The document discusses first order active RC filter sections based on inverting operational amplifier configurations. It describes how a bilinear transfer function can be realized using such a configuration, with the transfer function equal to the negative ratio of feedback and input impedances Z2 and Z1. Z1 and Z2 are formed using series RC networks, allowing the transfer function to take the standard bilinear form with poles and zeros defined by the RC component values.

1. Add Symbol
Analog Electronic Circuits
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0402xxx: Analog Electronic Circuits
Chapter 3: continued
Active RC Filters Design Techniques

2. 5. First Order Active RC Sections
▪ First order active RC section has transfer function in the s-plane of the following form:
T(S) is the transfer function of the first order filter (Bilinear Transfer Function) where N(S) and
D(S) are linear function with real constants (K, z1=ωz1, and p1=ωp1).
▪ Depending on the pole and the zero locations (p1,z1) of T(S), the magnitude
frequency response (|T(jω)|) can be a low pass response(|z1|>|p1| both located at
LHP(special case z1 at ∞ ideal integrator)), a high pass response (|z1|<|p1| both
located at LHP(special case z1 at → ideal differentiator)), or an all pass response
(|z1|=|p1| and z1>p1 → zero at RHP).
𝑇 𝑆 =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
=
𝑁(𝑆)
𝐷(𝑆)
= 𝐾
𝑆 + 𝑧1
𝑆 + 𝑝1
= 𝐾
𝑧1
𝑝1
∙
1 + Τ𝑆 𝑧1
1 + Τ𝑆 𝑝1
Analog Electronic Circuits
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Add your footnotes

3. Notes:
• In general, minimum phase transfer function has all zeros located at the LHP.
• Non-minimum phase transfer function has zeros in the RHP. If all zeros of transfer
function are all reflected about the jω-axis, there is no change in the magnitude transfer
function and only difference is in the phase-shift characteristics (All pass network).
• If the phase responses of two different transfer functions (two different systems) are
compared, the net phase shift over the frequency range from zero to infinity is less for the
transfer function with all its zeros in the LHP (minimum phase transfer function).
• The range of phase shift of a minimum phase curve is least possible or minimum
corresponding to a given magnitude curve, while the range of the non-minimum phase
curve is the largest possible for the given magnitude curve.
• All pass network is an example of non-minimum phase network which passes all
frequencies with equal gain while provides a phase (angle) delay for all frequencies.
• The phase lead network is a network that provides a positive phase angle over the
frequency range of interest, yielding a system to have adequate phase margin.
• The phase lag network is a network that provides a negative phase angle and a
significant attenuation over the frequency range of interest.
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4. What is the transfer function?
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▪ The transfer function (network function) of a linear, stationary (time-invariant
constant parameter) system is defined as the ratio of the Laplace transform of the
output variable signal to the Laplace transform of the input variable signal, with all
initial condition assumed to be zero. It gives input-output description of the
behaviour of a system without including any information about the internal structure
of the system and its behaviour. It is analytical tool for finding the frequency
response of the circuit.
▪ Transfer function representation ▪ Magnitude and phase transfer functions
𝑇 𝑗𝜔 = 𝑇(𝑗𝜔) 𝑒 𝑗𝜃 𝜔
𝑇 𝑗𝜔 =
ሽ𝑅𝑒 𝑁 𝑗𝜔 + 𝑗𝐼𝑚{𝑁(𝑗𝜔)
ሽ𝑅𝑒 𝐷 𝑗𝜔 + 𝑗𝐼𝑚{𝐷(𝑗𝜔) 𝑇 𝑗𝜔 =
𝑅𝑒 𝑁 𝑗𝜔 2 + ሽ𝐼𝑚{𝑁(𝑗𝜔) 2
𝑅𝑒 𝐷 𝑗𝜔 2 + ሽ𝐼𝑚{𝐷(𝑗𝜔) 2
𝜃 𝜔 = tan−1
𝐼𝑚 𝑁 𝑗𝜔
𝑅𝑒 𝑁 𝑗𝜔
− tan−1
𝐼𝑚 𝐷 𝑗𝜔
𝑅𝑒 𝐷 𝑗𝜔

5. Bilinear Transfer Function
𝑇 𝑗𝜔 = 𝐾
𝑗𝜔 + 𝑧1
𝑗𝜔 + 𝑝1
= 𝐾
𝑧1
𝑝1
∙
1 + Τ𝑗𝜔 𝑧1
1 + Τ𝑗𝜔 𝑝1
𝑇 𝑗𝜔 = 𝐾
𝑧1
𝑝1
∙
1 +
𝜔
𝑧1
2
1 +
𝜔
𝑝1
2
𝜃 𝜔 = (0°
𝑓𝑜𝑟 𝐾 > 0 𝑜𝑟 180°
𝑓𝑜𝑟 𝐾 < 0) + tan−1
𝜔
𝑧1
− tan−1
𝜔
𝑝1
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6. Bode Plots for Bilinear Transfer Function
It is a simple technique exists for obtaining an approximate plot of the magnitude and
phase of the transfer function given its poles and zeros. The technique is useful
particularly in the case of real poles and zeros. The method was developed by H. Bode,
and the resulting diagram are called Bode plots. The plots of the magnitude transfer
function |T(jω)| (in dB) and the phase transfer function θ(ω) (in degree) versus ω in
logarithmic representation is based on the asymptotic behaviour of |T(jω)| and θ(ω).
In case bilinear transfer function, define A(jω)=20log(|T(jω)|) and θ(ω)
𝐴 𝑗𝜔 = 20 log 𝐾
𝑧1
𝑝1
∙
1 +
𝜔
𝑧1
2
1 +
𝜔
𝑝1
2
𝜃 𝜔 = (0° 𝑜𝑟 180°) + tan−1
𝜔
𝑧1
− tan−1
𝜔
𝑝1
&
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7. 𝐴 𝜔 = 𝐴1 𝜔 + 𝐴2 𝜔 + 𝐴3 𝜔
𝐴 𝜔 = 20 log 𝐾
𝑧1
𝑝1
+ 20 log 1 +
𝜔
𝑧1
2
− 20 log 1 +
𝜔
𝑝1
2
A1(ω) is a constant term, A2(ω) provides a single zero at z1, A3(ω) provides a single pole at p1
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8. A2(ω) provides a single zero at z1A3(ω) provides a single pole at p1
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A2(ω) [dB]
ω<<z1 ω=z1 ω=10z1 ω>>z1
0 3 20 20log(ω/z1)
A3(ω) [dB]
ω<<p1 ω=p1 ω=10p1 ω>>p1
0 -3 -20 -20log(ω/p1)
The overall plot of A(ω) is shown for two cases: z1<p1 and
z1>p1, respectively and for simplicity: K= p1/z1. The break
frequencies occur at z1 and p1. Note that when |z1|=|p1| and
z1>p1, A(ω)=A1(ω) (all pass filter).

9. P1(ω) is a straight line (0ᵒ (case I) or 180ᵒ (case II)
P2(ω) provides a single zero at z1 (provides 45ᵒ/decade)
P3(ω) provides a single pole at p1 (provides -45ᵒ/decade)
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𝜃 𝜔 = (0°
𝑜𝑟 180°
) + tan−1
𝜔
𝑧1
− tan−1
𝜔
𝑝1
𝑃 𝜔 = 𝑃1 𝜔 + 𝑃2 𝜔 + 𝑃3 𝜔
P2(ω)
Case I | Case II
ω=0.1z1 ω=z1 ω=10z1
0ᵒ -180ᵒ 45ᵒ -135ᵒ 90ᵒ -90ᵒ
P3(ω)
ω=0.1p1 ω=p1 ω=10p1
0ᵒ -45ᵒ - 90ᵒ
P2(ω) Case I
P3(ω)
P2(ω) Case II

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Case a. The overall θ(ω) for |z1|<|p1| under case I, and for |z1|<|p1| under case II (both z1 and
p1 located at LHP and provide minimum phase transfer function). For case I, it is obvious that
the phase shift ranges over less than 80ᵒ). The network representing this feature is called
phase-lead network. It will exhibits like a differentiator if |z1|<<|p1|.
P(ω) Case I P(ω) Case II

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Case b. In the case |z1|>|p1| under K>0 (case I), and for |z1|>|p1| under K<0 (case II) both z1
and p1 located at LHP, the network representing this feature is called phase-lag network. It will
exhibits like a integrator if |z1|>>|p1|.
P(ω) Case I P(ω) Case II

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Case c. In the case |z1|<|p1| and z1 located at RHP, the complete θ(ω) will be
P(ω) Case I P(ω) Case II

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Case d. When |z1|=|p1| and z1 located at RHP (all pass filter), it provides non-minimum phase
transfer function. This is due to z1 at RHP. The phase shift ranges over 180ᵒ.
P(ω) Case I P(ω) Case II

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Notes:
• In case a. and case c., p1 has same location but z1 location is changed with same
magnitude. The only difference is in the phase-shift characteristics. If the phase
characteristics of the two system functions are compared, it can be readily shown that
the net phase shift over the frequency range from zero to infinity is less for the system
with all its zero in the LHP (the phase shift ranges over less than 80ᵒ). Hence, the
transfer function with all its zeros in the LHP is called minimum phase transfer function.
Case a. P(ω) Case I Case c. P(ω) Case II
Minimum
Phase Transfer
Function
Non-minimum
Phase
Transfer
Function

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Notes:
• The range of phase shift of a minimum phase transfer function is the least possible or
minimum corresponding to a given amplitude curve, whereas the range of the non-
minimum phase curve is the greatest possible for the given amplitude curve, as shown
below.
The zero is reflected about jω-axis and there is no
change in the magnitude transfer function. The only
difference is in the phase-shift characteristics.

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Exercises:
1. Obtain the Bode plot for the transfer function 𝑇 𝑆 = 6
𝑆+0.5
𝑆+3
2. Obtain the Bode plot for the 𝑇 𝑆 = 𝐾 𝑑 𝑆 (ideal differentiator)
3. Obtain the Bode plot for the 𝑇 𝑆 =
𝐾 𝑖
𝑆
(ideal integrator)

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Example:
Given the below Bode plot (magnitude
frequency response), find the magnitude
transfer function.
Solution:
It is obvious that the plot represents high order
magnitude transfer function. The three break
frequencies (poles and zeros) are
a. Double zero at the origin
b. Double pole at ωp1 = ω2
c. A single pole at ωp2 = ω3
Therefore, A1(ω) is given by
𝑇 𝑗𝜔 =
𝑗
𝜔
𝜔1
2
1 + 𝑗
𝜔
𝜔2
2
1 + 𝑗
𝜔
𝜔3
𝑇 𝑆 = 𝜔3
𝜔2
𝜔1
2
𝑆2
𝑆 + 𝜔2
2 𝑆 + 𝜔3
𝐴1 𝜔 = 20 log 𝑇 𝑗𝜔

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𝑇 𝑗𝜔 =
𝑗
𝜔
𝜔1
2
1 + 𝑗
𝜔
𝜔2
2
1 + 𝑗
𝜔
𝜔3
𝑇 𝑆 = 𝜔3
𝜔2
𝜔1
2
𝑆2
𝑆 + 𝜔2
2 𝑆 + 𝜔3
𝐴1 𝜔 = 20 log
𝜔
𝜔1
2
1 −
𝜔
𝜔2
2 2
+ 2
𝜔
𝜔2
2
∙ 1 +
𝜔
𝜔3
2
𝜃 𝜔 = 2 tan−1
𝜔
𝜔1
− 2 tan−1
𝜔
𝜔2
− tan−1
𝜔
𝜔3

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Exercise:
For the Bode plot (magnitude frequency response), find out the magnitude transfer function.

20. 5. First Order Active RC Sections
▪ First order sections based on inverting op-amp configuration
(Realization of the bilinear transfer function using the inverting op-amp configuration)
Analog Electronic Circuits
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Add your footnotes
The aim is to realize the bilinear transfer function using the active RC sections based on
inverting operational amplifier.
𝑇 𝑆 =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
= −
𝑍2
𝑍1
𝑍2
𝑍1
= 𝐾
𝑆 + 𝑧1
𝑆 + 𝑝1
𝑇 𝑆 =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
= −𝐾
𝑆 + 𝑧1
𝑆 + 𝑝1
If Z1 and Z2 are both series RC network, then Z1 = R1+(1/SC1),
Z2 = R2+R2/(SC2), and K= R2.

21. 5. First Order Active RC Sections
▪ First order sections based on inverting op-amp configuration
(Realization of the bilinear transfer function using the inverting op-amp configuration)
Analog Electronic Circuits
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Add your footnotes
𝑇 𝑆 =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
= −
𝑍2
𝑍1
= −
𝑅2 +
1
𝑆𝐶2
𝑅1 +
1
𝑆𝐶1
= −
𝑅1
𝑅2
∙
1 +
1
𝑆𝑅2 𝐶2
1 +
1
𝑆𝑅1 𝐶1
where Kz1=1/C2 and p1=1/C1.
Alternative form, in term
of the admittances
−
𝑍2
𝑍1
=
𝑌1
𝑌2

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Design Example:
Based on the following Bode plot (magnitude frequency response), design a practical first
order active RC op-amp section that meets the given characteristics of Bode plot.
Solution:
The two break frequencies (pols and zero) are
a. A single zero at ω=0.5rad/sec
b. A single pole at ω =6rad/sec
c. The constant term is -6dB
(20 log 𝐾
𝑧1
𝑝1
=-6dB → K=6)
𝑇 𝑆 = −6
0.5
6
∙
1 +
𝑆
0.5
1 +
𝑆
6
= −6 ∙
𝑆 + 0.5
𝑆 + 6
= −
𝑅2 +
1
𝑆𝐶2
𝑅1 +
1
𝑆𝐶1
Z1=R1+(1/SC1)=1+(1/6S),
Z2=R2+(1/SC2)=6+(1/3S)

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R1=1Ω, C1=(1/6)F, R2=6 Ω, C2=(1/3)F
To get the practical design, we need to do the frequency
scaling and the magnitude scaling.
Frequency Scaling:
Assume the circuit is designed for a certain frequency band and it is desired that the frequency
band be changed, keeping the same magnitude characteristics (frequency scaling kf =
1000rad/sec (frequency shift by 1000)). Thus, z1 = 500rad/sec and p1 = 6000rad/sec.
How this will affect the original circuit?
New C = C/kf while all R do not change by frequency scaling since they impedances do not
depend on the frequency. New C1=(1/6)mF and new C2=(1/3)mF.
Magnitude Scaling:
The above transfer function contains RC as ratio of R and as
product of RC. Then multiplying all R by scaling factor (km)
and dividing all C by the same quantity will not change the
transfer function.

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The magnitude of the elements (R and C) are scaled yielding to a practical
design. Let km=10,000; thus practical values of R and C are:
R1=10kΩ, R2=60kΩ, C1≈16nF, C2≈33nF
Element values before frequency
scaling and magnitude scaling
Element values after frequency
scaling and magnitude scaling
R kmR
C C/(kmkf)

25. 5. First Order Active RC Sections
▪ First order sections based on non-inverting op-amp configuration
(Realization of the bilinear transfer function using the non-inverting op-amp configuration)
Analog Electronic Circuits
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Add your footnotes
The aim is to realize the bilinear transfer function using the active RC sections based on
non-inverting operational amplifier.
𝑇 𝑆 =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
= 1 +
𝑍2
𝑍1
𝑇 𝑆 =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
= 𝐾
𝑆 + 𝑧1
𝑆 + 𝑝1
1 +
𝑍2
𝑍1
= 𝐾
𝑆 + 𝑧1
𝑆 + 𝑝1
𝑍2
𝑍1
=
𝐾 − 1 𝑆 + 𝐾𝑧1 − 𝑝1
𝑆 + 𝑝1
Since Z1 and Z2 are impedances, there are positive real values. K, z1, and
p1 must satisfy the following conditions:
𝐾 ≥ 1 and 𝐾𝑧1 > 𝑝1

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Two cases: K=1 and K=p1/z1>1
Case I: K=1
This implies z1>p1, that is, if K is unity, the non-inverting configuration can only realize a low
pass transfer function, and
𝑍2
𝑍1
=
𝑧1 − 𝑝1
𝑆 + 𝑝1
=
𝑧1 − 𝑝1
𝑆
1 +
𝑝1
𝑆
Therefore, the following assignments for Z1 and Z2 can be made:
Z1 = 1+(p1/S) = R1+(1/SC1), which yields, R1 = 1 and C1 = 1/p1. And Z2 = (z1-p1)/S = 1/SC2,
which yields, C2 = 1/( z1-p1).

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Two cases: K=1 and K=p1/z1>1
Case II: K=p1/z1>1
This implies z1<p1, that is the non-inverting configuration realizes a high pass transfer function.
One possible realization is
Z1 = 1+(1/(S/p1)) = R1+(1/SC1), and Z2 = (p1-z1)/z1 = R2.
𝑍2
𝑍1
=
𝑝1
𝑧1
− 1 𝑆
𝑆 + 𝑝1
=
𝑝1 − 𝑧1
𝑧1
1 +
1
𝑆
𝑝1

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▪ Cascading Connection:
5. First Order Active RC Sections
It is required to design high order filters to achieve new filters (Band Pass filters and Band
Reject filters) which cannot be obtained in case the first order filter. Also, high order filters
have better bandwidth and selectivity. The figure below demonstrate a cascade of N circuits
with bilinear transfer functions of T1, T2, and TN.

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Loading Effect:
Having minimum loading effect is considered in the cascaded connections based on the
Thevenin’s equivalent circuit. The Thevenin’s impedance taken from the output section needs
to be very smaller than the input impedance of the next cascaded section. Thus, the
Thevenin’s voltage will be the input voltage to the next cascaded section. Using op-amp in
designing RC filter is a good choice since it has very high input impedance and having very
low output impedance. Non-inverting op-amp is preferable since input voltage is applied
directly to non-inverting input. In case using inverting op-amp, then voltage buffer can be a
good choice to connect at the output of each section.
Zin of non-inverting op-amp Zout of non-inverting op-amp

30. 5. First Order Active RC Sections
▪ First order all pass filter
(lattice network=phase correction circuit=phase shaping)
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Add your footnotes
Circuit with all pass magnitude response is called phase correcting circuit, since the
magnitude of the output is constant for all frequency and only the phase of the output is
function of frequency. With this property, these circuits can be connected in cascade with other
circuits to correct for any desired phase (without changing the magnitude response). Also, it
can be used to design high order filter (such as notch filter or comb filter), by using cascaded
all pass filters and adder connected in feedforward connection.
First order all pass filter is characterized by the following transfer function:
1-LH pole and 1-RH zero at |a|.𝑇 𝑆 =
𝑉out
𝑉𝑖𝑛
= 𝐾
𝑆 − 𝑎
𝑆 + 𝑎
= 𝐾
𝑆 − 𝑧1
𝑆 + 𝑝1

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To have zero at RHP, it can be realized by a voltage difference (difference op-amp with one
input), shown below, that generates negative numerator coefficient.
By direct analysis, the transfer function of voltage difference
is given by:
where k = RF/R, z1 = kGG, and p1 = GG.
One possibility is by replacing Ga with SC (a capacitor).
where k = RF/R=1, z1 = k/(RGC), and p1 = -(1/RGC).
𝑇 𝑆 =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
=
𝐺 𝑎 − 𝑘𝐺 𝐺
𝐺 𝑎 + 𝐺 𝐺
𝑇 𝑆 =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
=
𝑆𝐶 − 𝑘𝐺 𝐺
𝑆𝐶 + 𝐺 𝐺
=
𝑆 −
𝑘
𝑅 𝐺 𝐶
𝑆 +
1
𝑅 𝐺 𝐶
= −
1 − 𝑆𝐶𝑅 𝐺
1 + 𝑆𝐶𝑅 𝐺

32. Analog Electronics Circuits
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The phase transfer function is given by
And the following design equation holds for desired θd
at a given frequency ωd
Design the previous first order all pass filter to provide
phase shift of 135ᵒ at 100rad/sec frequency.
Solution:
𝜃 𝜔 = 180° − 2 tan−1 𝜔𝑅 𝐺 𝐶 ; 𝑤ℎ𝑒𝑟𝑒 0° ≤ 𝜃 ≤ 180°
𝑅 𝐺 𝐶 =
cot
𝜃 𝑑
2
𝜔 𝑑
Example:
𝑅 𝐺 𝐶 =
cot
𝜃 𝑑
2
𝜔 𝑑
=
cot
135°
2
100
=
0.4142
100
= 4.142𝑚𝑠𝑒𝑐
C=0.1µF
RG=R=RF=41.42kΩ

33. Analog Electronics Circuits
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LTspice Simulation
Magnitude and phase responses of the all
pass filter:
θd = 135ᵒ at ωd = 100rad/sec= 2π*(15.9Hz)
This APF is considered as lead filter
providing a positive phase angle over the
frequency range of interest.
Phase response --- dashed line
Magnitude response ꟷ solid line

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If we need a lag filter; i.e. negative phase, it can be achieved by replacing GG with SC instead
of Ga.
One possibility is by replacing GG with SC (a capacitor).
where k = RF/R=1, z1 = 1/(kRaC), and p1 = -(1/RaC).
The phase transfer function is given by
𝑇 𝑆 =
𝐺 𝑎 − 𝑘𝑆𝐶
𝐺 𝑎 + 𝑆𝐶
= −𝑘
𝑆 −
1
𝑘𝑅 𝑎 𝐶
𝑆 +
1
𝑅 𝑎 𝐶
= −
𝑆 −
1
𝑅 𝑎 𝐶
𝑆 +
1
𝑅 𝑎 𝐶
=
1 − 𝑆𝐶𝑅 𝑎
1 + 𝑆𝐶𝑅 𝑎
𝜃 𝜔 = −2 tan−1 𝜔𝑅 𝑎 𝐶 ; 𝑤ℎ𝑒𝑟𝑒 − 180° ≤ 𝜃 ≤ 0°

35. Analog Electronics Circuits
35
The following design equation holds for desired θd at a given frequency ωd:
Design the previous first order all pass filter that has constant magnitude response and
provides phase shift of -30ᵒ at 100rad/sec frequency.
NB: Both the phase lead network and the phase lag network can be used as a phase
correction network depending on the system needs.
𝑅 𝑎 𝐶 =
− tan
𝜃 𝑑
2
𝜔 𝑑
Example:
𝑅 𝑎 𝐶 =
tan
−𝜃 𝑑
2
𝜔 𝑑
=
tan
−30°
2
100
=
0.2679
100
= 2.679𝑚𝑠𝑒𝑐
C=0.1µF
Ra=R=RF=26.8kΩ

36. Analog Electronics Circuits
36
▪ (1) R. Schaumann, H. Xiao, and M. E. Valkenburg, Design of Analog Filters, 2nd ed.,
Oxford University Press, 2010.
▪ (2) C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric Circuits, 3rd ed.,
McGraw-Hill Companies, Inc., 2007.
References