EE40Lecture 12Venkat AnantharamFirst Order CircuitsResponse of a CircuitNatural Response of an RC CircuitSolving for the Voltage (t 0)Solving for the Current (t > 0)Solving for Power and Energy Delivered (t > 0)Natural Response of an RL CircuitSolving for the Current (t 0)Solving for the Voltage (t > 0)Solving for Power and Energy Delivered (t > 0)Natural Response SummaryDigital SignalsPulse DistortionExampleFirst Order Circuits: Forced ResponseComplete SolutionThe Time ConstantWhat Does Xc(t) Look Like?Particular SolutionParticular Solution: F(t) ConstantThe Particular Solution: F(t) SinusoidThe Particular Solution: F(t) Exp.The Total Solution: F(t) SinusoidExampleSlide 1EE40 Spring 2008 Venkat AnantharamEE40Lecture 12Venkat Anantharam2/20/08 Reading: Chap. 4: first order circuitsSlide 2EE40 Spring 2008 Venkat AnantharamFirst Order CircuitsR+-Cvs(t)+-vc(t)+-vr(t)ic(t)vL(t)is(t)R L+-iL(t)KVL around the loop: KCL at the node:()() ()ccsdv tRCvtvtdt+=()() ()LLsdi tLit itRdt+=Slide 3EE40 Spring 2008 Venkat AnantharamResponse of a Circuit• Transient response of a circuit is the portion of the total response that dies to zero as time goes to infinity.• Steady-state response is what remains of the total response after the transient response is removed.• Natural response of a circuit is the portion of the response due to the initial conditions. This is generally part of the transient response, except in cases where there is some energy in the initial conditions that cannot dissipate through resistors (e.g. a pure LC oscillator).• Forced response is the portion of the response due to the forcing function (the right hand side of the differential equation that comes from the sources). In general this will also have a transient part. • We will later be mostly interested in the steady state response of circuits to a DC forcing function or more generally a sinusoidal forcing function• Note: some of this terminology is at variance with your book. Your book’s definition of ``forced response”. for instance, is inconsistent.Slide 4EE40 Spring 2008 Venkat AnantharamNatural Response of an RC Circuit• Consider the following circuit, for which the switch is closed for t < 0, and then opened at t = 0:Notation:0–is used to denote the time just prior to switching0+ is used to denote the time immediately after switching• The voltage on the capacitor at t = 0–is VoCRoVoRt = 0+−+v–Slide 5EE40 Spring 2008 Venkat AnantharamSolving for the Voltage (t ≥ 0)• For t > 0, the circuit reduces to• Applying KCL to the RC circuit:• Solution:+v–RCtetv/o V)(−=CRoVoR+−iSlide 6EE40 Spring 2008 Venkat AnantharamSolving for the Current (t > 0)• Note that the current changes abruptly:RCtoeVtv/)(−=RVieRVRvtitioRCto=⇒==>=+−−)0( )( 0,for 0)0(/+v–CRoVoR+−iSlide 7EE40 Spring 2008 Venkat AnantharamSolving for Power and Energy Delivered (t > 0)()RCtotRCxotRCtoeCVdxeRVdxxpweRVRvp/220/220/222121 )(−−−−=====∫∫RCtoeVtv/)(−=+v–CRoVoR+−iSlide 8EE40 Spring 2008 Venkat AnantharamNatural Response of an RL Circuit• Consider the following circuit, for which the switch is closed for t < 0, and then opened at t = 0:Notation:0–is used to denote the time just prior to switching0+ is used to denote the time immediately after switching• t<0 the entire system is at steady-state; and the inductor is Æ like short circuit• The current flowing in the inductor at t = 0–is Io and V across is 0.LRot = 0i+v–IoRSlide 9EE40 Spring 2008 Venkat AnantharamSolving for the Current (t ≥ 0)• For t > 0, the circuit reduces to• Applying KVL to the LR circuit:• v(t)=i(t)R•At t=0+, i=I0, • At arbitrary t>0, i=i(t) and• Solution:()()di tvt Ldt=LRoi +v–IoR-= I0e-(R/L)ttLReiti)/()0()(−=Slide 10EE40 Spring 2008 Venkat AnantharamSolving for the Voltage (t > 0)tLRoeIti)/()(−=• Note that the voltage changes abruptly:LRo+v–IoRI0RvReIiRtvtvtLRo=⇒==>=+−−)0()(0,for 0)0()/(Slide 11EE40 Spring 2008 Venkat AnantharamSolving for Power and Energy Delivered (t > 0)tLRoeIti)/()(−=LRo+v–IoR()tLRotxLRottLRoeLIdxReIdxxpwReIRip)/(220)/(220)/(222121 )(−−−−=====∫∫Slide 12EE40 Spring 2008 Venkat AnantharamNatural Response SummaryRC Circuit• Capacitor voltage cannot change instantaneously• time constantRL Circuit• Inductor current cannot change instantaneously• time constantRL=ττ/)0()()0()0(teitiii−+−==RiL+v–RCτ/)0()()0()0(tevtvvv−+−==RC=τSlide 13EE40 Spring 2008 Venkat AnantharamDigital SignalsWe compute with pulses. We send beautiful pulses in:timevoltagetimevoltageBut we receive lousy-looking pulses at the output:Capacitor charging effects are responsible!• Every node in a real circuit has capacitance; it’s the charging of these capacitances that limits circuit performance (speed)Slide 14EE40 Spring 2008 Venkat AnantharamPulse DistortionRThe input voltage pulse width must be large enough; otherwise the output pulse is distorted.(We need to wait for the output to reach a recognizable logic level, before changing the input again.)0123456012345TimeVoutPulse width = 0.1RC0123456012345TimeVout01234560 5 10 15 20 25TimeVout+Vout–Vin(t)C+–Pulse width = RC Pulse width = 10RCSlide 15EE40 Spring 2008 Venkat AnantharamExampleVinRVoutCSuppose a voltage pulse of width5 µs and height 4 V is applied to theinput of this circuit beginning at t = 0:R = 2.5 kΩC = 1 nFτ = RC = 2.5 µs• First, Voutwill increase exponentially toward 4 V.• When Vingoes back down, Voutwill decrease exponentially back down to 0 V.What is the peak value of Vout?The output increases for 5 µs, or 2 time constants.Æ It reaches 1-e-2or 86% of the final value.0.86 x 4 V = 3.44 V is the peak valueSlide 16EE40 Spring 2008 Venkat AnantharamFirst Order Circuits: Forced ResponseR+-Cvs(t)+-vc(t)+-vr(t)ic(t)vL(t)is(t)R L+-iL(t)KVL around the loop:vr(t) + vc(t) = vs(t))()(1)(tidxxvLRtvst=+∫∞−KCL at the node:()() ()LLsdi tLit itRdt+=()() ()ccsdv tRCvtvtdt+=Slide 17EE40 Spring 2008 Venkat AnantharamComplete 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 any particular solution and the associated complementary solution– Particular solution is any solution that satisfies the original equation.– Complementary solution is a solution of the homogeneous equation (the one with zero forcing function) used to satisfy the initial
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