Sheet Pile Wall Design and Construction: A Practical Guide for Civil Engineer...
Β
First Order Active RC Sections
1. 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:
π( π) =
πππ’π‘
πππ
=
π(π)
π·(π)
= πΎ
π + π§1
π + π1
= πΎ
π§1
π1
β
1 + ( π π§1β )
1 + ( π π1β )
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=Οz1, and p1=Οp1). 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(jΟ)|)canbe alow pass response (|z1|>|p1|both
locatedat Left-Half plane (special case 0z1 at β ο ideal integrator)), ahighpass response (|z1|<|p1|
both located at Left-Half plane (special case z1 at 0 ο ideal differentiator)), or an all pass response
(|z1|=|p1| and z1>p1 ο zero at Right-Half plane).
Note:
ο· Ingeneral,minimumphase transferfunction hasall the zerosof atransferfunction locatedat
the Left-Half of the s-plane.
ο· Non-minimumphasetransferfunction haszerosinthe Right-Halfof the s-plane.If all zerosof
transfer function are all reflected about the jΟ-axis, there is no change in the magnitude of
the transfer function and only difference is in the phase-shift characteristics (All pass
network).
ο· 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).
ο· 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.
ο· All pass networkisan example of non-minimumphase networkwhichpassesall frequencies
with equal gain while provides a phase (angle) delay for all frequencies.
ο· The phase leadnetworkisanetworkthat providesapositive phase angleoverthe frequency
range of interest, yielding a system to have adequate phase margin.
ο· 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):
π( ππ) = π π{ π( ππ)} + π πΌπ{π(ππ)}
π( ππ) = | π( ππ)| π( π)
Re{T(jΟ)} isthe real partof T(jΟ) whileIm{T(jΟ)}isthe imaginarypartof T(jΟ),andtheyare relatedas:
π π{ π( ππ)} = | π( ππ)|cos π , πππ πΌπ{ π( ππ)} = | π( ππ)| sin π
2. |T(jΟ)| is the magnitude frequency response given as
| π( ππ)| = β(π π{π(ππ)})2 + ( πΌπ{π( ππ)})2
and ΞΈ(Ο) is the phase frequency response given as
π( π) = tanβ1 (
πΌπ{π( ππ)}
π π{π( ππ)}
)
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
π( ππ) =
π π{ π( ππ)}+ ππΌπ{π(ππ)}
π π{ π·( ππ)}+ ππΌπ{π·(ππ)}
The magnitude transfer function is as following
| π( ππ)| =
β[ π π{ π( ππ)}]2 + [ πΌπ{π(ππ)}]2
β[ π π{ π·( ππ)}]2 + [ πΌπ{π·(ππ)}]2
The phase transfer function is as following
π( π) = π π ( π) β π π·( π) = tanβ1[
πΌπ{ π( ππ)}
π π{ π( ππ)}
]β tanβ1[
πΌπ{ π·( ππ)}
π π{ π·( ππ)}
]
The phase of the bilinear transfer function π( π) = πΎ
π+π§1
π+π1
is obtained as
π( π) = πβππ π ππ πΎ + πβππ π ππ ( ππ + π§1) β πβππ π ππ ( ππ + π1)
π( π) = (0Β° πππ πΎ > 0 ππ 180Β° πππ πΎ < 0) + tanβ1 (
π
π§1
) β tanβ1 (
π
π1
)
Notes:
ο· If ΞΈ>0α΅ ο Vout leads Vin and circuit that realizes T(S) is called leading circuit.
ο· If ΞΈ<0α΅ ο Vout lags Vin and circuit that realizes T(S) is called lagging circuit.
ο· If ΞΈ=0α΅ ο Vout and Vin are in phase.
ο· If ΞΈ=Β±180α΅ ο Vout and Vin are out of phase.
Bode Plots for the bilinear transfer function
The plotsof the magnitudefrequencyresponse|T(jΟ)|(indB) andthe phasefrequencyresponseΞΈ(Ο)
(in degree) versus Ο in logarithmic representation are called the Bode plots. They are based on the
asymptoticbehaviourof |T(jΟ)|andΞΈ(Ο).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.
3. Define A(jΟ) = 20 log |T(jΟ)| = 20 log |πΎ
π§1
π1
β
1+( ππ π§1β )
1+( ππ π1β )
| in dB. Therefore, the general magnitude
transfer function (in dB) is given by
π΄( π) = 20log (| πΎ
π§1
π1
| β
|1 + π
π
π§1
|
|1 + π
π
π1
|
)
π΄( π) = 20 log
(
| πΎ
π§1
π1
| β
β1 + (
π
π§1
)
2
β1 + (
π
π1
)
2
)
π΄( π) = π΄1( π) + π΄2( π)+ π΄3( π)
π΄( π) = 20 log| πΎ
π§1
π1
| + 20 log|1 + π
π
π§1
| β 20 log|1 + π
π
π1
|
A1(Ο) is a constant term, and its magnitude frequency response is plotted as
A2(Ο) provides a single zero at z1. Let us test it for some points as
4. For Ο/z1 <<1 (Ο<<z1), A2(Ο)β0dB
For Ο/z1 =1 (Ο=z1), A2(Ο)β3dB
For Ο/z1 =10 (Ο=10z1), A2(Ο)=20dB
For Ο/z1 >>1 (Ο>>z1), A2(Ο)β20log(Ο/z1) dB
The maximum deviation between the actual plot and the Bode plot occurs at Ο=z1. At Ο=2z1 (an
octave), A2(Ο)β6dB, and at Ο=10z1 (a decade), A2(Ο)β20dB (a single zero provides 20dB/decade)
A3(Ο) provides a single pole at p1. Let us test it for some points as
For Ο/p1 <<1 (Ο<<p1), A3(Ο)β0dB
For Ο/p1 =1 (Ο=p1), A3(Ο)β-3dB
For Ο/p1 =10 (Ο=10p1), A3(Ο)=-20dB
For Ο/p1 >>1 (Ο>>p1), A3(Ο)β-20log(Ο/p1) dB
The maximum deviation betweenthe actual plot and the Bode plot occurs at Ο=p1. At Ο=2p1 (an
octave), A3(Ο)β-6dB, and at Ο=10z1 (a decade), A2(Ο)β-20dB (a single pole provides -20dB/decade)
The complete plotof A(Ο) 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(Ο)=A1(Ο) (all pass response).
5. The phase frequency response function (in degree) for T(jΟ) is given by
π( π) = (0Β° πππ πΎ > 0 ππ 180Β° πππ πΎ < 0) + tanβ1 (
π
π§1
) β tanβ1 (
π
π1
)
π( π) = π1( π) + π2( π) + π3( π)
P1(Ο) is a straight line for all frequencies with 0α΅
for K>0 (case I) or 180α΅
for K<0 (case II)
P2(Ο) provides a single zero at z1 (provides 45α΅
/decade). Let us test it for some points as
For Ο/z1 = 0 (Ο=0), P2(Ο)=0α΅
for case I, or P2(Ο)=-180α΅
for case II
For Ο/z1 = 0.1 (Ο=0.1z1), P2(Ο)β0α΅
for case I, or P2(Ο)β-180α΅
for case II
For Ο/z1 = 1 (Ο=z1), P2(Ο)=45α΅
for case I, or P2(Ο)=-135α΅
for case II
For Ο/z1 = 10 (Ο=10z1), P2(Ο)β90α΅
for case I, or P2(Ο)β-90α΅
for case II
For Ο/z1 ο β (Οο β), P2(Ο)=90α΅
for case I, or P2(Ο)=-90α΅
for case II
case I (rad/sec) case II ..
P3(Ο) provides a single pole at p1 (provides -45α΅
/decade). Let us test it for some points as (since
conditiononKisconsideredinthe previouscase. The effectof Kisaddedtothe P2(Ο) phasefunction.)
For Ο/p1 = 0 (Ο=0), P3(Ο)=0α΅
For Ο/p1 = 0.1 (Ο=0.1p1), P3(Ο)β0α΅
For Ο/p1 = 1 (Ο=p1), P3(Ο)=-45α΅
For Ο/p1 β10 (Ο=10p1), P2(Ο)β-90α΅
For Ο/p1 ο β (Οο β), P2(Ο)=-90α΅
6. (rad/sec) .
Case a. Add the curves to get the complete ΞΈ(Ο) 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 lessthan80α΅
). 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 ΞΈ(Ο) will be
case I (rad/sec) case II .
Case c. In the case |z1|<|p1| and z1 located at Right-Half plane, the complete ΞΈ(Ο) will be
7. case I (rad/sec) case II .
Case d. When |z1|=|p1| and z1 located at Right-Half plane (all pass transfer function):
π( π) = (0Β° πππ πΎ > 0 ππ 180Β° πππ πΎ < 0) β 2tanβ1 (
π
π1
)
ΞΈ(Ο) for the two casesis givenbelow (providingnon-minimumphase transferfunction.Thisisdue to
z1 at Right-Half plane. The phase shift ranges over 180α΅
).
case I (rad/sec) case II .
Note:
ο· In case a. and case c., p1 has same location but z1 location is changed withsame magnitude.
The zero isreflectedaboutjΟ-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 shiftrangesoverlessthan80α΅
).Hence,the transferfunctionwithallitszeros
in the Left-Half plane is called a minimum phase transfer function.
ο· 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.
8. Exercises
(a) Obtain the Bode plot for the transfer function π( π) = 6
π+0.5
π+3
(b) Obtain the Bode plot for the π( π) = πΎπ π (ideal differentiator)
(c) Obtain the Bode plot for the π( π) =
πΎπ
π
(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
π+0.5
π+3
, and the results will give the Bode plot.
β
9. 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 Οp1 = Ο2
(c) A single pole at Οp2 = Ο3
π( ππ) =
(π
π
π1
)
2
(1 + π
π
π2
)
2
(1 + π
π
π3
)
π( π) = π3 (
π2
π1
)
2 π2
( π + π2)2( π+ π3)
Therefore, A1(Ο) is given by
π΄1( π) = 20log| π( ππ)|
10. π΄1( π) = 20 log
(
π
π1
)
2
β(1 β (
π
π2
)
2
)
2
+ (2
π
π2
)
2
β β1 + (
π
π3
)
2
And ΞΈ(Ο) is given by
π( π) = 2tanβ1(
π
π1
) β 2tanβ1 (
π
π2
) β tanβ1 (
π
π3
)
Exercise
For the Bode plot (magnitude frequency response), find out the magnitude transfer function.
11. 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 π( π) =
πππ’π‘
πππ
= βπΎ
π+π§1
π+π1
usingthe active RCsections
based on inverting operational amplifier
The transfer function of the above circuit is obtained as
π( π) =
πππ’π‘
πππ
= β
π2
π1
Thus, the following realization must be satisfied
π2
π1
= πΎ
π + π§1
π + π1
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.
14. If Z1 and Z2 are both series RC network, then Z1 = R1+(1/SC1), Z2 = R2+R2/(SC2), and K= R2.
π( π) =
πππ’π‘
πππ
= β
π2
π1
= β
π 2 +
1
ππΆ2
π 1 +
1
ππΆ1
π( π) = βπΎ
1 + ( π§1 πβ )
1 + ( π1 πβ )
= β
πΎ + πΎ( π§1 πβ )
1 + ( π1 πβ )
= β
π 2 +
1
ππΆ2
π 1 +
1
ππΆ1
= β
π 1
π 2
β
1 +
1
ππ 2 πΆ2
1 +
1
ππ 1 πΆ1
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 Ο=0.5rad/sec
17. π2
π1
=
( πΎ β 1) π + ( πΎπ§1 β π1)
π + π1
Since Z1 and Z2 are impedances,there are positivereal values. K,z1,and p1 must satisfythe following
conditions:
πΎ β₯ 1 and πΎπ§1 > π1
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
π2
π1
=
π§1 β π1
π + π1
=
( π§1 β π1)
π
1 +
π1
π
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
18. Thisimpliesz1<p1 thatisthe realizationof ahighpassresponse using the non-invertingconfiguration.
In this case we have:
π2
π1
=
(
π1
π§1
β 1) π
π + π1
=
(
π1 β π§1
π§1
)
1 +
1
π
π1
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
π2
π1
=
(
π1 β π§1
π§1
) π
π + π1
=
π1
π2
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.
19. One mainpointwhichneedstobe consideredisthe loadingeffect.We needtohave minimumloading
effect.Thisisdesignedbasedonthe Theveninβsequivalent.The Theveninβsimpedance takenfromthe
outputsectionneedstobe verysmallerthanthe inputimpedance of the nextcascadedsection.Thus,
the Theveninβsvoltage 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=βandZo=0). The below figuresshow the equivalent
models of the input and the output impedances of the non-ideal non-inverting op-amp.
21. 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
π( π) =
πππ’π‘
πππ
=
πΊπ β ππΊ πΊ
πΊπ + πΊ πΊ
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
π( π) =
πππ’π‘
πππ
= πΎ
π β π
π + π
where a = |z1|= |p1|with z1 locatedat Right-Half plane. There are several differentrealizations.One
possibility is by replacing Ga with SC (a capacitor).
π( π) =
πππ’π‘
πππ
=
ππΆ β ππΊ πΊ
ππΆ + πΊ πΊ
=
π β
π
π πΊ πΆ
π +
1
π πΊ πΆ
23. 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.
π( π) =
πΊπ β πππΆ
πΊπ + ππΆ
= βπ
π β
1
ππ π πΆ
π +
1
π π πΆ
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.
π( π) = β
π β
1
π π πΆ
π +
1
π π πΆ
=
1 β ππΆπ π
1 + ππΆπ π
The magnitude transfer function is unity for all frequencies.
| π( ππ)| = |
1 β πππΆπ π
1 + πππΆπ π
| = β
1 + ( ππ π πΆ)2
1 + ( ππ π πΆ)2 = 1
The phase transfer function is given by
π( π) = β2tanβ1( ππ π πΆ); π€βπππ β 180Β° β€ π β€ 0Β°