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ISU EE 475 - Lecture06

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Modeling of SystemsTable 2.3Voltage-current, voltage-charge, and impedance relationships for capacitors, resistors, and inductorsRLC network dttdiL)()(tRi∫dttiC)(1∫=++ )()(1)()(tvdttiCtRidttdiLLCsLRsLCsGRCsLCssVssGsVssRCssLCstvtdttdRCdttdLCtoutputtvtqCdttdqRdttqdLastvdttiCtRidttdiL11)(11)()(V)()()(V)(V)(V)()(v)(v)(v)(Cvq(t) , v)()(1)()(i(t)dtq(t) )()(1)()(22CCCC2CC2C2CC22++=++===++=++==++==++∫∫Block diagram of series RLC electrical networkMesh analysisMesh 1Mesh 201)()(012 )(1 221211122222111=⎟⎠⎞⎜⎝⎛+++−=−+=−++=−+ICsRLsLsIsVLsIILsRLsIICsIRLsImeshsVLsILsIIRmeshSum of impedance around mesh 1Sum of impedance around mesh 2Sum of impedance common to two meshesSum of applied voltages around the meshWrite equations around the meshes∆=∆⎥⎦⎤⎢⎣⎡−+=⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡⎥⎥⎦⎤⎢⎢⎣⎡++−−+)(0)( '0)(1122121sLsVLssVLsRIRulesCramersVIICsRLsLsLsLsRDeterminant() ()()()()()1212212212122132221221)()()(11RsLCRRsRRLCsVLCsICsRsLCRRsRRLCCsCsLCsRLCsLsRLsCsRLsLsR++++==++++=−+++=−⎟⎠⎞⎜⎝⎛+++=∆Kirchhoff current law at these two nodesi1i3i2i1+ i2 +i3=0i4i3+ i4 =0Nodal analysis()0)()()()()()/1( /1 ,/10)()()( vas marked node At the0)()()()()( vas marked node At the22122122112C21L=++−=−++===−+=−++−sVCsGsVGGsVsVGsVLsGGRGRGRsVsVsCsVRsVsVLssVRsVsVCLCLLCCCLLLconductanceKirchhoff current law()()()()(){}()()(){}()LCGsLCCLGGsGGCGsGsVsVLCGsCGLCGGGsGGCGsGsVsVCsGCGsLsGGCGsGsVsVGCsGLsGGGGsVsVGCsGLsGGGGsVLsGGsVGsVsVsVCsGGGLsGGCCCCCCL//)()(///1//)()(///1/)()(/1)()(/10)(/1)(0)()()(/1221221212222212212122221212222121222212121122221++++=+−++++=−+++=−+++=−+++⎥⎦⎤⎢⎣⎡−++=⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡+−−++Figure 2.7Block diagram of the network of Figure 2.6Make sure you can obtain thisReal part of admittance is conductance, imaginary part is susceptance; 1/R =G has only real partExample: car suspensionSuppose y(t) is measuredfrom equilibrium position when gravity has set in. So gravity is canceled by spring force at eq. pos.∴There are two forces on m:y )ry(bf y(y-r)kelongationkfdampingspring&&&opposing opposing −⋅=⋅=⋅=Newton’s Law:or num=den=T.F.=H(s)=orkrrbkyybymrybrykym+=++−−−−==∑&&&&&&&&)()( force↑↑↑↑⎥⎦⎤⎢⎣⎡+=⎥⎦⎤⎢⎣⎡++ rmkrmbymkymby&&&&LL0101 bbaa⎥⎦⎤⎢⎣⎡mkmb⎥⎦⎤⎢⎣⎡mkmb1012012)()(asasbsbmksmbsmksmbsRsY+++=+++=)()()()(2sRsHsRmksmbsmksmbsY


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ISU EE 475 - Lecture06

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