Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7EE201 Lecture 18 P. 1Source-free and step response20 8HStep1 : Compute RTH for t 0.4sRTH = 5 Step 2: Evaluate iL(t) for t 0.4s5+vL_0.4sExample: Find iL(t) and vL(t) for t 0 if iL(0-) = 10A, and the switch closes at t = 0.4s4 8H+vL(t)-iL(t) = [e ] iL(t0)-(RTH/L)(t-t0)EE201 Lecture 18 P. 2Step 3: Evaluate iL (t) for 0 t 0.4 s. RTH= 20 iL (t) = [e ] iL(0+) iL (t) = 10 e-2.5t A At t = 0.4 s, iL(0.4) = 10e-1 = 3.68 AWhere t0 = 0.4 s. But iL(0.4) is unknown-(20/8)(t-0)Step 4: Put solution for iL (0.4) into equation for iL (t) t 0.4 s.iL (t) = [e ] iL(0.4) iL (t) = 3.68 e-0.5(t-0.4) A-(4/8)(t-t0)EE201 Lecture 18 P. 3Step 5: Compute vL (t) from Ohm’s Law When 0 t 0.4 s : vL (t) = -RTH iL (t) vL (t)= -(20)(10)[e-2.5t ] = -200 e-2.5t V When t 0.4 s vL (t) = - RTH iL (t) vL (t) = -(4)3.68 e (-0.5)(t-0.4) vL (t) = -14.72 e (-0.5)(t-0.4) Vu(t) =Unit step functionsu(t - t0) =1 t 00 t 01 t t00 t t0Examples: v(t) = 10 u(t) Vv(t)=10V if t 00 if t 0iL (t) = 3.68 e-0.5(t-0.4) A iL (t) = 3.68 e-0.5(t-0.4) u(t-0.4) AStep ResponseAdd a constant independent source to RL or RC circuitEE201 Lecture 18 P. 4LFrom KVL vL (t)= vs - R iL (t) _R+_R++_C+_+_vL(t)vC(t)iL(t)iC(t) RL Circuit RC CircuitStep response considers circuits with constant current or voltage sources.Derivation of differential equation for RL circuit (iL (t) response )vL (t)= L d iL (t) dtvSvSEE201 Lecture 18 P. 5 d iL (t) R iL (t) vS dt L L(18.1)Derivation of differential equation for RC circuit (vc(t) response) dtd vc (t)ic(t) = CFrom Ohm’s Lawic(t) = vs(t) - vc(t) R+= d vC (t) vC (t) vS dt RC RC(18.2)+=EE201 Lecture 18 P. 6General Differential Eqn. for (18.1) and (18.2) dt d x(t) 1 x (t) = FF – a forcing function (constant)RL RC L/RTHRTHC+Solution form for general differential equation:x(t) = F + [x(t0+) - F] exp{-(t-t0)/}(18.3)Interpretation of solution is obtained by examining each term (see next page).General strategy: reduce complicated circuit (attached to inductor or capacitor) to its Thevenin equivalent, attach inductor or capacitor to external terminals of network, and use equations derived on Page 7.EE201 Lecture 18 P. 7iL(t)= iL() + [iL(t0+) - iL()] exp {-(RTH/L)(t-t0)}iL(t) = (vs/L)(L/RTH) + [iL(t0+)- (vs/L)(L/RTH)] exp{-(RTH/L)(t-t0)}RL Circuit Caselim t iL(t) = (vs/L)(L/RTH) = F = final valueiL(t) = (final iL) + [(initial iL) - (final iL)] exp{-(RTH/L)(t-t0)}vc(t) = vc() + [vc(t0+) - vc()] exp{-(t-t0)/(RTHC)}vc(t) = (Vs / RTHC)(RTHC) + [vc(t0+)- (vs/ RTHC)(RTHC) ] exp{-(t-t0)/(RTHC)}RC Circuit Caselim t vc(t)= (vs/ RTHC)(RTHC )= F = final valuevc(t) = (final vc) + [(initial vc) - (final vc]
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