Assembling of the SC Circuit Matrix Based on the Status of Switches

Transcription

1 WSS RNSINS on IRUIS and SYSMS ssembling of the S ircuit Matrix Based on the Status of Switches BHUMIL BRNÍK Department of lectronics and Informatics ollege of Polytechnics Jihlava olstého 6 86 Jihlava ZH RPUBLI brtnikb@gmailcom http//wwwvspjc bstract - his paper deals with assembling of the S circuit matrix based on the status of switches It is well known matrix assembly process using two-aphs or transformation aphs However the matrix can be built only on the basis of the status switches in the S circuit his procedure is somewhat simpler than the method of two-aphs Described method is compared with other methods o Key-Words - S circuit status of switches nodal charge method capacitance matrix four phases of switching Introduction It is well known matrix assembly process using two-aphs [] [] [] or transformation aphs [] [] [6] describing the circuit at all four stages of switching he phases are marked as even with the letters and odd and the nodes are numbered avoid confusion Problem Formulation he two-aph method uses a separate denomination of node of voltages and currents voltage triangle the square of the current his method can be simplified as follows ircuit with passive elements only ircuit containing capacir and a switched capacir whose circuit diaam is shown in Fig can be described by the set of equations for both phases Fig ircuit containing a switched capacir t t t t t t t fter the Z-transform a system of equations can be rewritten in the form his system of equations can be rewritten in next step in the matrix form where the matrix of the system is ie generally his matrix consists of four partial submatrix generally and By comparing this matrix with partial schematic diaams in Fig for both phases the submatrix can be obtained easy as follows as we can see he submatrix which are describing this circuit only in the even phase and in the odd phase are follows -ISSN -66X 7 Issue 6 olume June

2 WSS RNSINS on IRUIS and SYSMS Fig Diaams of circuit for even and odd phases [ ] Because charge is given by equation and charge is in first row which corresponds the line and voltage is in first column which corresponds the column member is in row and in the same column Similarly can be assembled a second matrix 6 [ ] 6 Similarly we can also assemble the remaining matrix 7 o However rows are of opposite phase than columns and their elements must be multiplied by 7 In the row is in the column member ie member simultaneously occurring in nodes and In the row is in the column member ie member simultaneously occurring in nodes and Both multiplied by of course Sign of the is positive because indexes of nodes and one number ie at the least is the same ie number of node is the same no phase the phase may be different hus after multiplication by this member is Sign of the is negative because the indexes and indexes no number is the same hus after multiplication by this member will be Similarly can be assembled a second matrix 8 o 8 Rules for sign of the members are in able Sign of member which is connected between output node of and other node except ound is negative already able Rules for sign of Phase one number of nodes is the same no number of nodes is the same the same different ircuit with active elements Let we are considering circuit containing voltage-voltage transducer [] [] three capacirs and the source of charge whose circuit diaam is shown in Fig Its equation is [7] where is voltage gain of the Fig ircuit with a his circuit can be described by nodal voltage method [] [] by the set of equations 9 9 -ISSN -66X 7 Issue 6 olume June

3 fter overwriting set of equations 9 in following form the matrix of the system is where matrix contains matrix of the voltage-voltage transducer only and is the remaining part he resulting matrix consists of partial submatrixes [8] he resulting capacitance matrix consists of six following partial submatrixes - - Because operational amplifier has not memory are submatrixes in positions and only In other positions and these submatrixes are not Since an operational amplifier in a switched circuit processes high frequencies the frequency characteristics of its amplification is taken in account by considering either one frequency s or both frequencies s s of its diffraction where is the maximum value of amplification in an open loop of feedback he frequency response can be now calculated by substituting or able Partial schematic diaams and its matrix representation Schematic diaam orresponding matrix WSS RNSINS on IRUIS and SYSMS -ISSN -66X 7 Issue 6 olume June

4 WSS RNSINS on IRUIS and SYSMS summary of relations between the matrix and the partial schematic diaam allow implementation of symbolic analysis [9] [] is given in able for an overview he capacir connected output node of the operational amplifier ie node in a different node than the common has in a column corresponding the index node always a negative sign his node is always outside the main diagonal of matrix as is shown Fig where is schematic diaam of simple circuit and corresponding matrix Fig Sign of is negative in matrix if index of column is the same as index of output node xamples his method whose principle was explained above will now be illustrated by solved two examples ommented solutions in detail onsider the circuit Fig containing of four nodes two capacirs and one operation amplifier It is be compiled his matrix Fig ircuit for example First it is necessary draw the schematic diaams separately for each phase as is shown in Fig6 In the even phase the circuit has three nodes in odd phase two nodes herefore the capacitance matrix will have five rows and five columns o he output of the operational amplifier is connected in node four therefore in row four ie and is necessary write the equation of the amplifier ie but in phases and only therefore operational amplifier has not memory Fig6 Schematic diaams separately for both phases hus parts of capacitance matrix will be following his part will be added submatrices for and phases in next steps he members of the matrix are given by indexes of the rows and columns For example in position row column is member because in the node capacir is connected 6 [ ] In position row column is member - because capacir is connected between nodes and as we can see hus will be [ ] 7 Similarly we can also assemble the remaining matrix in position row column is member because in node capacirs and are connected -ISSN -66X 7 Issue 6 olume June

5 WSS RNSINS on IRUIS and SYSMS [ ] 8 In position row column is the member - because capacir is connected between nodes and hus will be [ ] 9 Similarly in the matrix in position row column is the member because capacir is connected between nodes and he numbers and are identical therefore sign is positive in this case ie as we can see ie this member is finally negative In position row column is the member because capacir is connected between nodes and he numbers and are identical therefore sign is positive ie ie this member is finally negative But in position row column is the member because capacir is connected between nodes and Sign is negative in this case ie - because the indexes and no number is the same as we can see ie this member is finally ie this submatrix will be in following form Similarly the remaining matrix is following where member in row and column is because capacir is connected between nodes and s we can see the numbers in members and are identical therefore sign is positive ie ie this member will be negative finally hus the resulting capacitance matrix will be in following form -ISSN -66X 7 Issue 6 olume June

6 ommon solution without comment Previous solution has been very extensively commented his can lead the assumption that the described method is o laborious hus the next example will be solved without comment onsider the circuit Fig7 containing five nodes two capacirs two switched capacirs and one operational amplifier with finite amplification Schematic diaams separately for each phase are shown in Fig8 Fig7 Schematic diaam example -phase -phase Fig 8 Partial schematic diaams Partial matrixes are following WSS RNSINS on IRUIS and SYSMS -ISSN -66X 76 Issue 6 olume June

7 ompared with others methods Method described above will now be compared with two common methods ompared with two-aphs method For the purpose of comparison with the two-aph method [] [] will now solved the same example In this case the circuit is described by the matrix 6 B B 6 where and are submatrices onsider the same circuit ie circuit Fig9 with two capacirs two switched capacirs and operational amplifier with finite amplification he circuit has five nodes the numbers of nodes are in the circle Fig9 Schematic diaam for comparison Schematic diaams separately for each phase of this circuit are shown in Fig he numbers of nodes in square are for the charge in the triangle for the voltage -phase -phase Fig Partial schematic diaams WSS RNSINS on IRUIS and SYSMS -ISSN -66X 77 Issue 6 olume June

8 In the two-aph method two tables must be now prepared herefore matrix formulation required assembling following four tables for even and odd phases able for matrix able for matrix out inp lement out inp lement o he matrixes based on these tables are able for B matrix able for B matrix out inp lement out inp lement o he matrixes based on these tables are 9 9 he resulting matrix is after substituting in 6 in following form WSS RNSINS on IRUIS and SYSMS -ISSN -66X 78 Issue 6 olume June

9 ompared with full matrix method For the purpose of comparison with the full matrix method [] [7] will now solved the same example the Fig9 he circuit has five nodes and so its capacitance matrix will have them o But in matrix last row is replaced by equation of the ie hus the capacitance matrix is in form hus the resulting capacitance matrix containing all phases of switching will then have the following form matrix consists of ten rows and ten columns his matrix is somewhat unclear as we can see hus matrix is in last step reduced by closing the switches in its final form Final matrix consists six rows and six columns as we can see he last rows in and matrix consists of all eros because an operational amplifier has not memory ie transfer between phases is equal ero o s we can see the resulting matrices and are identical WSS RNSINS on IRUIS and SYSMS -ISSN -66X 79 Issue 6 olume June

10 WSS RNSINS on IRUIS and SYSMS able omparison table Step he two-aphs valuation state Full-matrix method method of switches Drawing a circuit diaam Drawing a circuit diaam Drawing a circuit diaam Drawing a circuit diaam for Drawing a circuit diaam Writing of the capacitance each phase with dual for each phase matrix numbering nodes ssembly four tables for each of the four phases ssembly of four matrices for each of the four phases Replacing row by equation of the ssembly of four matrices for each of the four phases ssembly resulting matrix ssembly resulting capacitance matrix ssembly resulting matrix - Reduction by closing switch Note Nodes have two different numbers in each phase one Nodes have one number Nodes have one number for the charge other for voltage G G onclusions comparison he comparison of the three methods is illustrated in able Individual steps of solution are described in this table in three columns s we can see Fig the method of two aphs requires two different type numbers of nodes in each phase o one for the charge in square and other for voltage in triangle herefore this way becomes these schemas a somewhat complicated he method of evaluation state of switches requires only one type of node numbers Fig8 While the two-aph method requires redrawing schematic diaams full-matrix then repeated rewriting matrices 6 onclusion Proposed method described above ie evaluation of the status of switches based on the general coordinate transformation method described in [7] etc In this case the parameters of the elements appear in the resulting matrix on the positions given by identification of the nodes of the elements with the nodes of the circuit as is shown in Fig ircuit consists of two elements G whose matrix is G G G G and resulting matrix of this circuit which is obtained G Fig oordinate transformation method through transformations of coordinates is G G G G G where index includes an element G and index includes an element G he numbers are indexes of nodes his general method is applied the switching circuits In position ig in above described method can be member ig cknowledgment his work was supported by the Faculty of lectrical ngineering and Informatics Department of lectrical ngineering University of Pardubice ech Republic References [] lach J Singhal K omputer Methods for ircuit nalysis and Design an Nostrand Reynhold New York 99 -ISSN -66X 8 Issue 6 olume June

11 WSS RNSINS on IRUIS and SYSMS [] lach J Basic Network heory with omputer pplication an Nostrand Reynhold New York 99 [] Brtník B Full aph Solution of Switched apacir ircuits by Means of wo-graphs WSS ransactions on ircuits and Systems Iss8 ol ugust pp67-77 [] Biolek D Biolkova nalysis of ircuits ontaining ctive lements by Using Modified -aphs dvances in Systems Science Measurement ircuits and ontrol WSS Press lectrical and omputer ngineering Series pp 79-8 ISBN [] Dostál he nalysis of the ctive omponents ontaining Switched apacirs by Nodal oltage Method lectronics horiont ol NoI 98 pp -6 [6] Biolek D Biolkova Flow Graphs Suitable for eaching ircuit nalysis Proceedings of the th WSS International onference on pplications of lectrical ngineering Praque ech Republic March - [7] Čajka J Kvasil J heory of iruits Linear ircuit SNL/LF Prague 979 [8] Biolek D Solving lectronic ircuits BN Publisher Prague [9] Hospodka J Bicak J pplication for Symbolic nalysis of Linear ircuits Including Switched ircuits Proceedings of the 6th WSS International onference System Science and Simulation in ngineering enice Italy November - 7 pp -8 [] Braun Z Žila Z Berka Z opological nalysis of Nonreciprocial lectrical Network with Help of Singular lements Proc of the th WSS Int onf on pplications of lectrical ngineering Prague ech Republic March - ISBN [] Martinek P Boreš P Hospodka J lectrics Filters U Publisher Prague -ISSN -66X 8 Issue 6 olume June