《信号与线性系统分析基础 》 课件 第5章 连续时间信号与系统的复频域分析.ppt

TheZero-InputResponseCharacteristicequation?j,(j=1,2,3,?,n)CharacteristicrootZero-InputResponseyzi(0)，yzi(1)，…，yzi(n-1)y(-1)，y(-2)，…，y(-n)yzi(k)=y(k)-yzs(k)=y(k)，k0InitialconditionCharacteristicequation:Characteristicroots:Zero-InputResponse:Example:TheZero-InputResponsey(k)+3y(k-1)+2y(k-2)=f(k),f(k)=2kε(k),y(-1)=0,y(-2)=1/2.Findyzi(k),k≥0.yzi(k)+3yzi(k-1)+2yzi(k-2)=0TheZero-StateResponseCharacteristicequation?j (j=1,2,3,?,n)Characteristicroot(distinctroots?j)Zero-StateResponseyzs(-1)=yzs(-2)=…=yzs(-n)=0Initialstateyzs(0)，yzs(1)，…，yzs(n-1)Initialcondition吉林大学TheUnitSampleResponse

and

TheUnitStepResponseTheUnitSampleResponseDefinitionTheunitsampleresponseisthezero-stateresponseofthesystemresultingfromtheapplicationoftheunitpulse?(k).Denotedh(k)Initialstateh(-1)=h(-2)=…=h(-n)=0Initialconditionh(0),h(1),h(2),…,h(n-1)HowtofindSolvingadifferenceequationZ-transformTheUnitSampleResponseDeterminationk0：?(k)=0，h(k)=0k=0：?(k)=1，h(0)——recursionk0：?(k)=0，h(k)——solutionofahomogeneousequationLTIsystem:LetthenCi：determinedbyh1(1),h1(2),…,h1(n)吉林大学Time-DomainAnalysisofDiscrete-TimeSystemsDiscrete-TimeSignalsDiscrete-TimeSignalsAdiscrete-timesignalf(k)hasvaluesforsomediscontinuouspointwhilehasnotdefinitionforotherpoints.k—integerDefinitionDiscrete-TimeSignalsAnalyticalmethod:Graphicalmethod:Sequencemethod:k=0RepresentationDiscrete-TimeSignalEnergyandPowerEnergy:Power:OperationofDiscrete-TimeSignalsAddition：Multiplication：Difference：forwarddifference:backwarddifferenceRunningsum：OperationofDiscrete-TimeSignalsTimeshift(m0)RightshiftLeftshiftTransformationsoftheIndependentVariableOperationofDiscrete-TimeSignalsTimereversalTransformationsoftheIndepend