] → ( 𝑠 )
𝑤
𝑤
𝑐
𝑐
= 10−𝐺𝑠,𝑑𝐵/10 − 1 …... equation (2)
Dividing equations (1) & (2) by each other yields
2𝑛
𝑤𝑠
( )
𝑤𝑝
=
10−𝐺𝑠,𝑑𝐵/10 − 1
10−𝐺𝑝,𝑑𝐵/10 − 1
This equation can easily be solved for the Filter Order n to yield the equation [4]
𝐥𝐨𝐠 𝟏𝟎 [(𝟏𝟎−𝑮𝒔,𝒅𝑩/𝟏𝟎 − 𝟏)/(𝟏𝟎−𝑮𝒑,𝒅𝑩/𝟏𝟎 − 𝟏)]
𝒏=
𝟐 𝐥𝐨𝐠 𝟏𝟎 (𝒘𝑺 /𝒘𝒑 )
We can also get equations for 𝑤𝑐 yields by [4]:
Solving Equation (1) ⇒ 𝒘𝒄 =
Solving Equation (2) ⇒ 𝒘𝒄 =
𝒘𝒑
(𝟏𝟎
−𝑮𝒑,𝒅𝑩 /𝟏𝟎
−𝟏)
𝟏/𝟐𝒏
𝒘𝒔
(𝟏𝟎−𝑮𝒔,𝒅𝑩 /𝟏𝟎 −𝟏)
𝟏/𝟐𝒏
……... equation (3)
……... equation (4)
Example/ Design a Butterworth filter with the following specifications [6]
{𝐺𝑝,𝑑𝐵 = −3, 𝑤𝑝 = 20, 𝐺𝑠,𝑑𝐵 = −25, 𝑤𝑠 = 50}, Find the transfer function 𝐻(𝑠)
Page | 4
Name: Ismaeel Laith Fahmi
Class: 3rd – Morning – A
Design steps:
Step 1: we compute the Filter Order As
25
𝐥𝐨𝐠 𝟏𝟎 [(𝟏𝟎−𝑮𝒔,𝒅𝑩/𝟏𝟎 − 𝟏)/(𝟏𝟎−𝑮𝒑,𝒅𝑩/𝟏𝟎 − 𝟏)]
𝒏=
→ 𝑛=
𝟐 𝐥𝐨𝐠 𝟏𝟎 (𝒘𝑺 /𝒘𝒑 )
log10 [
1010 − 1
3
1010
]
−1
50
2 log10 ( )
20
316.7
→ 𝑛 = 3.142 ⇒ 𝒏 = 𝟒
2 log10 (2.5)
→ 𝑛 = log10
Step 2: we compute the Cutoff Frequency 𝑤𝑐 As
𝒘𝒄 =
𝒘𝒄 =
𝒘𝒑
𝑮𝒑,𝒅𝑩
−
(𝟏𝟎 𝟏𝟎 −𝟏)
𝟏/𝟐𝒏
→ 𝒘𝒄 =
𝒘𝒔
(𝟏𝟎−𝑮𝒔,𝒅𝑩 /𝟏𝟎−𝟏)
𝟏/𝟐𝒏
20
1
3
8
(1010 −1)
→ 𝑤𝑐 =
→ 𝒘𝒄 = 𝟐𝟎. 𝟎𝟏
50
1
25
8
(1010 −1)
→ 𝒘𝒄 = 𝟐𝟒. 𝟑𝟔
Two different answers since n was rounded from 3.14 to 4, we choose 20.01.
Step 3: we compute the Transfer function 𝐻(𝑠) As
𝑗𝜋
For 𝑛 = 4 we can lookup coefficients in a table or compute by 𝑠𝑘 = 𝑒 2𝑛(2𝑘+𝑛−1)
𝐻𝑛 (𝑠) =
1
, this is the normalized transfer function.
𝑠 4 +2.6131𝑠 3 +3.4142𝑠 2 +2.6131𝑠+1
We obtain final Transfer function by replacing s in 𝐻𝑛 (𝑠) with 𝑠/𝑤𝑐 (easier way)
[6].
1
𝐻(𝑠) =
(
Page | 5
4
𝑠
)
20.01
+2.6131(
𝑠
)
20.01
3
+3.4142(
2
s
s
) +2.6131(
)+1
20.01
20.01
Name: Abdullah Khalid Abdalwahab
Class: 3rd – Morning – B
First Order Low Pass Butterworth Filter:
The below circuit shows the low pass Butterworth filter:
The required pass band gain of the Butterworth filter will mainly depends on the
resistor values of ‘R1’ and ‘Rf’ and the cut off frequency of the filter will depend
on R and C elements in the above circuit.
The gain of the filter is given as Amax = 1 + (R1 / Rf)
The impedance of the capacitor ‘C’ is given by the -jXC and the voltage across the
capacitor is given as,
Vc = - jXC / (R - jXC) * Vin
Where XC = 1 / (2πfc), capacitive Reactance.
The transfer function of the filter in polar form is given as
H(jω) = |Vout/Vin| ∟ø
Where gain of the filter Vout / Vin = Amax / √{1 + (f/fH)²}
And phase angle Ø = - tan-1 ( f/fH )
At lower frequencies means when the operating frequency is lower than the cut-off
frequency, the pass band gain is equal to maximum gain.
Vout / Vin = Amax i.e. constant.
At higher frequencies means when the operating frequency is higher than the cutoff frequency, then the gain is less than the maximum gain.
Vout / Vin < Amax
Page | 6
Name: Abdullah Khalid Abdalwahab
Class: 3rd – Morning – B
When operating frequency is equal to the cut-off frequency the transfer function is
equal to Amax /√2. The rate of decrease in the gain is 20dB/decade or 6dB/octave
and can be represented in the response slope as -20dB/decade.
Second Order Low Pass Butterworth Filter:
An additional RC network connected to the first order Butterworth filter gives us a
second order low pass filter. This second order low pass filter has an advantage
that the gain rolls-off very fast after the cut-off frequency, in the stop band.
In this second order filter, the cut-off frequency value depends on the resistor and
capacitor values of two RC sections. The cut-off frequency is calculated using the
below formula.
fc = 1 / (2π√R2C2)
The gain rolls off at a rate of 40dB/decade and this response is shown in slope 40dB/decade. The transfer function of the filter can be given as:
Vout / Vin = Amax / √{1 + (f/fc)4}
The standard form of transfer function of the second order filter is given as
Vout / Vin = Amax /s2 + 2εωns + ωn2
Where ωn = natural frequency of oscillations = 1/R2C2
ε = Damping factor = (3 - Amax ) / 2
For second order Butterworth filter, the middle term required is sqrt(2) = 1.414,
from the normalized Butterworth polynomial is
3 - Amax = √2 = 1.414
Page | 7
Name: Abdullah Khalid Abdalwahab
Class: 3rd – Morning – A
In order to have secured output filter response, it is necessary that the gain Amax is
1.586.
Higher order Butterworth filters are obtained by cascading first and second order
Butterworth filters. This can be shown as follows:
Where an and bn are pre-determined filter coefficients and these are used to
generate the required transfer functions.
Ideal Frequency Response of the Butterworth Filter:
The flatness of the output response increases as the order of the filter increases.
The gain and normalized response of the Butterworth filter for different orders are
given below:
Normalized Low Pass Butterworth Filter Polynomials:
Normalization is a process in which voltage, current or impedance is divided by
the quantity of the same unit of measure. This process is used to make a
dimensionless range or level of particular value.
The denominator polynomial of the filter transfer function gives us the Butterworth
polynomial. If we consider the s-plane on a circle with equal radius whose center is
Page | 8
Name: Abdullah Khalid Abdalwahab
Class: 3rd – Morning – B
at origin, then all the poles of the Butterworth filter are located in the left half of
that s-plane.
For any order filter the co-efficient of the highest power of ‘s’ should be always 1
and for any order filter the constant term is always 1. For even order filters all the
polynomial factors are quadratic in nature. For odd order filters all the polynomials
are quadratic except 1st order, for 1st order filter the polynomial is 1+s.
Butterworth polynomials in coefficients form is tabulated as given below:
The transfer function of the nth order Butterworth filter is given as follows:
H(jω) = 1/√{1 + ε² (ω/ωc)2n}
Where n is the order of the filter
ω is the radian frequency and it is equal to 2πf
And ε is the maximum pass band gain, Amax
Butterworth Low Pass Filter Example:
Let us consider the Butterworth low pass filter with cut-off frequency 15.9 kHz
and with the pass band gain 1.5 and capacitor C = 0.001µF.
fc = 1/2πRC
Page | 9
Name: Abdullah Khalid Abdalwahab
Class: 3rd – Morning – B
15.9 * 10³ = 1 / {2πR1 * 0.001 * 10-6}
R = 10kΩ
Amax = 1.5 and assume R1 as 10 kΩ
Amax = 1 + {Rf / R1}
Rf = 5 kΩ
Third Order Butterworth Low Pass Filter:
The cascade connection of 1st order and 2nd order Butterworth filters gives the
third order Butterworth filter. Third order Butterworth filter circuit is shown below.
For third order low pass filter the polynomial from the given normalized low pass
Butterworth polynomials is (1+s) (1+s+s²). This filter contains three unknown
coefficients and they are a0 a1 a2. The coefficient values for these are a0 = 1, a1 =
2 and a2 = 2. The flatness of the curve increases for this third order Butterworth
filter as compared with the first order filter.
Page | 10
Name: Ahmed Hashim Khaleal
Class: 3rd – Morning – A
HPF design &examples
Theoretical background and circuit diagrams of the filters
First order high pass Butterworth filter
High pass filter are often formed simply by interchanging frequency
Determining resistor and capacitors in low pass filter. That is a first order
high pass filter is formed from a first order low pass type by interchanging
components R and C. Figure- 1 shows a first order high pass Butterworth
filter with a low cut off frequency of . This is the frequency at which the
magnitude of the gain is 0.707 times its pass band value. Obviously, all
frequencies higher than
are pass band frequencies with the
highest frequency determined by the close loop bandwidth of the OPAMP The output voltage is given by:
or
f =frequency of the input signal
And magnitude of the voltage gain is:
Second order high-pass Butterworth filter
As in the case of the first order filter, a second order high pass filter can be
formed from a second order low pass filter simply by interchanging the
Page | 11
Name: Ahmed Hashim Khaleal
Class: 3rd – Morning – A
frequency determining resistors and capacitors. Figure 4 shows a second
order high pass filter.
The voltage gain magnitude equation of the second order high pass filter is
as follows:
f =frequency of the input signal
fig 2
fig 1
Calculation of circuit components
First order high pass Butterworth filter
For a high pass filter of cut off frequency 1KHz with a pass band gain 2
1KHz, C= 0.01
and hence,
Resistance has been taken as, R= (15+1) K
Since passband gain is 2, Rf=R1=10 K
Page | 12
Name: Ahmed Hashim Khaleal
Class: 3rd – Morning – A
Second order high-pass Butterworth filter
For designing a 2nd order high pass Butterworth filter of higher cut off frequency
𝑓𝑙 = 1𝐾ℎ𝑧 𝑡ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑐𝑜𝑚𝑝𝑛𝑒𝑛𝑡𝑠 𝑎𝑟𝑒 𝑡𝑎𝑘𝑒𝑛 𝑎𝑠 𝑅2 = 𝑅3 = 𝑅 𝑎𝑛𝑑 𝐶2 = 𝐶3
= 𝐶 = 0.01𝜇𝐹
𝑅=
1
2𝜋(108 )(10−8 )
To guarantee Butterworth response gain must be equel to
1.586 d hence Rf=0.586𝑅1
Chosen values are Rf=3.3k and 𝑅1 = 5.6𝑘
Frequency Response Data and Results
First order high pass Butterworth filter
The data for frequency response of the 1st order high pass filter is shown in
Table-1.
Input
frequency (
in (Hz)
10
100
200
500
700
800
1000
Page | 13
Input
voltage
in volt
1.00
1.00
1.00
1.00
1.00
1.00
1.00
Output
voltage
Gain
magnitude
(dB)=
in volt
0.08
0.75
1.25
1.80
1.94
2.00
2.00
Magnitude
0.08
0.75
1.25
1.80
1.94
2.00
2.00
-21.40
-02.50
01.94
05.10
05.74
06.02
06.02
Name: Ahmed Hashim Khaleal
Class: 3rd – Morning – A
3000
1.00
2.00
2.00
06.02
7000
1.00
2.00
2.00
06.02
10000
1.00
1.90
1.90
05.58
100000
1.00
1.95
1.95
05.80
The frequency response curve for 1st order high pass filter is shown in figure-3.
Fig 3: Frequency Response curve of 1st order Butterworth filter.
Second order high pass Butterworth filter
3.2Second order high pass Butterworth filter
st
The data for frequency response of the 1 order high pass filter is shown in Table-2
Input frequency
(
Page | 14
in (Hz)
Input voltage
in volt
Output voltage
Gainmagnitude
in volt
Magnitude
(dB)=
30
1.00
0.01
0.01
-38.42
100
1.00
0.03
0.03
-30.46
200
1.00
0.06
0.06
-24.44
500
1.00
0.43
0.43
-07.33
600
1.00
0.54
0.54
-05.35
700
1.00
0.70
0.70
-03.09
800
1.00
0.80
0.80
-01.94
900
1.00
0.95
0.95
-0.44
1000
1.00
1.05
1.05
0.42
Name: Ahmed Hashim Khaleal
Class: 3rd – Morning – A
1200
1.00
1.25
1.25
01.94
1500
1.00
1.38
1.38
02.79
2000
1.00
1.45
1.45
03.23
3000
1.00
1.55
1.55
03.81
7000
1.00
1.60
1.60
04.08
10000
1.00
1.60
1.60
04.08
100000
1.00
1.45
1.45
03.23
The frequency response curve for 2nd order high pass filter is shown in figure4.
Fig 4: Frequency Response curve of 2nd order Butterworth filter
Page | 15
Name: Haytham Ali Ibraheem
Class: 3rd – Morning – B
BPF design &examples
1. BUTTERWORTH BP-FILTER
To illustrate the utility of the proposed design method we have used it to design a
4th order band pass filter of the Butterworth type. The structure is an extended
Sallen- Key filter applying only one operational amplifier and is shown in Fig.1.
The operational amplifier is coupled as a unity gain amplifier since unity gain
amplifiers have the advantage of providing low power consumption, yielding a
large dynamic range, simplifying the amplifier design and being usable over a
larger frequency range than more conventional constant gain amplifiers.
The conventional way of designing such a filter would be either a cascade
approach of a second order high-pass filter and a second order low pass or by
cascading two second order band pass sections. Such an approach would require
two operational amplifiers with increased power consumption as a result. All
thought the classical cascade approach would be preferable in the high-Q case
due to lower sensitivities. Other possibilities with one opera- tional
Page | 16
Name: Haytham Ali Ibraheem
Class: 3rd – Morning – B
TABLE1
Normalized circuit components with ideal operational amplifier and with
normalized operational amplifier model t =62.5 rad/sec
C1 (F)
R1 (Ÿ)
C2 (F)
R2 (Ÿ)
R3 (Ÿ)
C3 (F)
R4 (Ÿ)
C4 (F)
Ideal
Op.Amp.
1.70000
0.8677687
0.991614
1.28364
5.44854
1.70000
1.97381
0.02880
Op.Amp.
model
1.70000
0.808192
1.23019
1.12370
5.96647
1.70000
1.24352
0.02880
Rel.Dev.
in %
0
í7.9
24
í12
9.5
0
í37
0
TABLE 2,
Normalized poles of the circuit response with operational amplifier
model Ȧ t =62.5 rad/sec
Poles
(rad/sec)
Real part
Imaginary part
í0.923877
í0.382684
í90.5906
0.382685
0.923879
0.0
amplifier also exist but this will be treated in another publication.
Page | 17
Name: Haytham Ali Ibraheem
Class: 3rd – Morning – B
The operational amplifier model is shown in figure 2 with the unity gain
frequency :𝐶𝑟 =
1
𝐶
, In this case we will design a Butterworth band pass
filter with a center frequency of 16 kHz and we will use an operational
amplifier with a unity gain frequency of 1MHz
We have normalized the whole design such that the center frequency is
1.0rad/sec. This gives a normal- ized unity gain frequency of t=
1MHz/16kHz = 62.5 and a normalized model capacitor with the value of
C = 1/ t
= 16mF. This corresponds to apply the low cost opera- tional amplifier
LM307.
2.SENSITIVITY ANALYSIS
Manufactured filters cannot guarantee to correspond ex- actly to the
designed filter performances. The effect of component tolerances should
be analyzed. The simplest way of predicting the effect of component
tolerances is to use the concept of network sensitivity assuming that the
component changes are small. The magnitude variability with respect to
the passive components and the unity gain frequency is given by,
Page | 18
Name: Haytham Ali Ibraheem
Class: 3rd – Morning – B
We have chosen to subdivide the Schoeffler criterion in two parts a passive part
with contribution only from the passive components:
and an active part with contribution only from the unity gain frequency of the
operational amplifier :
A sensitivity analysis was performed assuming the
relative changes of the resistors, capacitors and the unity gain
frequency to be uncorrelated random variables with a zero
mean Gaussian distribution and 1% standard deviation.
his corresponds to gain slopes of 40dB/decade. The
passive sensitivity index of the gain function approaches
Page | 19
Name: Mohammed Sattar Jaber
Class: 3rd – Morning – B
band stop filter design & examples
We have seen above that simple band stop filters can be made using first or second
order low and high pass filters along with a non-inverting summing op-amp circuit
to
reject a wide band of frequencies. But we can also design and construct band stop
filters to produce a much narrower frequency response to eliminate specific
frequencies by increasing the selectivity of the filter. This type of filter design is
called a “Notch Filter”.
Notch filters are a highly selective, high-Q, form of the band stop filter which can
be used to
reject a single or very small band of frequencies rather than a whole bandwidth of
different
frequencies. For example, it may be necessary to reject or attenuate a specific
frequency
generating electrical noise (such as mains hum) which has been induced into a circuit
from inductive loads such as motors or ballast lighting, or the removal of harmonics,
etc.
But as well as filtering, variable notch filters are also used by musicians in sound
equipment such as graphic equalizers, synthesizers and electronic crossovers to deal
with narrow peaks in the acoustic response of the music. Then we can see that notch
filters are widely used in much the same way as low-pass and high-pass filters.
Page | 20
Name: Mohammed Sattar Jaber
Class: 3rd – Morning – B
Notch filters by design have a very narrow and very deep stop band around their
center frequency with the width of the notch being described by its selectivity Q in
exactly the same way as resonance frequency peaks in RLC circuits.
The most common notch filter design is the twin-T notch filter network. In its basic
form, the
twin-T, also called a parallel-tee, configuration consists of two RC branches in the
form of two tee sections, that use three resistors and three capacitors with opposite
and opposing R and C elements in the tee part of its design as shown, creating a
deeper notch.
Basic Twin-T Notch Filter Design
The upper T-pad configuration of resistors 2R and capacitor 2C form the low-pass
filter section of the design, while the lower T-pad configuration of capacitors C and
resistor R form the high- pass filter section. The frequency at which this basic twinT notch filter design offers maximum attenuation is called the “notch
frequency”, ƒN and is given as:
Twin-T Notch Filter Equation
Being a passive RC network, one of the disadvantages of this basic twin-T notch
filter design is
Page | 21
Name: Mohammed Sattar Jaber
Class: 3rd – Morning – B
that the maximum value of the output (Vout) below the notch frequency is generally
less than the maximum value of output above the notch frequency due in part to the
two series resistances (2R) in the low-pass filter section having greater losses than
the reactance’s of the two series capacitors (C) in the high-pass section. As well as
uneven gains either side of the notch frequency, another disadvantage of this basic
design is that it has a fixed Q value of 0.25, in the order of -12dB. This is because at
the notch frequency, the reactance’s of the two series capacitors equals the
resistances of the two series resistors, resulting in the currents flowing in each branch
being out-of-phase by 180o. the notch. Single Op-amp Twin-T Notch Filter
Here the output from the twin-T notch filter section is isolated from the voltage divider
by a single non-inverting op-amp buffer. The output from the voltage divider is fed back
to “ground” point of R and 2C. The amount of signal feedback, known as the feedback
fraction k, is set by the resistor
ratio and is given as:
The value of Q is determined by the R3 and R4 resistor ratio, but if we wanted to
make Q fully adjustable, we could replace these two feedback resistors with a single
potentiometer and feed it into another op-amp buffer for increased negative gain. Also,
to obtain the maximum notch depth at the given frequency, resistors R3 and R4 could
be eliminated and the junction of R and 2C connected directly to the output.
Page | 22
Name: Mohammed Sattar Jaber
Class: 3rd – Morning – B
Band Stop Filter Example No2
Design a two op-amp narrow-band, RC notch filter with a center notch frequency, ƒN of
1kHz and a -3dB bandwidth of 100 Hz. Use 0.1uF capacitors in your design and
calculate the expected notch depth in decibels.
Data given: ƒN = 1000Hz, BW = 100Hz and C = 0.1uF.
1. Calculate value of R for the given capacitance of 0.1uF
2. Calculate value of Q
3. Calculate value of feedback fraction k
4. Calculate the values of resistors R3 and R4
5. Calculate expected notch depth in decibels, dB
Page | 23
Name: Mohammed Sattar Jaber
Class: 3rd – Morning – B
Notch Filter Design
Page | 24
Name: Jaffar Mohammed Jaber
Class: 3rd – Morning – A
Electronic Design for Analog Filters
Standard Filter Blocks:
The generic filter structure (Figure 1a) lets you realize a highpass or lowpass
implementation by substituting capacitors or resistors in place of components G1
through G4. Considering the effect of these components on the opamp feedback
network, one can easily derive a lowpass filter by making G2/G4 into capacitors
and G1/G3 into resistors. (Doing the opposite yields the highpass implementation.)
Figure 1. By substituting for G1 through G4 in the generic filter block (a), you can
implement a lowpass filter (b) or a highpass
filter (c).
The transfer function for the lowpass filter
(Figure 1b) is:
This equation is simpler with conductances.
Replace the capacitors with a conductance of
sC, and the resistors with a conductance of G.
If this looks complicated, you can "normalize"
the equation. Set the resistors equal to 1Ω or
the capacitors equal to 1F, and change the surrounding components to fit the
response. Thus, with all resistor values equal to 1Ω, the lowpass transfer function
is:
This transfer function describes the response of a generic, second-order lowpass
filter. We now take the theoretical tables of poles that describe the three main filter
responses, and translate them into real component values.
Page | 25
Name: Jaffar Mohammed Jaber
Class: 3rd – Morning – A
The Design Process To determine the type of filter required, you should use the
above descriptions to select the passband performance needed. The simplest way to
determine filter order is to design a 2nd-order filter stage, and then cascade
multiple versions of it as required. Check to see if the result gives the desired
stopband rejection, and then proceed with correct pole locations as shown in the
tables in the Appendix. Once pole locations are established, the
component values can soon be calculated.
First, transform each pole location into a quadratic expression similar to that in the
denominator of our generic 2nd-order filter. If a quadratic equation has poles of (a
± jb), then it has roots of (s - a - jb) and (s - a + jb). When these roots are
multiplied together, the resulting quadratic expression is s² - 2as + a² + b².
In the pole tables, "a" is always negative, so for convenience we declare s² + 2as +
a² + b² and use the magnitude of "a," regardless of its sign. To put this into
practice, consider a 4th-order Butterworth filter. The poles and the quadratic
expression corresponding to each pole location are as follows:
Poles (a ± jb)
Quadratic Expression
-0.9239 ± j0.3827 s² + 1.8478s + 1
-0.3827 ± j0.9239 s² + 0.7654s + 1
You can design a 4th-order Butterworth lowpass filter with this information.
Simply substitute values from the above quadratic expressions into the
denominator of equation 4. Thus, C2C4 = 1 and 2C4 = 1.8478 in the first filter,
implying that C4 = 0.9239F and C2 = 1.08F. For the second filter, C2C4 = 1 and
2C4 = 0.7654, implying that C4 = 0.3827F and C2 = 2.61F. All resistors in both
filters equal 1Ω. Cascading these two 2nd-order filters yields a 4th-order
Page | 26
Name: Jaffar Mohammed Jaber
Class: 3rd – Morning – A
Butterworth response with rolloff frequency of 1rad/s, but the component values
are impossible to find. If the frequency or component values above are not suitable,
read on.
It so happens that if you maintain the ratio of the reactances to the resistors, the
circuit response remains unchanged. You might therefore choose 1kΩ resistors. To
ensure that the reactances increase in the same proportion as the resistances, divide
the capacitor values by 1000.
We still have the perfect Butterworth response, but unfortunately the rolloff
frequency is 1rad/s. To change the circuit's frequency response, we must maintain
the ratio of reactances to resistances, but simply at a different frequency. For a
rolloff of 1kHz rather than 1rad/s, the
capacitor value must be
further reduced by a factor of
2π × 1000. Thus, the
capacitor's reactance does not
reach the original
(normalized) value until the higher frequency. The resulting 4th-order Butterworth
lowpass filter with 1kHz rolloff takes the form of Figure 2.
Figure 2. These two nonidentical 2nd-order filter sections form a 4th-order
Butterworth lowpass filter.
Using the above technique, you can obtain any even-order filter response by
cascading 2nd-order filters. Note, however, that a 4th-order Butterworth filter is
not obtained simply by calculating the components for a 2nd-order filter and then
cascading two such stages. Two 2nd-order filters must be designed, each with
different pole locations. If the filter has an odd order, you can simply cascade 2ndorder filters and add an RC network to gain the extra pole. For example, a 5thorder Chebyshev filter with 1dB ripple has the following poles:
Page | 27
Name: Jaffar Mohammed Jaber
Class: 3rd – Morning – A
Poles Quadratic Expression
-0.2265 ± j0.5918 s² + 0.453s + 0.402 2.488s² + 1.127s + 1
-0.08652 ± j0.9575 s² + 0.173s + 0.924 1.08s² + 0.187s + 1
-0.2800
see text
To ensure conformance with the generic filter described by equation 4, and to
ensure that the last term equals unity, the first two quadratics have been multiplied
by a constant. Thus, in the first filter C2C4 = 2.488 and 2C4 = 1.127, implying that
C4 = 0.5635F and C2 = 4.41F. For the second filter, C2C4 = 1.08 and 2C4 = 0.187,
implying that C4 = 0.0935F and C2 = 11.55F.
Earlier, it was shown that an RC circuit has a pole when 1 + sCR = 0: s = -1/CR. If
R = 1, then to obtain the
final pole at s = -0.28 you
must set C = 3.57F. Using
1kΩ resistors, you can
normalize for a 1kHz rolloff frequency as shown in Figure 3. Thus, designers can
boldly go and design lowpass filters of any order at any frequency.
Figure 3. A 5th-order, 1dB-ripple Chebyshev lowpass filter is constructed from
two non-identical 2nd-order sections and an output RC network.
All of this theory applies also to the design of highpass filters. It has been shown
that a simple RC lowpass filter has the transfer function:
Similarly, a simple RC highpass filter has the transfer function:
Normalizing these functions to correspond with the normalized pole tables gives
TF = 1/(1 + s) for lowpass and TF = 1/(1 + 1/s) for highpass.
Page | 28
Name: Jaffar Mohammed Jaber
Class: 3rd – Morning – A
Note that the highpass pole positions "s" can be obtained by inverting the lowpass
pole positions. Inserting those values into the highpass filter block ensures the
correct frequency response. To obtain the transfer function for the highpass filter
block, we need to go back to the transfer function of the lowpass filter block. Thus,
from
we obtain the transfer function of the equivalent highpass filter block by
interchanging capacitors and resistors:
Again, life is much simpler if capacitors are normalized instead of resistors:
Equation 9 is the transfer
function of the highpass filter block. This time, we calculate resistor values instead
of capacitor values. Given the general highpass filter response, we can derive the
highpass pole positions by inverting the lowpass pole positions and continuing as
before. Inverting a complex-pole location is easier said than done, however. As an
example, consider the 5th-order, 1dB-ripple Chebyshev filter discussed earlier. It
has two pole positions at (-0.2265 ± j0.5918).
The easiest way to invert a complex number is to multiply and divide by the
complex conjugate, thereby obtaining a real number in the numerator. You then
find the reciprocal by inverting the fraction as follows:
Inverting this expression yields pole positions that can then be converted to the
corresponding quadratic expression, and values calculated as before. The result is:
Poles Quadratic Expression
-0.564 ± j1.474
Page | 29
s² + 1.128s + 2.490 0.401s² + 0.453s + 1
References
References:
[1] ALL ABOUT ELECTRONICS. (2017, October 8). Filter Approximations:
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[Video]. YouTube. Available from
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[2] ALL ABOUT ELECTRONICS. (2017, October 16). Butterworth Filter :
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K92vdh3gAXoaOXXQDu&index=6
[3] Adam Panagos. (2014, July 16). Butterworth Filter - 02 - Normalized Filter
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