COMMUNICATIONCIRCUITS

Assoc.Prof.Zlatka Valkova‐Jarvis,PhD

TECHNICAL UNIVERSITY OF SOFIAFACULTY OF TELECOMMUNICATIONS

COMMUNICATIONCIRCUITSBachelorofEngineeringCourse

CompulsorySubject,CodeBTCEe25,Credits6,SemesterIVLectures:45 hours(15three‐hourlectures);LaboratoryWork:15 hours(5three‐hourlabs)

ElectiveSemesterProject:CodeBTCEe34,Credit1,SemesterVLECTURECONTENT1. Analogue linear one- and two-port circuits– basic parameters and descriptive equations, properties.

Analysis of analogue circuits using generalized matrices. Basic elements used in analogue linear circuits

2. Representation of analogue immittance functions – analytical representation, pole-zero diagrams, frequency responses. The Hurwitz test. Frequency and impedance scaling

3. Passive linear one-port circuits – canonic schemes. LC and RC one-port circuits - basic properties, synthesis of Foster and Cauer canonic one-port schemes

4. Description of analogue linear two-port circuits in the s-domain – transfer function (TF), pole-zero diagrams. Description of analogue linear two-port circuits in the frequency domain – frequency responses, polar diagrams

5. The transfer function obtained from a given magnitude or phase response. Frequency responses conditioned by basic type poles and zeros. Pole-zero diagrams and transfer functions obtained from analogue frequency responses. Asymptotic Bode diagrams

6. Description of analogue linear circuits in the time-domain - impulse and step responses, analogue convolution. Analogue impulse and step responses due to different transfer function’s poles. Step responses of general bilinear and biquadratic analogue transfer functions

LECTURE CONTENT7. An introduction to two-port passive circuit synthesis - terms of realisability. Properties and limitations of

structures with different elements and topology. Different types of two-port passive circuits. Synthesis of non-terminated L-shaped and lattice schemes

8. Amplitude correction – basic considerations, TF, cascade realisation, specifics of approximation. Passive and active amplitude-correction sections of first and second order. Attenuators

9. Phase correctors – basic principles of phase correction, transfer functions, cascade realisation. Delay circuits

10. Electrical filters- basic types, specifications. Approximation based on given requirements to the Magnitude Response. Frequency transformation for analogue filters. Basic filter transfer functions of n-th, first and second order

11. Polynomial (Butterworth, Chebyshev, Legendre, Bessel) and non-polynomial (Cauer, Inverse Chebyshev) classic approximations. Comparison of different approximations. Computer-based approximations –generalised equal-ripple and maximally-flat approximations

12. Synthesis of RC and single-terminated LC two-port ladder circuits with polynomial transfer functions. Analysis of ladder LC filter structures

13. Active circuits - basic principles of synthesis and design. Properties of cascade realisations. Low pole-quality factor (single-amplifier) first- and second-order polynomial and non-polynomial active filter sections – transfer functions and active models

14. Medium and large pole-quality factor active filter sections – basic models and realisations. Multi-amplifier universal biquadratic active filter sections. Switched-Capacitor active filters (SC-filters)

15. Direct active realisation methods by passive reactive schemes imitation. Gyrator models. Simulation of double-terminated LC-ladder circuits using frequency-dependent negative resistance and conductivity –Bruton transformations. Sensitivity – types and evaluation

LECTURE CONTENT16. Linear Time-Invariant (LTI) systems. Description of digital linear circuits in the time- and z-domains.

The interdependence of the s-plane and z-plane

17. Description of digital linear circuits in the frequency domain. Аnalogue to digital transformation. Digital frequency response in relation to the basic type of poles and zeros. Pole-Quality Factor

18. Basic elements used in digital LTI systems. Recursive (IIR – Infinite Impulse Response) and non-recursive (FIR - Finite Impulse Response) digital circuits. Stability considerations

19. Analysis of digital LTI systems using Signal Flow-Graph (SFG) transforms and simple circuit descriptions in time- and z-domain

20. Non-recursive (FIR) digital circuit realisations. Direct, predictive direct, cascade, lattice, cascade lattice and linear-phase structures

21. Design of non-recursive digital systems using window functions. Mini-max design methods – Parks-McClellan algorithm. Design of FIR systems with Matlab

22. IIR digital circuit realisations. Direct, cascade, parallel, leader and lattice realisations. Realisations based on parallel all-pass structures

23. Design of IIR digital systems. Design based on the bilinear z-transformation, invariant Impulse Response method. Direct methods. Digital frequency transforms. Design of IIR systems with Matlab

INFORMATION RESOURCES • Primary Sources1. Lecture slides (available at - http://sopko-tu-sofia.bg).2. LUTOVAC, M, D. TOSIC AND B. EVANS, Filter design for signal processing using Matlab and Mathematica,

Prentice Hall, 2001.3. SCHAUMANN, R AND M. E. VAN VALKENBURG, Design of analog filters, 2nd edition, The Oxford Series

in Electrical and Computer Engineering, 2009. 4. MITRA, S., Digital signal processing. 4-rd Edition, McGraw-Hill, 2010. 5. ADYA ALKA, Circuits and Systems, Alpha Science Intl Ltd, 2010.6. WING OMAR, Classical Circuit Theory, Springer, 2010.7. COUCH L., Digital & Analog Communication Systems, 7th edition, Prentice Hall, 2006.

• Supplementary Sources 8. MATLAB – R2010b + Toolboxes. MathWorks, 2010 and any other versions of MATLAB.9. CHEN, W.K., The Circuits and filters handbook, CRC Press & IEEE Press, 2009.

• Useful WWW addressesGeneralmaterialsonCC http://sopko.tu‐sofia.bg.RecordedseminarsonMATLAB http://www.mathworks.com/company/events/archived_webinars.jspExchangeoffilesforMATLAB http://www.mathworks.com/matlabcentral/fileexchange/loadCategory.doInteractivefilterdesign http://scpd.stanford.edu/SCPD/js/brandingFrame/externalURL2.htmCCcoursesinGeorgiaTechU,USA http://csip.ece.gatech.edu/csipindex.htmlMaterialsonCircuitsandDSP http://www.yov408.com/html/tutorials.php?&s=83CCcoursesinTampere,Finland http://www.tut.fi/public/oppaat/opas2004‐2005/kv/laitokset/InstituteofSignalProcessing/On‐linecoursesandmaterials http://www.techonline.com/community/homeEducationalmaterials http://www.onesmartclick.com/engineering/digital‐signal‐processing.htmlMITOpenCourses http://ocw.mit.edu/OcwWeb/Electrical‐Engineering‐and‐Computer‐Science/index.htm

METHOD OF ASSESSMENT

Assessment Two-hour written exam with 12 problems to be solved. Unlimited use of books and sources. The final grade

includes the assessments from the Lab works and from participation in the discussions during the lectures, with weighting coefficients as given in the table below.

Standards of AssessmentExcellent (6) – a very comprehensive knowledge of the entire course-material and of the main additional

sources. A deep understanding, perfect reasoning and sound creativity in solving complex problems and applying the knowledge. Well-demonstrated original, deductive thinking, together with the capability to argue, define and defend a position connected with the subject.

Very Good (5) – a comprehensive knowledge of the entire material with deep understanding, very good reasoning and manifest creativity in solving problems and applying the knowledge.

Good (4) – more than basic knowledge with the ability to solve simple and more difficult problems but showing some lack of creativity in solving problems and applying the knowledge.

Fair (3) – a basic knowledge and ability to solve simple problems.

Final Assessment Components

Component Weight

1 Written Exam 0,75

2 Lab Work Assessment 0.15

3 Participation in Lecture Discussions 0,10

LABORATORY WORK

1. Investigation of time-, frequency- and s-domain characteristics and descriptions of analogue circuits and transfer functions using MATLAB

2. Investigation of time-, frequency- and z-domain characteristics and descriptions of digital circuits and transfer functions using MATLAB

3. Magnitude approximation of given specifications. Realization and comparative studies of Butterworth, Chebyshev, Cauer, Inverse Chebyshev and Bessel approximations using MATLAB

4. Magnitude approximation of FIR and IIR digital with MATLAB. Design of non-recursive (FIR) digital filters using window functions. Mini-max methods of design. Design of IIR filters -Butterworth, Chebyshev, Cauer, Inverse Chebyshev and Bessel digital circuit approximations

5. Design of FIR and IIR filter structures with MATLAB. Basic digital circuit realizations - Direct, Cascade, Parallel, Ladder-lattice, and Lattice realizations.

Analogue linear one- and two-port circuits– basic parameters and descriptive equations, properties.

Analysis of analogue circuits using generalized matrices. Basic elements used in analogue linear circuits

Linear circuits can be described in the complex s-domain (the Laplace domain), frequency domain (the Fourier domain), and time domain.

Let's recap:The Laplace Transform is a powerful tool that is very useful in Electrical

Engineering. The transform allows equations in the "time domain" to be transformed into an equivalent equation in the complex s-domain (s = σ + jω) and vice-versa.

The Fourier Transform describes the system in the frequency domain and can be a specific case of the Laplace transform. The complex Laplace variable s can be separated into its real and imaginary parts: s = σ + jω (σ is the real part of s, ω is the imaginary part of s, and j is the imaginary number). For σ = 0, i.e. s = jω, the Laplace transform is the same as the Fourier transform if the signal is causal.

The variable ω is known as the "radial frequency”, and is given the units rad/s(radians per second)

1. ANALOGUE LINEAR ONE-PORT CIRCUITS

The descriptive parameter/function of a linear one-port circuit is:

Impedance: Complex Laplace Impedance Complex Fourier Impedance

Z(s)=R + jX (sjω) Z(j)= R(jω) + jX(jω)R – resistance; X - reactance

Admittance: Complex Laplace Admittance Complex Fourier Admittance

Y(s) =1/Z(s) =G+ j X -1 (sjω) Y(j)= G(jω) + jS(jω)G =1/R– conductance; 1/X- susceptance

Z(s) and Y(s) F(s) - Immitance function

The descriptive equation of a one-port circuit is Ohm‘s law: )().()()().()(sUsYsIsIsZsU

U

I F(s)

2. ANALOGUE LINEAR TWO-PORT CIRCUITS

Two-port passive circuit parametersThese consider the internal state of the circuit rather

than its working behaviour and are also called static, primary, transducer matrices, internal.

y-parameters - facilitate admittance matching calculations

z-parameters - facilitate impedance matching calculations

h-parameters (hybrid)

f-parameters

a-parameters (inverse transmission), providing electrical inputs

b-parameters (transmission), providing electrical outputs

-

+U1

I1

U2-

+ I2

2

1

2221

1211

2

1

UU

yyyy

II

2

1

2221

1211

2

1

II

zzzz

UU

2

1

2221

1211

2

1

UI

hhhh

IU

2

1

2221

1211

2

1

IU

ffff

UI

2

2

2221

1211

1

1

IU

aaaa

IU

1

1

2221

1211

2

2

IU

bbbb

IU

Two-port circuit transfer functionsThese consider the behaviour of the circuit and not the

unnecessary internal detail and are also called secondary parameters, dynamic, external, working

Tabl. 1-1

-

+U1

I1

U2-

+ I2

Dynamic parameter The Laplace domain

The Fourier domain

1 Voltage transfer function

2 Currency transfer function

3 Transfer Impedance

4 Transfer Admittance

5 Input Impedance

6 Output Impedance

)()()(

1

2

jUjUjTU

)()()(

1

2

jIjIjTI

)()()(

1

221

jIjUjZ

)()()(

1

221

jUjIjY

)()()(

1

1

jIjUjZin

)()()(

2

2

jIjUjZout

)()()(

1

2

sUsUsTU

)()()(

1

2

sIsIsTI

)()()(

1

221

sIsUsZ

)()()(

1

221

sUsIsY

)()()(

1

1

sIsUsZin

)()()(

2

2

sIsUsZout

An interaction between two-port passive circuit parameters and circuit transfer functions exists. Tabl. 1-2

y z h f a

TUT22

T21

Zy1Zy

T11z

T21

ZzZz

Th21

T21

ZhZh

T22

T21

ZfZf

T1112

T

ZaaZ

TU/ZT=22

21

yy

11

21

zz

h

21h 21f

11a1

TITy11

21

Zyy

T22

21

Zzz

T22

21

Zh1h

T11f

21

Zff

T2122 Zaa1

Z21T11y

21

Yyy

T22

21

Yz1z

T22

21

Yhh

Tf11

21

Yff T2221 Yaa

1

Y21T22

21

Zy1y

T11z

21

Zzz

Th11

21

Zhh T22

21

Zff

T1112 Zaa1

ZinTy11

T22

ZyZy1

T22

T11z

ZzZz

T22

Th11

Zh1Zh

T11f

T22

ZfZf

T2122

T1112

ZaaZaa

ZoutИy22

И11

ZyZy1

И11

И22z

ZzZz

И22h

И11

ZhZh

И11

Иf22

Zf1Zf

И2111

И2212

ZaaZaa

3. ANALYSIS OF ANALOGUE CIRCUITS USING GENERALIZED MATRICES

I. The method applies to analogue circuits having direct connection between the input and the output - this node is numbered as “zero”. The other nodes have the numbers from 1 to m starting from the input and moving towards the output.

II. A square matrix Y (m × m) is formed. Each element of the matrix Y is an admittance formed as follows:

- in the main diagonal are the admittances with identical indexes Yii - the sum of admittances connected to the node i in the circuit;

- the remaining admittances Yij are the sum of the admittances placed between node i and node j, with sign minus.

III. The necessary adjugates Δij and the matrix’s determinant Δ are calculated and used to determine the required dynamic parameter via the formulae in Table 1-3 (the second column – for the matrix of admittances Y).

Δij = (-1)i+j Mij, where Mij (referred to as the minor of Y) is the determinant of the (m − 1)×(m − 1) matrix that results from deleting row i and column j of Y.

Dynamic parameter

Connection with the matrix of:admittances impedances

1

2

U

UTU bbaaTaa

ab

Y ,

Tbb

ab

Y

021

2

IU

UTUaa

ab

bb

ab

1

2

I

ITI

Tbb

ab

Z bbaaTaa

ab

Z ,

1

221

I

UZ bbT

ab

Y

aaTbbaa

ab

Y

,

1

221

U

IY aaTbbaa

ab

Z

, bbT

ab

Z

1

1

I

UZin

Tbb

Taabbaa

Z

Z,

bbaaTaa

Tbb

Z

Z

,

2

2out

I

UZ

saa

sbbbbaa

Z

Z,

bbaasbb

saa

Z

Z

,

ZT is the impedance of the load (YT – the admittance);Zs is the impedance of the source.

Tabl. 1-3

NUMERICAL EXAMPLE 1:Determine the voltage transfer function,

when ZT=∞, of the following analogue circuit:Solution:

The admittances matrix [Y] is 3 × 3. The formulae for the voltage transfer function (Table 1-3) is:“a” denotes the number of the input node, i.e. a=1, while “b” denotes the number of the output node, i.e. b=3.For ZT= ∞, i.e. YT=0, the voltage transfer function will be: We calculate the adjugates Δ13 and Δ11 :

Hence, the voltage transfer function is:

R L

C

sLY

sCYR

Y

1;

;1

3

2

1

sLsL

sLsLsC

RR

RR

YYYYYYY

YY

110

1111

011

0

0Y

33

33211

11

bbaaTaa

abU

YsT

,

)(

11

13)(

aa

abU sT

sRLsLR

YYab1111 31

3113

sRL

sRCsCRsL

YYYYYYYYaa1111 213

1123321311

1

11

111

1

)()(

213

31

sRC

sRL

sRCsRL

sCRsL

sRL

YYY

YYsTU

4. BASIC ELEMENTS USED IN ANALOGUE LINEAR CIRCUITS

Active elements - the elements which are capable of delivering energy -Transistors, Triacs, Varistors, Vacuum Tubes, Relays, Solenoids and Piezo-Electric Devices;

Passive elements - the elements which will receive the energy and dissipate or store energy - Resistor, Capacitor and Inductor (linear passive elements) ; Diodes, Switches and Spark Gaps (non-linear passive devices)

Resistor - R, (or Conductance - G =1/R, S)Resistors are circuit elements that allow current to pass through them, but restrict the flow according to a specific ratio called "Resistance". Every resistor has a given resistance. Resistors are commonly used as heating elements.

Potentiometers - resistors that have a variable resistance as a function of position.

Thermisters - resistors that have a variable resistance as a function of temperature.

R

Capacitor - C, F ; ZC(s)=1/sC

Inductor - L, H ; ZL(s)=sL

Ideal Transformer – n (U2=nU1) coefficient of transformation

Operational Amplifiers (Op Amps) - active circuit components; have 2 input terminals and 1 output terminal .

Uout = A(U2 - U1) –descriptive equationAn Ideal Op Amp has: • infinite impedance, bandwidth, voltage gain • zero: output impedance, offset (error)

C

L

SourcesIndependent Sources - produce current/voltage at a particular rate that

is dependent only on time. • Current sources - sources that output a specified amount of current.

The voltage produced by the current source will be dependent on the current output, and the resistance of the load (Ohm's law). An ideal current source has Rin = , . Iout = const in any kind of load

A diagram of an ideal current source, I, driving a resistor, R, and creating a voltage V

• Voltage sources - produce a specified amount of voltage. The amount of current that flows out of the source is dependent on the voltage and the resistance of the load (again, Ohm's law). An ideal voltage source has Rin = 0, . Uout = const in any kind of load.

A diagram of an ideal voltage source, V, driving a resistor, R, and creating a current I

Dependent Sources - these are current or voltage sources whose output value is based on time or another value from the circuit. The following sources are possible:

• Current-controlled current source - Zin = 0; Zout = Descriptive parameter: Descriptive equation: I2 = I1

• Current-controlled voltage source - Zin = 0; Zout = 0Descriptive parameter: z11

Descriptive equation: U2 = z11I1

• Voltage-controlled current source – Zin = ; Zout = Descriptive parameter: y21 = gm.Descriptive equation: I2= gm (U2 - U1)

• Voltage-controlled voltage source - Zin = ; Zout = 0Descriptive parameter: Descriptive equation: U2 = U1

Impedance Converters• NIC-Negative Impedance ConverterDescriptive parameter: Descriptive equation: Z1 =-Z2

One Op Amps realization: =R1/R2

Application: negative element circuits; active analogue circuits• GIC-Generalized Impedance Converter Descriptive parameter: f(s) Descriptive equation:

Application: active analogue circuits, active inductors, capacitor multipliers, FDNC, FDNR, etc.

)()()(

)()()(

42

31 sZsZsZ

sZsZsZ Tin

NICZ1=- Z2

GICZin=f(s)ZT

OpAmp1

OpAmp2

Frequency-dependent negative resistors and conductors

•FDNC – Frequency Dependent Negative Conductivity – (D-element)

•FDNR – Frequency Dependent Negative Resistance – (N-element)

Descriptive parameter: D, F.s Descriptive equation: Realization: 1 GIC, 3 resistors and 2 capacitors.

For example:

CjjDjYDssY DD 22 )(;)(

Descriptive parameter: N, H.s Descriptive equation:Realization: 1 GIC, 3 resistors and 2 capacitors.

For example:

LjjNjZNssZ NN 22 )(;)(

Application: Active analoguecircuitsbasedonBruton transformations

223142

42

3111

)( NssRRRCCR

sCsC

RRsZ TTin

2

22423142

31 1

11

)( DssYDssRRCC

RRRR

sCsCsZ in

TTin

Gyrator – a passive linear two-port entirely non-reciprocal device

Descriptive parameter: Rgy – gyration resistance Descriptive equation: Z1 (s) Z2(s) =Rgy=const (Z1 (s) and Z2(s) are reverse impedances)

An important property of a gyrator is that it inverts the current-voltagecharacteristic of an electrical component or network.

In the case of linear elements, the impedance is also inverted, i. e. a gyrator can make a capacitive circuit behave inductively and vice versa

Realization:1 NIC, 2 resistors and 1 D-element (FDNC):

RgyRgy

Z1=

NIC

Application of a gyrator: It is primarily used in active filter design and miniaturisation.•to transform a load capacitance into an inductance

•to reduce the size and cost of a system by removing the need for heavy and expensive inductors

Representation of analogue immittance functions –analytical representation, pole-zero diagrams, frequency

responses. The Hurwitz test. Frequency and impedance scaling

1. REPRESENTATION OF ANALOGUE IMMITTANCE FUNCTIONS 1.1. Analytical representation in s-domain A circuit transfer function is the ratio of two polynomials N(s) (Numerator) and D(s)

(Denominator):

impedance Z(s)

admittanceY(s)

Limitations of impedance Z(s) and admittance Y(s): The coefficients bi and ak must be positive real numbers. The highest (m and n) and the lowest powers of the numerator and denomi-nator

polynomials (D(s) and N(s)) must not differ by more than one D(s) and N(s) are Hurwitz polynomials. For LC-one-port circuits they are modified

Hurwitz polynomials.Hence, when H(s) is an immitance function (Z(s) or Y(s)), both N(s) and D(s) must

be Hurwitz polynomials.

n

j

jj

m

i

ii

nn

nn

mm

mm

sa

sb

asasasabsbsbsb

sDsNsH

0

0

011

1

011

1

......

)()()(

.

1.2. Pole-zero Diagram

, where

•The poles and zeros can be only be either real or complex-conjugated.•The poles and zeros of the immittances are located in the left half-plane of the complex s-plane and/or on its boundary, the imaginary axis

n

jj

m

i

nn

m

ss

ssH

ssssssa

ssssssb

sD

sNsHi

m

1

10

21

000

)(

)(

))....()((

))...()((

)(

)()( 21

n

m

a

bH

An analogue immittance function can be described in the frequency domain (theFourier domain) by applying the substitution s⟶jω on the analytical expression in thecomplex s-domain (the Laplace domain):

1.3. One-port circuit representation in the frequency domain

jZ

jZ

j

Z1

2

jZ

1

2

magnetostrictive resonator

224

222

223

221

55

331

44

220 )()(

j

HjZsbsbsb

sasaasZ js

)(

)()(22

2

221

j

HjZ

Z

0

)(

)()(22

1

222

HjjZ

N S

L0 CL

C0

L C

The frequency responses Z(jω)/j=Z(ω) of an inductor, capacitor, parallel and serial resonant one-port circuits are:

piezoresonator

Z

0

2. HURWITZ POLYNOMIAL2.1. Strict Hurwitz Polynomial A polynomial P(s) is a strict Hurwitz polynomial whose

• Coefficients are positive real numbers(a necessary but not sufficient condition)

• Roots are located in the left half-plane of the complex s-plane.(a necessary and sufficient condition )A strict Hurwitz polynomial has no any missing term of 's'

A given polynomial can be tested to be Hurwitz or not by using the continued fraction expansion technique.

The Hurwitz strict polynomial test - strict Hurwitz polynomial verification1. The polynomial P(s) is split into a sum of two polynomials A(s) – even, and

B(s) – odd, i.e. P(s)=A(s)+B(s).

nnsasasaasP ...)( 2

210

...)( 44

220 sasaasA

...)( 55

331 sasasasB

2.The ratio or (the polynomial in the numerator is the

one with the higher degree).

3. (s) expands in continued fraction using Euclid’s algorithm:

4. If a) the coefficients Ci (i=1÷m), derived after the continued fraction expansion, are positive real numbers, and b) the total number of these coefficients is equal to the degree of the polynomial, i.e. m = n, the polynomial P(s) is a strict Hurwitz polynomial.

.

sC

sCsC

sCs

m

1...

11

1)(

32

1

)()()(sBsAs

)()()(sAsBs

NUMERICAL EXAMPLE 1:Check if the polynomial is a strict Hurwitz polynomial

SolutionWe determine the polynomials A(s) and B(s):

- even polynomial of 4th order- odd polynomial of 3rd order

Then we define the ratio: ,

and expand it in continued fraction:

Clearly, a) all coefficients C1 ÷C4, derived after the

continued fraction expansion are positive real numbers, andb) the total number of the coefficients is 4,

and the polynomial P(s) is of 4th degree as well.

Therefore, P(s) is a strict Hurwitz polynomial

13)( 234 sssssP

13)( 24 sssA

sssB 3)(

ssss

sBsAs

3

24 13)()()(

.

2.2. Modified Hurwitz Polynomial

A polynomial P(s) is a modified Hurwitz polynomial whose roots are located in the left half-plane of the complex s-plane and/or on the imaginary axis which is its boundary (this is a necessary and sufficient condition )

A modified Hurwitz polynomial always has some missing terms of 's’. Only when the roots of the polynomial are located on the imaginary axis

will the odd or even terms of 's' be missing. The Hurwitz modified polynomial test - modified Hurwitz

polynomial verificationA given polynomial can be determined to be modified Hurwitz or not by

using the continued fraction expansion technique of the ratio:

, where is the first derivative of the polynomial P(s)

The same conditions as those of the strict Hurwitz polynomial test a) andb) must be met to conclude that the polynomial is a modified Hurwitz polynomial.

dssdPsPs )()()(

dssdP )(

NUMERICAL EXAMPLE 2:Check if the polynomial is a modified Hurwitz polynomial

SolutionThe odd terms of s are missing,

so we apply the modified Hurwitz test:

Evidently, a) all coefficients C1 ÷C4, derived after the

continued fraction expansion are positive real numbers, andb) the total number of the coefficients is 4,

and the polynomial P(s) is of 4th degree as well.

Therefore, P(s) is a modified Hurwitz polynomial

86)( 24 sssP

3. FREQUENCY AND IMPEDANCE SCALING

The main dimensions of the electronic elements are not chosen in the best possible way:

Resistor - 1; in practice - 1k=1000, 1M=1000000 are used;Capacitor - 1F, in practice - 1nF=0,000000001F or 10pF=0,00000000001F;Inductor - 1Н, in practice - 1 mH (0,001H) or 1Н (0,000001Н).

The frequency range used in communications is wide – from a few Hz to tens of GHz. This makes the description of the circuits more difficult.

110.210.41010.2

111)( 8216

26

2

sss

sRCLCsRsL

sLRsC

sZ

sradLCp /10.5,01 8

C L

R 100

2H200pF

The main aim is to transform the value of the electronic elements and frequencies so that they are of the order of 1. In this way the coefficients in the transfer function will have values of the same order.

Impedance Scaling

When the impedance is divided by a coefficient kr,

The nature of the impedance doesn’t change :sC

sLRsZ 1)(

'''' 11

1)(

nnn

rrrrrn sC

sLRsCkk

LskR

ksC

sLR

ksZsZ

CkC

kLL

kRR

rn

rn

rn

'

'

'

;

;

Impedance scaling

kr is chosen to be around the average value of the resistors.

Frequency ScalingThe frequency range is scaled by a coefficient kf so that the most

important frequencies of the circuit become close to 1.

To keep the analytic expression of the circuit functions unchanged a multiplication and a division of the coefficient kf is carried out.

fn

fn k

fFfk

;

"

"" 11

n

n

ff

fff

nCj

LjRCk

kj

Lkk

jRk

jZZ

jkj

ksS

CkC

LkL

ff

fn

fn

;

;"

"

Frequency scaling

The scaled impedance is easier to work with because it has better coefficient numbers than the unscaled impedance.

Simultaneous Frequency and Impedance Scaling

fffrn

r

fn

rn

k

sSk

CkkCk

kLL

k

RR ;;;;

NUMERICAL EXAMPLE 3:Let us choose kr = R = 100 and kf = p = 0,5.108 and scale the elements:

./110.5,010.5,0

;110.2.10.5,0.10

;1100

10.5,010.2

;1100100

8

8

1082

86

sradk

FkCkC

Hkk

LL

kRR

f

p

frn

r

fn

rn

C L

R 100

2H200pF

1

1)(2

SS

SSZ110.210.4

1010.2)(8216

26

ss

ssZHzFkf

sradk

Fkk

CC

HLk

kL

RkR

f

f

fr

n

nf

r

nr

,

;/,

;,

;,

;,

Descaling

Passive linear one-port circuits – canonic schemes. LC and RC one-port circuits - basic properties, synthesis

of Foster and Cauer canonic one-port schemes 1. CANONIC REACTIVE (LC) ONE-PORT CIRCUITS

C2

C1

C4

C3

L1

oneDCcircuit

onecircuitforfrequencyω=∞

5inductorsand5capacitors

twoDCcircuits

twocircuitsforfrequencyω=∞

3inductorsand3capacitors

1)1()1(2

adac

aab

aZ1 Z1

Z2

bZ1 cZ2

dZ1

1

)1(

1

2

2

2

aad

aac

aab Z1 Z2

aZ1

dZ1

bZ1

cZ2

1;

)1)(1()(

1;

)1)(1()(

2

2

2

2

bbf

abbad

aae

babac Z1 Z2

aZ1 bZ2

fZ2

cZ1

dZ2

eZ1

abABbaA

bABbBf

bABBABe

bABbBd

bBABABc

4;1

22

;2)(

22

;2)(

2

bZ2

Z1

Z2

aZ1 cZ1 dZ2

eZ1 fZ2

Table 3‐1

No equivalent one-port circuits conditions of equivalence

1.

2.

3.

4.

2. BASIC PROPERTIES OF PASSIVE LC ONE-PORT CIRCUITSEXAMPLE 1:

21

2

22

2

2

221

212

12

22121

3

22

21 11

)(1

)()()(21

ss

sH

CLs

CLLLL

ssL

CLsLLsCLLs

CLssL

sLsZsZsZ CLL

12

0221

210

21

21

22

2

21

1;;1;13,23,2

jCL

jsjCLLLL

jsC

LLLLCL

PROPERTIES WHICH AFFECT CANONIC CIRCUITS1. The difference between the number of inductors and capacitors is no more than one. 2. There is no more than one DC circuit and no more than one circuit for frequency ω=∞.

PROPERTIES WHICH AFFECT RESONANCE1. The number of resonances is one less than the number of elements. 2. The first resonance (in the lowest frequency) is of:

• parallel type if there is a DC circuit;• serial type if there is no DC circuit.

3. When the frequency increases, the resonances alternate (parallel, then serial, then parallel again and so on).

PROPERTIES RELATING TO POLES AND ZEROS1. Each resonance frequency has a corresponding pair of “inner” poles or zeros located on the imaginary axis.2. For s=0 and s=∞ there is always a pole or zero called “outer”.

Characteristic row: s=

s=0 j

‐j

jj

‐j‐j

s=-‐j

j

1 2 3... ..... ...

s=s=0

PROPERTIES RELATING TO Z(j) AND Z(s)1. Analytical expression of an LC impedance consists of its numerator, the squares of the

frequencies of the serial type resonances and the parallel frequencies in the denominator.

22

322

1

224

222)(

HjjZ

‐ полюсвначалото;‐ полюсвначалото;‐ полюсвначалото;

5

53

31

44

220)(

sbsbsbsasaa

sZ ‐ polefors=0

4

42

20

55

331)(

sbsbbsasasa

sZ ‐ zerofors=0

PROPERTIES RELATING TO FREQUENCY RESPONSE

1. Since Z(jω)=jX(ω) and Y(jω)=jS(ω), the frequency responses of a reactance function are

or .

They have a sign and are always increasing functions located in 1 and 4 quadrants.jjZX )()(

j

jYS )()(

22

422

2

223

221)(

jHjZ

squares of the frequencies of the serial type resonances

squares of the frequencies of the parallel type resonances

2. The multiplier H/j indicates the presence of a DC circuit and the first resonance in the frequency ω1 is of serial type. In contrast, the multiplier jH specifies the absence of a DC circuit, and the first resonance ω1 is of parallel type.

3. Analytical expressions of LC-impedances in Laplace domain Z(s) – reactance functions:

squares of the frequencies of the serial type resonances

squares of the frequencies of the parallel type resonances

‐ полюсвначалото;

EXAMPLE 2:

22

322

1

224

222

1)(

LjjZ

2

321

23

21

44

24

22

24

22

35

123

221

2

24

222

2

1)(

ss

sssL

ssss

sLsZ

23

23

12

21

1;1CLCL

3. SYNTHESIS OF LC-DRIVING-POINT FUNCTIONS3.1. PARTIAL-FRACTION EXPANSION OF AN LC-DRIVING-POINT FUNCTION

)(lim);(lim;)(lim

)(

22

00

220

22sF

ss

AssFAssFA

ssA

sA

sAsF

l

slss

l l

l

l

LC impedance describes a one-port circuit comprised of inductors and capacitors.- First LC-form of Foster - partial-fraction expansion of an LC impedance

ll

l

o

ll

l l

l

CLs

sC

sA

sL

sZsZsZ

ssA

sA

sAsZ

1

1

11

)()()(

)(

2

0

220

C0L LlCl

L1C1

HA

CLF

AC

FA

CHAL

l

l

lll

ll

o

,1;,1

,1;,

22

0

EXAMPLE 3:Synthesise the first form of Foster using the following normalised LC impedance:Apply impedance and frequency descaling by coefficients kr=103 and kf=106

respectively.Solution:

2

31)( 2

22

sssssZ

2131lim)(lim

23

2

31lim)(lim

12

31lim)(lim

2

22

2

22

2

22

000

22

22

222

s

sssZs

sA

s

ssssZA

ss

ssssZA

s

l

sl

ss

ss

l

HALFA

C

FA

CHAL o

41

221

;221

1132

2311

;1

21

11

11

0

C0L L1C1

Descaling:

.210

2

1010

2

;25.041

10

10

;32

10

32

1010

32

;110

10

9631

1

6

3

11

9630

6

3

nFkk

CC

mHLkkL

nFkk

CC

mHLkkL

fr

f

r

fro

f

r

- Second LC-form of Foster - partial-fraction expansion of an LC admittance

l

ll

l

ll

l l

l

CLs

Ls

sLsC

sYsYsY

s

sAs

AsAsY

1

11

)()()(

)(

20

0

220

C L0

C1 Cl

L1 Ll

FA

LCH

AL

HA

LFAC

l

l

lll

ll

o

,1;,1

,1;,

22

0

EXAMPLE 4:Synthesise the second form of Foster using the following normalised LC admittance:Apply impedance and frequency descaling by coefficients kr=103 and kf=106

respectively.Solution:

HALFA

CFA

CHAL o 41

221;2

2111;

32

2311;1 2

1

11

11

0

The impedance and frequency descaling:

.61

10

61

1010

61;22

10

10

;21

10

21

1010

21;22

10

10

9632

26

3

22

9631

16

3

11

nFkk

CCmHL

kkL

nFkk

CCmHL

kkL

frf

r

frf

r

2

1

31

23lim)(lim

21

31

21lim)(lim

031

2lim)(lim;031

2lim)(lim

22

22

3

22

2

2

22

22

1

21

2

1

22

22

00022

2

222

2

221

2

ss

sss

ssYs

sA

ss

sss

ssYs

sA

ss

ssssYAsss

ssssYA

ss

ss

ssss

C1 C2

L1 L2

31

2)( 22

2

ss

sssY

3.2. CONTINUED-FRACTION EXPANSION OF AN LC-DRIVING-POINT FUNCTIONEvery LC-driving-point function can be represented as a sum of two functions:

)()()( 21 sZsZsZ )()()( 21 sYsYsY

If Z(s) has a pole for s=∞, then

Now, Z1(s) has a zero for s=∞, i.e. Y1(s)=1/Z1(s) has a pole for s=∞, which can be extracted:

)(1)(33

2 sZsLsY

Here Y2(s) has zero

for s=∞, but Z2(s)=1/Y2(s) has a pole for s=∞ and so on.

)()( 11 sZsLsZ

)(1)(22

1 sYsCsZ

The input impedance is described by the following continued-fraction expansion:The obtained one-port circuit is the first (LP) form of Cauer:

If Z(s) has a pole for s=0, that pole is extracted. The second (HP) form of Cauer is synthesised after

the following continued-fraction expansion is done:

Hence, when an input function (impedance or admittance) is expanded as a continued-fraction:

The following two schemes are derived:

As a rule: In order to synthesise the first (LP) form of Cauer, the input function(impedance Z(s) or admittance Y(s), the one which has pole for s=∞) is expandedas a continued-fraction as the polynomials N(s) and D(s) are presented indescending order of s. To synthesise the second (HP) form of Cauer, the input function having polefor s=0 (impedance Z(s) or admittance Y(s)) is expanded as a continued-fraction asthe polynomials N(s) and D(s) are presented in ascending order of s.

EXAMPLE 5:Synthesise the first and second form of Cauer using

the following normalised LC impedance:Apply impedance and frequency descaling by coefficients kr=103 and kf=106

respectively.

ss

sssssssZ

234

231)( 3

24

2

22

Solution:The characteristic order of Z(s):

a) First form of Cauer:Z(s) has pole for s=∞ and is expanded as a continued-fraction when the polynomialsN(s) and D(s) are in descending order of s:

.61

1061

101061;44

1010

;21

1021

101021;1

1010

9632

26

3

22

9631

16

3

11

nFkkC

CmHLkkL

nFkkC

CmHLkkL

frf

r

frf

r

Descaling:

b) Second form of Cauer:Z(s) has pole for s=0 and is expanded as a continued-fraction when the polynomials

N(s) and D(s) are in ascending order of s:

.551010

;252

10252

1010252

;245

1010

;32

1032

101032

6

3

24

9632

3

6

3

12

9631

1

mHLkkL

nFkk

CC

mHLkkL

nFkk

CC

f

r

fr

f

r

fr

Descaling:

OUTERPOLEANDZEROCOMBINATIONS

CHARACTERISTICROW

IMPEDANCEZ(jω)

FREQUENCYCHARACTERISTICX()

IMPEDANCEZ(jω)

FIRSTFORMOFFOSTERCANONICONE‐PORTCIRCUIT(SERIALTYPE)

SECONDFORMOFFOSTERCANONICONE‐PORTCIRCUIT(PARALLELTYPE)

FIRSTFORMOFCAUERCANONICONE‐PORTCIRCUIT(LOW‐PASSTYPE)

SECONDFORMOFCAUERCANONICONE‐PORTCIRCUIT(HIGH‐PASSTYPE)

4. BASIC PROPERTIES OF PASSIVE RC ONE-PORT CIRCUITS

01

01

2

2121

2

1 111

)(asabsb

CsRRRCRsR

sCR

RsZ RC

222

2

222

21

2221

1

)()(

CR

CRRRRZ RC

011

1

011

1

......)(

asasasabsbsbsbsF n

nn

n

mm

mm

RC

PROPERTIES RELATING TO ANALYTICAL EXPRESSIONS ZRC(s) AND YRC(s)1. The degrees m and n of the polynomials N(s) and D(s)

-for ZRC(s) m=n or m=n-1 (m>n is not possible);-for YRC(s) m=n or m=n+1 (m<n is not possible).

2. The polynomials N(s) and D(s) must be strict Hurwitz polynomials.

PROPERTIES RELATING TO POLES AND ZEROS1. Poles and zeros are real numbers, located on the negative real semi-axis and alternate.2. There is always a pole of impedance ZRC(s) (a zero of the admittanceYRC(s)) which is located

next to or exactly in the coordinate home (s=0). 3. For the frequency closest to infinity or infinity itself (s=) the impedance ZRC(s) may have

zero only, whereas the admittance YRC(s) may have only a pole.4. The first and the last special points (poles and zeros) are always of opposite type.

PROPERTIES WHICH CONCERN FREQUENCY RESPONSES1. Frequency responses are limited functions of the frequency (0<АЧХ<).2. Frequency responses ZRC() and YRC() are purely theoretical in nature.

5. SYNTHESIS OF RC-DRIVING-POINT FUNCTIONS5.1. PARTIAL-FRACTION EXPANSION OF AN LC-DRIVING-POINT FUNCTION

l l

ls

AsC

RsZ0

1)(

C0R

- First RC-form of Foster

l l

lRC s

sAsCR

sY0

1)(

CR

- Second RC-form of Foster

,;F,1)().(lim);(lim1);(lim l

ls

l00s ll

lRClRCs

RCA

RA

CsZsAssZC

sZRl

FA

CA

RsYs

sA

ssY

CsYR l

ll

llRC

ls

lRC

sRC

s l,;,1)(.

)(lim;

)(lim);(lim1

00

EXAMPLE 6:Synthesise the first and second

forms of Foster using the following normalised impedance:Solution:

Check the realisability:

First form of Foster:

Second form of Foster:

312)(

ssssZ

,21

15.0;F21

21

312)3(lim)().(lim

;,21

15.0;F21

21

312)1(lim)().(lim

;031

2lim)(lim1

;031

2lim)(lim

33

33

3s3

s3

11

11

1s1

s1

000

ss

3

1

lRC

lRC

sRC

s

RC

AR

AC

sssssZsA

AR

AC

sssssZsA

ssssssZ

C

ssssZR

FCRRFC

sss

sss

zZsY

,41;,2;,

32;,1

25.0

23

2311

220

Only ZRC(s) may have a pole for s=0. In this case, the impedance is expanded as a continued-fraction provided thatthe polynomials N(s) and D(s) are presented in ascending order of s. As a result the realization called second (HP) RC-form of Cauer is derived, consisting of alternating longitudinal capacitors and transversal resistors, starting with a capacitor. The same canonic realization will be obtained in case of YRC(s) expansion when its

first pole is next to s=0. The first element derived during the synthesis will be a transversal resistor.

5.2. CONTINUED-FRACTION EXPANSION OF AN RC-DRIVING-POINT FUNCTIONOnly YRC(s) may have a pole for s=∞ and if this pole exists, we expand the RC

admittance in a continued-fraction asthe polynomials N(s) and D(s) arepresented in descending order of s. Thus transversal capacitors and longitudinal resistors are derived. Each capacitor corresponds to a pole for s=∞. This leads to a chain-based realisation called first (LP) RC-form of Cauer.Another way to synthesize the LP form of Cauer is to expand the RC impedance in

descending order of s, in this case the lastpole of ZRC(s) is real and is located next to s=∞. The first obtained element now is a longitudinal resistor followed by a transversal capacitor, etc.

The rules for synthesis of the canonic first and second RC-forms of Cauer are summarised in Table 3-2:

the last pole for s=∞ the first pole for s=0 the last pole next to s=∞ the first pole next to s=0

ZRC not possible

ZRC – in ascending order of s

HP form of Cauer

not possible

YRC – in ascending order of s

HP form of Cauer

YRC

YRC – in descending order of s

LP form of Cauer

not possible

ZRC – in descending order of s

LP form of Cauer

not possible

js1= 1=0

j

js1= 1 1

j

Table 3‐2

EXAMPLE 7:Synthesise the first and second form of Cauer

using the following normalised impedance:Solution:

Check the realisability:

First (LP) form of Cauer:Since ZRC(s) has a zero for s=∞, then YRC (s) has a pole for s=∞. Hence, YRC (s)

will be expanded in a continued-fraction when the polynomials N(s) and D(s) are in descending order of s:

312)(

ssssZ

2

34

34

231

2)(

2

2

ssssY

ss

sss

ssZ

Second (HP) form of Cauer:The first pole of ZRC(s) is next to s=0, therefore YRC (s) must be expanded as a

continued-fraction to derive the HP RC-form of Cauer. The polynomials N(s) and D(s) are in ascending order of s:

2

34

34

2

31

2)(2

2

s

sssYss

s

ss

ssZ

Description of analogue linear two-port circuits in thes-domain – transfer function (TF), pole-zero diagrams.Description of analogue linear two-port circuits in the

frequency domain – frequency responses, polar diagrams

1. DESCRIPTION OF ANALOGUE LINEAR CIRCUITS

FOURIERTRANSFORM

dejHjHFthF

dteththFjHF

tj

tj

)(2

1)]([)(

)()]([)(

11

0

LAPLACETRANSFORM

dsesHj

sHLthL

dteththLsHL

st

st

)(2

1)]([)(

)()]([)(

11

jHsHth jsLaplas

Time domain Complex Laplace domain (s-domain) Complex Fourier domain

h(t)– ImpulseResponse;H(s)– SchematicFunction;H(jω)– SchematicFunctioninfrequencydomain.

A linear analogue system (one-, two- or many-port circuit) is fully described if the output signal y(t) (in time domain) or Y(s) (in s-domain), can be calculated for any given input signal x(t) (in the time domain) or X(s) (in the s-domain)

ANALOGUELINEAR SYSTEM

h(t),H(s),H(jω)

x(t) y(t)

X(s) Y(s)

If we have a circuit with impulse response h(t) in the time domain, with input x(t) and output y(t), we can find the Schematic Function H(s) of the circuit, in the Laplace domain, by transforming all three elements:

Therefore, the Schematic Function denoted with H(s), can be defined as either the Laplace-transformed representation of the impulse response,

)().()()()()( sHsXsYthtxty L

)]([)()( thLdtethsH st

or the ratio of the circuit output to its input in the Laplace domain:InputOutput

sXsYsH )()()(

• In case of one-port circuits H(s) is an impedance Z(s) or admittance Y(s);• The Schematic Function H(s) of a two-port circuit is a “transfer” (a ratio of the circuit

input over the circuit output) function Т(s) which can be: a voltage transfer function ТU(s), a currency transfer function ТI(s), a transfer impedance Z21(s), and a transfer admittance Y21(s).

The transfer function can be obtained by one of two methods: Transform the impulse response Transform the circuit, and solve.

When the Laplace transform is applied on the differential equation of the analogue system the transfer function is obtained. H(s) is the ratio of two polynomials N(s) and D(s) having the same coefficients as those in the differential equation below.

N-Numerator; D-Denominator

Limitations of transfer functions T(s) The coefficients ak must be positive real numbers, while bi can also be negative; The degree of the numerator m and denominator n may differ arbitrarily but must always be m ≤ n; D(s) must be a Hurwitz polynomial, but this is not essential for N(s).

n

j

jj

m

i

ii

nn

nn

mm

mm

sa

sb

asasasa

bsbsbsb

sD

sNsT

0

0

011

1

011

1

...

...

)(

)()(

m

i i

i

ij

jn

jj

dttxdb

dttyda

00

)()(

2. DESCRIPTION OF ANALOGUE LINEAR TWO-PORT CIRCUITS 2.1. TRANSFER FUNCTION – ANALYTICAL EXPRESSION

Polynomial TF

011

1 ...)()()()(

asasasa

sb

sD

sb

sDsNsT

nn

nn

mm

mm

Non-polynomial TF

011

1

2222

221

2

...).(..))((

)()()(

asasasa

sssHs

sDsNsT

nn

nn

lm

ReactionAction

)()(

)(1)(

sNsD

sTsG ‐ TransmissionConstant

Transfer functions are powerful tools for analysing circuits. If we know the transfer function of a circuit, we have all the information we need to understand the circuit, and we have it in a form that is easy to work with. When the transfer function is obtained, we can say that the circuit has been "solved" completely.

There are two types of Transfer Functions T(s):

The power of the denominator n is the order of the Transfer Function T(s).

Minimum phase TF – the zeros are on the left half-plane onlyNon-minimum phase TF – some of the zeros are on the right half-planeMaximum phase TF– zeros are positioned on the right half-plane

•The poles and zeros can only be either real or complex-conjugated.•The poles of the transfer functions are located in the left half-plane of the complex s-plane, while zeros can be anywhere in the s-plane

,)(

)(

))....()((

))...()((

)(

)()(

1

10

21

000 21

n

jj

m

i

nn

m

ss

ssH

ssssssa

ssssssb

sD

sNsTi

m

n

m

a

bH where

2.2. POLE-ZERO DIAGRAM

2.3. FREQUENCY RESPONSESA system function can be represented in both the Laplace s-domain and the

frequency domain:

)()()(

)()()(

0

0

jDjNjT

sa

sb

sDsNsT

jsn

k

kk

m

i

ii

)()(

)()(

......

...)(

2'1

21

12

1

54

32

144

22

0

44

33

22

10

BjA

jBA

bbbjbbb

bbjbbjbjN

)()(

)()(

......

...)(

2'2

22

22

2

54

32

144

22

0

44

33

22

10

BjA

jBA

aaajaaa

aajaajajD

Then,

)()()()(

)()()(

212'2

2'2

222

'212

'1

2'2

222

'2

'1

221

'22

'22

'22

'11

'22

'11

jTTBjABA

BAABjBA

BBAA

BjA

BjA

BjA

BjA

BjA

BjA

jDjNjT

evenfunction

oddfunction

The numerator and denominator polynomials are respectively:

2'2

222

2'1

2212

221 )()()(

BA

BATTjTT

MagnitudeResponsethemoduleof T(jω)isanevenfrequencyfunction:

)(21 )()()()()(

j

jseTjTTjTsTHence,

)()(ln)(ln)(ln )( jTeTjT j

dBTT

NpTT

dB

Np

),(lg20)(

),(ln)(

dBNp

NpdB

TT

TT

)(115.0)(

)(686.8)(

dBTTa dBdB ),(lg20)(

LogarithmicFrequencyResponses(forTransferFunctionsТ(s)only)

Attenuation

0 100 200 300 400 5000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

MagnitudeResponse

Frequency

ANALOGUEBAND‐PASSFILTER‐CAUER

0 100 200 300 400 500‐60

‐50

‐40

‐30

‐20

‐10

0

MagnitudeResponseindB

Frequency

ANALOGUEBAND‐PASSFILTER‐CAUER

0 50 100 150 200 250 300 350 400 450 5000

10

20

30

40

50

60

AttenuationindB

Frequency

ANALOGUEBAND‐PASSFILTER‐CAUER

'2

'1

221

'212

'1

1

2)()(

BBAABAABarctg

TTarctg

.

sddtdelay ,)()(

)(b

GroupDelay

PhaseConstant

PhaseResponsetheargumentofT(jω)isanoddfrequencyfunction:

2.4. POLAR DIAGRAM

MATLAB EXAMPLE:Analogue 6 order BP filter has transfer function:

Draw the pole-zero diagram and the frequency responses of the filter.

Solution:Ns=[11.88356.0943e‐13788874.34492.7785e‐84753403994.80182.0515];Ds=[158.70269709.98222661613.66191394.199644.e+6234.81e+8799.9e+10];pzmap(Ns,Ds);

Zero‐poleplot

1082354356

728334135

10.9,79910.81,23419,139410.6,2610.7,69702,580515,210.34,47510.778,210.87,78810.094,6883,11)(

ssssssssssssT

-18 -16 -14 -12 -10 -8 -6 -4 -2 0-300

-200

-100

0

100

200

300Pole-Zero Map

Real Axis (seconds-1)

Imag

inar

y A

xis

(sec

onds

-1)

Magnitude Response in relative units [T,w]=freqs(Ns,Ds,9000);m=abs(T);figure(2); plot(w,m);ylabel('Magnitude Response'); xlabel('Frequency'); title('ANALOGUE BAND-PASS FILTER - CAUER'); axis([0 500 0 1]); grid

Magnitude Response

Logarithmic Magnitude Responsem1=20*log10(m); figure(3); plot(w,m1); ylabel('Magnitude Response in dB'); xlabel('Frequency'); title('ANALOGUE BAND-PASS FILTER - CAUER'); axis([0 500 -60 0]); grid;

Magnitude Response in dB

0 100 200 300 400 500‐60

‐50

‐40

‐30

‐20

‐10

0

MagnitudeResponseindB

Frequency

ANALOGUEBAND‐PASSFILTER‐CAUER

0 100 200 300 400 5000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

MagnitudeResponse

Frequency

ANALOGUEBAND‐PASSFILTER‐CAUER

Logarithmic Attenuationa=-20*log10(m);figure(4); plot(w,a);ylabel('Attenuation in dB'); xlabel('Frequency'); title('ANALOGUE BAND-PASS FILTER - CAUER')axis([0 500 0 60]); grid;

Attenuation in dBPhase Responsefi=angle(T); figure(5); subplot(211),plot(w,fi*180/pi); ylabel('Phase Response in deg'); xlabel('Frequency'); title('ANALOGUE BAND-PASS FILTER - CAUER');axis([0 500 -180 180]); grid;subplot(212), plot(w, unwrap(fi)*180/pi);grid; axis([0 500 -300 300]); ylabel('Phase Response in deg'); xlabel('Frequency');

Phase Response

0 50 100 150 200 250 300 350 400 450 5000

10

20

30

40

50

60

AttenuationindB

Frequency

ANALOGUEBAND‐PASSFILTER‐CAUER

0 100 200 300 400 500

‐100

0

100

ANALOGUEBAND‐PASSFILTER‐CAUER

PhaseResponseindeg

Frequency

0 100 200 300 400 500

‐200

0

200

PhaseResponseindeg

Frequency

1. THE TRANSFER FUNCTION OBTAINED FROM A GIVEN MAGNITUDE RESPONSEVery often the Magnitude Response T() is known and the Transfer

Function T(s) is needed in order to synthesise the circuit. To obtain T(s) from a given magnitude response T() we apply the substi-

tution →s/j in the Magnitude Response power of two T 2()=T(j)2.

Let’s recall from mathematics that:A complex function can be represented: T 2()=T(j)2=T(j)T*(j)

where “*” denotes complex-conjugated; T*(j)=T(-j) for real coefficient functions as all transfer functions are.

Then )()()()()()()()( *2 sRsTsTjTjTjTjTjTsj

The transfer function obtained from a given magnitude or phase response. Frequency responses conditioned by

basic type poles and zeros. Pole-zero diagrams and transfer functions obtained from analogue frequency

responses. Asymptotic Bode diagrams

How to dissociate T(s) from R(s)?

It is known that , then .

D(s) must be a strict Hurwitz polynomial, i.e. the poles of the transfer function are located on the left half-plane of the complex s-plane.

Therefore, the roots of R(s), which are placed there, are roots of D(s), while those on the right half-plane of the complex s-plane are roots of D(-s).

In the case of a minimum phase TF derivation, the roots of the numerator of the R(s) in the left half-plane are roots of N(s), while those on the right half-plane are roots of N(-s).

When the derived TF is a non-minimum phase TF, more than one solution for N(s) and T(s) is possible, providing the same Magnitude Response but different Phase Responses.

)()()(sDsNsT

)()(

)()()()(

sDsN

sDsNsTsT

EXAMPLE 1:Derive a minimum phase and a non-minimum

phase TF having the following Magnitude Response:

Solution:The substitution jω⟶s leadsto:

In the square of the Magnitude Response analytical expression we substitute 2 with -s2 and 6 with ‐s6 :

Then calculate the roots of the numerator and the denominator of R(s):

)()()(

)()(

14

14)( 6

2

6

2222 sR

sDsN

sDsN

ssjT

s

6

2

14)()(

TjT

...;;

;

44

33

22

sjs

sjs

)1)(1)(1)(1()2)(2(

)1)(1()2)(2(

14)( 22336

2

ssssssss

ssss

sssR

The poles in the left half-plane set up the polynomial D(s)=(s+1)(s2+s+1).A minimum phase TF will have

the zero s01 placed on the left side of the imaginary axis:

The only possible non-minimum phase TF having the zero s02on the right half-plane is:

)1)(1)(1)(1()2)(2(

)1)(1()2)(2(

14

)( 22336

2

ssssssss

ssss

sssR

75,05,03,2 js

11 s

201 s

75,05,06,5 js

14 s

202 s

1222

)1)(1(2)( 232

ssss

sssssH MF

1222

)1)(1(2)( 232

ssss

sssssH NMF

2. THE TRANSFER FUNCTION OBTAINED FROM A GIVEN PHASE RESPONSE

Applying the substitution ω=s/j in the analytical expression of the Phase

Response:

we obtain:

The polynomials L(s) and M(s) are related as follows: L(s)+M(s)=N(s)D(-s) Hence, the procedure for obtaining a TF T(s) from a given Phase Response φ(ω) is as follows:

1. Substitute ω=s/j in the analytical expression of φ(ω) and derive the polynomials L(s) and М(s);

2. Calculate the roots of L(s) + М(s); 3. For a minimum phase TF T(s): the roots of L(s) + M(s) placed in the left

half-space are roots of N(s) and those in the right half-space, roots of D(-s)(their mirror images with respect to the jω axis are the roots of D(s));

4. When it comes to a non-minimum TF T(s), there are many polynomials N(s) and D(s) with a different distribution of the poles of L(s)+M(s).

'2

'1

221

'212

'1

1

2)()(

BBAABAABarctg

TTarctg

)(

)(/ sjM

sLarctgjs

EXAMPLE 2:Derive a minimum phase and a non-minimum phase

TF having the following Phase Response:

Solution:In the analytical expression of the Phase Response we substitute by s/j,

2 by -s2, 3 by –s3/j, etc.:

The polynomials L(s), M(s) and L(s)+M(s) =N(s)D(-s) are:

44

6)(

2

3

arctg

)()(

)44(6

446)( 2

3

2

3

sMjsLarctg

sjssarctgarctg j

s

)22)(2(464)().()()( 223 sssssssDsNsMsL

s’3

s1

s2 s’2

s’1

s3

j

1

-j

j

-1 -2 2

The roots of L(s)+M(s) are: s1=1+j; s2=2; s3=1‐j.The minimum phase and non-minimum phase

TF are, respectively:

;464

1)22)(2(

1)()()( 232

sssssssD

sNsTMP

222

)()()( 2

ss

ssDsNsTNMP

44)(

6)(2

3

ssM

sssL

3. MAGNITUDE AND PHASE RESPONSE DERIVATION FROM POLE-ZERO PLOT

n

mn

kk

m

ijsn

kk

m

i

nn

m

abH

sj

sjH

ss

ssH

ssssssassssssb

sTii

m

,)(

)(

)(

)(

))....()(())...()((

)(

1

10

1

10

21

000 21

.

;)(

)(00

0

k

iii

jkk

j

esjsj

esjsj

n

kk

m

iii

k

ii j

n

kk

m

in

k

jk

m

i

j

esj

sjH

esj

esjHjT 11

00

)()(

1

10

1

)(

1

)(0

)(

MagnitudeResponse

PhaseResponse

n

kk

m

ii

110 )()()(Phase Response:

dBsjsjHT kn

k

m

ii

,lg20lg20lg20)(1

01

dB

Magnitude Response:

EXAMPLE 3:

1

01)(

ss

sТ 1

01)(

jjjТ

0101 )(

arctg1

1 )(

arctg

22

22

1

01

1

01)(

jj

jТT101

)(

arctgarctg

2211 j

220101 j- pole - zero

4.FREQUENCY RESPONSES CONDITIONED BY BASIC TYPE POLES AND ZEROS4.1. FREQUENCY RESPONSES CONDITIONED BY

A NEGATIVE REAL POLE

11

11

1)(1)(

jjТ

ssТ js

Magnitude Logarithmic MagnitudeResponse Response

221

11)(

Т dBT ,lg20)( 2211

dB

TT

,62lg20lg202lg20lg20lg20)()(

2

11

121121

12

dBTTT ,lg20)0()(0 1011

dBTT ,32lg20lg202lg20)( 011111

dBT ,lg20)(11

1. 2.3.

To derive the graphic of T()dB we calculate it in several frequencies:

=straightlinewithslope–6dВ/oct

3dBerror

PhaseResponse:

We calculate φ1() in several frequencies:1

1 )( arctg

error

deg0,00)0(01

1

radarctg1.

2.

3.

deg45,4

1)(1

1111

radarctgarctg

deg90,2

)(1

11

radarctg

deg3.8447.11010)10(10

deg7.51.01.01.0)1.0(1.0

1

1111

1

1111

radarctgarctg

radarctgarctg

4.2. FREQUENCY RESPONSES CONDITIONED BY REAL ZERO

Real negative zero

jjTssT

sjs

010010

00

)()(

jjTssT

sjs

010010

00

)()(;Real positive zero

dBT ,lg20)( 22001

010 )(

arctg

Logarithmic Magnitude Response

Phase Response

010 )(

arctg

dBT ,lg20)( 22001

Logarithmic Magnitude Response

Phase Response

4.3. FREQUENCY RESPONSES CONDITIONED BY COMPLEX CONJUGATED POLES

222,1 js

22

220

21

012

2222212

2111)(

aa

asasjsjssssssT

22

220 ac

The denominator is a Hurwitz polynomial.

Frequency of the complex-conjugated poles

Quality factor of the poles1

022

22

2

22

22 1

21

2 aa

QП

Then the TF can be re-written as follows:

2200

0

120122

111)(C

П

C sQ

sasaaasasas

sT

ПQ5.0Small:QП<35Medium:QП=520Large:QП>20ExtraLarge: QП>100

2

221

02

411

lg20)0()(22

11

П

ПQ

QTTaaQ

ПmCm

lg40)()(lg20)( 221

22202 ТaaТ

QП determines the type of the Magnitude and Phase Responses

Always mC 2mb

1. – tгр(ω) – monotone decreasing curve

2. – tгр(ω) – the graphic has

a maximum for the frequency

2220

12 )(

C

П

CQarctg

aaarctg

s

QQd

dt

П

CC

C

П

Cгр ,)()(

2

22222

222

577.031 ПQ

577.031 ПQ

2max 41121П

QC

Qп Т2()max tГР uneven curve

4.4.FREQUENCY RESPONSES CONDITIONED BY COMPLEX CONJUGATED ZEROSComplex conjugated zeros

Frequency of the complex-conjugated zeros

Quality factor of the zeros1

0

0

20

20

0 2 bb

Q

20

2000

bc

2

0

2

012

0000

002

00

210)(

CC sQ

s

bsbsjsjs

sssssT

000 2,1 js

2

0

2

012

0000

002

00

210)(

CC sQ

s

bsbsjsjs

sssssT

000 2,1 js

Complex conjugated zeros on the imaginary axis

4.5. FREQUENCY RESPONSES CONDITIONED BY A ZERO AT S=0

20

2000020 21

)( sjsjssssssT

00 2,1 js

2)(;)(

j

jjTssT .2

)(;lg20)(;)( constdBTT dB

00 s

4.6. FREQUENCY RESPONSES CONDITIONED BY A POLE AT S=0

CONCLUSIONS1. Each zero of the transfer function, when the frequency is high enough,

increases the steepness of the Logarithmic Magnitude Response by 6 dB/oct(20 dB/dec) and raises the Phase Response by π/2 rad (90 deg).

2. Each pole of the transfer function, when the frequency is high enough, decreases the steepness of the Logarithmic Magnitude Response by -6 dB/oct(-20 dB/dec) and reduces the Phase Response by -π/2 rad (-90 deg).

3. When ω→∞ λ= –6(n-m), dB/oct; φ(ω)→ [(m-n) π]/2, rad

.2

)(;,lg20)(;)( constdBTT dB 211)(;/1)(

je

jjTssT

01 s

FREQUENCY RESPONSES CONDITIONED BY REAL POLES AND ZEROSNEGATIVE REAL POLE 1

b=1T(), dB

T0T0-3

-6dB/oct

b=1()

-45

-90

lg

-45/dec

0,1b

10b

NEGATIVE REAL ZERO 1

a=01

T() dB

T01

+6dB/oct

a=01

()90

45

lg

-45/dec

0,1a

10a

A POLE at s=0

()

lg -90

T(),dB

-6dB/oct

1

A ZERO at s=0 T(),

dB

+6dB/oct

1

()

lg

90

20 dB/dec = 6 dB/octThe jumps in the curve are not approximated

PROBLEM:

From a given TF or pole-zero diagram, plot the Logarithmic Magnitude Response and Phase Response via graphical summations.

012

2

21

221

22 )(

asass

sss

ssssssT

5. ASYMPTOTIC BODE DIAGRAMS

Plainly, s1 = –σ1 = –ωb1s2 = –σ2 = –ωb2

Only H is unknown. Using the expression:

we obtain:

012

21212 )(

asasH

ssH

ssssHsT

dBsjsjHT jn

j

m

idB i

,lg20lg20lg20)(1

01

dBTH bb ,lg20lg20lg20 210

INVERSE PROBLEM:

From a given Logarithmic Magnitude Response, determine:1. Poles and zeros of the TF2.Analytical expression of the TF 3.The graph of the Phase

Response

6. Asymptotic Bode diagrams – using MatlabBand-pass Transfer Function High-pass Transfer Function

34)()()(

2

sss

sDsNsT

Ns1=[010];Ds1=[143];bode(Ns1,Ds1);grid

34)()()( 2

2

sss

sDsNsT

Ns2=[100];Ds2=[143];figure(2);bode(Ns2,Ds2); grid

Low-pass Transfer Function

341

)()()( 2

sssDsNsT

Ns3=[001];Ds3=[143];figure(3);bode(Ns3,Ds3); grid

Description of analogue linear circuits in the time domain -impulse and step responses, analogue convolution.

Analogue impulse and step responses due to different transfer function poles. Step responses of general bilinear

and biquadratic analogue transfer functions

1. DESCRIPTION OF ANALOGUE LINEAR CIRCUITS IN THE TIME DOMAIN DIFFERENTIAL EQUATION

When the coefficients aj and bi are known, it is possible to calculate y(t)for a given x(t):

m

ii

iij

jn

jj dt

txdbdttyda

00

)()(

The coefficients aj and bi depend on the parameters of the circuit elements R, L, C, etc.;

n (the order of the derivative of the output signal y(t)) is the order of the differential equation and of the analogue system itself.

Descriptive parameters: the coefficients aj and bi.

Problem: differential equations of high order are difficult to solve.

STEADY-SPACE DESCRIPTIONSpace equation Output equation

IMPULSE AND STEP RESPONSES

The impulse and step responses are related as follows:The impulse response can be used to determine the outputfrom the input through the convolution operation:

dthxthtxty )()()()()(

)()()( ttdttxd BuAx )()()( ttty DuCx

x(t) - state vector, u(t) - input signal vector, y(t) - output signal vector,A, B, C and D - descriptive matrices

0100)( tза

tзаtu

u(t)

t

1 Linearsystem

h(t),g(t),T(s)

x(t) y(t)

The step response g(t) is the output that the circuit will produce when an unit step function u(t) is the input:

dttdgth )( ;)( dtthtg

Linearsystem

h(t),g(t),T(s)

x(t) y(t)(t)

t

0100)( tза

tзаt

The impulse response h(t) is the output that the circuit will produce when an ideal impulse function (Dirac impulse) (t) is the input:

2. ANALOGUE CIRCUITS – TIME-DOMAIN CONSIDERATIONSLAPLACE TRANSFORM INVERSE LAPLACE TRANSFORM

ssTLtg

sTLth)()(

)]([)(1

1

)]([)()]([)(tgsLsTthLsT

LAPLACE TRANSFORM OF SOME OFTEN-USED SIGNALS

0),( ttx )(txLsX 0),( ttx )(txLsX

1s1 t0sin 2

02

0

s

)(t 1 t0cos 20

2 ss

te s1 te t

0sin 20

20

s

t 21s te t

0cos 20

2

ss

tte 22 21

ss tsh 0 2

02

0

s

3. NATURAL AND UNNATURAL PARTS OF THE OUTPUT SIGNAL3.1 . NATURAL AND UNNATURAL PARTS OF AN OUTPUT SIGNAL

Transfer Function:

Input Signal:)()()(sDsNsT

)()()(sDsNsX

x

x

The natural part of Y(s) is determined only by the TF T(s) , while the unnatural part ofY(s) is set by the input signal X(s).

Output Signal

UnnaturalpartNaturalpart

n

jjx

m

ix

n

jj

m

i

x

x

ss

ss

ss

ssH

sDsN

sDsNsXsTsY

ii

1

10

1

10

)(

)(

)(

)(

)()(

)()()().()(

3.2. AN ALTERNATIVE DERIVATION OF THE TIME-DOMAIN CHARACTERISTICS Y(s) expands into a sum of partial fractions:

Apply the Inverse Laplace Transform:

xn

j jx

jxn

j j

jss

Ass

AsY

11)( )().(lim sTssA j

ssj

j

)().(lim sXssA xj

ssxj

xj

)()()()(11

tytyeAeAtysY unnatn

jnat

tsxj

n

j

tsj

ILT x xjj

When x(t)= δ(t), then y(t) has no unnatural part and y(t)=h(t).Therefore h(t) can be derived from the TF T(s), when only the natural part is considered.

4. IMPULSE RESPONSE CONDITIONED BY DIFFERENT TYPES OF POLES Practical (in the left half-plane) and purely theoretical (in the right half-plane)

poles are considered:

Y(s) expands into a sum of partial fractions:

where

The Inverse Laplace Transform is applied:

Grouping the complex-conjugated pairs А3 and А*3, А4 and А*

4, А5 and А*5,

corresponding to the complex conjugated poles according to the rule “the sum of a complex function and its complex-conjugated pair is equal to twice the real part of the complex function“:

01

201

223

2

444422223351

)(

asasasass

jsjsjsjsjsjsssssNsY

44

5

44

5

22

4

22

4

3

3

3

3

5

2

1

10)(

jsA

jsA

jsA

jsA

jsA

jsA

sA

sA

sA

sY

sYssA jssjj

lim

tjttjttjttjttjtjtt eeAeeAeeAeeAeAeAeAeAAty 444422223351554433210)(

tjttjttjtt eeAeeAeAeAeAAty 4422351543210 Re2Re2Re2)(

The complex residiuums are presented as follows:

then

Hence,

,;; 543554433

jjj eee AAAAAA

5444223351 Re2Re2Re2)( 543210

tjttjttjtt eeeeeeAeAAty AAA

544233210 cos2cos2cos2)(

cosRe

4251543

teteteAeAAty

etttt

j

AAA

5. STEP RESPONSE OF A FIRST ORDER TRANSFER FUNCTION5.1. FIRST ORDER TRANSFER FUNCTION (BILINEAR TF):

Step Response is derived

where

Then:

For t=0, t= и t=1/a0=1/σ1, the Step Response is:

i.е. after a time period of t=1/a0, 63% of the step process is complete.

0

01

1

01)(asbsb

s

sHsT

0

10)(1)().()(asA

sA

sTs

sTsXsY

NaturalpartUnnaturalpart

)0()(.lim0

000 T

ab

sYsAs

0

01

0

00101 )().(lim

0 ab

ba

babsYasA

as

0

0

01

0

0

0

10)(asab

b

sab

asA

sA

sY

0,)()( 0

0

01

0

01

teab

bab

sYLtg ta

)0()(63.0)0(37.0)()0()(11

;)(

;)0(0

1

0

01

0

0

00

0

0

1

ggggggeab

bab

ag

at

ab

gt

bgt

τ =1/a0 - time-constant of the Step ResponseRepresented by τ the Step Response will be:The transition process concludes after a time of 5τ, i.e. g(5τ) g(().

Important Result:It was shown that:

From the other side:

Therefore, g(0)= T(∞) and g(∞)=T(0), i.e. if the transfer function Т(s) is known, two values of the Step response g(t) are known as well: g(0) and g(∞).

This applies to every TF and Step Response.

/)()0()()( tegggtg

;)(;)0(0

01 a

bgbg

10

0

0

01 )(;)0(;)( bTabT

asbsbsT

g(t)g()

g(τ)

g(0)

0 5t

0.63[g()-g(0)]

g(0)<g(), т.е. b1<b0/a0 g(t)g(0)

g(τ)

g()

0 5t

0.63[g(0)-g()]

g(0)>g(), т.е. b1>b0/a0

EXAMPLE1:DrawtheStepResponseofatwo‐portcircuithavingTF.Calculateg(0),

g(∞), thetimeτ,ofthetransitionprocess;g(τ), thetimeforthetransitionprocessto

complete.Solution:

Then: g(0)=b1=1g(∞)=b0/a0=1/2τ=1/a0=1/2 – time‐constantofStepResponse

Thetransitionprocesscompletesin5τ=5/2=2.5s

TheStepResponseis:

685.012163.01

)0()(63.0)0()(

gggg

21

sssT

211

21

0

0

1

0

01

abb

asbsb

sssT

g(t)

g(0)=1

g(τ)=0.685

g()=0.5

0 0.5 5=2.5 t

0.63[g(0)-g()]

g(0)>g(), т.е. b1>b0/a0

5.2 STEP RESPONSE OF A SECOND-ORDER TRANSFER FUNCTION (BIQUADRATIC TF)

22

20

0

02

012

2

012

2)(c

c sQ

s

sQ

sH

asasabsbsbsT

HabT

abT

abH

c

2

22

20

0

0

2

2 )(;)0(;

)0()();()0(21

02 TbgTbg

tt ebbbebbbbssTLtg 21

2

021222

1

011212

1221

01 1)()(

Realpoles: ))((

)(21

012

2

ss

bsbsbsT

g(t)g()

g(τ)

g(0)

0 5t

g(t)g(0)

g(τ)

g()

0 5t

QО – zeros quality factorQП – poles quality factor ωО – natural frequency of zerosωC – natural frequency of poles

Complex conjugated poles (s1,2=-2 j2): 2

22

2

012

2

2222

012

2)())((

)(

sbsbsb

jsjsbsbsbsT

tebbbtebbbssTLtg tt

22222

22

201

222

222

022

222

01 sin1cos)()( 22

For t=0 and t= the Step Response is:

)0()(

);()0(

20

22

22

0

222

22

022

222

0

Tbbg

Tbbbbg

c

When Т(s) is of I order, g(t) increases or decreases fluently from g(0) to g(∞).

When Т(s)isofIIorder:А) for two real poles – as for the TF Т(s)

of I order;B) for complex-conjugated poles - g(t)

increases or decreases from g(0) to g(∞), but with an additional harmonic process in frequency ω2.

te 2

2

EXAMPLE2:DrawtheStepResponseofatwo‐portcircuithavingTF.Calculate

g(0)andg(∞).Solution:

Then:g(0)=b2=1g(∞)=b0/a0=1/2

Wecalculatethediscriminantofthedenominatorpolynomiala2s2+a1s+a0=0.

TheTFhasapairofcomplex‐conjugatedpoles,i.e.theStepResponseis:

24.0

12

2

ss

ssT

211

24.01

0

0

2

012

2

012

22

2

abb

asasabsbsb

ssssT

02.1.44.04 202

21 aaaD

g(t)

g(0)=1

g()=0.5

0 t

6. IMPULSE AND STEP RESPONSE OF AN LP FILTER

τИ

h(t)1

0.5

0 τЗ t

H

0 τЗ τУ t

0.01g(t)1

0.9

0.5

0.1

Delaytime τз – thetimetakenforh(t)toreachmaximumvalue orthetimefor g(t)to increasestoone‐halfofitssetvalueinthefilteroutput;WidthoftheImpulseResponse τИ– thewidthof h(t) at0.5ofthemaximumvalue;RisetimeoftheImpulseResponse τН– thetimethatg(t) increasesfrom 0.1to 0.9of itssetvalue;ThemaximumreboundoftheStepResponse ‐ γ Settlingtimeofthetransitionprocess τУ – thetimeforreboundsoftheStepResponsetoreduceto 1%ofitssetvalue.

Elmerformula

0

2

0

220

21

20

21

0

1

0

13 22;

aa

bb

b

b

a

abb

aa

H

ThelinkbetweenthebandwidthBWofthefilteratthe3dB level and thewidthoftheImpulseResponse:τИ fГ ≥0.318, fГ ‐ cut‐offfrequencyinHz.

WhenthereboundsoftheStepResponseofanLPfilterare γ< 5%, then:τН ωГ ≈ 2.2,respectively τН fГ ≈ 0.35,whereωГ = 2fГ,rad/s.

AnLPfiltershouldhaveacut‐offfrequency fГ notlessthan0.35τ,inordertopassanimpulsewhichisτwide.

7. STEP RESPONSE OF SECOND-ORDER SECTION TFReal poles

Complex-conjugated poles

EXAMPLE 3:AnanalogueBPfilterof6thorderhasthefollowingtransferfunction:

Drawthetimeresponsesofthefilter.Solution:

Ns=[0.0340897.74e‐180.063161‐6.019e‐180.012909‐3.6517e‐19];Ds=[10.279872.0005‐0.365671.2311‐0.105990.23305];

23305,010599,02311,136567,00005,227987,010.6517,3012909,010.019,6063161,010.748,7034089,0)( 23456

1921834185

ssssssssssssT

ImpulseResponseimpulse(Ns,Ds); grid;ylabel('amplitude(volts)');xlabel('time');title('IMPULSERESPONSEOFANANALOGUEBPFILTER');

StepResponsefigure(2);step(Ns,Ds);grid;ylabel('amplitude(volts)');xlabel('time');title('STEPRESPONSEOFANANALOGUEBPFILTER');

0 0.2 0.4 0.6 0.8 1‐0.2

‐0.15

‐0.1

‐0.05

0

0.05

0.1

0.15

0.2STEP RESPONSE OF AN ANALOGUE BP FILTER

time (seconds)

ampl

itude

(vol

ts)

0 0.2 0.4 0.6 0.8 1‐30

‐20

‐10

0

10

20

30IMPULSE RESPONSE OF AN ANALOGUE BP FILTER

time (seconds)

ampl

itude

(vol

ts)

An introduction to two-port passive circuit synthesis -terms of realisability. Properties and limitations of

structures with different elements and topology.Different types of two-port passive circuits. Synthesis of

non-terminated L-shaped and lattice schemes

Thetwo‐portcircuitsynthesisperformsthefollowingproceduresandintheorderspecified:1.Checkthetermsofrealisability – toassesswhetheritispossibletosynthesisetheschemeusingknownmethodsprovidedthattherequirementsoftheschemetosynthesiseareset.

2.Approximation – specifiedrequirementsaresubmittedbyfunctionsdescribingthetimeandfrequencycharacteristics.Forexample, toderivethetransferfunctionfromgivenrequirementsforMagnitudeResponse.

3.Synthesis – schemederivationanddimensioning.Checkingthetermsofrealisabilityisthefirstprocedure‐ thisisvery

importantbecauseitverifiestherealisabilityoftheschemesynthesis.

1. TERMS OF REALISABILITY Checkingthetermsofrealisabilityisthefirstprocedure;thisisveryimportantasit

verifiestherealisabilityofthecircuit.Themostcommonly‐usedtermsofrealisabilityareasfollows:I.Theprincipleofcausality – systemreactioncannotprecedetheactionthatcausesthatreaction.Mathematically,theprincipleofcausalityisdescribedbythePaley‐Wienertheorem, whereТ() istheMagnitudeResponse:

II. Conditionforstability – whenthedenominatorpolynomialD(s)isaHurwitzpolynomialwhoseroots(polesoftheTF)arelocatedinthelefthalf‐planeofthecomplexs‐plane,theTFoffersastablecircuit. Otherwise,thesynthesiswillresultinanunstableschemewithoutactualpracticalapplicability.

III. Requirementforthecoefficients (bi and ak)andthedegreeofthepolynomials(m and n):

• Thecoefficients akmustbepositiverealnumbers(D(s)mustbeastrictHurwitzpolynomial),whilebi canalsobenegative (N(s)maynotbeaHurwitzpolynomial).

• Thedegreeofthenumeratorm anddenominatornmaydifferarbitrarilybutmustalwaysbem ≤n.

dT

021)(ln

n

k

kk

m

i

ii

nn

nn

mm

mm

sa

sb

asasasabsbsbsb

sDsNsT

0

0

011

1

011

1......

)()()(

2.BASICPASSIVESTRUCTURES2.1.SYMMETRICANDASYMMETRICSTRUCTURESParametersmeasuredattheinputterminalscoincidewiththosemeasuredattheoutput

terminals.Obviously,thesymmetryisatthetransverseaxis ofthetwo‐portcircuit.

symmetric asymmetric

2.2.BALANCEDANDUNBALANCEDSTRUCTURESThesymmetryisatthelongitudinalaxisofthestructure.

balanced unbalanced

2.3.GROUNDEDCIRCUITSOftenunbalancedstructureshaveadirectconnectionbetweentheinputandoutput

terminals,thustransformingthetwo‐port(four‐terminal)circuitintoathree‐terminalcircuit(groundedcircuit).

Whenbalancedandgroundedcircuitsareincorrectlyconnectedbypassingofelementsispossible,suchastheZa(s)intheschemebelow (betweenthegroundedterminalsofthefirstandthethirdcircuitsthereisashortcircuit).Incaseslikethis, anoptocoupler ordecouplingtransformer(idealtransformerwitha1:1

ratio) isused tоconnectbalancedandunbalancedcircuits.

3.PROPERTIESANDLIMITATIONSOFSTRUCTURESWITHDIFFERENTELEMENTSANDTOPOLOGY

4.DIFFERENTTYPESOFTWO‐PORTPASSIVECIRCUITREALISATIONS4.1.DIRECTREALISATIONS

ThegivenTFissubjectedtosynthesisprocedures, leadingdirectlytoacircuitrealisation.I.DirectLatticeRealisation

ТF ofidlingmode

TFinactiveloadmode RTн=1:

II.DirectT‐shaped realisation with bilateralcoherentloadsТ‐shaped realisation is used in the synthesis of amplitude

and phase correctors.When Z1(s)Z2(s)=R

2=const,thecharacteristicimpedanceswillbeactiveandidentical.TheTFwillbeasfollows:

.. 2121 constRZZZZZ Тcc )(

)()(

)(2

2

1 sZRsZ

sZRRsT

constRZZZZZ abMcc 021 .

ab

abПХU ZZ

ZZzzsT

11

21)(

HUba

ba

T

TU sТ

YY

YY

y

y

RyRysТ

HH

HH

H

H )(211

)(22

21

22

21

011

1

2222

221

2

...).(..))((

)()()(

asasasasssHs

sDsNsT n

nn

n

lm

III.DirectLadderRealizationŻi canbeeitherasingleelement(inductororcapacitor)oraparallel/serialoscillatingcircuit.

PolynomialRealisation(Żi isasingleelement)

• Lowpass (LP) TF:

• Highpass (HP) TF:

• Bandpass (BP) TF:

Non‐polynomialRealisation(Żi isaparallelorserialoscillatingcircuit)

• Lowpass (LP)TF –m=0;0≤2l<n generallimitation: m+2l≤n• Highpass (HP)TF –m+2l=n; m>0• Bandpass (BP)TF– 0<m+2l<n; m>0• Bandstop (BS)TF –

011

1 ...)()()()(

asasasasb

sDsb

sDsNsT n

nn

n

mm

mm

011

1

0...

)(asasasa

bsT nn

nn

011

1 ...)(

asasasasbsT n

nn

n

nn

011

1 ...)(

asasasasbsT n

nn

n

mm

022

221 ...;2;0 anlm l

4.2.CASCADEREALISATIONSTheTFfactorises intofirstandsecondordermultipliers(factors):

Eachfactorissynthesised andthederivedsectionsareconnectedinacascade.I.Cascade‐coherentimplementation

All k sectionshavethe sameinputandoutputimpedancesR0,andthelastsectionisloadedcoherentlyRТ=R0.

II.Cascade‐untiedimplementation“Untying”meanstoeliminatetheinterferencebetweentwoormorecircuits.

EachsectioninaCascade‐untiedimplementationistheloadforthenextoneandaccomplishesitsownTF,MagnitudeandPhaseResponseastheywouldhavebeenrealizediftheothersectionweremissing. Theaimistoprovideextremeload: shortcircuit(ZТ≈0)oranidlingmode(ZТ =∞)

)(...)()()()()( 21 sTsTsHTsDsNsT k

buffer stage

highimpedanceinputtooutputlowresistance– eachsectioninthecascaderealisation operatesinidlingmode.Thismodeisoftenused,sincetheactiveschemeswithOpAmpseasily achieveZin→∞and Zout≈0.

lowresistanceinputtooutputhighimpedance – shortcircuitattheoutput

Inordertosatisfytheuntyingconditions,usually“bufferstages” areused.Theycanberesistorgroups,dependentsourcesoractiveschemes.

)()();()( 21 jZjZjZjZ outinoutin

)()();()( 21 jZjZjZjZ outinoutin

5.SYNTHESISOFNON‐TERMINATEDL‐SHAPEDANDLATTICESCHEMES5.1.SYNTHESISOFNON‐TERMINATEDL‐SHAPEDREALISATIONS

Non‐terminatedschemesusuallyworkinidlingmode.FortheTF:

themathematicallycorrectsolution:

isanelectrotechnically‐absurdsolutionsinceZ1(s)andZ2(s)mustbefractional‐rationalfunctions.This introducesauxiliarypolynomialL(s),suchthatbothimpedancesareofRC‐type:

,)()(

)()()(

)()()(

21

2

1

2sDsN

sZsZsZ

sUsUsT ПХ

)()()()()()()()(

)()(21

21

2 sNsDsZsDsZsDsZsZ

sNsZ

)()()()(

)()()(

1

2

sLsNsDsZ

sLsNsZ

EXAMPLE1:Synthesise anon‐terminatedL‐shaped two‐portcircuithavingthefollowingTF:

Solution:

CheckingthetermsofrealisabilityshowsthatthezerosandpolesoftheTFT(s)arelocatedcorrectlyforRCladderschemes: zeros: s01=0,s02= ∞,

poles:p1=‐0.36,p2=‐1.39.

Then:

)()(

274)( 2 sD

sNss

ssT

)()5,0)(1(4

)(264

)()()()(

)()()()(

21

2

sLss

sLss

sLsNsDsZ

sLs

sLsNsZ

-1

j

-0.5

j

0 Z2(s)

Z1(s)

-0.7

j

0Z2(s)

-1

j

-0.5Z1(s)

-0.7

)7,0()5,0)(1(4

)()5,0)(1(4)(

7,01

)7,0()()(

1

2

ssss

sLsssZ

ssss

sLssZ

Z1(s) issynthesised byfirstformofFoster:

1

'

'0

'1

1)7,0(

)5,0)(1(4)(l l

lsA

sCR

sssssZ

;35,086,2186,2

)7,0(5,0)(1(4lim)(.lim1

;4)7,0(

)5,0)(1(4lim)(lim

'0

01

0'0

s1

s'

FCss

ssssZsC

sssssZR

ss

49,07,034,0

94,234,01

34,0)5,0)(1(4lim

)()7,0(lim

'1

'1

0,7s

10,7s

'1

R

FC

sss

sZsA

Theone‐portcircuitZ1(s)is:

R’=4

C0’=0.35F C1’=2.94F

R1’=0.49

Z2(s)issynthesisedbyfirstformofFosteraswell:

1

"

"0

"2

17,0

1)(l l

lsA

sCR

ssZ

;07,0

lim)(.lim1;07,0

1lim)(lim0

20"

0s2

s"

sssZs

CssZR

ss

43,17,01;1

17,07,0lim

)()7,0(lim

"1

"1

0,7s

20,7s

"1

R

FC

ss

sZsA

C1”=1F

R1”=1,43

Theone‐portRC‐circuitZ2(s)is:

Thus,thesynthesised L‐shapedRCtwo‐portcircuithavingTFT(s) willbe:

R’=4

C0’=0,35F C1’=2,94F

R1’=0,49

C1”=1F

R1”

1,43

Optimalsynthesiswithcutsofzerosandpoles: )1()( sssL

-1

j

0 Z2(s)

-1

j

-0.5 Z1(s)

ss

ssss

sLsssZ

ssss

sLssZ

)5,0(4)1(

)5,0)(1(4)(

)5,0)(1(4)(

11

)1()()(

1

2

R’=4 C0’=0.5F

C1”=1F

R1”

1

5.2.SYNTHESISOFNON‐TERMINATEDLATTICESCHEMESThetransferfunctionidlingexpressedbyz‐parametersand

impedancesZa(s)and Zb(s)is:

1. AnauxiliarypolynomialL(s)isintroduced.ItdividesthenumeratorN(s) anddenominatorD(s) polynomials:

2. AnalyticalexpressionsofZa(s),Zb(s),z11 and z21 aredetermined:

If T(s)iseligibleforRC‐realization,whichisthemostcommoncase,auxiliarypolynomial L(s)ischosensothatZa(s),Zb(s) and z11 areRC‐impedances.3. Za(s) and Zb(s) aresynthesisedbyanyofthemethodsforthesynthesisofRC‐

impedances.

)()()(

11

21пх sD

sNZZZZ

zz

sTab

abU

ab

abU ZZ

ZZ

sLsDsLsN

sT

)()()()(

)( пх

.)()(;

)()()()(

;)()(;

)()()()(

21

11

sLsNz

sLsNsDsZ

sLsDz

sLsNsDsZ

b

a

EXAMPLE2:Synthesiseanon‐terminatedlatticetwo‐portcircuithavingthefollowingidling

transferfunction:

Solution:CheckingthetermsofrealisabilityshowsthatthezerosandpolesoftheTFT(s)are

locatedinplacesappropriateforRC‐latticeschemes. zeros:s01,02= 1± j (intherighthalf‐plane), poles:p1=‐0.32,p2=‐4.68(onthenegativerealsemi‐axis).Then,Za(s),Zb(s)and z11 willbeasfollows:

310222)( 2

2пх

sssssT

9,11;1,0zeros)(

)9,11)(1,0()(112

)()()( '

0'0

2

21 sssLss

sLss

sLsNsDZa

67.1;1zeros)(

)67,1)(1(3)(583

)()()( ''

0''0

2

21 sssLss

sLss

sLsNsDZb

68,4;32,0zeros)(

)68,4)(32,0()(

3102)()( '''

0'''0

211 21 ss

sLss

sLss

sLsDz

L(s) can be of second- or third-order. We choose the lowest possible order, i.e. second.

Thepole‐zerodiagramshelptochooseL(s) correctly.

)()68,4)(32,0(

11 sLssz

)()67,1)(1(3

sLssZb

)()9,11)(1,0(

sLssZa

-1.67 ‐1

j

j

j

-1

-1

-0.32

-0.1Za

Zb

z11

-4.68

-11.9

Apossiblechoiceis: ,.

Then:

)5.1())(()( 21 sssssL

)5.1(583)(;

)5.1(112)(

22

sssssZ

sssssZ ba

The following scheme is derived after the synthesis on the first form of Foster.

Clearly, there are too many elements for second order realisation.

SupposingtherootsofthepolynomialL(s)areselectedsothatsomeofthezerosofZa(s)or Zb(s) arecancelledout,theschemerealisation willbesimplified.WhenL(s)=s(s+1),thezeros01=–1 of Zb(s) andtheroot‐1ofL(s)willcanceleachother.

Za(s)and Zb(s)aresynthesizedviafirstformofFoster:

For Za(s)wederive:

.)67,1(3)1(

)67,1)(1(3583

;)1(

)9,11)(1,0(112

2

2

2

2

ss

ssss

ssssZ

ssss

ssssZ

b

a

;84,019,1)1(

)9,11)(1,0(lim)(.lim1

;1)1(

)9,11)(1,0(lim)(lim

0000

ss

FCssssssZs

C

sssssZR

sa

s

a

81,9181,9

1,081,9

81,9)9,11)(1,0(lim

)()1(lim

1

1

1s

1s1

R

FC

sss

sZsA a 1

For Zb(s)wederive:;3)67,1(3lim)(lim

ss'

sssZR b

.2,001,5)67,1(3lim)(.lim1 '0

00'0

FCssZsC s

bs

C 0=0,84FR 1=9,81

R’1=3C 1=0,1F

R=1

C’0=0,2FTheschemeofthenon‐terminatedlatticetwo‐portrealisation willbeasfollows:

Amplitude correction – basic considerations, TF, cascade realisation, specifics of approximation. Passive and active

amplitude-correction sections of first- andsecond- order. Attenuators

1.AMPLITUDECORRECTION– BASICCONSIDERATIONSOnpassingthroughelectricalcircuits,signalssufferfromlineardistortionsconsistingof

spectralcomponentratiochanges,calledamplitude‐frequencydistortions.WhentheMagnitudeResponseisafrequency‐independentconstantТ(ω)=T0=const.;

a(ω)=a0=const suchdistortionsaremissing.Thisconditionofundistortedtransmissionisusuallylimitedtoacertainrangeoffrequencies– theoneinwhichthespectrumofthetransmittedsignalislocated.Fig. 8‐1 showsthecorrectionofaband‐pass‐typecircuitwithattenuation аb(ω).Tomakethisattenuationa0=const.,itisnecessaryforeachfrequencytoаb(ω) toadd

attenuationequaltothedifferencebetweenthedesireda0 andthepresentаb(ω).

Thetwo‐portcircuitrealisesthatattenuation:аk(ω)=a0‐аb(ω) is calledanamplitudecorrector.

circuit amplitudecorrector

Fig.8‐1

ω

a(ω)

ak(ω)ab(ω)

a=ab+ak=a0

2.AMPLITUDE‐CORRECTIONANALOGUETRANSFERFUNCTIONS

j

jj

j

jj

jj

ji

ii

sQ

s

sQ

hsH

bscsHsT

i 20

02

20

02

0

0

)(

Amplitudecorrectorscanbesynthesisedasdirect orcascade structures. Inpractice,cascade‐coherentimplementationisthemostoftenused.

Forthispurpose,theamplitudecorrectorTF,whichisalwaysofpolynomialtype, factorisesintofirst‐ andsecond‐ordertransferfunctions,whicharesynthesisedasfirst‐ andsecond‐ordersectionsconnectedcascade‐coherently.

2.1.FIRST‐ORDERAMPLITUDE‐CORRECTIONTRANSFERFUNCTION

00 cb 00 bc 0

0)(bscsHsTI

a∞

a(ω)

a∞/2

H=b0/c0

ωω0ω

T(ω)

c0/b0

1

σ

jω

‐b0 ‐c0

HP type LPtype

σ

jω

‐b0‐c0

ω

1T(ω)

c0/b0

ω

a0

a(ω)

a0/2

H=1

ω0

20

02

20

02

200

2

200

2)(

s

Qs

sQ

hsH

bsshbssHsTII

Second‐orderTFofamplitudecorrectorsaretypicalbiquadraticfunctionsasfollows:2.2.SECOND‐ORDERAMPLITUDE‐CORRECTIONTRANSFERFUNCTION

h>1–band‐passtype;TheMagnitudeResponseT(ω)hasamaximum Hh forω=ω0.

Twocasesarepossibledependingonthevalueofthecoefficienth:

h<1–band‐stoptype;TheMagnitudeResponseT(ω)hasaminimum Hh forω=ω0.

h

T(ω)

1

ωω0 ωω0

a(ω),Np

a0/2

a0

ω’ ω”

h>1H=1

h>1H=1/h

h

1T(ω)

a0/2

a(ω),Np

a0

h<1H=1

h<1H=1

ω’ω0ω”ωω0ω

3.AMPLITUDE‐CORRECTIONCIRCUITAPPROXIMATIONSincethedesignoftheamplitudecorrectorsuses the knownattenuation

curve aв(ω) orMagnituderesponseTв(ω) ofthecorrectedcircuit,approximatingtheamplitudecorrectordoesn’t pose a serious mathematical problem. The curve of the corrector aк(ω) or Tк(ω), canbeapproximatedbystandardfunctions.

In practice, it is simply the aggregation of the attenuation curves of the first- and second-order sections. By summation of the four curves, given on slides 2 and 3, arbitrarily complicated curves for the amplitude corrector’s attenuation can be obtained. Thus a cascade realisation transfer function is derived and the sections in the cascade are synthesised.

Amplitudecorrectorsandtheir constituent sectionsmusthaveconstantcharacteristic impedance,inordertobeconnectedcoherently to each other and to the corrected circuit.

Thelatticerealisation(fig.8‐2) hasthesepropertieswhenŻа and Żb arereciprocalimpedances,i.e.describereverseone‐portcircuits.

LatticerealisationscaledTFis:

11

11

)(

н

н

н

н

b

b

a

aнU Z

ZZZ

sТ Fig.8‐2

4.PASSIVEAMPLITUDE‐CORRECTIONCIRCUITSТ‐shapedrealisation(fig.8‐3) isveryoftenusedas

anamplitude‐correctionsection.WhentheimpedancesZ1(jω)and Z2(jω)describe

reversetoRone‐portcircuits,theT‐shapedcircuitTFis:

ForagivenTF(first‐ orsecond‐order)whichisasectionofacascaderealisationamplitude‐correctionTF,andwhenRL=RS=R(L‐Load,S‐Supply),synthesising cascadesectioncircuits comesdowntodeterminingtheimpedancesZ1(s)andZ2(s).TheseimpedancesusuallyresultinLC or RLC one‐portcircuits:

L‐shapedrealisation(fig.8‐4)isalsosometimesused.Ż1and Ż2 againmustbereverseimpedances.

.)(

)()(

)(2

2

1 sZRsZ

sZRRsT

.)(1

)()(

)(;)()(1)(

1

221 R

sTsT

sZRsZ

sTsTRsZ

Fig.8‐3

Fig.8‐4

R R

Ż1

Ż2RRR

R

Ż1

Ż2 RR

4.1.PASSIVEAMPLITUDE‐CORRECTIONFIRST‐ORDERSECTIONS

000

0

0

0)( bcbscs

cbsTLP

.,;,

;,11;,1

00

002

00

02

001

0

01

FcRbbcC

bcRbR

Hcb

RLbcRR

.,;,

;,;,1

00

02

002

0

001

001

cbRcRH

cbRL

ccbRRF

cbRC

Т‐shapedrealisationsmayhavetheirzerosanywhereinthecomplexs‐planeexceptfortherealaxis.

;)( 000

0 bcbscssTHP

PassiveAmplitude‐correctionsectionoffirst‐orderandofHPtype

PassiveAmplitude‐correctionsectionoffirst‐orderandofLPtype

R R

R1

R2

L2

C1

R R

R1

R2

C1

L1

00 bc

00 bc

4.2.PASSIVEAMPLITUDE‐CORRECTIONSECOND‐ORDERSECTIONS

Qbh

bsshbss

sQ

s

sQ

hssTBS

1,1)( 200

2

200

2

20

02

20

02

HhRbLF

hRbC

hhRR ,1;,

11;,1

01

011

HhbRLF

RhbC

hRhR ,

1;,1;,

1 02

022

Passiveamplitude‐correctionsectionofsecond‐orderandofBPtype:

Qbh

bsshbss

hsQ

s

sQ

hs

hsTBP

1,111)( 200

2

200

2

20

02

20

02

HbhhRLF

hRhbChRR ,1;,

1;,1

01

011

HhRhbLF

RhbhC

hRR ,

1;,1;,

1 02

022

Passiveamplitude‐correctionsectionofsecond‐orderandofBStype:

R R

R1

R2

L2

C1 L1

C2

11

Hh

hHh11

R R

R1

R2

L2

C1

C2

L1

5.ACTIVEAMPLITUDE‐CORRECTIONSECTIONS5.1.ACTIVEAMPLITUDE‐CORRECTIONFIRST‐ORDERSECTIONS

The individual АRC‐sections connect on the principle „high impedance input to lowresistance output“, while only the input and the output of the complicated cascaderealisation ensure coherent connection. The easiest way to provide coherent connectionis the use of buffer stages before the first and after the last section in the cascaderealisation.

First‐order TFs have poles and zeros on the negative real axis and can besynthesised as RC L‐shaped structure after which a buffer stage with a variable gaincontrol OpAmp (Operational Amplifier) is connected.

000

0)( bcbscssTHP

.1,1, 200

10

001

R

cbC

ccbR .1,,

00202001 cbCbRbcR

000

0

0

0)( bcbscs

cbsTLP

Fig.8‐5 Fig.8‐6

5.2.ACTIVEAMPLITUDE‐CORRECTIONSECOND‐ORDERSECTIONS

Amplitude‐correctionTFof band‐pass andband‐stop typecanbesynthesisedviathesamebiquadraticactivescheme(fig.8‐7).Thetype(BPorBS)isdeterminedbythevalueofthecoefficienth.

TheTFofthesectioninfig.8‐7isscaledbyfrequencysothatω0=1.

21

54

21

615

1

4

213

2

21

543

62

54

1

1

213

2

2

6200

2

200

2

11

11

1)(

CCGG

CCGGG

CG

CCGss

CCGGG

GGGG

CG

CCGss

GG

bsshbssHsT

When we equalise the coefficients of the identical powers of s of the biquadratic TF andthe section TF, the elements of the scheme will be:

pp

pppnn

np

p

n

p

np hbCGGC

bCGGCGG

HHGG

HGGG

GCG

HGGG

GGhbCGCG

22

22

642

3522

1 22

;1;;;;2

where C1+C2=C and Gp arearbitrarilychosen.WhenanindependentadjustmentoftheTFparametersisneeded,schemeswith2

or4OpAmps areused.

Fig.8‐7

C1

R2R1

C2

R3

R4

R6

R5

6.ATTENUATORSAttenuators,aswellasamplitude‐correctioncircuits,impact on the amplitudes of the

frequency components of the signal.Attenuatorshavepermanentattenuationforallfrequenciesandpermanentinput

andoutputimpedances,usuallyidenticaltoeachother.Symmetricattenuatorsarecalledextenders becausetheycausetheworkingeffectof

longercircuit(circuitwithhigherattenuation).Dependingonthetypeofelements,theattenuatorsarereactive andresistive.

Intelecommunications,resistiveattenuatorsaremostoftenused.ThemostsimpleattenuatorisanL‐shapedvoltagedivider

(fig.8‐8а).Itcanberealisedasapotentiometer(Fig.8‐8b)whoseinputandoutputimpedancesshift as theattenuationchanges.

shaRRathRR T

T 21 ;2

222

;2 21acthRRshaRR П

П

T‐ andП‐shapedschemesaremoreoftenused.Theyaredepictedinfig. 8‐9аandb respectively,togetherwiththeformulaefortheirsynthesis(а,Npandtheinputimpedances RT or RП aregiven).Forhigherattenuations(over 5‐10Np)resistorshaveverysmallvalues.

Henceattenuatorswithhighattenuationaresynthesisedascascadeschemes.

Fig.8‐8(а) (b)

Fig.8‐9(а) (b)

R1/2

R2 ŻTŻTR1/2

ŻR1

2R2 2R2Ż

R1

R2ŻПŻT

R1R2

R

.1

;1 21

aa

eRRReR

ThecircuitsinFig.8‐9аand b cannotbeadjustable,becausechangingthevalueofoneoftheelementstochangetheattenuationwillalsochangetheinputimpedances.

Keepinginputimpedancesunchangedispossibleonlyifallthreeresistorsarechangedatonce,whichisdifficult.

Agoodsolutionforanadjustableattenuatorwithunchangeableinputimpedancesisdepictedinfig.8‐10.

Fig.8‐10

ForgiveninputresistorR andattenuation а :

thefollowinganalyticalexpressionsforsynthesisareobtained(whenR1R2=R2=const.,i.e.Zc1=Zc2=R=const):

NpRR

RRa ,1ln1ln

2

1

R RR1

R2R R

Phasecorrectors – basicprinciplesofphasecorrection,transferfunctions,cascaderealisation.Delaycircuits

1.PRINCIPLESOFPHASECORRECTIONPhase‐frequencydistortionschangetheinitialphaseratiosofthespectralcomponents

ofprocessedsignals,whichcanbecompensatedforbyso‐calledphase‐frequencycorrectors.

Thesecorrectorsmustinfluencetheinitialphasesofthespectralcomponentssignalsbutkeeptheamplitudesofthesecomponentsunchanged.

ThisimpliesthattheMagnitudeResponse(orAttenuation)ofaphase‐frequencycorrectormustbeafrequency‐independentconstantwithinthefrequencyrangeofthetransmittedsignal, i.e. .)(;)()( 00 constaaconstTTjT

Forthisreason,phase‐frequencycorrectorsarealsocalledall‐passcircuits.Aphase‐frequencycorrectoriscascadedtothecorrectedcircuit addingtoitsGroupDelay(GD)tГР.b valuetГР.ksuchthatthetotalGDtГР. ispermanent(fig.9‐1а).

Inregardtothephaseconstantb itshouldbealinearfrequencyfunction(fig.9‐1b). (а) (b)Fig.9‐ 1

Fig.9‐2

Forinstance,theGDcorrectionofaBPfifth‐orderButterworthfilter(fig.9‐2)consistsoffollowing:1. FromthetotalGD,whichisahorizontalline, theGDofthecorrectedcircuitisextractedthusdeliveringtheGDofthecorrector:

tгр. ‐ tгр.НЧФ =tгр.k2.Theobtainedcurve(theexpectedGDofthecorrectortгр.k)isapproximatedbydifferentGDcurvesofthefirst‐ andsecond‐orderphase‐correctionsections,thusspecifyingtheexactorderoftheall‐passTF.

3.Thecurvesofeachsectionareoptimisedsothattheoverallerrorisminimal.

TotalGD: tгр.

FilterGD:tгр.LPF

GDofthephasecorrector:tгр.k

tгр. Thisdescribedmethodologyrequirescascaderealisationofthephase‐frequencycorrector.

Inordertopreventadditionaldistortiondueto incoherentload,thephase‐frequencycorrectorshouldhavepermanentinputandoutputresistances.

Thephase‐frequencycorrectormustbesymmetric soastobeabletobeconnectedatanypointofthecorrectedcircuit.

2. ANALOGUEALL‐PASS(PHASE)TFThepolynomialinthedenominatorisastrictHurwitzpolynomial:

)(...)(1

2210

n

iin

nn ssasasasaasD

InorderfortheMagnitudeResponsetobeconstantatallfrequencies eitherN(s)=D(s),whichwouldmakethecircuitmeaningless,or N(s)andD(s) mustbeconjugatedpolynomials:

)()(‐)(1

33

2210

n

iin ssasasasaasDsN

Therefore, anall‐passTFwillbe:

)(

)()1(

)()(‐

)()()(

1

13

32

210

33

2210

n

ii

n

ii

ss

ss

sasasaasasasaa

sDsD

sDsNsT

All‐passTFaretypicalnon‐polynomial andnon‐minimumphase TF– theallzeroesarelocatedintheright‐halfofthes‐plane.SinceN(s)andD(s) areconjugatedpolynomials,thezeroesaremirror‐symmetricaltothepoles.

Cascaderealisationisthemostoftenusedandanall‐passcascadeTFwillbe:

l ll

ll

k k

kssss

sssTsT 2

2)()(

Thefollowingall‐passfirst‐ andsecond‐ordersectionsexist:

2.1.ALL‐PASSFIRST‐ORDERTF

sssT

)(I

Phaseconstant

arctgb 2II

dBTa 0II Attenuation

22

Iгр.I

2

ddbt

GroupDelay

σ

jω

‐σ σ

b1

=

/2

tГР.1

2/

1/

=

20

02

20

02

2

2II )(

sQ

s

sQ

s

sssssT

Phaseconstant

220

0

2II 22)(

QarctgarctgbII

dBHTa ,lg20IIII Attenuation

GroupDelay

1414222

220

222

2II

гр.IIQQQ

Qd

dbt

114 2Qm

2.1.ALL‐PASSSECOND‐ORDERTF

b2

ω =

2

tГР2.

ω

4Q/ω0

ω0

2Q/ω0

tmax

ωmTheGDcurvetГР.2hasamaximumwhenQ > 5.Inpractice,however,smallervaluesforQ areused.WhenQ < 0.5thetwopolesofTII(s) becomerealandtheTFcanbepresentedasaproductoftwoTFeachoffirst‐order.

SYNTHESISOF LATTICESCHEMESWITH PERMANENTINPUTRESISTANCE

TheserealisationsareusuallyLC or RLCcoherentlyloaded by,where Ża(s)and Żb(s)arereverseimpedances.

When Ża and Żb arescaledbycoefficientkr=RT=R0,sothatRTн = R0н =1,then:

Thus,foragivenTF:

Thetwosectionsaresynthesisedandtheelementsaredescaledbythesamecoefficientkr=RT=R0.OntheoutputthedescaledresistorRTispluggedin.

baT ZZRR 0

HHHHHHHH ababab YYRYYRZZ /11/11 2200

11

11

11

11

)1(

12

)( 2

2

H

H

H

H

H

H

H

H

H

H

HH

HH

b

b

a

a

b

b

a

a

a

a

ba

baHU Z

ZZZ

YY

YY

Y

YYY

YYsТ

НU

НU

ab

НU

НUa sT

sTY

sYsTsTsY

ННН )(1

)(11)(,)(1)(1)(

3.PASSIVEALL‐PASSSECTIONSOFFIRST‐ ANDSECOND‐ORDERThesynthesisoftheall‐passfirst‐ andsecond‐ordersectionsfollowsthesynthesis

procedureforsynthesisofcoherentlyloadedlatticestructures(fig.9‐3).

Fig.9‐3

Passiveall‐passI ordersections:scaledbyR; /1;/1 ba CL

Passiveall‐passII ordersections:0000

1;;1;

Q

CQLQ

LQC bbaa

sssTI

)(

ssssHsTII 2

2)(

For passiveTII(s),H=1 whileforactiverealisationsH hasanarbitraryvalue.

EXAMPLE 1:Synthesiseacoherentlyloaded(RT=R0=1)latticestructurehavingthefollowingall‐

passTF:

Solution:

Thepolesandzerosare:

Forа<2theyarecomplex‐conjugatedandarelocatedinaquadrantsymmetry(zeroesareontheright‐halfofthes‐plane) ‐ typicaloftheall‐passTFwhichisofnon‐minimum‐phasetype.

Table 7‐1 showsthattheTFcanbesynthesisedasalatticeorgroundedRLC‐structure.

HereitisrealisedasalatticeLC structure.

11)( 2

2

assasssT HU

41

2;

41

2

22,1

20 2,1

ajasajas

.11

11)(

;11)(1)(1)(

2

2

asass

asY

sY

asas

ass

sTsTsY

HH

H

ab

HU

HUa

4.ACTIVEALL‐PASSSECTIONSOFFIRST‐ ANDSECOND‐ORDER4.1.ACTIVEALL‐PASSSECTIONSOFFIRST‐ORDER

1;I

CG

sCGsCG

sssT

Awell‐knownactiveall‐passfirst‐ordersectionisdepictedinfig.9‐4. WhenwecomparetheTFrealisedbythesectiontoatypicalall‐passfirst‐orderTFTI(s) thefollowingscaledvaluesforthe conductivity G andthecapacitorС areobtained:

G G

GC

Fig.9‐4

21

543

21

615

1

4

213

2

21

543

62

54

1

1

213

2

2

62

2II 11

11

1

CCGGG

CCGGG

CG

CCGss

CCGGG

GGGG

CG

CCGss

GG

ssssHsT

C and Gp arearbitrarilychosentoensurethatGn and G4 willberealpositives.Then,when wederive:10

.22

;1G;;;;)2( 22

22

642

3522

1pp

ppnn

np

p

n

p

np CGGC

CGGCG

HHG

HGGG

GCG

HGGG

GGCGCG

4.2..ACTIVEALL‐PASSSECTIONSOFSECOND‐ORDER

Fig.9‐5

C1

R2R1

C2

R3

R4

R6

R5

All‐passsecond‐orderTFsaretypicalbiquadraticTFsandaresynthesisedasactiveamplitude‐correctionsecond‐ordersectionsbutthecoefficienth is h = ‐1.Thesectioninfig. 8‐7 isused,givenagainforconvenience(fig. 9‐5).ItsTFisasfollows:

5.DELAYCIRCUITSDelaycircuitsaretwo‐portcircuitsprovidingdelaysofsignalsbyaspecifiedtime

withoutchangingthesignalwaveform.DelaycircuitshavealinearPhaseResponseanda constantMagnitudeResponse

foraspecifiedbandoffrequencies,i.e.theyareall‐passcircuits.ThedelaycircuitTFisobtainedbyPadenapproximation,whichusesBessel

polynomials.

Forexample,thedelaycircuitsarepartofthedevicestoprotectHi‐Fistereo‐equipment.Beingdirectlyconnectedtothespeakers,thestereoamplifierscanbedamagedintheeventofpowersupplyinstability. Theprotectivedeviceprovidesthenecessarydelayofthesignalthusprotectingthestereo‐equipment.

Typical"popoff“whenturningstereo‐equipmentonoroffalsocanbeavoidedbyensuringthesignalsaredelayed.

Timeshift