AnnouncementsReviewLecture #5First Order CircuitsComplete SolutionThe Time ConstantWhat Does Xc(t) Look Like?The Particular SolutionExample2nd Order CircuitsA 2nd Order RLC CircuitThe Differential EquationSlide 13Slide 14The Complementary SolutionCharacteristic EquationDamping Ratio and Natural FrequencyOverdamped : Real Unequal RootsUnderdamped: Complex RootsCritically damped: Real Equal RootsSlide 21Slide 22Slightly Different ExampleSlide 24Why is Single-Frequency Excitation Important?Slide 26Steady-State Sinusoidal AnalysisThe Good News!PhasorsComplex ImpedanceSinusoidsPhaseSlide 33EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu1AnnouncementsLecture 3 updated on web. Slide 29 reversed dependent and independent sources. Solution to PS1 on web todayPS2 due next Tuesday at 6pmMidterm 1 Tuesday June18th 12:00-1:30pm. Location TBD.EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu2ReviewCapacitors/InductorsVoltage/current relationshipStored Energy1st Order CircuitsRL / RC circuitsSteady State / Transient responseNatural / Step responseEE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu3Lecture #5OUTLINEChap 4RC and RL Circuits with General SourcesParticular and complementary solutionsTime constantSecond Order CircuitsThe differential equationParticular and complementary solutionsThe natural frequency and the damping ratioChap 5Types of Circuit ExcitationWhy Sinusoidal Excitation?PhasorsComplex ImpedancesReadingChap 4, Chap 5 (skip 5.7)EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu4First Order Circuits( )( ) ( )cc sdv tRC v t v tdt+ =KVL around the loop:vr(t) + vc(t) = vs(t)R+-Cvs(t)+-vc(t)+ -vr(t)vL(t)is(t)R L+-)()(1)(tidxxvLRtvstKCL at the node:( )( ) ( )LL sdi tLi t i tR dt+ =ic(t)iL(t)EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu5Complete SolutionVoltages and currents in a 1st order circuit satisfy a differential equation of the formf(t) is called the forcing function.The complete solution is the sum of particular solution (forced response) and complementary solution (natural response).Particular solution satisfies the forcing functionComplementary solution is used to satisfy the initial conditions. The initial conditions determine the value of K.( )( ) ( )dx tx t f tdtt+ =/( )( ) 0( )cctcdx tx tdtx t Kett-+ ==( )( ) ( )ppdx tx t f tdtt+ =Homogeneous equation( ) ( ) ( )p cx t x t x t= +EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu6The Time ConstantThe complementary solution for any 1st order circuit isFor an RC circuit, = RCFor an RL circuit, = L/R/( )tcx t Ket-=EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu7What Does Xc(t) Look Like? = 10-4/( )tcx t et-=• is the amount of time necessary for an exponential to decay to 36.7% of its initial value.•-1/ is the initial slope of an exponential with an initial value of 1.EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu8The Particular SolutionThe particular solution xp(t) is usually a weighted sum of f(t) and its first derivative.If f(t) is constant, then xp(t) is constant.If f(t) is sinusoidal, then xp(t) is sinusoidal.EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu9ExampleKVL:R = 5kΩ+-C = 1uF+-vc(t)+ -vr(t)ic(t)t = 0vs(t) = 2sin(200t)EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu102nd Order CircuitsAny circuit with a single capacitor, a single inductor, an arbitrary number of sources, and an arbitrary number of resistors is a circuit of order 2.Any voltage or current in such a circuit is the solution to a 2nd order differential equation.EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu11A 2nd Order RLC CircuitR+-Cvs(t)i (t)LApplication: FiltersA bandpass filter such as the IF amp for the AM radio.A lowpass filter with a sharper cutoff than can be obtained with an RC circuit.EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu12The Differential EquationKVL around the loop:vr(t) + vc(t) + vl(t) = vs(t)i (t)R+-Cvs(t)+-vc(t)+-vr(t)L+-vl(t)1 ( )( ) ( ) ( )tsdi tRi t i x dx L v tC dt- �+ + =�22( )( ) 1 ( ) 1( )sdv tR di t d i ti tL dt LC dt L dt+ + =EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu13The Differential EquationThe voltage and current in a second order circuit is the solution to a differential equation of the following form:Xp(t) is the particular solution (forced response) and Xc(t) is the complementary solution (natural response).2202( ) ( )2 ( ) ( )d x t dx tx t f tdt dta w+ + =( ) ( ) ( )p cx t x t x t= +EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu14The Particular SolutionThe particular solution xp(t) is usually a weighted sum of f(t) and its first and second derivatives.If f(t) is constant, then xp(t) is constant.If f(t) is sinusoidal, then xp(t) is sinusoidal.EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu15The Complementary SolutionThe complementary solution has the following form:K is a constant determined by initial conditions.s is a constant determined by the coefficients of the differential equation.( )stcx t Ke=22022 0st ststd Ke dKeKedt dta w+ + =2 202 0st st sts Ke sKe Kea w+ + =2 202 0s sa w+ + =EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu16Characteristic EquationTo find the complementary solution, we need to solve the characteristic equation:The characteristic equation has two roots-call them s1 and s2.2 20 002 0s szw wa zw+ + ==1 21 2( )s t s tcx t K e K e= +21 0 01s zw w z=- + -22 0 01s zw w z=- - -EE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu17Damping Ratio and Natural FrequencyThe damping ratio determines what type of solution we will get:Exponentially decreasing ( >1)Exponentially decreasing sinusoid ( < 1)The natural frequency is 0It determines how fast sinusoids wiggle.0azw=21 0 01s zw w z=- + -22 0 01s zw w z=- - -damping ratioEE40 Summer 2006: Lecture 5 Instructor: Octavian Florescu18Overdamped : Real Unequal RootsIf > 1, s1 and s2 are real and not equal.ttceKeKti1211200200)(00.20.40.60.81-1.00E-06ti(t)-0.200.20.40.60.8-1.00E-06ti(t)EE40 Summer 2006: Lecture 5
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