Network analysis synthesis Lecture notes

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DEPARTMENT OF ELECTRICAL ENGINEERING THIRD SEMESTER ,(EE/EEE) SUBJECT:NETWORK THEORY SUBJECT CODE-1303 SYLLABUS :NETWORK THEORY (3-1-0) MODULE-I (10 HOURS) Coupled Circuits: Self-inductance and Mutual inductance, Coefficient of coupling, dot convention, Ideal Transformer, Analysis of multi-winding coupled circuits, Analysis of single tuned and double tuned coupled circuits. Transient study in RL, RC, and RLC networks by Laplace transform method with DC and AC excitation. Response to step, impulse and ramp inputs. Two Port networks: Two port parameters, short circuit admittance parameter, open circuit impedance parameters, Transmission parameters, Image parameters and Hybrid parameters. Ideal two port devices, ideal transformer. Tee and Pie circuit representation, Cascade and Parallel Connections. MODULE-II (10 HOURS) Network Functions & Responses: Concept of complex frequency, driving point and transfer functions for one port and two port network, poles & zeros of network functions, Restriction on Pole and Zero locations of network function. Impulse response and complete response. Time domain behavior form pole-zero plot. Three Phase Circuits: Analysis of unbalanced loads, Neutral shift, Symmetrical components, Analysis of unbalanced system, power in terms of symmetrical components MODULE-III (10 HOURS) Network Synthesis: Realizability concept, Hurwitz property, positive realness, properties of positive real functions, Synthesis of R-L, R-C and L-C driving point functions, Foster and Cauer forms MODULE-IV (10 HOURS) Graph theory: Introduction, Linear graph of a network, Tie-set and cut-set schedule, incidence matrix, Analysis of resistive network using cut-set and tie-set, Dual of a network. Filters: Classification of filters, Characteristics of ideal filters BOOKS 1. Mac.E Van Valkenburg, “Network Analysis”, 2. Franklin Fa-Kun. Kuo, “Network Analysis & Synthesis”, John Wiley & Sons. 3. M. L. Soni, J. C. Gupta, “A Course in Electrical Circuits and Analysis”, 4. Mac.E Van Valkenburg, “Network Synthesiss”, 5. Joseph A. Edminister, Mahmood Maqvi, “Theory and Problems of Electric Circuits”, Schaum's Outline Series, TMH MODULE- I (10 hrs) 1.Magnetic coupled circuits. (Lecture -1) 1.1.Self inductance When current changes in a circuit, the magnetic flux linking the same circuit changes and e.m.f is induced in the circuit. This is due to the self inductance, denoted by L. di VL dt FIG.1 1.2.Mutual Inductance The total magnetic flux linkage in a linear inductor made of a coil is proportional to the current passing through it; that is, Fig. 2  Li . By Faraday‟s law, the voltage across the inductor is equal to the time derivative of the total influx linkage; given by, di d LN dt dt 1.3. Coupling Coefficient A coil containing N turns with magnetic flux Ø_ linking each turn has total magnetic flux linkage λ=NØ . By Faraday‟s law, the induced emf (voltage) in the coil is dd   eN  dt dt  . A negative sign is frequently included in this equation to signal that the voltage polarity is established according to Lenz‟s law. By definition of self-inductance this voltage is also given by Ldi=dtÞ; hence, The unit of flux(Ø) being the weber, where 1 Wb = 1 V s, it follows from the above relation that 1 H = 1 Wb/A. Throughout this book it has been assumed that Ø and i are proportional to each other, making L = (NØ) /I = constant Fig.3 In Fig.3 , the total flux resulting from current i1 through the turns N1 consists of leakage flux, Ø11, and coupling or linking flux, Ø12. The induced emf in the coupled coil is given by N (dØ /dt). This same voltage can be written using the mutual inductance M: 2 12 d 21 MN 1 di 2 di d 1 12 e M N 2 dt dt or d 12 MN 2 di 1 Also, as the coupling is bilateral, d 21 MN 1 di 2 dd   2 12 21 M N N 21 dt di   2 kk  12 NN 21 di di  12  d d 2 2 1  k N N  12 di di 12 2  k L L 12 Hence,mutual inductance , M is given by M k L L 12 And the mutual reactance X is given by M X k X X M 12 The coupling coefficient, k, is defined as the ratio of linking flux to total flux:  12 21 k  12 1.4.Series connection of coupled circuit ( lecture 2) When inductors are connected together in series so that the magnetic field of one links with the other, the effect of mutual inductance either increases or decreases the total inductance depending upon the amount of magnetic coupling. The effect of this mutual inductance depends upon the distance apart of the coils and their orientation to each other. Mutually connected inductors in series can be classed as either “Aiding” or “Opposing” the total inductance. If the magnetic flux produced by the current flows through the coils in the same direction then the coils are said to be Cumulatively Coupled. If the current flows through the coils in opposite directions then the coils are said to be Differentially Coupled as shown below. 1.4.1.Cumulatively Coupled Series Inductors Fig.4 While the current flowing between points A and D through the two cumulatively coupled coils is in the same direction, the equation above for the voltage drops across each of the coils needs to be modified to take into account the interaction between the two coils due to the effect of mutual inductance. The self inductance of each individual coil, L and L respectively will be the same 1 2 as before but with the addition of M denoting the mutual inductance. Then the total emf induced into the cumulatively coupled coils is given as: Where: 2M represents the influence of coil L on L and likewise coil L on L . 1 2 2 1 By dividing through the above equation by di/dt we can reduce it to give a final expression for calculating the total inductance of a circuit when the inductors are cumulatively connected and this is given as: L = L + L + 2M total 1 2 If one of the coils is reversed so that the same current flows through each coil but in opposite directions, the mutual inductance, M that exists between the two coils will have a cancelling effect on each coil as shown below. 1.4.2.Differentially Coupled Series Inductors Fig.5 The emf that is induced into coil 1 by the effect of the mutual inductance of coil 2 is in opposition to the self-induced emf in coil 1 as now the same current passes through each coil in opposite directions. To take account of this cancelling effect a minus sign is used with M when the magnetic field of the two coils are differentially connected giving us the final equation for calculating the total inductance of a circuit when the inductors are differentially connected as: L = L + L – 2M total 1 2 Then the final equation for inductively coupled inductors in series is given as: Inductors in Series Example No2 Two inductors of 10mH respectively are connected together in a series combination so that their magnetic fields aid each other giving cumulative coupling. Their mutual inductance is given as 5mH. Calculate the total inductance of the series combination. Inductors in Series Example No3 Two coils connected in series have a self-inductance of 20mH and 60mH respectively. The total inductance of the combination was found to be 100mH. Determine the amount of mutual inductance that exists between the two coils assuming that they are aiding each other. 1.4.DOT RULE (lecture 3) Fig.6 Fig.7 The sign on a voltage of mutual inductance can be determined if the winding sense is shown on the circuit diagram, as in Fig. To simplify the problem of obtaining the correct sign, the coils are marked with dots at the terminals which are instantaneously of the same polarity. To assign the dots to a pair of coupled coils, select a current direction in one coil and place a dot at the terminal where this current enters the winding. Determine the corresponding flux by application of the right-hand rule see Fig. 14-7(a). The flux of the other winding, according to Lenz‟s law, opposes the first flux. Use the right-hand rule to find the natural current direction corresponding to this second flux see Fig. 14-7(b). Now place a dot at the terminal of the second winding where the natural current leaves the winding. This terminal is positive simultaneously with the terminal of the first coil where the initial current entered. With the instantaneous polarity of the coupled coils given by the dots, the pictorial representation of the core with its winding sense is no longer needed, and the coupled coils may be illustrated as in Fig. 14-7(c). The following dot rule may now be used: (1) when the assumed currents both enter or both leave a pair of coupled coils by the dotted terminals, the signs on the M-terms will be the same as the signs on the L-terms; but (2) if one current enters by a dotted terminal while the other leaves by a dotted terminal, the signs on the M-terms will be opposite to the signs on the L-terms 1.5.CONDUCTIVELY COUPLED EQUIVALENT CIRCUITS From the mesh current equations written for magnetically coupled coils, a conductively coupled equivalent circuit can be constructed. Consider the sinusoidal steady-state circuit of Fig. 14-9(a), with the mesh currents as shown. The corresponding equations in matrix form are Fig.8 R j L j M I V  1 1 1 1    j M R j M I V  2 2 2 2 1.6.Single tuned coupled circuit R0 I1 I2 Rs R1 i2 AC i1 Ls L1 Fig.9 Z I Z I E 11 1 12 2 Z I Z I 0 21 1 22 2 ZE  11  Z 0 EZ  21 21 I 2 ZZ  Z Z Z Z 11 12 11 22 12 21  ZZ  21 22 E() jM I 2 22 R() R jX M 1 2 2 For LR E() jM I 2  1  22 R R j Ls M 12  Cs   EM V O   1  22 C R R j Ls M  s 12  Cs     FIG. variation of V with for different values of K 0 V EM O A E  1  22 EC R R j Ls M  s 12  Cs    At resonance EM I 2r 22 R R M 12 r  EM() 1 r V 2r 22 R R M C  12rs  M 1 A r 22 R R M C  12rs RR 1 1 M opt  r 1.7.Double tuned coupled circuit.(lecture-4) Rs R1 R2 i2 L2 Cp i1 L1 AC Fig.10 In this circuit the tuning capacitors are placed both in primary as well as secondary side. From the figure the eq. impedance on primary side is, Z , 11 Given by  1 Z R R j L  11 0 p p p  C p   R jX 11  1 Z R jL 22ss C  s  R jX 22 Z  Z jM 12 21 Applying loop analysis EZ 1 12 I 2 2 Z Z Z 11 12 12 I E M 21 V 0 22 j C C R jX R jX M  ss 1 1 2 2 EM 1  22 C R jX R jX M   s 1 1 2 2 V EM 0 1 AE 1 22 E C R jX R jX M  1 s 1 1 2 2 EM 1 1  22 C R jX R jX M E   1 s 1 1 2 2 M  22 C R jX R jX M  s 1 1 2 2 At resonance EM 1 r I 2 22 res R R M 12 r EM 1 1 r V o 22 res R R M C 12 r r s  FIG. variation of V with for different values of K 0 Or, E M / C 1 s  22 R R M 12 r V E M / C o res 1 s AE res 1 22 E R R M 1 1 2 r EM 1 1  22 C R R M E  sr 12 1 M  22 C R R M  sr 12 For maximum out put voltage at resonance, The denominator should be minimum As RR 2 12  M r M RR 12 M K L L opt crit p s  r This value will give the optimum M M opt 2.Two-Port Networks 2.1.Terminals and Ports (lecture-5) In a two-terminal network, the terminal voltage is related to the terminal current by the impedance Z=V / I. Fig.1 In a four-terminal network, if each terminal pair (or port) is connected separately to another circuit as in Fig. , the four variables i1, i2, v1, and v2 are related by two equations called the terminal characteristics. These two equations, plus the terminal characteristics of the connected circuits, provide the necessary and sufficient number of equations to solve for the four variables. 2.2.Z-PARAMETERS (open circuit parameters) The terminal characteristics of a two-port network, having linear elements and dependent sources, may be written in the s-domain as V =Z I +Z I (1) 1 11 1 12 2 V =Z I +Z I (2) 2 21 1 22 2 Z = V / I (for I =0) 11 1 1 2Z = V / I (for I =0) 21 2 1 2 Z V /I (for I =0) 12= 1 2 1 Z = V /I (for I =0) 22 2 2 1 2.3.Y-PARAMETERS(short circuit parameters) The terminal characteristics may also be written as , where I1 and I2 are expressed in terms of V1 and V2. I1 = Y11V1 + Y12V2 (3) I2 = Y21V1 + Y22V2 (4) this yields Y =I /V forV =0 11 1 1 2 Y =I /V For V =0 11 1 2 1 Y =I /V for V =0 21 2 2 2 Y =I /V forV =0 22 2 22 1 5ohm 5 ohm 5 ohm c a 5 0hm d b Fig.2 Solution: We can make two separate neyworks one a T netwotk comprising R1,R2,R3 and anetwork containing R4 only. 5 ohm 5 ohm c a 5 0hm d b 1 Y mho 11 7.5 1 Y  mho 12 15 1 Y mho 21 15 1 Y mho 22 7.5 Similarly for the next network 5ohm c a d b 1 Y mho 11 5 1 Y  mho 21 5 1 Y  mho 12 5 1 Y mho 22 5 Y  Y Y  net T B 1/ 34 /15  Y net 4 /15 1/ 3  (ans) Example.Find the Y parameters for the given circuit 1 ohm 2 ohm c a I1 I2 V2 V1 2 0hm 4 ohm d 1 ohm 2 ohm c a I1 I2=0 V1 2 0hm d I 1 Y 11 V 1 V0 2 11 Y mho 11 22 2 1 22 V 1  2 IV 2 11 2 I 2 2 2 2 2 4 I 1 2 Y mho 21 V 4 1 I0 2 1 ohm 2 ohm c a I1 I2 V2 2 0hm 4 ohm d b VV 5 22 I 2 2 12  8  24  12  12   24  12  5 Y mho 22 8 1 Y  mho 12 4 . 2.4.Transmission Parameters (ABCD parameters) The transmission parameters A, B, C, and D express the required source variables V1 and I1 in terms of the existing destination variables V2 and I2. They are called ABCD or T-parameters and are defined by V = AV – BI 1 2 2 I = CV – DI 1 2 2

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