Inductance and CapacitanceSerial-parallel combination of inductance and capacitanceCp-7 Response of First Order RL and RC CircuitsThe natural response of an RC circuitThe step response of an RL CircuitThe step response of an RC circuitNatural and Step Response of RLC CircuitsThe Overdamped
Voltage responseThe Step Response of a Parallel RLC CircuitThe Natural Response of a Serial RLC CircuitInductance and Capacitance Inductors and capacitors cannot generate energy, they are classified as passive elements. Inductor: The symbol for impedance is L and measured as henrys. The relationship between the voltage and current at the terminals of an inductor: Lv dtdiLv = )(1)(00∫+=tttidvLtiτ The power is: vip = The energy is : 221Liw = Capacitor: It is represented by C and is measured in farads(F) The relationship between the voltage and current at the terminals of a capacitor: Cv dtdvCi = ∫+=ttvdiCtv0)0(1)(τ The power is: dtdvCvvip == The energy is : 221Cvw = i + - + - iSerial-parallel combination of inductance and capacitance Inductor is serial L2v2L1v1L3v3 321LLLL ++= dtdiLLLvvvv )(321321++=++= Inductors in parallel L2i2L3i3L1i1 Total Inductance value: 3211111LLLL++= )()()()111(0302013213210titititvLLLiiiitt+++++=++=∫τ Capacitor: Capacitors in serial: C3v3C1v1C2v2 Total capacitor value: 3211111CCCCT++= Capacitors in parallel C3i3C1i1C2i2 v i v iTotal capacitor value: 321CCCCT++= Example: iC=2F≥<≤<=202000)(tfortforttforti For 0<t 0021)(1)( ===∫∫∞−∞−τττddiCtvtt For 00 <≤ t 4)2(21210)(1)0()(20200ttddiCvtvttt==+=+=∫∫ττττ For 0≥t 102142)(1)2()(222=+=+=∫∫τττddiCvtvtt ≥<≤<=2120400)(2tfortforttfortv Remainder: aedteatat=∫ )1(2−=∫ataedtteatat 11+=+∫ntdttnn for all n accept for n=-1 xdtt log1=∫− (Reference: Standard Mathematical Tables CRC Press.) + -Cp-7 Response of First Order RL and RC Circuits The natural response of an RL circuit: All the source current Is appears in the inductance before t=0. Switch is opened at t=0, then inductance begins releasing energy. We find the v(t) for t>=0. RoLiI2IsRv LiRv Using KVL to obtain an expression 0=+ RidtdiL This is a first-order differential equation. The solution of it tLReIti−=0)( ,t>=0 where 0I denotes the initial current in the inductor The voltage on the resistor is tLRIv−= Re0, +≥ 0t 0)0( =−v and RIv0)0( = The on the resistor RvRivip22=== or tLRIp220Re−= for +≥ 0t Energy tLRteLIpdxw22001(21−−==∫, 0≥t We call time constant RL=τ t=0The natural response of an RC circuit VgRoCRv C Rv 0=+RvdtdvC τ/0)(teVtv−=, 0≥t where τ is the time constant is RC=τ. τ/0)(teRVti−=, +≥ 0t τ/220teRVvip−==, +≥ 0t )1(1/2200τtteVCdxpp−−==∫, 0≥t The step response of an RL Circuit VsRLv(t) After switch is closed. KVL dtdiLRiVs+= t=0 Vg + - itLRsseRVIRVti)/(0)()(−−+= If the initial energy is zero I0=0, then tLRsseRVRVti)/()(−−= The voltage on the inductor is tLRstLRsseVeRVRVdtdLdtdiLv)/()/( −−=−== The step response of an RC circuit IsRC Using KCL , the differential equation: sccIvdtdvC =+ The solution of the differential equation gives the voltage in the capacitor: RCtssceRIVRIv/0)(−−+=, 0≥t The current in the capacitor yields the differential equation 01=+ iRCdtdi ! RCtseRVIi/0−−=, +≥ 0t vc iThe Integrating Amplifier cc-00V-0ccCVV1Vs++Rs Let assume op amp is ideal 0=+sfii 0=pv, so that sssRvi = and dtdvCiof= There fore ssovCRdtdv1−= ∫+=ttossotvdvCRtv0)(1)(0τ vo vp vn is ifNatural and Step Response of RLC Circuits 8.1 The Natural Response of a Parallel RLC Circuit Rv+-CVoL Applying KCL to the circuit 0100=+−+∫dtdvCIdvLRvtτ I0 is constant, we differentiate one to get 0122=++LCvdtdvRCdtvd This is second-order differentiation equation. The characteristic equation is 012=++LCRCss The solution of the differential equation is tstseAeAv2121+= Characteristic roots of the solution are 2021ωαα−+−=s 2021ωαα−−−=s where RC21=α and LC10=ω There are three possible outcomes 1. If 220αω< roots are will be real. The voltage response call it overdamped. 2. If 220αω> roots are will be complex. The voltage response call it underdamped. 3. If 220αω= roots are will be real and equal. The voltage response call it critically damped. ic iL iRThe Overdamped 220αω< Voltage response tstseAeAv2121+= 1. Using R, L, C, find the roots s1 and s2 of the characteristic equation 2. Using circuit analysis, find )0(+v 21)0( AAv +=+ 3. Using circuit analysis, find dtdv /)0(+ 2211)0()0(AsAsCidtdvC+==++ 4. Find the values of A1 and A2 from these two equation 5. Substitute the values s1, s2, A1 and A2 into the solution of the differential equation The Underdamped 220αω>Voltage response The voltage is tweBtweBtvdtdtsincos)(21αα−−+= where 220αω−=dw B1 and B2 values can be found by solving the following equations 10)0( BVv ==+ 21)0()0(BBCidtdvdCωα+−==++ The critically 220αω=Voltage response The voltage response is tteDteDtvαα−−+=21)( D1 and D2 values can be found by solving the following equations 20)0( DVv ==+ 21)0()0(DDCidtdvCα−==++The Step Response of a Parallel RLC Circuit Rvt=0+-CVoLI Using KCL, we have IdtdvCRviL=++, and dtdiLvL= or LCILCidtdiRCdtidLLL=++122 Solution: +=responsenaturaltheasformsametheoffunctionfIi +=responsenaturaltheasformsametheoffunctionfVv The Natural Response of a Serial RLC Circuit LRC Applying KVL into the circuit 0100=+++∫tVidCdtdiLRiτ which can be arranged as V0 I0 i022=++LCidtdiLRdtid This is second order differential equation. The characteristic equation is 012=++LCsLRs The roots of characteristic equation is 2022,1ωαα−±−=s where LR2=αrad/s and LC10=ωrad/s Thus possible solution 1. tstseAeAti2121)( += if 220αω< 2. teBteBtidtdtωωααsincos)(21−−+= if 220αω> 3. tteDteDtiαα−−+=21)( if 220αω= The Step Response of a Serial RLC Circuit LVt=0RC Applying KVL to the serial RLC circuit cvdtdiLRiV ++= dtdvCic=, and 22dtvdCdtdic=, substitute them into the equation LCVLCvdtdvLRdtvdCCC=++22 vC iThree possible solution 1. tstsfCeAeAVv21'2'1++= if 220αω< 2. teBteBVvdtdtfCωωααsincos'2'1−−++= if 220αω> 3.