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Berkeley ELENG 40 - Lecture Notes

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1EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu1Review Capacitors/Inductors Voltage/current relationship Stored Energy 1stOrder Circuits RL / RC circuits Steady State / Transient response Natural / Step responseEE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu2Lecture #5OUTLINE Chap 4 RC and RL Circuits with General Sources Particular and complementary solutions Time constant Second Order Circuits The differential equation Particular and complementary solutions The natural frequency and the damping ratio Chap 5 Types of Circuit Excitation Why Sinusoidal Excitation? Phasors Complex ImpedancesReadingChap 4, Chap 5 (skip 5.7)2EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu3First Order Circuits()() ()ccsdv tRCvtvtdt+=KVL around the loop:vr(t) + vc(t) = vs(t)R+-Cvs(t)+-vc(t)+-vr(t)vL(t)is(t)R L+-)()(1)(tidxxvLRtvst=+∫∞−KCL at the node:()() ()LLsdi tLit itRdt+=ic(t)iL(t)EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu4Complete Solution Voltages and currents in a 1st order circuit satisfy a differential equation of the form f(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 function Complementary solution is used to satisfy the initial conditions.  The initial conditions determine the value of K.()() ()dx txtftdtτ+=/()() 0()cctcdx txtdtxt Keττ−+==()() ()ppdx txtftdtτ+=Homogeneous equation() () ()pcxtxtxt=+3EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu5The Time Constant The complementary solution for any 1st order circuit is For an RC circuit, τ = RC For an RL circuit, τ = L/R/()tcxt Keτ−=EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu6What Does Xc(t) Look Like?τ = 10-4/()tcxteτ−=• τ is the amount of time necessary for an exponential to decay to 36.7% of its initial value.•-1/τ is the initial slope of an exponential with an initial value of 1.4EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu7The Particular Solution The 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 2005: Lecture 5 Instructor: Octavian Florescu82nd Order Circuits Any 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.5EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu9A 2nd Order RLC CircuitR+-Cvs(t)i(t)L Application: FiltersA 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 2005: Lecture 5 Instructor: Octavian Florescu10The 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 tCdt−∞++=∫22()() 1 () 1()sdv tRdit dititLdt LC dt L dt++=6EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu11The 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()()dxt dxtxtftdt dtαω++=() () ()pcxtxtxt=+EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu12The Particular Solution The 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.7EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu13The 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.()stcxt Ke=220220st ststdKe dKeKedt dtαω++=22020st st sts Ke sKe Keαω++=22020ssαω++=EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu14Characteristic Equation To find the complementary solution, we need to solve the characteristic equation: The characteristic equation has two roots-call them s1and s2.2200020ssζω ωαζω++==1212()ststcxtKe Ke=+21001sζω ω ζ=− + −22001sζω ω ζ=− − −8EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu15Damping Ratio and Natural Frequency The damping ratio determines what type of solution we will get: Exponentially decreasing (ζ >1) Exponentially decreasing sinusoid (ζ < 1) The natural frequency is ω0 It determines how fast sinusoids wiggle.0αζω=21001sζω ω ζ=−+ −22001sζω ω ζ=−− −damping ratioEE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu16Overdamped : Real Unequal Roots If ζ > 1, s1and s2are real and not equal.ttceKeKti⎟⎠⎞⎜⎝⎛−−−⎟⎠⎞⎜⎝⎛−+−+=1211200200)(ςωςωςωςω00.20.40.60.81-1.00E-06ti(t)-0.200.20.40.60.8-1.00E-06ti(t)9EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu17Underdamped: Complex Roots If ζ < 1, s1and s2are complex. Define the following constants:()12() cos sintcddxte A tA tαωω−=+0αζω=201dωωζ=−-1-0.8-0.6-0.4-0.200.20.40.60.81-1.00E-05 1.00E-05 3.00E-05ti(t)EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu18Critically damped: Real Equal Roots If ζ = 1, s1and s2are real and equal.0012()ttcxt Ke Kteςωςω−−=+10EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu19ExampleFor the example, what are ζ and ω0?dttdvLtiLCdttdiLRdttids)(1)(1)()(22=++22002() ()2()0cccdx t dx txtdt dtζω ω++=2001,2 ,2RRCLCL Lωζωζ===10Ω+-769pFi(t)159μHEE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu20Example ζ = 0.011 ω0= 2π455000 Is this system over damped, under damped, or critically damped? What will the current look like?-1-0.8-0.6-0.4-0.200.20.40.60.81-1.00E-05 1.00E-05 3.00E-05ti(t)11EE40 Summer 2005: Lecture 5 Instructor: Octavian Florescu21Slightly Different Example


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Berkeley ELENG 40 - Lecture Notes

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