Chapter LU Decomposition More Examples Electrical Engineering

Transcription

1 Chapter 4.7 LU Decomposition More Examples Electrical Engineering Example Three-phase loads e common in AC systems. When the system is balanced the analysis can be simplified to a single equivalent rcuit model. However, when it is unbalanced the only practical solution involves the solution of simultaneous line equations. n one model the following equations need to be solved Find the values of,,,,, and using LU decomposition. Solution u u2 u u4 u5 u6 2 u22 u2 u24 u25 u26 2 u u4 u5 u6 [ A] [ L][ U ] u44 u45 u u55 u u66 The [ U ] matrix is the same as the one found at the end of the forwd elimination steps of the naïve Gauss elimination method. Forwd Elimination of Unknowns Since there e six equations, there will be five steps of forwd elimination of unknowns. First step Divide Row by.746 and multiply it by.456, that is, multiply Row by Row.656 ( ) [ ] [ 72.64] Subtract the result from Row 2 to get 4.7.

2 4.7.2 Chapter Divide Row by.746 and multiply it by., that is, multiply Row by Row (.45) Subtract the result from Row to get Divide Row by.746 and multiply it by.8, that is, multiply Row by Row (.724) Subtract the result from Row 4 to get Divide Row by.746 and multiply it by., that is, multiply Row by Row (.45) Subtract the result from Row 5 to get [ ] [.686]

3 LU Decomposition-More Examples: Electrical Engineering Divide Row by.746 and multiply it by.8, that is, multiply Row by Row (.724) Subtract the result from Row 6 to get Second step Divide Row 2 by.94 and multiply it by.9464, that is, multiply Row 2 by Row 2 (.994) Subtract the result from Row to get Divide Row 2 by.94 and multiply it by.484, that is, multiply Row 2 by Row 2 (.456) Subtract the result from Row 4 to get

4 4.7.4 Chapter Divide Row 2 by.94 and multiply it by.9464, that is, multiply Row 2 by Row 2 (.994) Subtract the result from Row 5 to get Divide Row 2 by.94 and multiply it by.484, that is, multiply Row 2 by Row 2 (.456) 5 5 [ ] [.577] Subtract the result from Row 6 to get Third step Divide Row by and multiply it by.526, that is, multiply Row by Row (.6686) [ ] [ 4.269] Subtract the result from Row 4 to get

5 LU Decomposition-More Examples: Electrical Engineering Divide Row by and multiply it by.98697, that is, multiply Row by Row (.2677) 5 [ ] [.78275] Subtract the result from Row 5 to get Divide Row by and multiply it by.78644, that is, multiply Row by Row (.) 5 5 [ ] [.6272] Subtract the result from Row 6 to get Fourth step Divide Row 4 by.264 and multiply it by.2679, that is, multiply Row 4 by Row 4 (.257)

6 4.7.6 Chapter 4.7 Subtract the result from Row 5 to get Divide Row 4 by.264 and multiply it by.526, that is, multiply Row 4 by Row 4 (.429) 5 [ ] [.678] Subtract the result from Row 6 to get Fifth step Divide Row 5 by.8775 and multiply it by.675, that is, multiply Row 5 by Row 5 (.7474) [ ] [ 45.62] Subtract the result from Row 6 to get The coeffient matrix after the completion of the forwd elimination steps is the [ ] U matrix.

7 LU Decomposition-More Examples: Electrical Engineering [ U ].746 L. Now find [ ] [ L ] From Step of the forwd elimination process From Step 2 of the forwd elimination process From Step of the forwd elimination process

8 4.7.8 Chapter From Step 4 of the forwd elimination process From Step 5 of the forwd elimination process Hence [ L ] L Z C Now that [ L ] and [ U ] e known, solve [ ] This provides the six equations z 2.656z + z2..45z z + z ( ) z z z z z z 2.724z +.456z z + z4.9.45z z z z4 + z5.724z +.456z2 +.z +.429z z5 + z6 ( ) ( ) 6. Forwd substitution stting from the first equation gives z 2 Substituting the value of z into the second equation, z2. 656z ( )

9 LU Decomposition-More Examples: Electrical Engineering Substituting the values of z and z 2 into the third equation, z 6..45z (. 994) z ( 2) (.994)( ) Substituting the values of z, z 2, and z into the fourth equation, z z.456z z.9.724( 2).456( ).6686( 6.747) Substituting the values of z, z 2, z, and z 4 into the fifth equation, z z.994 z z. 257 z 6..45( 2) (.994)( ).2677( 6.747) (.257)( 62.86) 6.5 Substituting the values of z, z 2, z, z 4, and z 5 into the sixth equation, z z.456z2.z.429z z ( 2).456( ).( 6.747).429( 62.86).74745( 6.5) 5.76 Hence z 2 z2 z [ Z ] z z z U Z. Now solve [ ] ( ) ( ) This provides the six equations for [ ] (.456) +. + (.8) +. + (.8)

10 4.7. Chapter 4.7 (.526) (.78644) (.675) From the sixth equation Substituting the value of into the fifth equation, (.675) (.675) Substituting the values of and into the fourth equation, Substituting the values of,, and into the third equation, (.526) (.78644) (.526) (.78644) Substituting the values of,,, and into the second equation, Substituting the values of,,,, and into the first equation, (.456). (.8). (.8) 9. The solution vector is ( ) ( ) ( ) 2.746

11 LU Decomposition-More Examples: Electrical Engineering SMULTANEOUS LNEAR EQUATONS Topic LU Decomposition More Examples Summy Examples of LU decomposition Major Electrical Engineering Authors Aut Kaw Date August 8, 29 Web Site