First Order Active RC Sections

Manuscript 1
First Order Active RC Sections
Part 2
First order active RC section has transfer function in the s-plane of the following form:
𝑇( 𝑆) =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
=
𝑁(𝑆)
𝐷(𝑆)
= 𝐾
𝑆 + 𝑧1
𝑆 + 𝑝1
= 𝐾
𝑧1
𝑝1
∙
1 + ( 𝑆 𝑧1⁄ )
1 + ( 𝑆 𝑝1⁄ )
T(S) is the transfer function of the first order filter (Bilinear Transfer Function),where it can be seen
that the numerator and the denominator (N(S) and D(S); respectively) are linear function with real
constants (K, z1=ωz1, and p1=ωp1). Circuits that realize T(S) are called first order circuits (First Order
Sections).z1 isthe zeroof T(S) andp1 isthe pole of T(S). Dependingonthe pole andthe zerolocations
(p1,z1) of T(S),the magnitudefrequencyresponse(|T(jω)|)canbe alow pass response (|z1|>|p1|both
locatedat Left-Half plane (special case 0z1 at ∞  ideal integrator)), ahighpass response (|z1|<|p1|
both located at Left-Half plane (special case z1 at 0  ideal differentiator)), or an all pass response
(|z1|=|p1| and z1>p1  zero at Right-Half plane).
Note:
 Ingeneral,minimumphase transferfunction hasall the zerosof atransferfunction locatedat
the Left-Half of the s-plane.
 Non-minimumphasetransferfunction haszerosinthe Right-Halfof the s-plane.If all zerosof
transfer function are all reflected about the jω-axis, there is no change in the magnitude of
the 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 left-hand s-plane (the minimum phase transfer
function).
 The range of phase shift of a minimum phase transfer functionis least possible or minimum
correspondingtoa givenamplitude curve,while the range of the non-minimumphase curve
is the largest possible for the given amplitude curve.
 All pass networkisan example of non-minimumphase networkwhichpassesall frequencies
with equal gain while provides a phase (angle) delay for all frequencies.
 The phase leadnetworkisanetworkthat providesapositive phase angleoverthe 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.
In general,the transferfunctioncan be expressedasthe sum of a real part and an imaginarypart as
following (rectangular form and polar form, respectively):
𝑇( 𝑗𝜔) = 𝑅𝑒{ 𝑇( 𝑗𝜔)} + 𝑗 𝐼𝑚{𝑇(𝑗𝜔)}
𝑇( 𝑗𝜔) = | 𝑇( 𝑗𝜔)| 𝜃( 𝜔)
Re{T(jω)} isthe real partof T(jω) whileIm{T(jω)}isthe imaginarypartof T(jω),andtheyare relatedas:
𝑅𝑒{ 𝑇( 𝑗𝜔)} = | 𝑇( 𝑗𝜔)|cos 𝜃 , 𝑎𝑛𝑑 𝐼𝑚{ 𝑇( 𝑗𝜔)} = | 𝑇( 𝑗𝜔)| sin 𝜃

|T(jω)| is the magnitude frequency response given as
| 𝑇( 𝑗𝜔)| = √(𝑅𝑒{𝑇(𝑗𝜔)})2 + ( 𝐼𝑚{𝑇( 𝑗𝜔)})2
and θ(ω) is the phase frequency response given as
𝜃( 𝜔) = tan−1 (
𝐼𝑚{𝑇( 𝑗𝜔)}
𝑅𝑒{𝑇( 𝑗𝜔)}
)
The previous equations are correct, but in our study are often inconvenient to evaluate transfer
function usingthem.Since the transferfunctionis a ratioof two complex number,togetconvenient
form, the transfer function can be written as
𝑇( 𝑗𝜔) =
𝑅𝑒{ 𝑁( 𝑗𝜔)}+ 𝑗𝐼𝑚{𝑁(𝑗𝜔)}
𝑅𝑒{ 𝐷( 𝑗𝜔)}+ 𝑗𝐼𝑚{𝐷(𝑗𝜔)}
The magnitude transfer function is as following
| 𝑇( 𝑗𝜔)| =
√[ 𝑅𝑒{ 𝑁( 𝑗𝜔)}]2 + [ 𝐼𝑚{𝑁(𝑗𝜔)}]2
√[ 𝑅𝑒{ 𝐷( 𝑗𝜔)}]2 + [ 𝐼𝑚{𝐷(𝑗𝜔)}]2
The phase transfer function is as following
𝜃( 𝜔) = 𝜃 𝑁 ( 𝜔) − 𝜃 𝐷( 𝜔) = tan−1[
𝐼𝑚{ 𝑁( 𝑗𝜔)}
𝑅𝑒{ 𝑁( 𝑗𝜔)}
]− tan−1[
𝐼𝑚{ 𝐷( 𝑗𝜔)}
𝑅𝑒{ 𝐷( 𝑗𝜔)}
]
The phase of the bilinear transfer function 𝑇( 𝑆) = 𝐾
𝑆+𝑧1
𝑆+𝑝1
is obtained as
𝜃( 𝜔) = 𝑝ℎ𝑎𝑠𝑒 𝑜𝑓 𝐾 + 𝑝ℎ𝑎𝑠𝑒 𝑜𝑓 ( 𝑗𝜔 + 𝑧1) − 𝑝ℎ𝑎𝑠𝑒 𝑜𝑓 ( 𝑗𝜔 + 𝑝1)
𝜃( 𝜔) = (0° 𝑓𝑜𝑟 𝐾 > 0 𝑜𝑟 180° 𝑓𝑜𝑟 𝐾 < 0) + tan−1 (
𝜔
𝑧1
) − tan−1 (
𝜔
𝑝1
)
Notes:
 If θ>0ᵒ  Vout leads Vin and circuit that realizes T(S) is called leading circuit.
 If θ<0ᵒ  Vout lags Vin and circuit that realizes T(S) is called lagging circuit.
 If θ=0ᵒ  Vout and Vin are in phase.
 If θ=±180ᵒ  Vout and Vin are out of phase.
Bode Plots for the bilinear transfer function
The plotsof the magnitudefrequencyresponse|T(jω)|(indB) andthe phasefrequencyresponseθ(ω)
(in degree) versus ω in logarithmic representation are called the Bode plots. They are based on the
asymptoticbehaviourof |T(jω)|andθ(ω).To show how the Bode plotscan be constructed,we begin
with the magnitude transfer function
The customary magnitude and phase coordinates for Bode plots are shown below.

Define A(jω) = 20 log |T(jω)| = 20 log |𝐾
𝑧1
𝑝1
∙
1+( 𝑗𝜔 𝑧1⁄ )
1+( 𝑗𝜔 𝑝1⁄ )
| in dB. Therefore, the general magnitude
transfer function (in dB) is given by
𝐴( 𝜔) = 20log (| 𝐾
𝑧1
𝑝1
| ∙
|1 + 𝑗
𝜔
𝑧1
|
|1 + 𝑗
𝜔
𝑝1
|
)
𝐴( 𝜔) = 20 log
(
| 𝐾
𝑧1
𝑝1
| ∙
√1 + (
𝜔
𝑧1
)
2
√1 + (
𝜔
𝑝1
)
2
)
𝐴( 𝜔) = 𝐴1( 𝜔) + 𝐴2( 𝜔)+ 𝐴3( 𝜔)
𝐴( 𝜔) = 20 log| 𝐾
𝑧1
𝑝1
| + 20 log|1 + 𝑗
𝜔
𝑧1
| − 20 log|1 + 𝑗
𝜔
𝑝1
|
A1(ω) is a constant term, and its magnitude frequency response is plotted as
A2(ω) provides a single zero at z1. Let us test it for some points as

For ω/z1 <<1 (ω<<z1), A2(ω)≈0dB
For ω/z1 =1 (ω=z1), A2(ω)≈3dB
For ω/z1 =10 (ω=10z1), A2(ω)=20dB
For ω/z1 >>1 (ω>>z1), A2(ω)≈20log(ω/z1) dB
The maximum deviation between the actual plot and the Bode plot occurs at ω=z1. At ω=2z1 (an
octave), A2(ω)≈6dB, and at ω=10z1 (a decade), A2(ω)≈20dB (a single zero provides 20dB/decade)
A3(ω) provides a single pole at p1. Let us test it for some points as
For ω/p1 <<1 (ω<<p1), A3(ω)≈0dB
For ω/p1 =1 (ω=p1), A3(ω)≈-3dB
For ω/p1 =10 (ω=10p1), A3(ω)=-20dB
For ω/p1 >>1 (ω>>p1), A3(ω)≈-20log(ω/p1) dB
The maximum deviation betweenthe actual plot and the Bode plot occurs at ω=p1. At ω=2p1 (an
octave), A3(ω)≈-6dB, and at ω=10z1 (a decade), A2(ω)≈-20dB (a single pole provides -20dB/decade)
The complete plotof A(ω) isshownfor twocases: z1<p1 and z1>p1, respectively andfor simplicity: K=
p1/z1 (scaling factor). The break frequencies occur at z1 and p1. Note that when |z1|=|p1| and z1>p1,
A(ω)=A1(ω) (all pass response).

The phase frequency response function (in degree) for T(jω) is given by
𝜃( 𝜔) = (0° 𝑓𝑜𝑟 𝐾 > 0 𝑜𝑟 180° 𝑓𝑜𝑟 𝐾 < 0) + tan−1 (
𝜔
𝑧1
) − tan−1 (
𝜔
𝑝1
)
𝜃( 𝜔) = 𝑃1( 𝜔) + 𝑃2( 𝜔) + 𝑃3( 𝜔)
P1(ω) is a straight line for all frequencies with 0ᵒ
for K>0 (case I) or 180ᵒ
for K<0 (case II)
P2(ω) provides a single zero at z1 (provides 45ᵒ
/decade). Let us test it for some points as
For ω/z1 = 0 (ω=0), P2(ω)=0ᵒ
for case I, or P2(ω)=-180ᵒ
for case II
For ω/z1 = 0.1 (ω=0.1z1), P2(ω)≈0ᵒ
for case I, or P2(ω)≈-180ᵒ
for case II
For ω/z1 = 1 (ω=z1), P2(ω)=45ᵒ
for case II
For ω/z1 = 10 (ω=10z1), P2(ω)≈90ᵒ
for case I, or P2(ω)≈-90ᵒ
for case II
For ω/z1 ∞ (ω∞), P2(ω)=90ᵒ
for case II
case I (rad/sec) case II ..
P3(ω) provides a single pole at p1 (provides -45ᵒ
/decade). Let us test it for some points as (since
conditiononKisconsideredinthe previouscase. The effectof Kisaddedtothe P2(ω) phasefunction.)
For ω/p1 = 0 (ω=0), P3(ω)=0ᵒ
For ω/p1 = 0.1 (ω=0.1p1), P3(ω)≈0ᵒ
For ω/p1 = 1 (ω=p1), P3(ω)=-45ᵒ
For ω/p1 ≈10 (ω=10p1), P2(ω)≈-90ᵒ
For ω/p1 ∞ (ω∞), P2(ω)=-90ᵒ

(rad/sec) .
Case a. Add the curves to get the complete θ(ω) for |z1|<|p1|under K>0 (case I), and for
|z1|<|p1|under K<0 (case II) (both z1 and p1 located at Left-Half plane and provide minimum phase
transfer function. This is due to zero at Left-Half plane. For case I, it is obvious that the phase shift
rangesover lessthan80ᵒ
). The networkrepresentingthisfeature iscalledphase-leadnetwork.Itwill
exhibits like a differentiator if |z1|<<|p1|
case I (rad/sec) case II .
Case b. In the case |z1|>|p1|underK>0 (case I),and for |z1|>|p1| under K<0 (case II) bothz1 and p1
located at Left-Half plane, the network representing this feature is called phase-lag network. It will
exhibits like a integrator if |z1|>>|p1|. The complete θ(ω) will be
Case c. In the case |z1|<|p1| and z1 located at Right-Half plane, the complete θ(ω) will be

Case d. When |z1|=|p1| and z1 located at Right-Half plane (all pass transfer function):
𝜃( 𝜔) = (0° 𝑓𝑜𝑟 𝐾 > 0 𝑜𝑟 180° 𝑓𝑜𝑟 𝐾 < 0) − 2tan−1 (
𝜔
𝑝1
)
θ(ω) for the two casesis givenbelow (providingnon-minimumphase transferfunction.Thisisdue to
z1 at Right-Half plane. The phase shift ranges over 180ᵒ
).
Note:
 In case a. and case c., p1 has same location but z1 location is changed withsame magnitude.
The zero isreflectedaboutjω-axisandthere isnochange inthe magnitude transferfunction.
The onlydifference isinthe phase-shiftcharacteristics.If the phase characteristicsof 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 Left-Half
plane (the phase shiftrangesoverlessthan80ᵒ
).Hence,the transferfunctionwithallitszeros
in the Left-Half plane is called a minimum phase transfer function.
 The range of phase shift of a minimum phase transfer function is the least possible or
minimumcorrespondingtoagivenamplitude curve,whereasthe range of the non-minimum
phase curve is the greatest possible for the given amplitude curve, as shown below.

Exercises
(a) Obtain the Bode plot for the transfer function 𝑇( 𝑆) = 6
𝑆+0.5
𝑆+3
(b) Obtain the Bode plot for the 𝑇( 𝑆) = 𝐾𝑑 𝑆 (ideal differentiator)
(c) Obtain the Bode plot for the 𝑇( 𝑆) =
𝐾𝑖
𝑆
(ideal integrator)
*Note: To check your solution, you can use wolframAlpha website (www.wolframalpha.com)
In the search engine field write: Bode plot of 6
𝑆+0.5
𝑆+3
, and the results will give the Bode plot.
∞

Example: Given the below Bode plot (magnitude frequency response), find the magnitude transfer
function.
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
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)
Therefore, A1(ω) is given by
𝐴1( 𝜔) = 20log| 𝑇( 𝑗𝜔)|

𝐴1( 𝜔) = 20 log
(
𝜔
𝜔1
)
2
√(1 − (
𝜔
𝜔2
)
2
)
2
+ (2
𝜔
𝜔2
)
2
∙ √1 + (
𝜔
𝜔3
)
2
And θ(ω) is given by
𝜃( 𝜔) = 2tan−1(
𝜔
𝜔1
) − 2tan−1 (
𝜔
𝜔2
) − tan−1 (
𝜔
𝜔3
)
Exercise
For the Bode plot (magnitude frequency response), find out the magnitude transfer function.

First order sections based on inverting op-amp configuration
(Realization of the bilinear transfer function using the inverting op-amp configuration)
The aimisto realize the bilineartransferfunction 𝑇( 𝑆) =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
= −𝐾
𝑆+𝑧1
𝑆+𝑝1
usingthe active RCsections
based on inverting operational amplifier
The transfer function of the above circuit is obtained as
𝑇( 𝑆) =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
= −
𝑍2
𝑍1
Thus, the following realization must be satisfied
𝑍2
𝑍1
= 𝐾
𝑆 + 𝑧1
𝑆 + 𝑝1
Noted that Z1 and Z2 are either impedances of R, C, or combination of them satisfying that only one
capacitoris usedineach. Since we are giventhe transferfunction,we needtofindasuitable network
that representsit (network synthesis). There are many realizations obtained for this configuration
using the above transfer function that can be noticed from the below table.

-------------------------------------------------------------------------------------------------------------------------------------

First Order Active RC Sections

If Z1 and Z2 are both series RC network, then Z1 = R1+(1/SC1), Z2 = R2+R2/(SC2), and K= R2.
𝑇( 𝑆) =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
= −
𝑍2
𝑍1
= −
𝑅2 +
1
𝑆𝐶2
𝑅1 +
1
𝑆𝐶1
𝑇( 𝑆) = −𝐾
1 + ( 𝑧1 𝑆⁄ )
1 + ( 𝑝1 𝑆⁄ )
= −
𝐾 + 𝐾( 𝑧1 𝑆⁄ )
1 + ( 𝑝1 𝑆⁄ )
= −
𝑅2 +
1
𝑆𝐶2
𝑅1 +
1
𝑆𝐶1
= −
𝑅1
𝑅2
∙
1 +
1
𝑆𝑅2 𝐶2
1 +
1
𝑆𝑅1 𝐶1
Kz1=1/C2 and p1=1/C1.
This RC op-amp realization of T(S) is shown below in term of the impedances.
Alternative form, in term of the admittances (duality RC configuration), is given as (-Z2/Z1=-Y1/Y2)
Design Example:
Basedonthe followingBode plot(magnitude frequencyresponse), designapractical firstorderactive
RC op-amp section that meets the given characteristics of Bode plot.
It is obvious that the plot represents first order magnitude transfer function. 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)
Therefore,
𝑇( 𝑆) = −𝐾
𝑧1
𝑝1
∙
1 + ( 𝑆 𝑧1⁄ )
1 + ( 𝑆 𝑝1⁄ )
= −𝐾
𝑆 + 𝑧1
𝑆 + 𝑝1
𝑇( 𝑆) = −6 (
0.5
6
) ∙
(1 +
𝑆
0.5
)
(1 +
𝑆
6
)
= −6 ∙
( 𝑆 + 0.5)
( 𝑆 + 6)
𝑇( 𝑆) = −𝐾
1 + ( 𝑧1 𝑆⁄ )
1 + ( 𝑝1 𝑆⁄ )
= −
𝐾 + ( 𝐾𝑧1 𝑆⁄ )
1 + ( 𝑝1 𝑆⁄ )
= −
𝑅2 +
1
𝑆𝐶2
𝑅1 +
1
𝑆𝐶1
Z1=R1+(1/SC1)=1+(1/6S), and Z2=R2+(1/SC2)=6+(1/3S)
Using the previous impedance realization
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 circuitis designedfora certainfrequencybandandit is desiredthatthe frequencyband
be changed, keeping the same magnitude characteristics. For example, we would like to realize the
circuit with frequency scaling kf = 1000rad/sec (frequency shift by 1000). Therefore, z1 = 500rad/sec
and p1 = 6000rad/sec. How this will affect the original circuit?
NewC = C/kf while all Rdonot change by frequencyscalingsincetheyimpedancesdonotdependon
the frequency. New C1=(1/6)mF and new C2=(1/3)mF.

Magnitude Scaling
The above transferfunctioncontainsRandC as ratioof R and as productof RC. Thenmultiplying all R
byscalingfactor(km) and dividingall Cbythe same quantitywill notchange the transferfunction;i.e.,
the circuit shown below has the same transfer function as the previous one.
.
The magnitude of the elements (R and C) are scaled yielding to a practical design.
Let km=10000, thempractical valuesof R and C are (after frequency scaling and magnitude scaling):
R1=10kΩ, R2=60kΩ, C1≈16nF, C2≈33nF
First order sections based on non-inverting op-amp configuration
(Realization of the bilinear transfer function using the non-inverting op-amp configuration)
The aim isto realize the bilineartransferfunction 𝑇( 𝑆) =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
= 𝐾
𝑆+𝑧1
𝑆+𝑝1
usingthe active RC sections
based on non-inverting operational amplifier
The transfer function of the above circuit is obtained as
𝑇( 𝑆) =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
= 1 +
𝑍2
𝑍1
Thus, the following realization must be satisfied
1 +
𝑍2
𝑍1
= 𝐾
𝑆 + 𝑧1
𝑆 + 𝑝1
Elementvaluesbefore frequencyscalingand
magnitude scaling
Element values after frequency scaling and
magnitude scaling
R kmR
C C/(kmkf)

𝑍2
𝑍1
=
( 𝐾 − 1) 𝑆 + ( 𝐾𝑧1 − 𝑝1)
𝑆 + 𝑝1
Since Z1 and Z2 are impedances,there are positivereal values. K,z1,and p1 must satisfythe following
conditions:
𝐾 ≥ 1 and 𝐾𝑧1 > 𝑝1
Let us consider two cases: K=1 and K=p1/z1>1
Case I: K=1
Thisimpliesz1>p1 (and|z1|>|p1|excludez1<p1 leadstoinvertingamplifier)thatis,if Kisunity,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). This RC op-amp realizationis shown below in term
of the impedances.
In term of the admittances (duality RC configuration), it is given as (Z2/Z1=Y1/Y2)
Case II: K=(p1/z1)>1

Thisimpliesz1<p1 thatisthe realizationof ahighpassresponse using the non-invertingconfiguration.
In this case we have:
𝑍2
𝑍1
=
(
𝑝1
𝑧1
− 1) 𝑆
𝑆 + 𝑝1
=
(
𝑝1 − 𝑧1
𝑧1
)
1 +
1
𝑆
𝑝1
One possiblerealizationcanbe obtainedbythe followingassignmentandshownbelowconfiguration:
Z1 = 1+(1/(S/p1)) = R1+(1/SC1), and Z2 = (p1-z1)/z1 = R2
In term of the admittances (duality RC configuration), it is given as
𝑍2
𝑍1
=
(
𝑝1 − 𝑧1
𝑧1
) 𝑆
𝑆 + 𝑝1
=
𝑌1
𝑌2
The resulting realization is shown below.
Cascading Connection:
Itisrequiredtodesignhighorderfilterstoachievenew filters(BandPassfiltersandBandRejectfilters)
whichcannot be obtainedincase the firstorder filter.Also,highorderfiltershave betterbandwidth
and selectivity. The figure below demonstrate acascade of N circuitswith bilineartransferfunctions
of T1, T2, and TN.

One mainpointwhichneedstobe consideredisthe loadingeffect.We needtohave minimumloading
effect.Thisisdesignedbasedonthe Thevenin’sequivalent.The Thevenin’simpedance takenfromthe
outputsectionneedstobe verysmallerthanthe inputimpedance of the nextcascadedsection.Thus,
the Thevenin’svoltage will be the inputvoltage tothe nextcascadedsection(withminimumlosses).
Since the active amplifierisan op-amp,thisisnot a significantissue.The feature of op-ampishaving
very high input impedance and having very low output impedance. The non-inverting op-amp
configurationispreferable since it has very low Zo at low frequencyandZin isinfinite underideal op-
amp. (the input voltage is applied directly to the non-inverting input of the ideal op-amp). In case
using the inverting amplifier, then voltage buffer can be a good choice to connect at the output of
eachsection(idealvoltagebufferop-amphasZin=∞andZo=0). The below figuresshow the equivalent
models of the input and the output impedances of the non-ideal non-inverting op-amp.

Summary of Transfer Function response of the first order section.

First order all pass filter (lattice network=phase correction circuit=phase shaping)
Circuitwithall pass magnitude response iscalledphase correctingcircuit,since the magnitudeof the
output is constant for all frequency and only the phase of the output is function of frequency. With
thisproperty,these circuitscanbe connectedincascade withothercircuitstocorrect forany desired
phase (withoutchangingthemagnituderesponse).Also,itcanbe usedtodesignhighorderfilter(such
as notch filterorcombfilter),byusingall passfilterandadderconnectedinfeedforwardconnection.
The inputs of the adder are output of all pass filter scaled by 0.5 and the input signal. Cascading all
pass filter and takes the last all pass filter output to be input of the adder with the input signal will
generate the response of comb filter (this method can be applied for both analog comb filter and
digital comb filter).
To have zero at Right-Half plane, it can be realized by a voltage difference (difference op-ampwith
one input), shown below, that generates negative numerator coefficient.
By directanalysis,the transferfunctionof the above circuitisgivenby
𝑇( 𝑆) =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
=
𝐺𝑎 − 𝑘𝐺 𝐺
𝐺𝑎 + 𝐺 𝐺
where k = RF/R, z1 = kGG, and p1 = GG.
And it is matched with the general transfer function of all pass filter is given as
𝑇( 𝑆) =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
= 𝐾
𝑆 − 𝑎
𝑆 + 𝑎
where a = |z1|= |p1|with z1 locatedat Right-Half plane. There are several differentrealizations.One
possibility is by replacing Ga with SC (a capacitor).
𝑇( 𝑆) =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
=
𝑆𝐶 − 𝑘𝐺 𝐺
𝑆𝐶 + 𝐺 𝐺
=
𝑆 −
𝑘
𝑅 𝐺 𝐶
𝑆 +
1
𝑅 𝐺 𝐶

where k = RF/R, z1 = k/(RGC), and p1 = -(1/RGC).
It can be notedthat z1 locationcan be smallerthan, greaterthan,or equal to |p1|,dependingonthe
choice of the constant k < 1, k > 1, or k = 1. Based on out particular interest, we choose k = RF/R =1.
𝑇( 𝑆) =
𝑆 −
1
𝑅 𝐺 𝐶
𝑆 +
1
𝑅 𝐺 𝐶
= −
1 − 𝑆𝐶𝑅 𝐺
1 + 𝑆𝐶𝑅 𝐺
The magnitude transfer function is unity for all frequencies.
| 𝑇( 𝑗𝜔)| = |−
1 − 𝑗𝜔𝐶𝑅 𝐺
1 + 𝑗𝜔𝐶𝑅 𝐺
| = √
1 + ( 𝜔𝑅 𝐺 𝐶)2
1 + ( 𝜔𝑅 𝐺 𝐶)2 = 1
In orderto have anglesexpressedineitherfirstorfourthquarter,the phase transferfunctionisgiven
by
𝜃( 𝜔) = 180° − 2tan−1( 𝜔𝑅 𝐺 𝐶); 𝑤ℎ𝑒𝑟𝑒 0° ≤ 𝜃 ≤ 180°
And the following design equation holds for desired θd at a given frequency ωd
𝜃𝑑 = 180° − 2tan−1( 𝜔 𝑑 𝑅 𝐺 𝐶)
180° = 𝜃𝑑 + 2tan−1( 𝜔 𝑑 𝑅 𝐺 𝐶)
90° =
𝜃𝑑
2
+ tan−1( 𝜔 𝑑 𝑅 𝐺 𝐶)
𝜔 𝑑 𝑅 𝐺 𝐶 = tan (90° −
𝜃𝑑
2
)
𝑅 𝐺 𝐶 =
cot(
𝜃𝑑
2
)
𝜔 𝑑
Example: design the previous first order all pass filter that has constant magnitude response and
provides phase shift of 135ᵒ at 100rad/sec frequency.
𝑅 𝐺 𝐶 =
cot(
𝜃𝑑
2
)
𝜔 𝑑
=
cot(
135°
2
)
100
𝑅 𝐺 𝐶 =
0.4142
100
= 4.142𝑚𝑠𝑒𝑐
Let C=0.1µF and RG=R=RF, yields R=41.42kΩ.

Design simulation
This all pass filter is considered as a lead filter providing a positive phase angle over the frequency
range of interest.If we needalagfilter;i.e.negativephase, itcanbe achievedbyreplacingGG withSC
instead of Ga.
𝑇( 𝑆) =
𝐺𝑎 − 𝑘𝑆𝐶
𝐺𝑎 + 𝑆𝐶
= −𝑘
𝑆 −
1
𝑘𝑅 𝑎 𝐶
𝑆 +
1
𝑅 𝑎 𝐶
where k = RF/R, z1 = 1/(kRaC), and p1 = -(1/RaC).
It can be notedthat z1 locationcan be smallerthan,greaterthan,or equal to |p1|,dependingonthe
choice of the constant k> 1, k < 1, or k = 1. Basedon outparticularinterest(all passfilter),we choose
k = RF/R =1.
𝑇( 𝑆) = −
𝑆 −
1
𝑅 𝑎 𝐶
𝑆 +
1
𝑅 𝑎 𝐶
=
1 − 𝑆𝐶𝑅 𝑎
1 + 𝑆𝐶𝑅 𝑎
The magnitude transfer function is unity for all frequencies.
| 𝑇( 𝑗𝜔)| = |
1 − 𝑗𝜔𝐶𝑅 𝑎
1 + 𝑗𝜔𝐶𝑅 𝑎
| = √
1 + ( 𝜔𝑅 𝑎 𝐶)2
1 + ( 𝜔𝑅 𝑎 𝐶)2 = 1
The phase transfer function is given by
𝜃( 𝜔) = −2tan−1( 𝜔𝑅 𝑎 𝐶); 𝑤ℎ𝑒𝑟𝑒 − 180° ≤ 𝜃 ≤ 0°

And the following design equation holds for desired θd at a given frequency ωd
𝜃𝑑 = −2tan−1( 𝜔 𝑑 𝑅 𝑎 𝐶)
𝑅 𝑎 𝐶 =
− tan (
𝜃𝑑
2
)
𝜔 𝑑
Example: design the previous first order all pass filter that has constant magnitude response and
provides phase shift of -30ᵒ at 100rad/sec frequency.
𝑅 𝑎 𝐶 =
tan (
−𝜃𝑑
2
)
𝜔 𝑑
=
tan(
30°
2
)
100
𝑅 𝑎 𝐶 =
0.2679
100
= 2.679𝑚𝑠𝑒𝑐
Let C=0.1µF and Ra=RF=R, yields R=26.79kΩ≈26.8kΩ.

First Order Active RC Sections

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