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Syllabus
Active Filters
Introduction, Active versus Passive Filters, Types of Active Filters,
First-Order Filters, The Biquadratic Function, Butterworth Filters,
Transfer Function Realizations, Low pass Filters, High-Pass Filters,
Band-Pass Filters, Band-Reject Filters, All-Pass Filters, Switched
Capacitor Filters, Filter Design Guide Lines.
Filter Basics
• A filter is a frequency-selective circuit that passes a specified
band of frequencies and blocks or attenuates signals of
frequencies outside this band.
• A filter is used to remove (or attenuate) unwanted frequencies
in an audio signal
• “Stop Band” – the part of the frequency spectrum that is
attenuated by a filter.
• “Pass Band” – part of the frequency spectrum that is
unaffected by a filter.
• Filters are usually described in terms of their “frequency
responses,” e.g. low pass, high pass, band pass, band reject
(or notch)
Advantages of Active Filters over
Passive Filters
(i) The maximum value of the transfer function or gain may be greater
than unity,
(ii) The loading effect is minimal, which means that the output
response of the filter is essentially independent of the load driven
by the filter.
(iii) The active filters do not exhibit insertion loss. Hence, the
passband gain is equal to 0 dB.
(iv) Complex filters can be realized without the use of inductors,
(v) The passive filters using R, L and C components are realizable
only for radio frequencies. Because, the inductors become very
large, bulky and expensive at audio frequencies. Due to low Q at
low frequency applications, high power dissipation is incurred.
The active filters
overcome these problems,
(vi) Rapid, stable and economical design of filters for variety of
applications is possible.
(vii) The active filters are easily tunable due to flexibility in gain and
frequency adjustments.
(viii) The op-amp has high input impedance and low output impedance.
Hence, the active filters using op-amp do not cause loading effect
on the source and load. Therefore, cascading of networks does
not need buffer amplifier.
(ix) Active filters for fixed frequency and variable frequency can be
designed easily. The adjustable frequency response is obtained
by varying an external voltage signal.
(x) There is no restriction in realizing rational function using active
networks.
(xi) Use of active elements eliminates the two fundamental restrictions
of passivity and reciprocity of RLC networks.
Limitations of Active Filters over
Passive Filters
(i) The high frequency response is limited by the gain-bandwidth
product and slew rate of the practical op-amps, leading to
comparatively lower bandwidth than the designed bandwidth.
(ii) The design of active filters becomes costly for high frequencies.
(iii) Active filters require dual polarity dc power supply whereas
passive filters do not.
(iv) The active element is prone to the process parameter variations
and they are sensitive to ambient conditions like temperature.
Hence, the performance of the active filter deviates from the
ideal response.
Ideal Filter Characteristics
Filter Characteristics
CLASSIFICATION OF FILTERS
Types of Filters
• Butterworth – Flat response in the pass band & stop band and
called flat-flat filter.
• Chebyshev – steeper roll-off but exhibits pass band ripple
(making it unsuitable for audio systems) & flat stopband.
• Cauer – It has equiripple both in pass & stop band.
9
passband
passband
stopband
Butterworth filter magnitude
response
stopband
Chebyshev filter magnitude
response
passband
stopband
Cauer filter magnitude response
FILTERS BASED ON FREQUENCY
Low pass filter (LPF)
High pass filter (LPF)
20db/decade
20db/decade
Understanding Poles and Zeros
The transfer function provides a basis for determining important
system response characteristics
The transfer function is a rational
function in the complex variable s = σ + jω, that is
zi’s are the roots of the equation N(s) = 0, and are defined to be the
system zeros, and the pi’s are the roots of the equation D(s) = 0,
and are defined to be the system poles.
Example
A linear system is described by the differential equation
Find the system poles and zeros. Solution: From the differential
equation the transfer function is
The system therefore has a single real zero at s = −1/2, and a pair of
real poles at s = −3 and s = −2.
Consider a Pole at Zero. Its response is constant.
Consider Poles at +a and –a. The exponential responses are
shown, for a function k/s+a, and k/s-a
Consider conjugate poles +jω & -jω & their mirror image on the
right side, along with their responces which is decaying sine wave
and increasing sine wave.
The equation shown has 3 poles & one Zero at -1. Zeros show how
fast the amplitudes vary.
Frequency Response of filters
•Ideal
•Practical
• Filters are often described in terms of poles and zeros
– A pole is a peak produced in the output spectrum
– A zero is a valley (not really zero)a
Order of the Filter
Comparison of FIR & IIR Filter
1.
2.
3.
4.
5.
6.
7.
8.
9.
FIR (Finite Impulse Response) (non-recursive)
filters produce zeros.
In signal processing, a finite impulse response
(FIR) filter is a filter whose
impulse
response (or response to any finite length input)
is of finite duration, because it settles to zero in
finite time.
Filters combining both past inputs and past
outputs can produce both poles and zeros.
FIR filters can be discrete- time or continuoustime, and digital or analog.
FIR is always stable
FIR has no limited cycles.
FIR has no analog history.
FIR filters are dependent upon linear-phase
characteristics.
FIR is dependent upon i/p only.
10. FIR’s delay characteristics is much better, but
they require more memory.
11. FIR filters are used for tapping of a higherorder.
1.
IIR (infinite Impulse Response) (recursive) filters
produce poles.
2. This is in contrast to infinite impulse
response (IIR) filters, which may have internal
feedback and may continue to respond
indefinitely (usually decaying).
3. IIR filters are difficult to control and have no
particular phase.
4. IIR can be unstable
5. IIR is derived from analog.
6. IIR filters make polyphase implementation
possible.
7. IIR filters are used for applications which are
not linear.
8. IIR filters are dependent on both i/p and o/p.
9. IIR filters consist of zeros and poles, and
require less memory than FIR filters.
10. IIR filters can become difficult to implement,
and also delay and distortion adjustments can
alter the poles & zeroes, which make the filters
unstable.
11. IIR filters are better for tapping of lower-orders,
since IIR filters may become unstable with
tapping higher-orders.
ACTIVE FILTERS USING OP-AMP:
Filters are frequency selective circuits. They are required to pass a
specific band of frequencies and attenuate frequencies outside the
band. Filters using an active device like OPAMP are called active
filters. Other way to design filters is using passive components like
resistor, capacitor and inductor.
ADVANTAGES OF ACTIVE FILTERS:
Possible to incorporate variable gain
Due to high Zi & Z0 of the OPAMP, active filters donot load the input
source or load.
Flexible design.
FREQUENCY RESPONSE OF
FILTERS:
Gain of a filter is given as, G=Vo/Vin
Ideal & practical frequency responses of different types of filters are
shown below.
First Order Low-Pass Butterworth Filter
1st-ORDER Butterworth LPF:
Because of simplicity, Butterworth filters are considered.
In 1st. order LPF which is also known as one pole LPF. Butterworth
filter and it’s frequency response are shown above.
• RC & values decide the cut-off frequency of the filter.
• Resistors R1 & RF will decide it’s gain in pass band. As the OPAMP is used in the non-inverting configuration, the closed loop
gain of the filter is given by
RF
AVF  1 
R1
EXPRESSION FOR THE GAIN OF
THE FILTER:
 jX C
Voltage across the capacitor is, V1 
Vin          (1)
R  jX C
Reactance of the capacitor is, X C 
Equation (1) becomes
Output of the filter is,
 R 
Vin
V0  AVF  V1  1  F 
R1  1  j 2fRC

V
AVF
 0 
Vin
 f 

1  j 
f
 H
1
2fC
 1 
 j
Vin

2fC 
 jVin
Vin

V1 


   (2)
2

fRC
 1  2fRC  j 1 
R  j

j
2

fC


Vin=
Vin
1 j 2fRC 
f = frequency of input signal.
Where fH= 1/2πRC cut-off frequency.
f = frequency of the input signal
The gain magnitude and phase angle equations of the low-pass filter
can be obtained by converting Equation (7-lb) into its equivalent
polar form, as follows:
The operation of the low-pass filter can be verified from the gain
magnitude equation, (7-2a):
1. At very low frequencies, that is, f < fH,
2. At f = fH,
3. f < fH,
DESIGN PROCEDURE:
Step1: Choose the cut-off frequency fH
Step2: Select a value of ‘C’ ≤ 1µF (Approximately
between .001 & 0.1µF)
Step3: Calculate the value of R using
Step4: Select resistors R1 & R2 depending on the
desired pass band gain.
=2. So RF=R1
For a first order Butterworth LPF, calculate the cut –off frequency if
R=10K & C=0.001µF.Also calculate the pass band voltage gain if
R1=10K RF =100K
1
1
fH 

 15.915KHz
3
6
2RC 2 10 10  0.00110
1+100K/10K =11
Design a I order LPF for the following specification
Pass band voltage gain = 2. Cut off frequency, fC = 10KHz.
AVF = 2; Let RF = 10K
RF/R1=1 Let C = 0.001µF
1
1
1
fH 
&R 

2RC
2f H C 2 10 103  0.00110 6
R=15.9K
1st ORDER HPF:
fL is shown for HPF
Circuit diagram & frequency response are shown above.
Again RC components decide the cut off frequency of the HPF
where as RF & R1 decide the closed loop gain.
EXPRESSION FOR THE GAIN:
Voltage V1 
R
Vin
R  jX C
1
WhereX C 
2fC
 f 

R
R
R  j 2fC 
j 
V1 
Vin 

Vin
j
1
1  j 2fRC
 fL  V
R
R
in
2fC
j 2fC
f 
1  j 
 fl 
Output voltage =
 jf 
AVF  
V0
 fL 

Gain = V
 f 
in
1  j  
 fL 
 jf 
AVF  
 fL  V
V0  AVF .V1 
f in
1 j
fL
Magnitude=
SECOND-ORDER LOW-PASS BUTTERWORTH FILTER
The gain of the second-order filter is set by R1, and RF, while the high
cutoff frequency fH is determined by R2, C2, R3, and C3, as follows:
High Cutoff frequency,
Filter Design
1. Choose a value for the high cutoff frequency fH
2. To simplify the design calculations, set R2 = R3 = R and C2 = C3 = C.
Then choose a value of C ≤ 1µF
3. Calculate the value of R using Equation for fH:
4. 4. Finally, because of the equal resistor (R2 = R3) and capacitor (C2
= C3) values, the pass band voltage gain AF = (1 + RF/R1) of the
second-order low-pass filter has to be equal to 1.586. That is, RF =
0.586/R1 This gain is necessary to guarantee Butterworth
response. Hence choose a value of R1 < 100 kΩ and calculate the
value of RF .
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 frequency determining resistors and capacitors.
Figure 7-8(a) shows the second-order high-pass filter.
SECOND-ORDER HIGH-PASS BUTTERWORTH FILTER
7.8 (a)
Second-order Hi pass-pass Filter Analysis
i1
Replace VC
i2
i3
4th Order Filter
Q=fc/BW = fC/(fH-fL)
Narrow Band-Pass Filter
DESIGN EQUATIONS:
Q
Select C1 = C2 =C R1 
2f C CAF
Q
Q
R2 
RB 
2
f C C
2f C C 2Q  AF


R3
A is the gain at f =fC AF 
2R1
Condition on gain AF<2Q2
Notch filterj
45
Shunted Twin T Filter
with swapped R & C
ALL PASS FILTER:
It is a special type of filter which passes all the frequency
components of the input signal to output without any
attenuation. But it introduces a predictable phase shift for
different frequency of the input signal.
The all pass
filters are also
called as delay
equalizers or
phase correctors.
Switched-Capacitor Filters
•
•
•
•
Active RC filters are difficult to implement totally on an IC due to the
requirements of large valued capacitors and accurate RC time
constants
The switched capacitor filter technique is based on the realization that
a capacitor switched between two circuit nodes at a sufficiently high
rate is equivalent to a resistor connecting these two nodes.
Switched capacitor filter ICs offer a low cost high order filter on a single
IC.
Can be easily programmed by changing the clock frequency.
Switched-Capacitor Filters
S1
R
+
v1
i
S2
+
+
v2
v1
-
-
+
CR
-
v1  v 2
i R
v1  v 2
R
i
q1 = CRv1
q2 = CRv2
Dq = q1-q2 = CR(v1-v2)
v2
-
the value of R is a function of CR and fC. For a fixed value of
C, the value of R can be adjusted by adjusting fC
Copyright © S.Witthayapradit.2009
q
i
 qfC  fC C R v1  v2 
T
T
Requ
1
fC 
T
v1  v2
1


i
fC C R
R equ
+
+
v1
v2
-
-
jω
X
X
Imaginary
X
-jω
X
-a
X
X
X
X
+a
Real
X
Zeros: roots of N(s)
• Poles: roots of D(s)
• Poles must be in the left half plane for the system to be stable
• As the poles get closer to the boundary, the system becomes less stable
• Pole-Zero Plot: plot of the zeros and poles on the complex s plane