Summary from previous lectureGoals for todayElectrical dynamical variables: charge, current, voltageElectrical resistanceCapacitanceCapacitanceInductanceSummary: passive electrical elements; SourcesCombining electrical elements: networksImpedances in series and in parallelThe voltage dividerExample: the RC circuitInterpretation of the RC step responseExample: RLC circuit with voltage sourceExample: two-loop networkThe operational amplifier (op-amp)Lecture 04 – Wednesday, Sept. 122.004 Fall ’07 Summary from previous lecture• Laplace transform• Transfer functions and impedancesL [f(t)] ≡ F (s)=Z+∞0−f(t)e−stdt.L [u(t)] ≡ U( s )=1s.L£e−at¤=1s + a.Lh˙f(t)i= sF (s) − f(0−).L·Zt0−f(ξ)dξ¸=F (s)s.⇔f(t)x(t)F (s)X(s)TF(s)=X(s)F (s)Z(s)=F (s)X(s)Ts(s)Ω(s)JbTF(s):=Ω(s)Ts(s)=1Js+ b.ZJ= Js; Zb= b;TF(s)=1ZJ+ ZbLecture 04 – Wednesday, Sept. 122.004 Fall ’07 Goals for today• Dynamical variables in electrical systems:– charge,– current,–voltage.• Electrical elements: –resistors, – capacitors, – inductors, –amplifiers.• Transfer Functions of electrical systems (networks)• Next lecture (Friday):– DC motor (electro-mechanical element) model – DC motor Transfer FunctionLecture 04 – Wednesday, Sept. 122.004 Fall ’07 Electrical dynamical variables: charge, current, voltage+ –+ +– –++++++++v(t)i(t):=dq( t )dtcharge qcharge flow ≡ current i(t)voltage (aka potential) v( t)Coulomb [Cb]Amp´ere [A]=[Cb]/[sec]Volt [V]+−Lecture 04 – Wednesday, Sept. 122.004 Fall ’07 v(t)=Ri(t) ⇒ V (s)=RI(s) ⇒V (s)I(s)= R ≡ ZR• Collisions between the mobile charges and the material fabric (ions, generally disordered) lead to energy dissipationElectrical resistance(loss). As result, energy must be expended to generate current along the resistor;i.e., the current flow requires application of potential across the resistor• The quantity ZR=R is called the resistance (unit: Ohms, or Ω)• The quantity GR=1/R is called the conductance (unit: Mhos or Ω-1)++++++++v( t )+i(t)+−Lecture 04 – Wednesday, Sept. 122.004 Fall ’07 Capacitance• Since similar charges repel, the potential v is necessary to prevent the charges from flowing away from the electrodes (discharge)• Each change in potential v(t+Δt)=v(t)+Δv results in change of the energy stored in the capacitor, in the form of charges moving to/away from the electrodes (↔ change in electric field)++ ++++++++++++++++++++++++ +−−−−−−−−−−−−−−−−−−−−−v( t )electrode(conductor)electrode(conductor)dielectric(insulator)+−i(t)i(t)E(t)Lecture 04 – Wednesday, Sept. 122.004 Fall ’07 Capacitance• Capacitance C:• in Laplace domain:q(t)=Cv(t) ⇒dq(t)dt≡ i(t)=Cdv(t)dt++ ++++++++++++++++++++++++ +−−−−−−−−−−−−−−−−−−−−−v( t )electrode(conductor)electrode(conductor)dielectric(insulator)+−i(t)i(t)E(t)I(s )=CsV (s) ⇒V (s)I(s )≡ ZC(s)=1CsLecture 04 – Wednesday, Sept. 122.004 Fall ’07 Inductance• Current flow i around a loop results in magnetic field B pointing normal to the loop plane. The magnetic field counteracts changes in current; therefore, to effect a change in current i(t+Δt)=i(t)+Δi a potential v must be applied (i.e., energy expended)• Inductance L:• in Laplace domain:v( t )B(t)i(t)v(t)=Ldi(t )dtV (s)=LsI(s) ⇒V (s)I( s)≡ ZL(s)=Ls+−Lecture 04 – Wednesday, Sept. 122.004 Fall ’07 Summary: passive electrical elements; Sources+−Voltage source:v(t) independentof current through.Electrical inputs: voltage source, current sourceCurrent source:i(t) independentof voltage across.Ground:potential reference v(t)=0alwaysTable removed due to copyright restrictions.Please see: Table 2.3 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.Lecture 04 – Wednesday, Sept. 122.004 Fall ’07 Combining electrical elements: networks• Kirchhoff Current Law (KCL)– charge conservation• Kirchhoff Voltage Law (KVL)– energy conservationv(t)vC(t)V (s)VC(s)Network analysis relies on two physical principlesi1ik···+−Pik(t)=0−+−+v1vk···Pvk(t)=0PVk(s )=0PIk(s)=0-+ΩΩCourtesy of Prof. David Trumper. Used with permission.Lecture 04 – Wednesday, Sept. 122.004 Fall ’07 Impedances in series and in parallel+−+−I1I2V2V1V+−V+−Z1Z2IZV+−IZ+−Impedances in parallelKCL: I = I1+ I2.KVL: V1+ V2≡ V .From definition of impedances:Z1=V1I1; Z2=V2I2.Therefore, equivalent circuit has1Z=1Z1+1Z2³⇔ G = G1+ G2.´Impedances in seriesKCL: I1= I2≡ I.KVL: V = V1+ V2.From definition of impedances:Z1=V1I1; Z2=V2I2.Therefore, equivalent circuit hasZ = Z1+ Z2µ⇔1G=1G1+1G2.¶I1I2V2V1IZ1Z2+−Lecture 04 – Wednesday, Sept. 122.004 Fall ’07 The voltage divider−+V2ViZ1Z2+−+−IViV2Since the two impedances are in series, they combine to an equivalent imp edanceZ = Z1+ Z2.The curre nt flowing through the combined impedance isI =VZ.Therefore, the voltage drop across Z2isV2= Z2I = Z2VZ⇒V2Vi=Z2Z1+ Z2.Z2Z1+ Z2ViZ+−+−IEquivalent circuit for computing the current I.Block diagram & Transfer FunctionLecture 04 – Wednesday, Sept. 122.004 Fall ’07 Example: the RC circuit+−Vi+−+−ViVCZ2=1CsVCWe recognize the voltage divider configuration, with the voltage across the ca-pacitor as output. The transfer function is obtained asTF(s)=VC(s)Vi(s)=1/CsR +1/Cs=11+RCs=11+τ s,where τ ≡ RC. Further, we note the similarity to the transfer function of therotational mechanical system c onsisting of a motor, inertia J and viscous frictioncoefficient b that we saw in L ecture 3. [The transfer function was 1/b(1 + τs),i.e. identical within a multiplicative constant, and the time constant τ wasdefined as J/b.] We can use the analogy to establish properties of the RCsystem without re—deriving them: e.g., the respo nse to a step input Vi= V0u(t)(step response) isVC(t)=V0³1 − e−t/τ´u(t), where now τ = RC .11+RCsBlock diagram & Transfer FunctionZ1= RLecture 04 – Wednesday, Sept. 122.004 Fall ’07 Interpretation of the RC step response++++++++++++++++++++++++++ +−−−−−−−−−−−−−−−−−−−−−+−Vi+−+−Z2=1CsVCZ1= RVC(t)=V0³1 − e−t/τ´u(t), τ = RC.VC(t)[Volts]t [msec]V0=1Volt R =2kΩ C =1μFCharging of a capacitor:becomes progressively moredifficult as charges accumulate.Capacity (steady-state) is reachedasymptotically (VC→V0as t→∞)CLecture 04 – Wednesday, Sept. 122.004 Fall ’07 Example: RLC
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