Lecture Notes for EE-226 Circuit Analysis-II

Dr. Ghulam Mustafa

Professor, Department of Electrical Engineering

Pakistan Institute of Engineering & Applied Sciences
Email: gm@pieas.edu.pk, Homepage: faculty.pieas.edu.pk/gm
 Lecture-7
Filter Circuits

What are filter circuits?

7-1
 What are filter circuits?
[center]
We saw in previous chapter that impedance of capacitors and inductors changes with
frequency. When a sinusoidal signal with varying frequency is applied to circuits
containing capacitors and inductors, these circuits pass to the output only those signals
that reside in a certain range of frequencies. Such circuits are called
frequency-selective circuits or filters.

Many devices that communicate via electric signals, such as telephones, radios,
televisions, and satellites, employ frequency-selective circuits. A home stereo system
with a graphic equalizer is an excellent example of a collection of filter circuits. Each
band in the graphic equalizer is a filter that amplifies sounds (audible frequencies) in
the frequency range of the band and attenuates frequencies outside of that band. Thus
the graphic equalizer enables you to change the sound volume in each frequency band.

What are filter circuits? 7-2

 What are Filter Circuits?
[center]
Recall that the transfer function of a circuit H(s) provides an easy way to compute the
steady-state response to a sinusoidal input.
x(t) = A cos(ωt + ϕ)
yss (t) = A|H(jω)| cos[ωt + θ(ω) + ϕ], H(jω) = |H(jω)|ejθ(ω) .
To study the frequency response of a circuit, we replace a fixed-frequency sinusoidal
source with a varying-frequency sinusoidal source. The magnitude and phase of the
output signal depends only on the magnitude and phase of the transfer function
H(jω).

If we can vary the frequency of a sinusoidal source without changing its magnitude or
phase angle, the amplitude and phase of the output will vary only if those of the
transfer function vary as the frequency of the sinusoidal source is changed.
What are filter circuits? 7-3
 What are Filter Circuits?
[center]
Frequency response plots of
ideal filter circuits.
(a) Low-pass filter
(b) High-pass filter
(c) Bandpass filter
(d) Bandreject filter
ωc is called the cutoff frequency.
The range of frequencies of the
input signals passed from the
input to the output is called the
passband. Frequencies not in a
circuit’s passband are in its
stopband.
What are filter circuits? 7-4
 What are Filter Circuits?
[center]
There are many circuits that act as filters. We broadly classify into two types:

1. Passive Filters: Filter circuits that employ only the passive elements: resistors,
capacitors, and inductors. The largest output amplitude such filters can achieve is
usually 1.
2. Active Filters: Filter circuits that also employ operational amplifiers. These filter
circuits can achieve output amplitudes greater 1.

What are filter circuits? 7-5

 Part-1
Passive Filters

Low-Pass Filters

High-Pass Filters

Band-Pass Filters

Band-Reject Filters

Drawing Bode Diagrams using MATLAB

7-6
 Low-Pass Filters
[center]
A Series RL Circuit

At low frequencies (ω ≪ R), the inductor’s

impedance is very small compared with the
resistor’s impedance, and the inductor
effectively functions as a short circuit.
At high frequencies (ω ≫ R), the inductor’s
impedance is very large compared with the
resistor’s impedance, and the inductor thus
functions as an open circuit, effectively
blocking the flow of current in the circuit.

Low-Pass Filters 7-7

 Low-Pass Filters
[center]
This series RL circuit selectively passes low-frequency inputs to the output, and it
blocks high-frequency inputs from reaching the output. This circuit’s response to
varying input frequency thus has the shape shown in figure below:
The upper plot shows how
|H(jω)| varies with frequency.
The lower plot shows how θ(jω)
varies as a function of frequency.
Compare with the ideal magnitude
plot for a low-pass filter, the
frequency response for realistic
circuit varies gradually. How we
define cutoff frequency?

Low-Pass Filters 7-8

 Low-Pass Filters
[center]
The definition for cutoff frequency widely used by electrical engineers is√the frequency
for which the transfer function magnitude is decreased by the factor 1/ 2 from its
maximum value:
1
|H(jωc )| = √ Hmax ,
2
Where Hmax is the maximum magnitude of the transfer function.

The passband of a realizable filter is defined as the range of frequencies in which the
amplitude of the output voltage is at least 70.7% of the maximum possible amplitude.

Low-Pass Filters 7-9

 Low-Pass Filters
[center]
To quantitatively analyse the series RL circuit, we draw its s domain equivalent as
shown below.

The voltage transfer function for this circuit is

Vo (s) R/L
H(s) = =
Vi (s) s + R/L
Low-Pass Filters 7-10
 Low-Pass Filters
[center]
To study the frequency response, we make the substitution s = jω

R/L
H(jω) =
jω + R/L

The magnitude and phase of H(jω) will be

R/L
|H(jω)| = p ,
ω + (R/L)2
2
 
−1 ωL
θ(jω) = − tan .
R

Observe the behavior of |H(jω)| and θ(jω) when frequency increases from ω = 0 to
ω = ∞.
Low-Pass Filters 7-11
 Low-Pass Filters
[center]
We can compute the cutoff frequency, ωc , from |H(jω)|. Recall that
|H(jωc )| = √12 Hmax . For the low-pass filter Hmax = H(j0), therefore

1 R/L
|H(jωc )| = √ |1| = p
2 ω 2 + (R/L)2

Solving for ωc , we get

R
ωc =
L

The cutoff frequency, ωc , can be set to any desired value by appropriately selecting
values for R and L.

Low-Pass Filters 7-12

 Example
[center]
Electrocardiology is the study of the electric signals produced by the heart. These
signals maintain the heart’s rhythmic beat, and they are measured by an instrument
called an electrocardiograph. This instrument must be capable of detecting periodic
signals whose frequency is about 1 Hz (the normal heart rate is 72 beats per minute).
The instrument must operate in the presence of sinusoidal noise consisting of signals
from the surrounding electrical environment, whose fundamental frequency is 60
Hz—the frequency at which electric power is supplied.

Choose values for R and L in the series RL low-pass circuit such that the resulting
circuit could be used in an electrocardiograph to filter out any noise above 10 Hz and
pass the electric signals from the heart at or near 1 Hz. Then compute the magnitude
of Vo at 1 Hz, 10 Hz, and 60 Hz to see how well the filter performs.

Low-Pass Filters 7-13

 Example
[center]
The cutoff frequency of the desired low-pass filter is 10 Hz. Since
R
ωc =
L
We cannot choose R and L independently. Let’s choose a commonly available value of
L, 100 mH. Next, to use the above equation to calculate R, we need to convert the
cutoff frequency from hertz into radians per second:

ωc = 2π(10) = 20πrad/s.

The value of R is

R = ωc L = (20π)(100 × 10−3 ) = 6.28Ω.

Low-Pass Filters 7-14

 Example
[center]
We can compute the magnitude of Vo using |Vo (ω)| = |H(jω)||Vi (ω)|

R/L 20π
|Vo (ω)| = p |Vi | = √ |Vi |
2
ω + (R/L)2 ω + 400π 2
2

Table below summarizes the computed magnitude values for the frequencies 1 Hz, 10
Hz, and 60 Hz.

f(Hz) |Vi | (V) |Vo | (V)

1 1.0 0.995
10 1.0 0.707
60 1.0 0.164

Low-Pass Filters 7-15

 Low-Pass Filters
[center]
A Series RC Circuit
The series RC circuit also behaves as a low-pass filter.
1. At zero frequency (ω = 0): The impedance of the
capacitor is infinite, and the capacitor acts as an
open circuit. The input and output voltages are
thus the same.
2. At frequencies increasing from zero: The
impedance of the capacitor decreases relative to the
impedance of the resistor. The output voltage is
smaller than the source voltage.
3. At infinite frequency (ω = ∞): The impedance of
the capacitor is zero, and the capacitor acts as a
short circuit. The output voltage is thus zero.

Low-Pass Filters 7-16

 Exercise
[center]

For the series RC circuit show in figure.

1. Find the transfer function between the source
voltage and the output voltage.
2. Determine an equation for the cutoff frequency in
the series RC circuit.
3. Choose values for R and C that will yield a
low-pass filter with a cutoff frequency of 3 kHz.

Low-Pass Filters 7-17

 Exercise
[center]
1. The s-domain equivalent for the circuit is
shown. The transfer function is
1
RC
H(s) = 1
s+ RC

2. The magnitude of H(jω) is

1 1
|H(jω)| = q RC =⇒ |H(jωc )| = q RC
1 2 1 2


ω 2 + RC


ωc2 + RC

At ωc , |H(jωc )| = √12 Hmax and Solving, we get

Hmax = |H(j0)| = 1, so
1
ωc =
RC
Low-Pass Filters 7-18
 Exercise
[center]
3. Let’s choose C = 1µF, then R will be
1 1
R= = = 53.05Ω.
ωc C (2π)(3 × 103 )(1 × 10−6 )

Low-Pass Filters 7-19

 Low-Pass Filters (Summary)
[center]
The general form for the transfer functions
of the two low-pass filters is:

ωc
H(s) =
s + ωc

Recall that an important parameter for

first-order circuits is the time constant, τ ,
which characterizes the shape of the time
response. For the RL circuit, the time
constant has the value L/R; for the RC
circuit, the time constant is RC. Compare
the time constants to the cutoff frequencies
for these circuits and notice that
1
τ= .
Low-Pass Filters ωc 7-20
 High-Pass Filters
[center]
A Series RC Circuit

At ω = 0, the capacitor behaves like an open

circuit, so there is no current flowing in the
resistor. There is no voltage across the
resistor, and the circuit filters out the
low-frequency source voltage before it reaches
the circuit’s output.
At ω = ∞, the capacitor behaves as a short
circuit, and thus there is no voltage across the
capacitor. The input voltage and output
voltage are the same.

High-Pass Filters 7-21

 High-Pass Filters
[center]
Figure below shows the frequency response plot for the series RC high-pass filter. For
reference, the dashed lines indicate the magnitude plot for an ideal high-pass filter.

High-Pass Filters 7-22

 High-Pass Filters
[center]

The s-domain equivalent of the RC circuit

is shown. Applying s-domain voltage
division, the transfer function is
s
H(s) =
s + 1/RC

For frequency response, substitute s = jω

to get The magnitude and phase equations are
jω ω
H(jω) = |H(jω) = p
jω + 1/RC ω2 + (1/RC)2
θ(ω) = 90◦ − tan−1 ωRC.

High-Pass Filters 7-23

 High-Pass Filters
[center]
√
Now we calculate the cutoff frequency. Recall that at ωc , H(jωc ) = (1/ 2)Hmax . For
a high-pass filter,Hmax = |H(jω)|ω=∞ = |H(j∞)| = 1, therfore
1 ωc
√ =p
2 ωc2 + (1/RC)2

Solving, we get

1
ωc = .
RC

The cutoff frequency for the series RC circuit has the value 1/RC, whether the circuit
is configured as a low-pass filter or as a high-pass filter.

High-Pass Filters 7-24

 Exercise
[center]
Show that the series RL circuit in Figure
below also acts like a high-pass filter:
1. Derive an expression for the circuit’s
transfer function.
2. Use the result from (a) to determine
an equation for the cutoff frequency in
the series RL circuit.
3. Choose values for R and L that will
yield a high- pass filter with a cutoff
frequency of 15 kHz.

High-Pass Filters 7-25

 High-Pass Filters (Summary)
[center]
The general form for the transfer functions
of the two high-pass filters is:

s
H(s) =
s + ωc

Any circuit with transfer function as above

would behave as a high-pass filter with a
cutoff frequency of ωc .
Also, we have discovered that a series RC
circuit has the same cutoff frequency
whether it is configured as a low-pass filter
or as a high-pass filter. The same is true of
a series RL circuit.

High-Pass Filters 7-26

 Effect of Load on Filter Output
[center]
Examine the effect of placing a load
resistor in parallel with the inductor in the
RL high-pass filter shown in figure below.
1. Determine the transfer function for
the circuit
2. Sketch the magnitude plot for the
loaded RL high-pass filter, using the
values for R = 500Ω and L = 5.31mH
and letting RL = R. On the same
graph, sketch the magnitude plot for
the unloaded RL high-pass filter.

High-Pass Filters 7-27

 Effect of Load on Filter Output
[center]
RL sL
RL +sL
H(s) =
R + RRLL+sL
sL
 
RL
R+RL s
=  
RL R
s+ R+RL L
Ks
= ,
s + ωc
where
RL
K= , ωc = KR/L. 500 1
R + RL K= = , ωc = KR/L = 7.5kHz.
500 + 500 2
High-Pass Filters 7-28
 Band-Pass Filters
[center]
Now we consider two filters that pass signal with frequencies in a band while filtering
out all signal with frequencies not in the passband. We are already familiar with the
notion of the cutoff frequency. We now introduce three more notions that we need to
describe the characteristics of band-pass and band-reject filters.

Center Frequency or Resonant Frequency, ωo

The center frequency or the resonant frequency ωo is defined as the frequency for
which a circuit’s transfer function is purely real. When a circuit is driven at the
resonant frequency, we say that the circuit is in resonance, because the frequency of
the forcing function is the same as the natural frequency of the circuit. The center
frequency is the geometric center of the pass band:
√
ωo = ωc1 ωc2

Band-Pass Filters 7-29

 Band-Pass Filters
[center]
Bandwidth, β
Bandwidth is the width of the passband.

Quality Factor, Q
The quality factor is defined as the ratio of the center frequency to the bandwidth.
The quality factor gives a measure of the width of the passband, independent of its
location on the frequency axis.

Although there are five different parameters that characterize the bandpass filter —
ωc1 , ωc2 , ωo , β, and Q. Only two of the five can be specified independently.

Band-Pass Filters 7-30

 Band-Pass Filters
[center]
A Series RLC Circuit
At ω = 0, the capacitor behaves like an open
circuit, and the inductor behaves like a short circuit,
so there is no current flowing in the resistor. There
is no voltage across the resistor, and the circuit
filters out the low-frequency source voltage before it
reaches the circuit’s output.
At ω = ∞, the capacitor behaves as a short circuit,
and the inductor behaves like an open circuit. The
inductor now prevents current from reaching the
resistor, and again the output voltage is zero.
Between ω = 0 and ω = ∞, both the inductor and
capactor will have finite impedances, and some
voltage will drop across the resistor.
Band-Pass Filters 7-31
 Band-Pass Filters
[center]
Figure shows the frequency response plot
of the series RLC band-pass circuit. The
figure also shows the magnitude plot of the
ideal band-pass filter.
Notice how the phase angle of the output
voltage changes. The capacitor contributes
positive phase shift, therefore at very low
frequencies, the phase angle at the output
maximizes at +90◦ . The inductor
contributes a negative phase shift, at very
high frequencies, the phase angle at the
output reaches its negative maximum of
−90◦ .

Band-Pass Filters 7-32

 Band-Pass Filters
[center]
For quantitative analysis, we draw the s-domain
equivalent of the series RLC circuit as shown. Using
the s-domain voltage division, the transfer function
is
(R/L)s
H(s) =
s2 + (R/L)s + (1/LC)

The magnitude and phase angle will be

ω(R/L) We now calculate the five

|H(jω)| = p ,
[(1/LC) − ω 2 ]2 + [ω(R/L)]2 parameters that characterise the
band-pass filter.
 
◦ −1 ω(R/L)
θ(ω) = 90 − tan .
(1/LC) − ω 2

Band-Pass Filters 7-33

 Band-Pass Filters
[center]
Recall that center frequency is the frequency for the which the circuit’s transfer
function is real. This will occur when the sum of the impedance of the capacitor and
inductor is zero:
1
jωo L + = 0.
jωo C
Solving, we get
r
1
ωo = .
LC

Band-Pass Filters 7-34

 Band-Pass Filters
[center]
Next, we calculate the cutoff frequencies, ωc1and ωc2 . Recall that, at cutoff
frequencies the magnitude of the transfer function is √12 Hmax . What will be Hmax ?

Since Hmax = |H(jωo )|, so

ωo (R/L)
Hmax = |H(jωo )| = p
[(1/LC) − ωo2 ]2 + [ωo (R/L)]2
p
(1/LC)(R/L)
=r hp i2 = 1.
[(1/LC) − (1/LC)]2 + (1/LC)(R/L)

Now, from the magnitude equation

1 ωc (R/L)
√ =p
2 [(1/LC) − ωc2 ]2 + [ωc (R/L)]2
1 1
√ =p .
Band-Pass Filters
2 [(ωc L/R) − (1/ωc RC)]2 + 1 7-35
 Band-Pass Filters
[center]
Equating the denominators
L 1
±1 = ωc − =⇒ ωc2 L ± ωc R − 1/C = 0.
R ωc RC
The solution yields
s
R 2
  
R 1
ωc1 =− + + ,
2L 2L LC
s 
R 2
 
R 1
ωc2 = + + .
2L 2L LC

Band-Pass Filters 7-36

 Band-Pass Filters
[center]
We can also use the above equations to confirm that center frequency ωo is the
geometric mean of the two cutoff frequencies.
√
ωc = ωc1 ωc2
v   r
u s  s 
2   2  
R R 1  R R 1  1
u
= t− + + + + = .
u
2L 2L LC 2L 2L LC LC

Recall that the bandwidth of a bandpass filter is the difference between the two cutoff
frequencies. Therefore
β = ωc2 − ωc1
 s    s  
2   2  
R R 1   R R 1 
= + + − − + +
2L 2L LC 2L 2L LC
R
= .
Band-Pass Filters L 7-37
 Band-Pass Filters
[center]
The last parameter, the quality factor, is defined as the ratio of the center frequency to
the bandwidth.
p r
1/LC L
Q= =
(R/L) CR2
The expressions for the cutoff frequencies can be written in terms of the center
frequency, bandwidth and quality factor:
s  s 
2
β β 2
β β 2
ωc1 = − + + ωo , ωc2 = + + ωo2 .
2 2 2 2
 s   s 
 2  2
1 1  , ωc2 = ωo .  1 + 1 + 1
ωc1 = ωo . − + 1+ .
2Q 2Q 2Q 2Q

Band-Pass Filters 7-38

 Band-Pass Filters (Summary)
[center]
The general form of the transfer function is
βs
H(s) =
s2 + βs + ωo2

Any circuit with transfer function like above

will act as a bandpass filter. The transfer
function can also be written in the form
Kβs
H(s) =
s2 + βs + ωo2

where the values for K and β depend o

wether the series resistance of the voltage
source is zero or non-zero.

Band-Pass Filters 7-39

 Band-Pass Filters (Summary)
[center]
Recall that the natural response of the series RLC circuit is characterized by the neper
frequency (α) and the resonant frequency (ωo ), given as
r
R 1
α= rad/s and ωo = .
2L LC
We see that the same parameter ωo characterises both the time response and the
frequency response. That is why the center frequency is also called the resonant
frequency. The bandwidth and neper frequency are related by
β = 2α.
The natural response of a series RLC circuit may be under-damped, overdamped, or
critically damped. The transition from an overdamped to an underdamped response
occurs when Q = 1/2. A circuit whose frequency response contains a sharp peak at
ωo , indicating a high Q and a narrow bandwidth, will have an underdamped natural
response. A circuit whose frequency response has a broad bandwidth and a low Q will
have an overdamped natural response.
Band-Pass Filters 7-40
 Example
[center]
A graphic equalizer is an audio amplifier that allows you to select different levels of
amplification within different frequency regions. Using the series RLC circuit, choose
values for R, L, and C that yield a bandpass circuit able to select inputs within the
1 − 10 kHz frequency band. Such a circuit might be used in a graphic equalizer to
select this frequency band from the larger audio band (generally 0 − 20 kHz) prior to
amplification.

Band-Pass Filters 7-41

 Example
[center]
A graphic equalizer is an audio amplifier that allows you to select different levels of
amplification within different frequency regions. Using the series RLC circuit, choose
values for R, L, and C that yield a bandpass circuit able to select inputs within the
1 − 10 kHz frequency band. Such a circuit might be used in a graphic equalizer to
select this frequency band from the larger audio band (generally 0 − 20 kHz) prior to
amplification.

We need to compute values for R, L, and C that produce a bandpass filter with cutoff
frequencies of 1 kHz and 10 kHz. Any approach we choose will provide only two
equations—insufficient to solve for the three unknowns. We need to select a value for
either R, L, or C and use the equations to calculate the remaining component values.
We choose 1µF capacitor. Next, we compute the center frequency as
p p
fo = fc1 fc2 = (1000)(10, 000) = 3162.28Hz.
Band-Pass Filters 7-41
 Example
[center]
Next, compute the value of L using the computed center frequency and the selected
value for C.
1 1
L= 2 = = 2.533mH.
ωo C [2π(3162.28)]2 (10−6 )
Next, calculate Q
fo 3162.28
Q= = = 0.3514
fc1 − fc1 10, 000 − 1000
Now, R will be calculated as
s s
L 0.0025
R= = = 143.24Ω.
CQ2 (10−6 )(0.3514)2

Band-Pass Filters 7-42

 Exercise
[center]
1. Show that the RLC circuit in figure is
also a bandpass filter by deriving an
expression for the transfer function
H(s).
2. Compute the center frequency, ωo .
3. Calculate the cutoff frequencies, ωc1
and ωc2 , the bandwidth, β, and the
quality factor, Q.
4. Compute values for R and L to yield a
bandpass filter with a center frequency
of 5 kHz and a bandwidth of 200 Hz,
using a 5 mF capacitor.

Band-Pass Filters 7-43

 Exercise
[center]
Examine the effect of assuming a nonzero
source resistance, Ri , on the characteristics
of a series RLC bandpass filter.
1. Determine the transfer function for
the circuit in figure.
2. Sketch the magnitude plot for the
circuit using the values for
R = 143.24Ω, L = 2.533 mH, and
C = 1µF and setting Ri = R. On the
same graph, sketch the magnitude
plot for the circuit where Ri = 0.

Band-Pass Filters 7-44

 Band-Reject Filters
[center]
Band-reject filters pass source voltages outside the band between the two cutoff
frequencies to the output (the passband), and attenuate source voltages before they
reach the output at frequencies between the two cutoff frequencies (the stopband).

Band-reject filters are characterized by the same parameters as band- pass filters: the
two cutoff frequencies, the center frequency, the band- width, and the quality factor.
Again, only two of these five parameters can be specified independently.

We examine two circuits that function as band-reject filters and then compute
equations that relate the circuit component values to the characteristic parameters for
each circuit.

Band-Reject Filters 7-45

 Band-Reject Filters
[center]
We again consider the series RLC circuit
but this time the output voltage is
measured across the capacitor-inductor
pair.
At ω = 0, the inductor behaves like a short
circuit and the capacitor behaves like an
open circuit, but at ω = ∞, these roles
switch. The output voltage is defined over
an effective open circuit, and thus the
output and input voltages have the same
magnitude.
Between these two passbands, both the
inductor and the capacitor have finite
impedances of opposite signs.
Band-Reject Filters 7-46
 Band-Reject Filters
[center]
Figure below presents a sketch of the frequency response of the series RLC band-reject
filter. The magnitude plot is overlaid with that of the ideal band-reject filter.

Band-Reject Filters 7-47

 Band-Reject Filters
[center]
Quantitative Analysis of the Band-Reject Filter
After transforming to the s-domain, we use
voltage division to construct the transfer
function
1 1
sL + sC s2 + LC
H(s) = 1 = 2 .
R + sL + sC s +R 1
L s + LC

The magnitude and phase relations are

1
LC − ω2
|H(jω)| = q ,
ωR 2
1

2

LC − ω2 + L
!
ωR
θ(ω) = − tan−1 1
L
.
LC − ω2
Band-Reject Filters 7-48
 Band-Reject Filters
[center]
For the bandr-eject filter, the center frequency is still defined as the frequency for
which the sum of the impedances of the capacitor and inductor is zero. In the
bandpass filter, the magnitude at the center frequency was a maximum, but in the
band-reject filter, this magnitude is a minimum. This is because in the band-reject
filter, the center frequency is not in the pass-band; rather, it is in the stop-band. The
center frequency is given by
r
1
ωo = .
LC
It can be verified from the magnitude relation that |H(jωo )| = 0.

Band-Reject Filters 7-49

 Band-Reject Filters
[center]
The expressions for the cutoff frequencies are
s 
R 2
 
R 1
ωc1 = − + + ,
2L 2L LC
s 
R 2
 
R 1
ωc2 = + + .
2L 2L LC
We can use the expressions for the cutoff frequencies to generate the expressions for
the bandwidth and quality factors:
R
β= ,
L
and
r
L
Q= .
Band-Reject Filters R2 C 7-50
 Band-Reject Filters
[center]
Again, we can write the expressions for the cutoff frequencies cin terms of the center
frequency, bandwidth and quality factor:
s  s 
2
β β 2
β β 2
ωc1 = − + + ωo , ωc2 = + + ωo2 .
2 2 2 2

 s   s 
 2  2
1 1 1 1
ωc1 = ωo . − + 1+ , ωc2 = ωo .  + 1+ .
2Q 2Q 2Q 2Q

Although not given here, a parallel RLC circuit also produces a band-reject filter.

Band-Reject Filters 7-51

 Band-Reject Filters (Summary)
[center]
The general form of the transfer function is

s2 + ωo2
H(s) =
s2 + βs + ωo2

Any circuit with transfer function like

above will act as a band-reject filter.

Band-Reject Filters 7-52

 Exercise-I
[center]
A resistor, denoted as Rl , is added in
series with the inductor in the series RL
circuit. The new low-pass filter circuit is
shown in figure below.
1. Derive the expression for H(s) where
H(s) = Vo /Vi .
2. At what frequency will the magnitude
of H(jω) be maximum?
3. What is the maximum value of the
magnitude of H(jω)?
4. At what frequency will the magnitude
of H(jω) equal
√ its maximum value
divided by 2?
Band-Reject Filters 7-53
 Exercise-II
[center]
For the bandreject filter in figure, calculate

a) vo
b) fo
c) Q
d) ωc1
e) fc1
f) ωc1
g) fc2 .

Band-Reject Filters 7-54

 Drawing Bode Diagrams using MATLAB
[center]
We have seen that the frequency response plot is a very important tool for analyzing a
circuit’s behavior. Now, we learn how to create a type of the frequency response plot
of the transfer function of a circuit, the Bode diagram. A Bode diagram consists of a
magnitude plot and a phase plot. The most efficient method for generating and
plotting the amplitude and phase response is to use a digital computer and a numerical
computation and visualizaton software to give us accurate numerical plots of |H(jω)|
and θ(jω) versus ω.

MATLAB is a powerful numerical computation and visualization software developed by

Mathworks. We can use MATLAB to generate Bode diagrams.

Drawing Bode Diagrams using MATLAB 7-55

 Drawing Bode Diagrams using MATLAB
[center]
Recall the low-pass filter example where we designed a series RL low-pass circuit with
component values R = 6.28Ω and L = 100mH to give a cutoff frequency of 10 Hz or
20π rad/s. Also, recall that the transfer function of the low-pass filter is
ωc
H(s) = .
s + ωc

Drawing Bode Diagrams using MATLAB 7-56

 Drawing Bode Diagrams using MATLAB
[center]
Use the following MATLAB code to generate the Bode diagram.
% Drawing Bode Diagram Matlab.m
% Matlab script file to create Bode diagram of a transfer funcion
% Define component values
R = 6.28;
L = 100*10ˆ(-3);

% Compute the cuttoff frequency

wc = R/L;

% Define transfer function

Hs= tf([wc], [1 wc]);

% Call Matlab function to generate Bode diagram

bode(Hs);

% Turn the grid on for Bode plot

grid on;

Drawing Bode Diagrams using MATLAB 7-57

 Drawing Bode Diagrams using MATLAB
[center]
Bode Diagram
0

-5 System: Hs
Frequency (rad/s): 62.8
Magnitude (dB): -3.01

Magnitude (dB)
-10

-15

-20

-25
0
Phase (deg)

-45

-90
10 0 10 1 10 2 10 3
Frequency (rad/s)

Drawing Bode Diagrams using MATLAB 7-58

 Drawing Bode Diagrams using MATLAB
[center]
Note that a Bode diagram uses a logarithmic scale on the frequency axis to
accommodate a wide range of frequencies of interest. Normally we plot three or four
decades of frequencies, 1 Hz to 1 kHz in our example, choosing the frequency range
where the transfer function characteristics are changing. If we plot both the magnitude
and phase angle plots, they again share the frequency axis.

Second, instead of plotting the absolute magnitude of the transfer function versus
frequency, the Bode magnitude is plotted in decibels (dB) versus the log of the
frequency. If the magnitude of the transfer function is |H(jω)| , its value in dB is
given by

AdB = 20 log10 |H(jω)|.

Drawing Bode Diagrams using MATLAB 7-59

 Drawing Bode Diagrams using MATLAB
[center]
It is important to remember that while |H(jω)| is an unsigned quantity, AdB is a
signed quantity. When AdB = 0, the transfer function magnitude is 1, since
20 log10 (1) = 0. When AdB < 0, the transfer function magnitude is between 0 and 1,
and when AdB > 0, the transfer function magnitude is greater than 1. Finally, note
that
√
20 log10 (1/ 2) = −3dB.

Recall that we define the cutoff frequency of filters by determining the frequency√at
which the maximum magnitude of the transfer function has been reduced by 1/ 2.
For the Bode diagram, this corresponds to the frequency at which the maximum
magnitude of the transfer function in dB has been reduced by 3 dB. Look at the cutoff
frequency marked on the Bode diagram.

Drawing Bode Diagrams using MATLAB 7-60

 Part-2
Active Filters

Active Low-Pass and High-Pass Filters

Scaling

Op Amp Bandpass and Band-Reject Filters

Higher-Order Op Amp Filters

Narrowband Band-Pass and Band-Reject Filters

7-61
 Active Filters
[center]

Active filters have certain advantages over passive filters.

▶ Active circuits can produce bandpass and band-reject filters without using
inductors.
▶ Active filters provide a control over amplification not available in passive filter
circuits.
▶ Active filter work well when gain, load variation, and physical size are important
parameters in the design specifications.

7-62
 Active Low-Pass and High-Pass Filters
[center]
Qualitatively analyse this circuit!

Active Low-Pass and High-Pass Filters 7-63

 Active Low-Pass and High-Pass Filters
[center]
Qualitatively analyse this circuit!
When the frequency of the source is varied,
only the impedance of the capacitor is
affected. At very low frequencies, the
capacitor acts like an open circuit, and the
op amp circuit acts like an amplifier with a
gain of −R2 /R1 . At very high frequencies,
the capacitor acts like a short circuit,
thereby connecting the output of the op
amp circuit to ground. This circuitthus
functions as a low-pass filter with a
passband gain of −R2 /R1 .

Active Low-Pass and High-Pass Filters 7-63

 Active Low-Pass and High-Pass Filters
[center]
This circuit is analogous to the inverting
amplifier circuit, so its transfer function is
−Zf /Zi , where the impedance in the input
path (Zi ) is the resistor R1 , and the
impedance in the feedback path (Zf ) is the
parallel combination of the resistor R2 and
the capacitor C. The transfer function is
1


Zf R2 ∥ sC
H(s) = − =− The transfer function has the same form as
Zi R1
ωc the passive low-pass filter, with an
= −K exception: The gain in the pass-band, K,
s + ωc
is set by the ratio R2 /R1 . The passband
where K = R2 /R1 and ωc = 1/R2 C. gain and the cutoff frequency to be
specified independently.
Active Low-Pass and High-Pass Filters 7-64
 Example
[center]
For the active low-pass filter circuit, calculate values for C and R2 that, together with
R1 = 1Ω, produce a low-pass filter having a gain of 1 in the passband and a cutoff
frequency of 1 rad/s. Construct the transfer function for this filter and use it to sketch
a Bode magnitude plot of the filter’s frequency response.

Active Low-Pass and High-Pass Filters 7-65

 Example
[center]
For the active low-pass filter circuit, calculate values for C and R2 that, together with
R1 = 1Ω, produce a low-pass filter having a gain of 1 in the passband and a cutoff
frequency of 1 rad/s. Construct the transfer function for this filter and use it to sketch
a Bode magnitude plot of the filter’s frequency response.
R2 = KR1 = (1)(1) = 1Ω.

Active Low-Pass and High-Pass Filters 7-65

 Example
[center]
For the active low-pass filter circuit, calculate values for C and R2 that, together with
R1 = 1Ω, produce a low-pass filter having a gain of 1 in the passband and a cutoff
frequency of 1 rad/s. Construct the transfer function for this filter and use it to sketch
a Bode magnitude plot of the filter’s frequency response.
R2 = KR1 = (1)(1) = 1Ω.

1 1
C= = = 1F.
R1 ωc (1)(1)

Active Low-Pass and High-Pass Filters 7-65

 Example
[center]
For the active low-pass filter circuit, calculate values for C and R2 that, together with
R1 = 1Ω, produce a low-pass filter having a gain of 1 in the passband and a cutoff
frequency of 1 rad/s. Construct the transfer function for this filter and use it to sketch
a Bode magnitude plot of the filter’s frequency response.
R2 = KR1 = (1)(1) = 1Ω.

1 1
C= = = 1F.
R1 ωc (1)(1)

ωc 1
H(s) = −K =−
s + ωc s+1

Active Low-Pass and High-Pass Filters 7-65

 Example
[center]
We can use MATLAB create the Bode magnitude plot.
%Example15.1: Active Low-Pass Filter
R1=1; %1 Ohm
K=1;
wc = 1; % 1 rad/s

% Design R2 and C
R2=K*R1;
C = 1/(R2*wc);

% Plot Bode diagram

w = logspace(-1,1); % Generate frequencies from 10ˆ{-1} to 10ˆ{1} rad/s
[mag,phase] = bode(H,w); % Compute magnitude and phase response
semilogx(w,20*log10(mag(:)),'LineWidth',1.5); % Plot magnitude only in dBs

grid on;

Active Low-Pass and High-Pass Filters 7-66

 Example
[center]
% Set plot properties
ax=gca;
ax.YLim = [-20 10];
ax.GridColor = [0,0,0];
ax.MinorGridColor = [0,0,0];
ax.GridLineStyle = '-';
ax.MinorGridLineStyle = '-';
ax.GridAlpha = 1;
ax.MinorGridAlpha = 1;
ax.FontSize = 12;

xlabel('$\omega$ rad/s', 'interpreter', 'latex');

ylabel('$\vert H(j \omega) $ dB', 'interpreter', 'latex');

Active Low-Pass and High-Pass Filters 7-67

 Example
[center]
10

-5

-10

-15

-20
10 -1 10 0 10 1

Active Low-Pass and High-Pass Filters 7-68

 Active Low-Pass and High-Pass Filters
[center]
The circuit shown is a first-order high-pass
filter. The transfer function is
Zf R2
H(s) = − =− 1
Zi R1 + sC
s
= −K
s + ωc
where K = R2 /R1 and ωc = 1/R1 C.
Again, the transfer function has the same
form as the passive high-pass and the gain
in the pass-band, K, is set by the ratio
R2 /R1 . The passband gain and the cutoff
frequency can be specified independently.

Active Low-Pass and High-Pass Filters 7-69

 Exercise
[center]

30
X 10000
X 499.7 Y 19.99
Figure shows the Bode magnitude plot of a 20 Y 16.99

high-pass filter. For the active high-pass

10
filter circuit, calculate values of R1 and R2
that produce the desired magnitude 0

response. Use a 0.1µF capacitor. If a -10

10 kΩ load resistor is added to this filter,
how will the magnitude response change? -20

-30

-40
10 0 10 1 10 2 10 3 10 4

Active Low-Pass and High-Pass Filters 7-70

 Exercise
[center]
Note that the gain in the passband is 20 dB; therefore, K = 10. Also note that the 3
dB point is 500 rad/s. Therefore, the transfer function is
s
H(s) = −10
s + 500

R2 1
= K = 10 and 500 =
R1 R1 C
Since C = 0.1µF, we can find

R1 = 20kΩ, R2 = 200kΩ.

There will be no effect of load resistance because we assume an ideal op amp.

Active Low-Pass and High-Pass Filters 7-70
 Scaling
[center]
In the design and analysis of both passive and active filter circuits, working with
element values such as 1Ω, 1H, and 1F is convenient. Although these values are
unrealistic for specifying practical components, they greatly simplify computations.
After making computations using convenient values of R, L, and C, the designer can
transform the convenient values into realistic values using the process known as
scaling.

Magnitude Scaling
We scale a circuit in magnitude by multiplying the impedance at a given frequency by
the scale factor km . If we let unprimed variables represent the initial values of the
parameters, and we let primed variables represent the scaled values of the variables,
then
′ ′ ′
R = km R, L = km L, C = C/km .
Scaling 7-71
 Scaling
[center]
Frequency Scaling
In frequency scaling, we change the circuit parameters so that at the new frequency,
the impedance of each element is the same as it was at the original frequency. If we let
kf denote the frequency scale factor, both inductors and capacitors are multiplied by
1/kf .
′ ′ ′
R = R, L = L/kf , C = C/kf .
A circuit can be scaled simultaneously in both magnitude and frequency.
′
R = km R,
′ km
L = L,
kf
′ 1
C = C.
km kf
Scaling 7-72
 Exercise
[center]
The series RLCp circuit shown has a center
frequency of 1/LC = 1 rad/s, a
bandwidth of R/L = 1 rad/s, and thus a
quality factor of 1. Use scaling to compute
new values of R and L that yield a circuit
with the same quality factor but with a
center frequency of 500 Hz. Use a 2µF
capacitor.

Scaling 7-73
 Exercise
[center]

′
ωo 2π(500)
kf = = = 3141.59
ωo 1

1 C 1
km = ′ = = 159.155
kf C (3141.59)(2 × 10−6 )
Therefore
′
R = km R = 159.155Ω
′ km
L = L = 50.66mH.
kf

Scaling 7-73
 Op Amp Bandpass and Band-Reject Filters
[center]
This approach is motivated by the Bode
magnitude plot shown in figure. The
bandpass filter consists of three separate
components:
1. A unity-gain low-pass filter whose
cutoff frequency is ωc2 , the larger of
the two cutoff frequencies;
2. A unity-gain high-pass filter whose
cutoff frequency is ωc1 , the smaller of
the two cutoff frequencies; and
3. A gain component to provide the
desired level of gain in the passband.

Op Amp Bandpass and Band-Reject Filters 7-74

 Op Amp Bandpass and Band-Reject Filters
[center]
This method assumes that ωc1 is smaller than ωc2 . The resulting filter is called
ω
broadband bandpass filter. A filter is called broadband, if ωcc2 ≫ 2.
1

We can construct a circuit to give a broadband bandpass filter as shown.

Op Amp Bandpass and Band-Reject Filters 7-75

 Op Amp Bandpass and Band-Reject Filters
[center]
The transfer function of the cascaded bandpass filer is
   
Vo −ωc2 −s −Rf
H(s) = =
Vi s + ωc2 s + ωc1 Ri
This is not the standard form of the transfer function of a bandpass filter as we saw for
passive filters
βs
HBP = 2
s + βs + ωo2
If ωc2 ≫ ωc1 , then (ωc1 + ωc2 ) ≈ ωc2 . Then
−Kωc2 s
H(s) =
s2 + ωc2 s + ωc1 ωc2

Op Amp Bandpass and Band-Reject Filters 7-76

 Op Amp Bandpass and Band-Reject Filters
[center]
The components values can be computed as
1
ωc 2 = .
RL CL

1
ωc1 = .
R H CH
Note that
−Kωc2 (jωo )
|H(jωo )| =
(jωo )2 + ωc2 (jωo ) + ωc1 ωc2
Kωc2
= =K
ωc1
Therefore
Rf
|H(jωo )| = .
Op Amp Bandpass and Band-Reject Filters
Ri 7-77
 Example
[center]
Design a bandpass filter for a graphic equalizer to provide an amplification of 2 within
the band of frequencies between 100 and 10, 000 Hz. Use 0.2 mF capacitors.

Since ωc2 = 100ωc1 , we can say that ωc2 ≫ ωc1 .

1 1
ωc2 = = 2π(10000) =⇒ RL = ≈ 80Ω.
R L CL [2π(10000)](0.2 × 10−6 )

1 1
ωc1 = = 2π(100) =⇒ RH = ≈ 7958Ω.
RH CH [2π(100)](0.2 × 10−6 )
Finally, for the gain stage, let Ri = 1kΩ, then
Rf = 2(1000) = 2kΩ.

Op Amp Bandpass and Band-Reject Filters 7-78

 Example
[center]
The resulting circuit is

Op Amp Bandpass and Band-Reject Filters 7-79

 Op Amp Bandpass and Band-Reject Filters
[center]
Now we show that a band-reject filter can
be constructed by using a parallel
connection of low-pass and high-pass filters
and a summing amplifier. The transfer
function is
  
−Rf −ωc1 −s
H(s) = +
Ri s + ωc1 s + ωc2
 2 
Rf s + 2ωc1 s + ωc1 ωc2
=
Ri (s + ωc1 )(s + ωc2 )

1 1
Again ωc1 = , ωc2 = ,
RL CL R H CH
Rf
K= .
Op Amp Bandpass and Band-Reject Filters Ri 7-80
 Exercise
[center]

Design a circuit based on the parallel

band-reject op amp filter. The Bode
magnitude response of this filter is shown.
Use 0.5µF capacitors in your design.

Op Amp Bandpass and Band-Reject Filters 7-81

 Higher-Order Op Amp Filters
[center]
Remember that an ideal filter has a
discontinuity at the point of cutoff, which
sharply divides the passband and the
stop-band. Although we cannot hope to
construct a circuit with a discontinuous
frequency response, we can construct
circuits with a sharper, yet still continuous,
transition at the cutoff frequency.
An approach is suggested by the Bode
magnitude plot of cascaded identical
low-pass prototype filters. An n-element
cascade of identical low-pass filters will
transition from the passband to the
stop-band with a slope of 20n dB/dec.
Higher-Order Op Amp Filters 7-82
 Higher-Order Op Amp Filters
[center]
The block diagram and the circuit diagram
for such a cascade are shown. The transfer
function is
   
−1 −1 −1
H(s) =
s+1 s+1 s+1
(−1) n
= .
(s + 1)n

The order of the filter is determined by the

number of poles in the transfer function.
An issue with this approach is, as the order
of the low-pass filter is increased by adding
prototype low-pass filters to the cascade,
the cutoff frequency also changes.
Higher-Order Op Amp Filters 7-83
 Higher-Order Op Amp Filters
[center]
We can solve this issue if we are able to calculate the cutoff frequency of the higher
order filters formed in the cascade of first-order filters, we can use frequency scaling to
calculate component values that move the cutoff frequency to its specified location. If
we start with a cascade of n prototype√low-pass filters, we can compute the cutoff
frequency that results in |H(jω)| = 1/ 2.
(−1)n
H(s) =
(s + 1)n

(−1)n 1
|H(jωcn )| = n
=√
(jωcn + 1) 2

1 2/n
 
1 1 1
p = √ =⇒ 2 = √
2 + 1)n
(wcn 2 wcn + 1 2
√ √
q
n 2 n
2 = wcn + 1 =⇒ ωcn = 2 − 1.
Higher-Order Op Amp Filters 7-84
 Higher-Order Op Amp Filters
[center]
The cutoff frequency of a fourth-order unity-gain low-pass filter constructed from a
cascade of four prototype low-pass filters:
√
q
4
ωc4 = 2 − 1 = 0.435rad/s.
Thus, we can design a fourth-order low-pass filter with any arbitrary cutoff frequency
by starting with a fourth-order cascade consisting of prototype low-pass filters and then
scaling the components by kf = ωc /0.435 to place the cutoff frequency at any value of
ωc desired.

By cascading identical low-pass filters, we can increase the asymptotic slope in the
transition and control the location of the cutoff frequency, but the gain of the filter is
not constant between zero and the cutoff frequency ωc . Remember that in an ideal
low-pass filter, the passband magnitude is 1 for all frequencies below the cutoff
frequency.
Higher-Order Op Amp Filters 7-85
 Higher-Order Op Amp Filters
[center]
Butterworth Filters

A unity-gain Butterworth low-pass filter has a transfer function whose magnitude is

given by
1
|H(jω)| = p
1 + (ω/ωc )2n
where n is an integer that denotes the order of the filter. Note that

1. The cutoff frequency is ωc rad/s for all values of n.

2. If n is large enough, the denominator is always close to unity when ω < ωc .
3. In the expression for |H(jω)| , the exponent of ω/ωc is always even.

Given an equation for the magnitude of the transfer function, how do we find H(s)?
Higher-Order Op Amp Filters 7-86
 Higher-Order Op Amp Filters
[center]
Butterworth Filters
We set ωc equal to 1 rad/s to find H(s). We wil thenl use scaling to transform the
prototype filter to a filter that meets the given filtering specifications. Note that if N is
a complex quantity, then |N |2 = N N ∗ , where N ∗ is the conjugate of N . It follows that

|H(jω)|2 = H(jω)H(−jω) = H(s)H(−s) ∵ s = jω.

Now observe that s2 = −ω 2 .Thus

1 1 1 1
|H(jω)|2 = 2n
= 2 n
= 2 n
=
1+ω 1 + (ω ) 1 + (−s ) 1 + (−1)n s2n

Higher-Order Op Amp Filters 7-87

 Higher-Order Op Amp Filters
[center]
Butterworth Filters
The procedure for finding H(s) for a given value of n is as follows:

1. Find the roots of the polynomial

1
= 0.
1 + (−1)n s2n

2. Assign the left-half plane roots to H(s) and the right-half plane roots to H(−s).
3. Combine terms in the denominator of H(s) to form first- and second-order factors.

Higher-Order Op Amp Filters 7-88

 Example
[center]
Find the Butterworth transfer function for
n = 2.
For n = 2, find the roots of the polynomial
Roots s2 and s3 are in the left-half plane.
1 + (−1)2 s4 = 0
Thus,
Rearranging, we find 1
H(s) = √ √ √ √
(s + 1/ 2 − j/ 2)(s + 1/ 2 + j/ 2)
s4 = −1 = 1 180◦
1
= √ .
2
s + 2s + 1
The four roots are

s1 = 1 45◦ , s2 = 1 135◦ ,
s3 = 1 225◦ , s4 = 1 315◦ .

Higher-Order Op Amp Filters 7-89

 Higher-Order Op Amp Filters
[center]
Normalized ( ωc = 1 rad/s) Butterworth Polynomials up to the Eighth Order are

n nth-Order Butterworth Polynomial

1. (s + 1)
2
√
2. (s + 2s + 1)
√
3. (s + 1)(s2 + 2s + 1)
4. (s2 + 0.765s + 1)(s2 + 1.848s + 1)
5. (s + 1)(s2 + 0.618s + 1)(s2 + 1.618s + 1)
√
6. (s2 + 0.518s + 1)(s2 + 2s + 1)(s2 + 1.932s + 1)
7. (s + 1)(s2 + 0.445s + 1)(s2 + 1.247s + 1)(s2 + 1.802s + 1)
8. (s2 + 0.390s + 1)(s2 + 1.111s + 1)(s2 + 1.666s + 1)(s2 + 1.962s + 1)

Higher-Order Op Amp Filters 7-90

 Higher-Order Op Amp Filters
[center]
Butterworth Filter Circuits
Butterworth polynomials are the product of first- and second-order factors. We can
construct a circuit whose transfer function has a Butterworth polynomial in its
denominator by cascading op amp circuits. A block diagram of such a cascade is
shown, using a fifth-order Butterworth polynomial as an example.

All odd-order Butterworth polynomials include the factor (s + 1), so all odd-order
Butterworth filter circuits must have a subcircuit that provides the transfer function
H(s) = 1/(s + 1). This is the transfer function of the prototype low-pass op amp
filter. So what remains is to find a circuit that provides a transfer function of the form
H(s) = 1/(s2 + b1 s + 1).

Higher-Order Op Amp Filters 7-91

 Higher-Order Op Amp Filters
[center]
Butterworth Filter Circuits
A circuit to implement second-order factor
Va − Vi Va − Vo
+ (Va − Vo )sC1 + =0
R R
Vo − Va
Vo sC2 + = 0.
R
Simplifying

(2 + RC1 s)Va − (1 + RC1 s)Vo = Vi ,

−Va + (1 + RC2 s)Vo = 0. 1
Vo R2 C1 C2
H(s) = = 2 2 1 .
Vi s + RC1 s + R2 C1 C2
Solving using Crammer’r rule, we get

Higher-Order Op Amp Filters 7-92

 Higher-Order Op Amp Filters
[center]
Set R = 1Ω
1
Vo C1 C2
H(s) = = 2 2 1
Vi s + C1 s + C1 C2

To obtain transfer function

1
H(s) = ,
s2 + b1 s + 1
choose capacitor values so that
2 1
b1 = , 1= .
C1 C1 C2

Higher-Order Op Amp Filters 7-93

 Example
[center]
Design a fourth-order Butterworth low-pass filter with a cutoff frequency of 500 Hz
and a passband gain of 10. Use as many 1kΩ resistors as possible. Compare the Bode
magnitude plot for this Butterworth filter with that of the identical cascade filter.

The fourth-order Butterworth polynomial is

(s2 + 0.765s + 1)(s2 + 1.848s + 1).
Component values to implement the polynomial (s2 + 0.765s + 1) are
2 2
C1 = = = 2.61F,
b1 0.765
1
C2 = = 0.38F.
C1
Similarly, capacitor values for second cascade are found to be C3 = 1.08 F and
C4 = 0.924 F.
Higher-Order Op Amp Filters 7-94
 Example
[center]
These component values given fourth-order Butterworth filter with a cutoff frequency
of 1 rad/s. A frequency scale factor of kf = 3141.6 will move the cutoff frequency to
500 Hz. A magnitude scale factor of km = 1000 will permit the use of 1kΩ resistors in
place of 1Ω resistors.
R = 1kΩ, C1 = 831nF, C2 = 121nF, C3 = 344nF, and C4 = 294nF.
For the inverting stage, let R1 = 1kΩ; then
Rf = 10R1 = 10kΩ.
The resulting circuit is

Higher-Order Op Amp Filters 7-95

 Example
[center]

The Bode diagrams of the Buttherworth

and identical cascade filters are

Higher-Order Op Amp Filters 7-96

 Higher-Order Op Amp Filters
[center]
The Order of a Butterworth Filter
What is the smallest value of n that will
meet the filtering specifications?
In the design of a low-pass filter, the
filtering specifications are usually given in
terms of the abruptness of the transition
region, specified as Ap , ωp , As , and ωs .
The order can then be
1
Ap = 20 log10 q = −10 log10 (1 + ωp2n ),
2n
1 + ωp
1
As = 20 log10 p = −10 log10 (1 + ωs2n ).
1 + ωs2n

Higher-Order Op Amp Filters 7-97

 Higher-Order Op Amp Filters
[center]
The Order of a Butterworth Filter log10 (σs /σp )
or n= .
Taking anti-logarithm log10 (ωs /ωp )

10−0.1Ap = (1 + ωp2n ), If ωp is the√cutoff frequency, the Ap equals

−20 log10 2, and σp = 1. Therefore
10−0.1As = (1 + ωs2n ).
log10 (σs )
Solving for ωp and ωs , and dividing n= .
log10 (ωs /ωp )
 n √
ωs 10−0.1As − 1 σs
=√ = For steep transition region, 10−0.1As ≫ 1,
ωp 10−0.1Ap − 1 σp
thus σs ≈ 10−0.05As and
log10 σs ≈ −0.05As .
n log10 (ωs /ωp ) = log10 (σs /σp )
−0.05As
n= . (nearest integer)
log10 (ωs /ωp )
Higher-Order Op Amp Filters 7-98
 Higher-Order Op Amp Filters
[center]
Butterworth High-Pass, Band-Pass, and Band-Reject Filters
An nth-order Butterworth high-pass filter has a transfer function with the nth-order
Butterworth polynomial in the denominator, just like the nth-order Butterworth
low-pass filter. But in the high-pass filter, the numerator of the transfer function is sn .
Again, we can use the cascade approach. The first-order factor is achieved by including
a prototype high-pass filter, with R1 = R2 = 1Ω, and C = 1 F) in the cascade.

To implement second-order factor with

transfer function
s2
H(s) = ,
s2 + b1 s + 1
we can use circuit as shown.

Higher-Order Op Amp Filters 7-99

 Higher-Order Op Amp Filters
[center]
Butterworth High-Pass, Band-Pass, and Band-Reject Filters

Vo s2
H(s) = = 2 2 1
Vi s + R2 C s + R1 R2 C 2

Setting C = 1 yields

Vo s2
H(s) = = 2 2 1
Vi s + R2 s + R1 R2

Then, find R1 and R2 using

2 1
b1 = , 1= .
R2 R1 R2

Higher-Order Op Amp Filters 7-100

 Higher-Order Op Amp Filters
[center]
Butterworth High-Pass, Band-Pass, and Band-Reject Filters
Note that the high-pass circuit was obtained from the low-pass circuitby interchanging
resistors and capacitors. Second, the prototype transfer function of a high-pass filter
can be obtained from that of a low-pass filter by replacing s in the low-pass expression
with 1/s.

Use frequency and magnitude scaling to design a Butterworth high-pass filter with
practical component values and a cutoff frequency other than 1 rad/s. Adding an
inverting amplifier to the cascade will accommodate designs with nonunity passband
gains.

Combine low-pass and high-pass filters in cascade to produce nth-order Butterworth

bandpass filters, and in parallel with a summing amplifier to produce nth-order
Butterworth band-reject filters.

Higher-Order Op Amp Filters 7-101

 Narrowband Band-Pass and Band-Reject Filters
[center]
The cascade and parallel component designs for synthesizing bandpass and bandreject
filters from simpler low-pass and high-pass filters result in only broadband, or low-Q,
filters.
  
−ωc −s sωc 0.5βs
H(s) = = 2 2
= 2
s + ωc s + ωc s + 2ωc s + ωc s + βs + ωc2
From the rightmost expression, we can determine
β = 2ωc , and ωo2 = ωc2 .

ωo ωc 1
Now Q= = = .
β 2ωc 2
With discrete real poles, the highest quality bandpass filter (or band-reject filter) we
can achieve has Q = 1/2.
Narrowband Band-Pass and Band-Reject Filters 7-102
 Narrowband Band-Pass and Band-Reject Filters
[center]
A circuit to produce high-Q bandpass filter
is shown. Summing currents at inverting
terminal
Va Vo
=−
1/sC R3

Solving for Va gives

Vo
Va = −
sR3 C
At node a
Vi − Va Va − Vo Va Va
= + +
R1 1/sC 1/sC R2

Narrowband Band-Pass and Band-Reject Filters 7-103

 Narrowband Band-Pass and Band-Reject Filters
[center] Equating the terms, we can find values of
Solving for Vi components.
Vi = (1 + 2sR1 C + R1 /R2 )Va − sR1 CVo 2
β= ,
R3 C
The transfer function is then 1
−s
Kβ = ,
Vo R1 C
R1 C
H(s) = = 2 2 1 1
Vi s + R3 C s + ReqR ωo2 = .
3C
2
Req R3 C 2
where Req = R1 ∥R2 . The standard for of For a prototype version with ωo = 1 rad/s
the transfer function for bandpass filter is and C = 1F, we have
−Kβs
H(s) = R1 = Q/K, R2 = Q/(2Q2 − K),
s2 + βs + ωo2
R3 = 2Q.
Narrowband Band-Pass and Band-Reject Filters 7-104
 Example
[center]
Design a bandpass filter, which has a center frequency of 3000 Hz, a quality factor of
10, and a passband gain of 2. Use 0.01µF capacitors in your design. Compute the
transfer function of your circuit, and sketch a Bode plot of its magnitude response.

Narrowband Band-Pass and Band-Reject Filters 7-105

 Example
[center]
Design a bandpass filter, which has a center frequency of 3000 Hz, a quality factor of
10, and a passband gain of 2. Use 0.01µF capacitors in your design. Compute the
transfer function of your circuit, and sketch a Bode plot of its magnitude response.

Since Q = 10 and K = 2, the values for R1 , R2 , and R3 in the prototype circuit are

R1 = Q/K = 5, R2 = Q/(2Q2 − K) = 10(200 − 2) = 10/198, R3 = 2Q = 20.

The scaling factors are kf = 6000π. and km = 108 /kf . After scaling

R1 = 26.5kΩ, R2 = 268.0kΩ, R3 = 106.1kΩ.

Narrowband Band-Pass and Band-Reject Filters 7-105

 Example
[center] The Bode diagram is
The circuit is shown.

The transfer function is

−3770s
H(s) =
s2 + 1885.0s + 355 × 106
Narrowband Band-Pass and Band-Reject Filters 7-106
 Narrowband Band-Pass and Band-Reject Filters
[center]
The circuit shown in figure is an active
high-Q band-reject filter known as the
twin-T notch filter because of the two
T-shaped parts of the circuit at the nodes
labeled a and b.
Summing the currents away from node a:

2(Va − σVo )
(Va − Vi )sC + (Va − Vo )sC + =0
R

or Va [2sCR + 2] − Vo [sCR + 2σ] = sCRVi .

Narrowband Band-Pass and Band-Reject Filters 7-107

 Narrowband Band-Pass and Band-Reject Filters
[center]
Summing the currents away from node b: Use Cramer’s rule to solve for Vo
Vb − Vi Vb − Vo 2(RCs + 1) 0 sCRVi
+ + (Vb − σVo )2sC = 0
R R 0 2(RCs + 1) Vi
−RCs −1 0
Vo =
or Vb [2 + 2RCs] − Vo [1 + 2σRCs] = Vi . 2(RCs + 1) 0 −(RCs + 2σ)
0 2(RCs + 1) −(2σRCs + 1)
Summing the currents away from the −RCs −1 RCs + 1
non-inverting input terminal of the top op (R2 C 2 s2 + 1)Vi
amp gives =
R 2 C 2 s2
+ 4RC(1 − σ)s + 1
Vo − Vb
(Vo − Va )sC + =0 The transfer function is
R
s2 + R21C 2


Vo
H(s) = =
or − sRCVa − Vb + (sRC + 1)Vo = 0. Vi [s2 + 4(1−σ) 1
RC s + R2 C 2 ]
Narrowband Band-Pass and Band-Reject Filters 7-108
 Narrowband Band-Pass and Band-Reject Filters
[center] We have three parameters (R, C, and σ)
The standard form of the transfer function and two design constraints (ωo and β).
of a band-reject filter: Thus one parameter is chosen arbitrarily; it
is usually the capacitor value because this
s2 + ωo2 value typically provides the fewest
H(s) =
s2 + βs + ωo2 commercially available options. Once C is
Equating the two, gives chosen,

1 1
ωo2 = , R= ,
R C2
2 ωo C
4(1 − σ) β 1
β= . and σ = 1 − =1− .
RC 4ωo 4Q

Narrowband Band-Pass and Band-Reject Filters 7-109

 Exercise
[center]
Design a high-Q active band-reject filter, based on the circuit discussed, with a center
frequency of 5000 rad/s and a bandwidth of 1000 rad/s. Use 1µF capacitors in your
design.

Narrowband Band-Pass and Band-Reject Filters 7-110

 Exercise
[center]
Design a high-Q active band-reject filter, based on the circuit discussed, with a center
frequency of 5000 rad/s and a bandwidth of 1000 rad/s. Use 1µF capacitors in your
design.

In the bandreject prototype filter, ωo = 1 rad/s, R = 1Ω, and C = 1 F. As ωo and Q

are given, C can be chosen arbitrarily, and R and σ can be found. From the
specifications, Q = 5. We see that

R = 200Ω, σ = 0.95.

Therefore we need resistors with the values 200Ω (R), 100Ω (R/2), 190Ω (σ R), and
10Ω [(1 − σ) R].

Narrowband Band-Pass and Band-Reject Filters 7-110