chapter 09 - Chapter 9 - Complete Response of Circuits with...

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Unformatted text preview: Chapter 9 - Complete Response of Circuits with Two Energy Storage Elements Exercises Apply KVL to right mesh: Ex. 9.3-1 di ( t ) + v ( t ) + 1( i ( t ) −is ( t ) ) = 0 dt di ( t ) ⇒ v ( t ) = −2 − ( i ( t )−is ( t ) ) dt 2 The capacitor current and voltage are related by i (t ) = 1 dis ( t ) 1 di ( t ) d 2i ( t ) 1 dv ( t ) 1 d di ( t ) = −2 −i ( t )+ is ( t ) = − − dt dt 2 2 dt 2 dt 2 dt 2 dt ∴ d 2i ( t ) 1 di ( t ) 1 dis ( t ) + + i( t ) = 2 dt 2 dt 2 dt The inductor voltage is related to the inductor current by di ( t ) v (t ) = 1 dt Ex. 9.3-2 Apply KCL at the top node: is ( t ) = Using the operator s = v (t ) 1 dv ( t ) + i (t ) + 1 2 dt d we have dt v (t ) = s i (t ) v (t ) 1 + sv ( t ) ⇒ is ( t ) = v ( t ) + 1 s 2 is ( t ) = v ( t ) + i ( t ) + sv ( t ) 2 Therefore 2s is ( t ) = 2 s v ( t ) + 2 v ( t ) + s 2 v ( t ) ⇒ d 2v ( t ) dv ( t ) d i (t ) +2 + 2 v (t ) = 2 s 2 dt dt dt 9-1 Ex. 9.3-3 Using the operator s = d , apply KVL to the dt left mesh: i1 ( t ) + s ( i1 ( t ) −i 2 ( t ) ) = vs ( t ) Apply KVL to the right mesh: 1 2 i 2 ( t ) + 2 i 2 ( t ) + s ( i 2 ( t ) −i1 ( t ) ) = 0 s 1 2 i1 ( t ) = 2 i 2 ( t ) + 2 i 2 ( t ) + i 2 ( t ) s s Combining these equations gives: 3s i2 ( t ) + 4si2 ( t ) + 2i2 ( t ) = s vs ( t ) 2 2 or Ex. 9.4-1 d 2i2 ( t ) di2 ( t ) d 2 vs ( t ) 3 +4 + 2i2 ( t ) = dt 2 dt dt 2 Using the operator s = top node: i s (t ) = d , apply KCL at the dt v (t ) 1 + i (t ) + s v (t ) 4 4 Apply KVL to the right-most mesh: v (t ) − ( s i (t ) + 6 i (t )) = 0 Combining these equations gives: s 2 i ( t ) + 7 s i ( t ) + 10 i ( t ) = 4 i s ( t ) The characteristic equation is: s 2 + 7 s + 10 = 0 . The natural frequencies are: s = −2 and s = −5 . Ex. 9.4-2 Assume zero initial conditions . Write mesh d equations using the operator s = : dt 1 s i1 ( t ) − i 2 ( t ) + 7 + 10 i1 ( t ) − 10 = 0 2 and 1 v ( t ) − 7 − s i1 ( t ) − i 2 ( t ) = 0 2 9-2 Now 0.005 s v ( t ) = i 2 ( t ) ⇒ v ( t ) = 200 200 i2 (t ) s i2 (t ) s so the second mesh equation becomes: 1 − 7 − s i1 ( t ) − i 2 ( t ) = 0 2 Writing the mesh equation in matrix form: s 10 + 2 −1 s 2 1 s i1 ( t ) 3 2 = 1 200 i 2 ( t ) 7 s+ 2 s − Obtain the characteristic equation by calculating a determinant: 10 + − 1 s 2 = s 2 + 20s + 400 = 0 ⇒ s1, 2 = −10 ± j 17.3 1 200 s+ 2 s s 2 − 1 s 2 Ex. 9.5-1 After t = 0 , we have a parallel RLC circuit with 1 1 7 1 1 2 = = = =6 α= and ω o = 2 RC 2(6)(1 / 42) 2 LC (7)(1 / 42) 2 7 − 6 = −1, − 6 2 dv ( t ) ∴ vn (t ) = A1e −t + A2 e−6t . We need vn (0) and n dt 7 ∴ s 1 , s 2 = −α ± α −ω = − ± 2 2 2 o to evaluate A1 & A2 . t =0 At t = 0+ we have: iC ( 0+ ) = −10 A ⇒ dv ( t ) 10 = = 420 V s 1 dt t =0+ 42 Then vn (0+ ) = 0 = A1 + A2 A1 = −84 , A2 = 84 dvn =−420= − A1 − 6 A2 dt t =0+ 9-3 Finally ∴ vn (t ) = −84e −t + 84e −6t V 1 1 = 0 ⇒ s 2 + 40s + 100 = 0 s+ RC LC Therefore s 1 , s 2 = −2.68 , −37.3 Ex. 9.5-2 s2 + vn ( t ) = A1 e −2.68t + A2 e−37.3t , v(0) = 0 = A1 + A2 v (0 + ) 1 dv(0+ ) + KCL at t = 0 yields + i (0 ) + = 0 so 1 40 dt dv(0+ ) = − 40 v(0+ ) − 40 i (0+ ) = − 40(0) − 40(1) = − 2.7 A1 − 37.3 A2 dt Therefore: A1 = −1.16 , A2 = 1.16 ⇒ v(t ) = vn (t ) = −1.16e −2.68t + 1.16e−37.3t + Ex. 9.6-1 For parallel RLC circuits: α = 1 1 1 1 2 = = 50, ω o = = = 2500 −3 LC (0.4)(10−3 ) 2 RC 2(10)(10 ) The roots of the characteristic equations are:∴ s1, 2 = −50 ± (50) 2 − 2500 = −50, −50 The natural response is vn (t ) = A1 e−50t + A2 t e−50t . At t = 0+ we have: −v(0+ ) ic (0 ) = = − . 8V 10Ω dv( t ) i (0+ ) ∴ =c = −800 V / s dt c + t =0 + 9-4 So vn (0+ ) = 8 = A1 ⇒ vn (t ) = 8e−50t + A2 t e−50t dv(0+ ) = −800 = − 400 + A2 ⇒ A2 = − 400 dt ∴ vn (t ) = 8 e 50t − 400 t e−50t V 1 1 = = 8000 2 RC 2(62.5)(10−6 ) 1 1 ω o2 = = = 1 08 −6 LC (.01)(10 ) α= Ex. 9.7-1 2 ∴ s = − α ± α 2 −ω o = − 8000 ± (8000) 2 −108 = − 8000 ± j 6000 ∴ vn (t ) = e −8000t [ A1 cos 6000 t + A2 sin 6000 t ] at t = 0+ 10 + ic (0+ ) = 0 62.5 ⇒ ic (0+ ) = −.24 A 0.08 + ∴ dv(0+ ) ic (0+ ) = = −2.4 ×10+5 V/s dt C dvn (0+ ) vn (0 ) = 10 = A1 and = −2.4 × 105 = 6000 A2 − 8000 A1 ⇒ A2 = −26.7 dt + ∴ vn (t ) = e−8000t [10 cos 6000 t − 26.7 sin 6000 t ] V Ex. 9.8-1 d 2v ( t ) dv ( t ) +5 + 6 v ( t ) = vs ( t ) so the characteristic equation is 2 dt dt s 2 + 5 s + 6 = 0 . The roots are s 1 , s 2 = −2, − 3 . The differential equation is d 2v ( t ) dv ( t ) (a) +5 + 6 v ( t ) = 8 . Try v f ( t ) = B . Substituting into the differential equation gives 2 dt dt 6 B = 8 ∴ v f (t ) = 8 / 6 V . 9-5 (b) d 2v ( t ) dt 2 +5 dv ( t ) dt + 6 v ( t ) = 3 e − 4 t . Try v f ( t ) = B e− 4 t . Substituting into the differential 3 3 equation gives (−4) 2 B + 5(−4) B + 6 B = 3 ⇒ B = . ∴ v f ( t ) = e−4t . 2 2 2 d v (t ) dv ( t ) (c) +5 + 6 v ( t ) = 2 e − 2 t . Try v f ( t ) = B t e− 2 t because –2 is a natural frequency. 2 dt dt Substituting into the differential equation gives ( 4t − 4) B + 5 B (1 − 2t ) + 6 Bt = 2 ⇒ B = 2. ∴ v f ( t ) = 2 t e −2t . Ex. 9.8-2 d 2i ( t ) di ( t ) + 20 i ( t ) = 36 + 12 t . Try i f ( t ) = A + B t . Substituting into the differential dt 2 dt equation gives 0 + 9 B + 20( A + Bt ) = 36 + 12t ⇒ B = 0.6 and A = 1.53. +9 ∴ i f ( t ) = 1.53 + 0.6 t A Ex. 9.9-1 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = R2 R1 + R 2 1 Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives: vC ( t ) R2 vs ( t ) = L +C d vC ( t ) = iL ( t ) dt d iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get 9-6 vs ( t ) = L v (t ) d vC ( t ) d d + C vC ( t ) + R1 C + C vC ( t ) + vC ( t ) R2 dt R 2 dt dt = LC L d R1 d2 + R1 C vC ( t ) + 1 + vC ( t ) + vt R2 dt R2 C ( ) dt 2 (a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω Use the steady state response as the forced response: R2 1 1= V v f = vC ( ∞ ) = 2 R1 + R 2 The characteristic equation is R1 1+ 1 R1 R2 2 2 s + + s+ = s + 6s + 8 = ( s + 2 ) ( s + 4 ) = 0 R 2 C L LC so the natural response is vn = A1 e −2 t + A2 e −4 t V The complete response is vc ( t ) = iL ( t ) = + vC ( t ) 1.309 + 1 + A1 e −2 t + A2 e−4 t V 2 d vC ( t ) = −1.236 A1 e −2 t − 3.236 A2 e−4 t + 0.3819 dt At t = 0 () 0 = vc 0+ = A1 + A2 + 0.5 ( ) = −1.236 A − 3.236 A 0 = iL 0 + 1 2 + 0.3819 Solving these equations gives A1 = −1 and A2 = 0.5, so vc ( t ) = 1 −2 t 1 −4 t −e + e V 2 2 (b) C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω Use the steady state response as the forced response: R2 1 v f = vC ( ∞ ) = 1= V R1 + R 2 4 The characteristic equation is 9-7 R1 1+ 1 R1 R2 2 2 2 s + + s+ = s + 4s + 4 = ( s + 2 ) = 0 R 2 C L LC so the natural response is vn = ( A1 + A2 t ) e −2 t V The complete response is vc ( t ) = iL ( t ) = vC ( t ) + At t = 0+ 1 + ( A1 + A2 t ) e −2 t V 4 d 1 vC ( t ) = + dt 4 () (( A 0 = vc 0+ = A1 + () 0 = iL 0+ = 2 ) − A1 ) − A2 t e −2 t 1 4 1 + A2 − A1 4 Solving these equations gives A1 = −0.25 and A2 = −0.5, so vc ( t ) = 1 1 1 −2 t − + t e V 4 4 2 (c) C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω Use the steady state response as the forced response: R2 4 v f = vC ( ∞ ) = 1= V R1 + R 2 5 The characteristic equation is R1 1+ 1 R1 R2 2 2 s + + s+ = s + 4 s + 20 = ( s + 2 − j 4 ) ( s + 2 + j 4 ) = 0 R 2 C L LC so the natural response is vn = e −2 t ( A1 cos 4 t + A2 sin 4 t ) V The complete response is iL ( t ) = vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V vC ( t ) 4 + A 2 −2 t A1 1d vC ( t ) = 0.2 + e cos 4 t − e−2 t sin 4 t 8 dt 2 2 9-8 At t = 0+ () 0 = vc 0+ = 0.8 + A1 () 0 = iL 0+ = 0.2 + A2 2 Solving these equations gives A1 = -0.8 and A2 = -0.4, so vc ( t ) = 0.8 − e −2 t ( 0.8 cos 4 t + 0.4 sin 4 t ) V Ex 9.9-2 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = R2 R1 + R 2 1, iL ( ∞ ) = 1 R1 + R 2 and vo ( ∞ ) = R2 R1 + R 2 1 Next, represent the circuit by a 2nd order differential equation: KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives: d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L Use the substitution method to get vs ( t ) = R1 C d d d L iL ( t ) + R 2 iL ( t ) + L iL ( t ) + R 2 iL ( t ) + R1 iL ( t ) dt dt dt d2 d = R1 LC 2 iL ( t ) + ( L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) dt dt Using iL ( t ) = vo ( t ) gives R2 L d R1 + R 2 d2 vs ( t ) = LC 2 vo ( t ) + + R1 C vo ( t ) + vt R2 dt R2 o ( ) R2 dt R1 9-9 (a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω Use the steady state response as the forced response: R2 1 1= V v f = vo ( ∞ ) = 2 R1 + R 2 The characteristic equation is R2 1+ 1 R2 R1 2 2 s + + s+ = s + 6s + 8 = ( s + 2 ) ( s + 4 ) = 0 R1 C L LC so the natural response is vn = A1 e −2 t + A2 e −4 t V The complete response is 1 + A1 e −2 t + A2 e −4 t V 2 A1 −2 t A2 −4 t v (t ) 1 iL ( t ) = o e+ eV = + 1.309 2.618 1.309 1.309 vo ( t ) = vC ( t ) = 1.309 iL ( t ) + At t = 0+ 1d 1 iL ( t ) = + 0.6180 A1 e −2 t + 0.2361 A2 e −4 t 4 dt 2 () 0 = iL 0+ = () 0 = vC 0+ = A1 A2 1 + + 2.618 1.309 1.309 1 + 0.6180 A1 + 0.2361 A2 2 Solving these equations gives A1 = −1 and A2 = 0.5, so vo ( t ) = 1 −2 t 1 −4 t −e + e V 2 2 (b) C = 1 F, L = 1 H, R1 = 1 Ω, R2 = 3 Ω Use the steady-state response as the forced response: R2 3 v f = vo ( ∞ ) = 1= V R1 + R 2 4 The characteristic equation is R2 1+ 1 R2 R1 2 2 s2 + + s+ = s + 4s + 4 = ( s + 2 ) = 0 R1 C L LC 9-10 so the natural response is vn = ( A1 + A2 t ) e −2 t V The complete response is vo ( t ) = iL ( t ) = vo ( t ) 3 vC ( t ) = 3 iL ( t ) + At t = 0+ 3 + ( A1 + A2 t ) e −2 t V 4 = 1 A1 A2 −2 t t e V + + 4 3 3 3 A1 A2 A2 −2 t d iL ( t ) = + + t e + 4 3 3 3 dt () 0 = iL 0+ = () 0 = vC 0+ A1 + 1 4 3 3 A1 A2 =+ + 43 3 Solving these equations gives A1 = -0.75 and A2 = -1.5, so vo ( t ) = 3 3 3 −2 t − + t e V 4 4 2 (c) C = 0.125 F, L = 0.5 H, R1 = 4 Ω, R2 = 1 Ω Use the steady state response as the forced response: R2 1 v f = vo ( ∞ ) = 1= V R1 + R 2 5 The characteristic equation is R2 1+ 1 R2 R1 2 2 s + + s+ = s + 4 s + 20 = ( s + 2 − j 4 ) ( s + 2 + j 4 ) = 0 R1 C L LC so the natural response is vn = e −2 t ( A1 cos 4 t + A2 sin 4 t ) V The complete response is vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V iL ( t ) = vo ( t ) 1 = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V 9-11 vC ( t ) = iL ( t ) + At t = 0+ 1d iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e−2 t sin 4 t 2 dt () ( 0 ) = 0.2 + 2 A 0 = iL 0+ = 0.2 + A1 0 = vC + 2 Solving these equations gives A1 = -0.8 and A2 = -0.4, so vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) V Ex. 9.10-1 At t = 0+ no initial stored energy ⇒ v1 (0+ ) = v2 (0+ ) = i (0+ ) = 0 3 di (0+ ) di (0+ ) +0=0 ⇒ =0 10 dt dt 0 dv1 (0+ ) + ⇒ =0 KCL at A : + i1 (0 ) + 0 = 0 1 dt 5 dv2 (0+ ) = 10 ⇒ KCL at B : − 0 + i2 (0+ ) − 10 = 0 ⇒ i2 (0+ ) = 6 dt KVL : − 0 + For t > 0: dv2 (0+ ) = 12 V s dt v1 1 d v1 + +i = 0 1 12 dt 5 d v2 = 10 KCL at B : − i + 6 dt 3 di − v1 + + v2 = 0 KVL: 10 dt Eliminating i yields 1 d v1 5 d v2 v1 + + − 10 = 0 12 dt 6 dt 3 5 d 2 v2 −v1 + + v2 = 0 10 6 dt 2 KCL at A : Next 9-12 v1 = v2 + 1 d 2 v2 4 dt 2 ⇒ d v1 d 2 v2 1 d 3 v2 = + dt dt 2 4 dt 3 Now, eliminating v1 v2 + 1 d 2 v2 1 d v2 1 d 3 v2 5 d v2 + + = 10 + 4 dt 2 12 dt 4 dt 3 6 dt Finally, the circuit is represented by the differential equation: d 3 v2 d2 v dv + 12 2 2 + 44 2 + 48v2 = 480 3 dt dt dt The characteristic equation is s 3 + 12s 2 + 44s + 48 = 0 . It’s roots are s1, 2,3 = −2, −4, −6 . The natural response is vn = A1e −2t + A2 e −4t + A3e −6t Try v f = B as the forced response. Substitute into the differential equation and equate coefficients to get B = 10. Then v2 (t ) = vn (t ) + v f (t ) = A1e −2t + A2 e −4t + A3e−6t + 10 We have seen that v2 (0+ ) = 0 and dv2 (0+ ) d 2 v2 (0+ ) = 12 V/s . Also = 4[v1 (0+ ) − v2 (0+ )] = 0 . dt dt 2 Then v2 (0+ ) = 0 = A1 + A2 + A3 + 10 dv2 (0+ ) = 12 = −2 A1 − 4 A2 − 6 A3 dt d 2 v2 (0+ ) = 0 = 4 A1 + 16 A2 + 36 A3 dt 2 Solving these equations yields A1 = −15, A2 = 6, A3 = −1 so v2 (t ) = ( −15 e −2t + 6 e −4t − e −6t + 10 ) V Ex. 9.11-1 1 1 s2 + s+ =0 RC LC In our case L = 0.1, C = 0.1 so we have s 2 + 10 s + 100 = 0 R 9-13 a) R = 0.4 Ω ⇒ s 2 + 25s + 100 = 0 s1,2 = −5, − 20 b) R = 1 Ω ⇒ s 2 + 10 s + 100 = 0 s1,2 = − 5 ± j 5 3 9-14 Problems Section 9-3: Differential Equations for Circuits with Two Energy Storage Elements P9.3-1 KCL: iL ( t ) = v (t ) dv ( t ) +C R2 dt KVL: vs ( t ) = R1iL ( t ) + L diL ( t ) + v (t ) dt dv ( t ) L dv ( t ) d 2v ( t ) v (t ) vs ( t ) = R1 +C + LC + v (t ) + dt R2 dt dt 2 R2 d 2v ( t ) R L dv ( t ) = 1 + 1 v ( t ) + R1C + + [ LC ] vs ( t ) R2 dt dt 2 R2 In this circuit R1 = 2 Ω, R 2 = 100 Ω, L = 1 mH, C = 10 µ F so 2 dv ( t ) −8 d v ( t ) vs ( t ) = 1.02v ( t ) + .00003 + 10 dt dt 2 dv ( t ) d 2 v ( t ) 8 8 10 vs ( t ) = 1.02 × 10 v ( t ) + 3000 + dt dt 2 P9.3-2 Using the operator s = KCL: is ( t ) = d we have dt v (t ) + iL ( t ) + Csv ( t ) R1 KVL: v ( t ) = R2iL ( t ) + LsiL ( t ) Solving by usingCramer's rule for iL ( t ) : iL ( t ) = is ( t ) R2 Ls + + R2Cs + LCs 2 + 1 R1 R1 R2 L 2 1 + iL ( t ) + + R2C siL ( t ) + [ LC ] s iL ( t ) = is ( t ) R1 R1 9-15 In this circuit R1 = 100 Ω, R 2 = 10 Ω, L = 1 mH, C = 10 µ F so 1.1iL ( t ) + .00011siL ( t ) + 10−8 s 2iL ( t ) = is ( t ) diL ( t ) d 2iL ( t ) + = 108 is ( t ) 1.1× 10 iL ( t ) + 11000 2 dt dt 8 P9.3-3 After the switch closes, a source transformation gives: KCL: iL ( t ) + C dvc ( t ) vs ( t ) + vc ( t ) + =0 dt R2 KVL: diL ( t ) − vc ( t ) − vs ( t ) = 0 dt di ( t ) vc ( t ) = R1is ( t ) + R1iL ( t ) + L L − vs ( t ) dt Differentiating R1is ( t ) + R1iL ( t ) + L d vc ( t ) d is ( t ) d iL ( t ) d 2iL ( t ) d vs ( t ) = R1 + R1 +L − dt dt dt dt 2 dt Then d is ( t ) d iL ( t ) d 2iL ( t ) d vs ( t ) vs ( t ) + R1 +L − iL ( t ) + C R1 + dt dt dt 2 dt R2 di ( t ) 1 + R1is ( t ) + R1iL ( t ) + L L − vs ( t ) = 0 R2 dt Solving for i L ( t ) : d 2i L ( t ) R1 − R1 R1 di s ( t ) 1 dv s ( t ) 1 di L ( t ) R1 1 + + + + i s (t ) − + i L (t ) = 2 dt LC R 2 L dt L dt L R 2C dt LR 2 C LC 9-16 Section 9-4: Solution of the Second Order Differential Equation - The Natural Response P9.4-1 From Problem P 9.3-2 the characteristic equation is: 1.1×108 +11000 s + s 2 = 0 ⇒ s1 , s2 = −11000± (11000) 2 − 4(1.1×108 ) = −5500± j8930 2 P9.4-2 KVL: 40 ( is ( t ) − iL ( t ) ) = (100 × 10−3 ) diL ( t ) + vc ( t ) dt The current in the inductor is equal to the current in the capacitor so 1 dv ( t ) iL ( t ) = ×10−3 c 3 dt 2 40 dv ( t ) 100 d vC ( t ) vC ( t ) = 40 is ( t ) − × 10−3 C − × 10−6 2 3 dt 3 dt d 2 vC ( t ) dv ( t ) + 400 C + 30000 vC ( t ) = 40 is ( t ) 2 dt dt 2 s + 400s + 30000 = 0 ⇒ ( s + 100)( s + 300) = 0 ⇒ s1 = −100, s2 = −300 P9.4-3 v ( t ) − vs ( t ) dv ( t ) + iL ( t ) + (10 × 10−6 ) =0 1 dt di ( t ) KVL: v ( t ) = 2iL ( t ) + (1× 10−3 ) L dt KCL: 0 = 2 iL ( t ) + (1× 10−3 ) vs ( t ) = 3iL ( t ) + 0.00102 diL ( t ) dt diL ( t ) dt − vs ( t ) + iL ( t ) + (10 × 10−6 ) ( 2 ) + 10−8 d 2iL ( t ) dt 2 ⇒ d 2iL ( t ) dt diL ( t ) dt + (10 × 10−6 ) (10−3 ) + 102000 diL ( t ) dt d 2iL ( t ) dt + 3 × 108 iL ( t ) = 108 vs ( t ) s 2 + 102000s + 3 ×108 = 0 ⇒ s1 = 3031, s2 = −98969 9-17 Section 9.5: Natural Response of the Unforced Parallel RLC Circuit P9.5-1 dv( 0 ) d = −3000 V/s . Using the operator s = , the node dt dt − vs ( t ) v( t ) v( t ) L + = 0 or L C s 2 + s + 1 v ( t ) = v s ( t ) equation is Csv ( t ) + R sL R 1 1 s+ = 0 ⇒ s 2 + 500 s + 40, 000 = 0 The characteristic equation is: s 2 + RC LC The initial conditions are v ( 0 ) = 6 V, The natural frequencies are: s1,2 = −250 ± 2502 − 40, 000 = −100, −400 The natural response is of the form v ( t ) = Ae−100t + Be−400t . We will use the initial conditions to evaluate the constants A and B. v ( 0) = 6 = A + B ⇒ dv ( 0 ) = −3000 = −100 A − 400 B dt A = −2 and B = 8 Therefore, the natural response is v ( t ) = −2e−100t + 8e−400t t >0 P9.5-2 The initial conditions are v ( 0 ) = 2 V, i ( 0 ) = 0 . 1 1 s+ = 0 ⇒ s 2 + 4s + 3 = 0 RC LC The natural frequencies are: s1 , s2 = −1, − 3 The natural response is of the form v ( t ) = Ae − t + Be −3t . We will use the initial conditions to evaluate the constants A and B. dv ( t ) = − Ae − t − 3Be −3t . At t = 0 this becomes Differentiating the natural response gives dt dv ( t ) v ( t ) dv ( t ) v (t ) i (t ) dv ( 0 ) + + i ( t ) = 0 or =− − . = − A − 3B . Applying KCL gives C dt R dt RC C dt dv ( 0 ) v ( 0) i ( 0) =− At t = 0 this becomes − . Consequently dt RC C The characteristic equation is: s 2 + −1A − 3B = − v ( 0) i ( 0) 2 − =− −0 = − 8 RC C 14 9-18 Also, v ( 0 ) = 2 = A + B . Therefore A = −1 and B = 3. The natural response is v ( t ) = −e − t + 3e −3t V P9.5-3 di1 ( t ) di ( t ) −3 2 =0 dt dt di ( t ) di ( t ) KVL : − 3 1 +3 2 + 2i2 ( t ) = 0 dt dt KVL : i1 + 5 (1) ( 2) d , the KVL equations are dt (1+5s )i1 + ( −3s )i2 = 0 3s 2 i1 = 0 ⇒ (1 + 5s ) ( 3s + 2 ) − ( 3s ) i1 = 0 ⇒ (1 + 5s ) i1 − ( 3s ) 3s + 2 ( −3s ) i1 + ( 3s + 2 ) i2 = 0 1 The characteristic equation is (1+5s ) ( 3s + 2 ) − 9 s 2 = 6 s 2 + 13s + 2 = 0 ⇒ s1, 2 = − , −2 6 Using the operator s = −t −t The currents are i1 ( t ) = Ae 6 + Be−2t and i2 ( t ) = Ce 6 + De −2t , where the constants A, B, C and D must be evaluated using the initial conditions. Using the given initial values of the currents gives i1 ( 0 ) = 11 = A + B and i 2 ( 0 ) = 11 = C + D Let t = 0 in the KCL equations (1) and (2) to get di1 ( 0 ) dt =− di2 ( 0 ) A C 33 143 = − − 2 B and =− =− −D 2 6 6 6 dt So A = 3, B = 8, C = −1 and D = 12. Finally, i1 (t ) = 3e− t / 6 + 8e−2t A and i 2 (t ) = −e −t / 6 + 12e −2t A 9-19 Section 9.6: Natural Response of the Critically Damped Unforced Parallel RLC Circuit P9.6-1 After t = 0 Using KVL: 100 ic ( t ) + 0.025 dic ( t ) + vc ( t ) = 0 dt The capacitor current and voltage are related by: ic ( t ) = 10−5 ∴ d 2 vc ( t ) dt 2 + 4000 dvc ( t ) dt dvc ( t ) dt + 4 × 106 vc ( t ) = 0 The characteristic equation is: s 2 + 4000 s + 4 × 106 = 0 The natural frequencies are: s1,2 = − 2000, − 2000 The natural response is of the form: vc ( t ) = A1e−2000t + A2 t e−2000t (The capacitor current is continuous at t = 0 in this circuit because it is equal to the inductor current.) Before t = 0 the circuit is at steady state vc ( 0+ ) = 3 = A1 dvc ( 0+ ) = 0 = −2000 A1 + A2 ⇒ A2 = 6000 dt ∴ vc ( t ) = ( 3 + 6000t ) e−2000t V for t ≥ 0 P9.6-2 After t = 0 Using KCL: ∫ t −∞ 1 dvc ( t ) =0 4 dt d 2 vc ( t ) dv ( t ) ⇒ +4 c + 4vc ( t ) = 0 dt dt vc (τ ) dτ + vc ( t ) + The characteristic equation is: s 2 + 4 s + 4 = 0 9-20 The natural frequencies are: s1,2 = −2, −2 The natural response is of the form: vc ( t ) = A1e−2t + A2 t e−2 t Before t = 0 the circuit is at steady state i L ( 0+ ) = i L ( 0 − ) and vC ( 0+ ) = vC ( 0− ) i C ( 0 + ) = −i L ( 0 + ) = − 2 A dv ( 0+ ) dt vc ( 0 + )=0= A1 and dvc ( 0+ ) dt = −8 = A2 = i C ( 0+ ) 14 = −8 V ⇒ vc ( t ) = −8 t e−2 t V P9.6-3 Assume that the circuit is at steady state before t = 0. The initial conditions are vc ( 0− ) = 104 V & iL ( 0− ) = 0 A After t = 0 − vc ( t ) + .01 KVL: diL ( t ) + 106 iL ( t ) = 0 dt (1) KCL: d 2iL ( t ) dvC ( t ) di ( t ) = −C .01 + 106 L iL ( t ) = −C 2 dt dt dt ∴ 0.01 C ( 2) d 2iL ( t ) di ( t ) + 106 C L + iL ( t ) = 0 2 dt dt The characteristic equation is: ( 0.01 C ) s 2 + (106 C ) s + 1 = 0 The natural frequencies are: s1,2 = −106 C ± (10 C ) 6 2 − 4 ( 0.01C ) 2 ( 0.01C ) For critically-damped response: 1012 C 2 − .04C = 0 ⇒ C = 0.04 pF so s1,2 = −5 ×107 , −5 × 107 . 9-21 The natural response is of the form: iL ( t ) = A1e −5×10 t + A2 t e −5×10 7 diL + ( 0 ) = 100 dt di ( 0 ) = 106 So iL ( 0 ) = 0 = A1 and L dt Now from (1) ⇒ 7 t vc ( 0+ ) − 106 iL ( 0+ ) = 106 A s = A2 ∴ iL ( t ) = 106 t e−5×10 t A 7 Now v ( t ) = 106 iL ( t ) = 1012 t e−5×10 t V 7 P9.6-4 The characteristic equation can be shown to be: s 2 + The natural frequencies are: s1,2 = −250, − 250 1 1 = s 2 + 500 s + 62.5 × 103 = 0 s+ RC LC The natural response is of the form: v ( t ) = Ae −250t + B t e−250t dv ( 0 ) = −3000 = −250 A + B ⇒ B = −1500 dt − 1500 t e−250t V v ( 0 ) = 6 = A and ∴ v ( t ) = 6e−250t P9.6-5 After t=0, using KVL yields: di ( t ) t + Ri ( t ) + 2 + 4 ∫0 i (τ ) dτ = 6 (1) dt v( t ) Take the derivative with respect to t: d 2i ( t ) di ( t ) +R + 4i ( t ) = 0 dt 2 dt The characteristic equation is s 2 + Rs + 4 = 0 Let R = 4 for critical damping ⇒ ( s + 2) 2 =0 So the natural response is i ( t ) = A t e −2t + B e −2t i ( 0 ) = 0 ⇒ B = 0 and ∴ i ( t ) = 4 t e −2t A di ( 0 ) = 4 − R ( i( 0 ) ) = 4 − R ( 0 ) = 4 = A dt 9-22 Section 9-7: Natural Response of an Underdamped Unforced Parallel RLC Circuit P9.7-1 After t = 0 KCL: vc ( t ) dv ( t ) + iL ( t ) + 5 ×10−6 c =0 dt 250 KVL: vc ( t ) = 0.8 (1) diL ( t ) dt ( 2) d 2vc ( t ) dv ( t ) + 800 c + 2.5 ×105 v ( t ) = 0 ⇒ s 2 + 800 s + 250, 000 = 0, s1,2 = −400 ± j 300 c dt dt 2 The natural response is of the form v c (t ) Before t = 0 the circuit is at steady state: = e −400t A1 cos 300t + A 2sin 300t −6 A 500 vc ( 0+ ) = vc ( 0− ) = 250 −6 iL ( 0+ ) = iL ( 0− ) = ( 500 )+6 = 3 V From equation (1) : dvc ( 0+ ) dt = − 2 × 105 iL ( 0+ ) − 800vc ( 0+ ) = 0 vc ( 0+ ) = 3 = A1 dvc ( 0+ ) dt = 0 = −400 A1 + 300 A2 ⇒ A2 = 4 ∴ vc ( t ) = e −400t [3 cos 300t + 4 sin 300t ] V 9-23 P9.7-2 Before t = 0 v ( 0+ ) = v ( 0− ) = 0 V i ( 0+ ) = i ( 0− ) = 2 A After t = 0 KCL : t v (t ) 1 d v (t ) + + 2 ∫ v (τ ) dτ + i ( 0 ) = 0 0 1 4 dt Using the operator s = d we have dt 1 2 v (t ) + s v (t ) + v (t ) + i (0) = 0 ⇒ 4 s (s 2 + 4s + 8) v ( t ) = 0 The characteristic equation and natural frequencies are: s 2 + 4s + 8 = 0 The natural response is of the form: v ( t ) = e v ( 0 ) = 0 = B1 and −2 t B1 cos 2t + B 2 sin 2t dv ( 0 ) = 4 −i ( 0 ) − v ( 0 ) = −4 [ 2] = −8 = 2 B2 or B2 = −4 dt so v ( t ) = −4e −2t sin 2t V P9.7-3 After t = 0 1 dvc ( t ) vc ( t ) + + iL ( t ) = 0 4 dt 2 4 diL ( t ) KVL : vc ( t ) = + 8 iL ( t ) dt KCL : Characteristic Equation: ⇒ s = −2 ± j 2 d 2i L ( t ) dt 2 +4 di L ( t ) dt (1) ( 2) + 5 i L ( t ) = 0 ⇒ s 2 + 4 s + 5 = 0 ⇒ s1,2 = −2 ± i Natural Response: i L ( t ) = e −2t A1 cos t + A2 sin t 9-24 Before t = 0 vc ( 0− ) 48 + − = 7 4 8 + 2 ⇒ vc ( 0 ) = vc ( 0 ) = 8 V 2 8 iL ( 0+ ) = iL ( 0− ) = − = −4 A 2 diL ( 0+ ) dt = vc ( 0+ ) 4 − 2iL ( 0+ ) = 8 A − 2 ( −4 ) = 10 4 s iL ( 0+ ) = −4 = A1 diL ( 0+ ) dt ∴ iL ( t ) = e −2t [ −4 cos t + 2 sin t ] A = 10 = − 2 A1 + A2 ⇒ A2 = 2 P9.7-4 The plot shows an underdamped response, i.e. v ( t ) = e − α t [ k1 cos ω t + k2 sin ω t ] + k3 . Examining the plot shows v ( ∞ ) = 0 ⇒ k3 = 0, v ( 0 ) = 0 ⇒ k1 = 0 . Therefore, v ( t ) = k2 e − α t sin ω t . Again examining the plot we see that the maximum voltage is approximately 260 mV the time is approximately 5 ms and that the minimum voltage is approximately −200 mV the time is approximately 7.5 ms. The time between adjacent maximums is approximately 5 ms so 2π ω≈ = 1257 rad/s . Then 5 ×10−3 −α ( 0.005 ) 0.26 = k2 e sin (1257 (.005 ) ) (1) −0.2 = k2 e −α ( 0.0075 ) sin (1257 (.0075 ) ) ( 2) To find α we divide (1) by (2) to get ( 6.29 rad ) ⇒ sin ( 9.43 rad ) α ( 0.0025 ) sin − 1.3 = e e0.0025α = 1.95 ⇒ α = 267 From (1) we get k2 = 544 . Then v ( t ) = 544e −267t sin 1257t V 9-25 P9.7-5 After t = 0 The characteristic equation is: s2 + 1 1 s+ = 0 or s 2 + 2 s + 5 = 0 RC LC The natural frequencies are: s1,2 = −1 ± j 2 The natural response is of the form: v(t ) = e − t B1 cos 2t + B 2 sin 2t v ( 0+ ) 2 ) = − 5 − i ( 0 ) = − 5 − 110 = − 1 V 2s dv ( 0 ) 3 1 = 10 − = − B + 2 B ⇒ B = − v(0 ) = 2 = B1 . From KCL, ic ( 0 + + + so + dt 2 3 Finally, v ( t ) = 2e − t cos 2t − e− t sin 2t V 2 1 2 2 2 t≥0 9-26 Section 9-8: Forced Response of an RLC Circuit P9.8-1 After t = 0 KCL : is ( t ) = v (t ) R KVL : v ( t ) = L is ( t ) = 2 d 2iL ( t ) dt 2 dv ( t ) diL ( t ) dt dt d 2iL ( t ) L diL ( t ) + iL ( t ) + LC R dt dt 2 d 2iL ( t ) dt + iL ( t ) + C + 1 diL ( t ) 1 1 + iL ( t ) = is ( t ) RC dt LC LC + ( 650 ) diL ( t ) dt + (105 ) iL ( t ) = (105 ) is ( t ) (a) Try a forced response of the form i f ( t ) = A . Substituting into the differential equations gives 0+0+ A 1 1 = ⇒ −3 (.01) (1×10 ) (.01) (1×10−3 ) A = 1 . Therefore i f ( t ) = 1 A . (b) Try a forced response of the form i f ( t ) = A t + B . Substituting into the differential equations gives 0 + A 65 1 + ( A t + B) = 0.5 t . Therefore A = 0.5 and ( 0.01) ( 0.001) (100 ) ( 0.001) B = −3.25 × 10 −3 . Finally i f ( t ) = 5 t − 3.25 × 10−3 A . (c) Try a forced response of the form i f ( t ) = A e−250 t . It doesn’t work so try a forced response of the form i f ( t ) = B t e−250 t . Substituting into the differential equation gives ( −250 )2 B e−250t − 500 B e−250t + 650 ( −250 ) B t e−250t + B e−250t + 105 B t e−250t = 2 e−250t . Equating coefficients gives ( 250 ) and 2 2 B + 650 ( −250 ) B + 105 B = 0 ⇒ ( 250 ) + 650 ( −250 ) + 105 B = 0 ⇒ −500 B + 650 B = 2 ⇒ [ 0] B = 0 B = 0.0133 Finally i f ( t ) = 0.0133 t e −250t A . 9-27 P9.8-2 After t = 0 d 2v ( t ) dt d 2v ( t ) dt + v (t ) R dv ( t ) 1 + v (t ) = s L dt LC LC + 70 dv ( t ) dt + 12000v ( t ) = 12000 vs ( t ) (a) Try a forced response of the form v f ( t ) = A . Substituting into the differential equations gives 0 + 0 + 12000 A = 24000 ⇒ A = 2 . Therefore v f ( t ) = 2 V . (b) Try a forced response of the form v f ( t ) = A + B t . Substituting into the differential equations gives 70 A + 12000 A t + 12000 B = 2400 t . Therefore A = 0.2 and B = Finally v f ( t ) = ( −1.167 × 10−3 ) t + 0.2 V . −70 A = −1.167 × 10 −3 . 12000 (c) Try a forced response of the form v f ( t ) = A e −30 t . Substituting into the differential equations gives 900 Ae −30t − 2100 Ae −30 t + 12000 Ae −30 t = 12000 e −30t . Therefore A = v f ( t ) = 1.11e−250 t V . 12000 = 1.11 . Finally 10800 9-28 Section 9-9: Complete Response of an RLC Circuit P9.9-1 First, find the steady state response for t < 0, when the switch is open. Both inputs are constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. After a source transformation at the left of the circuit: i L ( 0) = and 11 − 4 = 2.33 mA 3000 v C ( 0) = 4 V After the switch closes Apply KCL at node a: vC R +C d vC + iL = 0 dt Apply KVL to the right mesh: L d d i L + Vs − vC = 0 ⇒ vC = L i L + Vs dt dt After some algebra: V 1d 1 d2 i+ iL + iL = − s 2L dt R C dt LC R LC ⇒ d2 d i + ( 500 ) i L + (1.6 × 105 ) i L = −320 2L dt dt The characteristic equation is s 2 + 500 s + 1.6 × 105 = 0 ⇒ s1,2 = −250 ± j 312 rad/s 9-29 After the switch closes the steady-state inductor current is iL(∞) = -2 mA so i L ( t ) = −0.002 + e −250 t ( A1 cos 312 t + A2 sin 312 t ) d i L (t ) + 4 dt = 6.25 e −250 t −250 ( A1 cos 312 t − A2 sin 312 t ) − 312 ( A1 sin 312 t − A2 cos 312 t ) + 4 v C ( t ) = 6.25 = 6.25 e −250 t ( 312 A2 − 250 A1 ) cos 312 t + ( 250 A2 + 312 A1 ) sin 312 t + 4 Let t = 0 and use the initial conditions: iL ( 0+ ) = 0.00233 = −0.002 + A1 ⇒ 0.00433 = A1 vC ( 0+ ) = 4 = 6.25 ( 312 A2 − 250 A1 ) + 4 ⇒ A2 = 250 250 A2 = ( 0.00433) = 0.00347 312 312 Then i L ( t ) = −0.002 + e −250 t ( 0.00433 cos 312 t + 0.00345 sin 312 t ) = −0.002 + 0.00555 e −250 t cos ( 312 t − 36.68° ) A v C ( t ) = 4 + 13.9 e −250 t sin ( 312 t ) V i (t ) = vC (t ) 2000 = 2 + 6.95 e −250 t sin ( 312 t ) mA (checked using LNAP on 7/22/03) P9.9-2 First, find the steady state response for t < 0. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. i ( 0) = −1 = −0.2 A 1+ 4 and v (0) = 4 ( −1) = −0.8 V 1+ 4 9-30 For t > 0 Apply KCL at node a: v − Vs d +C v+i = 0 R1 dt Apply KVL to the right mesh: R2 i + L d d i − v = 0 ⇒ v = R2 i + L iL dt dt After some algebra: L + R1 R 2C d R1 + R 2 d2 Vs i+ i+ i= 2 dt R1 L C dt R1 L C R1 L C The forced response will be a constant, if = B so 1 = ⇒ d2 d i +5 i +5i =1 2 dt dt d2 d B + 5 B + 5 B ⇒ B = 0.2 A . 2 dt dt To find the natural response, consider the characteristic equation: 0 = s 2 + 5 s + 5 = ( s + 3.62 ) ( s + 1.38 ) The natural response is in = A1 e −3.62 t + A2 e−1.38 t so i ( t ) = A1 e −3.62 t + A2 e−1.38 t + 0.2 Then d v ( t ) = 4 i ( t ) + 4 i ( t ) = −10.48 A1 e−3.62 t − 1.52 A2 e−1.38 t + 0.8 dt At t=0+ −0.2 = i ( 0 + ) = A1 + A2 + 0.2 −0.8 = v ( 0 + ) = −10.48 A1 − 1.52 A2 + 0.8 so A1 = 0.246 and A2 = −0.646. Finally i ( t ) = 0.2 + 0.246 e −3.62 t − 0.646 e−1.38 t A 9-31 P9.9-3 First, find the steady state response for t < 0. The input is constant so the capacitors will act like an open circuits at steady state. v1 ( 0 ) = and 1000 (10 ) = 5 V 1000 + 1000 v2 ( 0 ) = 0 V For t > 0, Node equations: v1 − 10 1 v −v d + × 10−6 v1 + 1 2 = 0 1000 6 1000 dt 1 d ⇒ 2 v1 + × 10−3 v1 − 10 = v2 6 dt v1 − v2 1 d = × 10−6 v2 1000 16 dt 1 d ⇒ v1 − v2 = × 10−3 v2 16 dt After some algebra: d2 d v + ( 2.8 × 104 ) v1 + ( 9.6 ×107 ) v1 = 9.6 × 108 21 dt dt The forced response will be a constant, vf = B so d2 d B + ( 2.8 × 104 ) B + ( 9.6 × 107 ) B = 9.6 × 108 2 dt dt ⇒ B = 10 V . To find the natural response, consider the characteristic equation: s 2 + ( 2.8 × 104 ) s + ( 9.6 × 107 ) = 0 ⇒ s1,2 = −4 × 103 , −2.4 × 104 The natural response is 3 vn = A1 e −4×10 t + A2 e−2.4×10 4t so 3 v1 ( t ) = A1 e−4×10 t + A2 e−2.4×10 4t + 10 At t = 0 9-32 5 = v1 ( 0 ) = A1 e −4×103 ( 0 ) + A2 e −2.4×104 ( 0 ) + 10 = A1 + A2 + 10 (1) Next 1 d 2 v1 + ×10−3 v1 − 10 = v2 6 dt ⇒ d v1 = −12000v1 + 6000 v2 + 6 × 104 dt At t = 0 d v1 ( 0 ) = −12000v1 ( 0 ) + 6000 v2 ( 0 ) + 6 × 104 = −12000 ( 5 ) + 6000 ( 0 ) + 6 × 104 = 0 dt so 3 4 d v1 ( t ) = A1 −4 × 103 e −4×10 t + A2 −2.4 × 104 e−2.4×10 t dt ( ) ( ) At t = 0+ 0= d −4×103 ( 0 ) −2.4×104 ( 0 ) v1 ( 0 ) = A1 −4 × 103 e + A2 −2.4 × 104 e = A1 −4 × 103 + A2 −2.4 × 104 dt ( ) ( ) ( ) ( ) so A1 = −6 and A2 = 1. Finally v1 ( t ) = 10 + e −2.4 ×10 4t P9.9-4 For t > 0 3t − 6 e−4 ×10 V for t > 0 KCL at top node: diL ( t ) 1 dv ( t ) − 5 cos t + iL ( t ) + =0 0.5 dt 12 dt (1) KVL for right mesh: 0.5 diL ( t ) 1 dv ( t ) = + v (t ) 12 dt dt ( 2) Taking the derivative of these equations gives: d 2iL ( t ) diL ( t ) 1 d 2 v ( t ) of (1) ⇒ 0.5 + + = −5 sin t dt 12 dt 2 dt 2 dt 2 2 d of ( 2 ) ⇒ 0.5 d iL ( t ) = 1 d v ( t ) + dv ( t ) dt 12 dt 2 dt 2 dt d (3) ( 4) 9-33 d 2iL ( t ) di ( t ) Solving for in ( 4 ) and L in ( 2 ) & plugging into ( 3) gives 2 dt dt d 2v ( t ) dv ( t ) +7 + 12 v ( t ) = − 30 s i n t dt 2 dt The characteristic equation is: s 2 + 7s+12 = 0 . The natural frequencies are s1,2 = −3, −4 . The natural response is of the form vn (t ) = A1e −3t + A2 e −4t . Try a forced response of the form v f ( t ) = B1 cos t + B 2 sin t . Substituting the forced response into the differential equation and equating like terms gives B1 = 21 33 and B 2 = − . 17 17 v ( t ) = vn (t ) + v f ( t ) = A1e−3t + A2 e−4t + 21 33 cos t − sin t 17 17 We will use the initial conditions to evaluate A1 and A2. We are given iL ( 0 ) = 0 and v ( 0 ) = 1 V . Apply KVL to the outside loop to get 1 iC ( t ) + iL ( t ) + 1( iC ( t ) ) + v ( t ) − 5 cos t = 0 At t = 0+ iC ( 0 ) = 5cos ( 0 ) + iL ( 0 ) − v ( 0 ) 5 + 0 − 1 = =2 A 2 2 dv ( 0 ) iC ( 0 ) 2 = = = 24 V/s dt 1 12 1 12 ⇒ + 33 dv(0 ) = 24 = −3 A1 − 4 A2 − dt 17 v(0+ ) = 1 = A1 + A2 + 21 17 A1 = 25 429 A2 = − 17 Finally, ∴ v(t ) = 25e −3t − 429e −4t − 21 cos t + 33 sin t V 17 9-34 P9.9-5 Use superposition. Find the response to inputs 2u(t) and –2u(t-2) and then add the two responses. First, consider the input 2u(t): For 0< t < 2 s Using the operator s = d we have dt KVL: vc ( t ) + siL ( t ) + 4 iL ( t ) − 2 = 0 (1) KCL: 1 3 s vc ( t ) ⇒ vc ( t ) = iL ( t ) (2) s 3 Plugging (2) into (1) yields the characteristic equation: ( s 2 + 4 s + 3) = 0 . The natural frequencies are s1,2 = −1 , −3 . The inductor current can be expressed as iL ( t ) = iL (t ) = in (t ) + i f (t ) = ( A1 e −t + A2 e−3t ) + 0 = A1 e− t + A2 e −3t . Assume that the circuit is at steady state before t = 0. Then vc (0+ ) = 0 and iL (0+ ) = 0 . Using KVL we see that (1) diL (0+ ) = 4 2 − iL (0+ ) − vC (0+ ) = 8 A/s . Then dt iL (0) = 0 = A1 + A2 A1 = 4 , A2 = −4 . diL (0) = 8 = − A1 − 3 A2 dt Therefore iL (t ) = 4e − t − 4e−3t A . The response to 2u(t) is 0 t<0 v1 (t ) = 8 − 4 iL (t ) = −t −3t 8 − 16 e + 16 e V t > 0 . = 8 − 16 e − t + 16 e −3t u ( t ) V The response to –2u(t-2) can be obtained from the response to 2u(t) by first replacing t by t-2 everywhere is appears and the multiplying by –1. Therefore, the response to –2u(t-2) is v2 (t ) = −8 + 16e − (t − 2 ) − 16 e−3(t − 2 ) u ( t - 2 ) V . By superposition, v(t ) = v1 (t ) + v2 (t ) . Therefore v(t ) = 8 − 16e −t + 16e−3t u (t ) + −8 + 16e− (t − 2) − 16 e −3(t − 2) u (t − 2) V 9-35 P9.9-6 First, find the steady state response for t < 0, when the switch is closed. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. i ( 0) = − and After the switch opens 5 = −1.25 A 4 v (0) = 5 V Apply KCL at node a: v d + 0.125 v = i 2 dt Apply KVL to the right mesh: −10 cos t + v + 4 d i+4i =0 dt After some algebra: d2 d v + 5 v + 6 v = 20 cos t 2 dt dt The characteristic equation is s 2 + 5 s + 6 = 0 ⇒ s1,2 = −2, − 3 Try vf = A cos t + B sin t d2 d A cos t + B sin t ) + 5 ( A cos t + B sin t ) + 6 ( A cos t + B sin t ) = 20 cos t 2( dt dt 9-36 ( − A cos t − B sin t ) + 5 ( − A sin t + B cos t ) + 6 ( A cos t + B sin t ) = 20 cos t ( − A + 5 B + 6 A) cos t + ( − B − 5 A + 6 B ) sin t = 20 cos t Equating the coefficients of the sine and cosine terms yields A =2 and B =2. Then vf = 2 cos t + 2 sin t v ( t ) = 2 cos t + 2 sin t + A1 e −2 t + A2 e −3 t Next v (t ) d + 0.125 v ( t ) = i ( t ) ⇒ 2 dt d v (t ) = 8 i (t ) − 4 v (t ) dt d V 5 v ( 0 ) = 8 i ( 0 ) − 4 v ( 0 ) = 8 − − 4 ( 5 ) = −30 dt s 4 Let t = 0 and use the initial conditions: 5 = v ( 0 ) = 2 cos 0 + 2 sin 0 + A1 e −0 + A2 e −0 = 2 + A1 + A2 d v ( t ) = −2 sin t + 2 cos t − 2 A1 e −2 t − 3 A2 e−3 t dt −30 = d v ( 0 ) = −2 sin 0 + 2 cos 0 − 2 A1 e −0 − 3 A2 e −0 = 2 − 2 A1 − 3 A2 dt So A1 = −23 and A2 = 26 and v ( t ) = 2 cos t + 2 sin t − 23 e −2 t + 26 e −3 t V for t > 0 9-37 P9.9-7 First, find the steady-state response for t < 0. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. i ( 0) = 0 A and After t = 0 Apply KCL at node a: v (0) = 0 V C d v=i dt Apply KVL to the right mesh: d 8 i + v + 2 i + 4 (2 + i) = 0 dt d 12 i + v + 2 i = −8 dt 2 1 d d 4 After some algebra: v + ( 6) v + v = − 2 dt dt C 2C The forced response will be a constant, vf = B so 1 d2 d 4 B + ( 6) B + B = − 2 dt dt C 2C (a) ⇒ B = −8 V d2 d v + ( 6 ) v + ( 9 ) v = −72 . 2 dt dt 2 The characteristic equation is s + 6 s + 9 = 0 ⇒ s1,2 = −3, −3 When C = 1/18 F the differential equation is Then v ( t ) = ( A1 + A2 t ) e −3t − 8 . Using the initial conditions: 9-38 0 = v ( 0 ) = ( A1 + A2 ( 0 ) ) e0 − 8 ⇒ 0 = i ( 0) = C So (b) A1 = 8 d v ( 0 ) = C −3 ( A1 + A2 ( 0 ) ) e0 + A2 e0 ⇒ dt A2 = 3 A1 = 24 v ( t ) = ( 8 + 24 t ) e −3t − 8 V for t > 0 d2 d v + ( 6 ) v + ( 5 ) v = −40 2 dt dt 2 The characteristic equation is s + 6 s + 5 = 0 ⇒ s1,2 = −1, −5 When C = 1/10 F the differential equation is Then v ( t ) = A1 e − t + A2 e −5 t − 8 . Using the initial conditions: 0 = v ( 0 ) = A1 e0 + A2 e0 − 8 ⇒ A1 + A2 = 8 ⇒ A1 = 10 and A2 = −2 d 0 0 0 = v ( 0 ) = − A1 e − 5 A2 e ⇒ − A1 − 5 A2 = 0 dt So v ( t ) = 10 e − t − 2 e −5 t − 8 V for t > 0 (c) d2 d v + ( 6 ) v + (10 ) v = −80 2 dt dt 2 The characteristic equation is s + 6 s + 10 = 0 ⇒ s1,2 = −3 ± j When C = 1/20 F the differential equation is Then v ( t ) = e −3 t ( A1 cos t + A2 sin t ) − 8 . Using the initial conditions: 0 = v ( 0 ) = e0 ( A1 cos 0 + A2 sin 0 ) − 8 ⇒ 0= So A1 = 8 d v ( 0 ) = −3 e0 ( A1 cos 0 + A2 sin 0 ) + e0 ( − A1 sin 0 + A2 cos 0 ) ⇒ dt A2 = 3 A1 = 24 v ( t ) = e −3 t ( 8 cos t + 24 sin t ) − 8 V for t > 0 9-39 P9.9-8 The circuit will be at steady state for t<0: so iL(0+) = iL(0−) = 0.5 A and vC(0+) = vC(0−) = 2 V. For t>0: Apply KCL at node b to get: 1 1d 1 1d = i (t ) + v (t ) ⇒ i (t ) = − v (t ) L C L 4 4 dt 4 4 dt C Apply KVL at the right-most mesh to get: 4i L( t) + 2 d 1 d i (t ) = 8 vc ( t ) + v ( t ) L dt 4 dt c Use the substitution method to get d 1 1 d 1 1 d 1 d 4 − vC ( t ) + 2 − vC ( t ) = 8 vc ( t ) + v ( t ) dt 4 4 dt 4 4 dt 4 dt c or d2 d v t + 6 v (t ) + 2 v (t ) 2 C( ) C dt dt C d2 d The forced response will be a constant, vC = B so 2 = B + 6 B + 2B ⇒ B = 1 V . dt dt 2 2= To find the natural response, consider the characteristic equation: 0 = s 2 + 6 s + 2 = ( s + 5.65 ) ( s + 0.35 ) The natural response is vn = A1 e −5.65 t + A2 e −0.35 t vC ( t ) = A1 e −5.65 t + A2 e −0.35 t so +1 Then iL ( t ) = 1 1d 1 −5.65 t −0.35 t + + 0.0875 A2 e vC ( t ) = + 1.41A1 e 4 4 dt 4 9-40 At t=0+ 2 = vC ( 0 + ) = A1 + A + 1 2 1 1 = iL ( 0+ ) = + 1.41A1 + 0.0875 A 2 2 4 so A1 = 0.123 and A2 = 0.877. Finally vC ( t ) = 0.123 e −5.65 t + 0.877 e −0.35 t +1 V P9.9-9 The inductor current and voltage are related be After t = 0 v (t ) = L di ( t ) dt (1) Apply KCL at the top node to get C dv( t ) v( t ) + i (t ) + =5 dt 2 (2) d , and substituting (1) into (2) yields ( s 2 + 4s + 29 ) i ( t ) = 5 . dt The characteristic equation is s 2 + 4 s + 29 = 0 . The characteristic roots are s1,2 = − 2 ± j 5 . Using the operator s = The natural response is of the form in ( t ) = e −2t [ A cos 5t + B sin 5t ] . Try a forced response of the form i f ( t ) = A . Substituting into the differential equation gives A = 5 . Therefore i f ( t ) = 5 A . The complete response is i(t ) = 5 + e−2t [ A cos 5t + B sin 5t ] where the constants A and B are yet to be evaluated using the initial condition: i (0) = 0 = A + 5 ⇒ A = −5 di (0) di (0) 2A 0 = v ( 0) = L ⇒ = 0 = −2 A + 5 B ⇒ B = = −2 dt dt 5 Finally, i (t ) = 5 + e −2t [ −5 cos 5t − 2 sin 5t ] A . P9.9-10 9-41 Assume that the circuit is at steady before t = 0. 2 ×9 = 6 A 2+1 1 v ( 0 + ) = v (0 − ) = × 9× 1.5 = 4.5 V 2+1 i (0 + ) = i (0 − ) = Apply KCL at the top node of the current source to get After t = 0: i ( t ) + 0.5 dv( t ) v( t ) + = is ( t ) dt 1.5 (1) Apply KVL and KCL to get dv( t ) v( t ) 5di( t ) v ( t ) + 0.5 + + i (t ) 0.5 = dt dt 1.5 (2) Solving for i(t) in (1) and plugging into (2) yields d 2 v( t ) 49 dv( t ) 4 di ( t ) 2 + + v ( t ) = is ( t ) + 2 s 2 dt dt 30 dt 5 5 where is ( t ) = 9 + 3e −2t A d 49 4 , the characteristic equation is s 2 + s + = 0 and the characteristic 30 5 dt roots are s1,2 =−.817 ± j.365 . The natural response has the form Using the operator s = vn (t ) = e−0.817t A1 cos (0.365 t ) + A2 sin (0.365 t ) Try a forced response of the form v f (t ) = B0 + B1e −2 t . Substituting into the differential equations gives B0 = 4.5 and B1 = −7.04 . The complete response has the form v(t ) = e−.817 t A1 cos(0.365 t ) + A2 sin (0.365 t ) + 4.5 − 7.04 e −2t Next, consider the initial conditions: v (0) = 4.5 = A1 + 4.5 − 7.04 ⇒ A1 = 7.04 9-42 d v ( 0) 4 4 = 2 is (0) - 2 i (0) − v(0) = 2(9 + 3) − 2(6) − (4.5) = 6 dt 3 3 6= d v( 0 ) = −0.817 A1 + 0.365 A2 + 14.08 ⇒ A2 = −6.38 dt So the voltage is given by v(t ) = e−0.817 t 7.04 cos(0.365 t ) + A2 sin (0.365 t ) + 4.5 − 7.04 e−2t Next the current given by i (t ) = is (t ) − Finally v(t ) d v(t ) − 0 .5 1.5 dt i (t ) = e −0.817 t [ 2.37 cos(0.365t ) + 7.14 sin(0.365t ) ] + 6 + 0.65e −2t A P9.9-11 First, find the steady state response for t < 0. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. va ( 0− ) = −4 i ( 0− ) ( ) i ( 0− ) = 2va ( 0− ) = 2 −4 i ( 0− ) ⇒ i ( 0− ) = 0 A and v ( 0− ) = 10 V 9-43 For t > 0 Apply KCL at node 2: va d + K va + C v=0 R dt KCL at node 1 and Ohm’s Law: va = − R i so d 1+ K R v= i dt C Apply KVL to the outside loop: L d i + R i + v − Vs = 0 dt After some algebra: d2 Rd 1+ K R 1+ K R v+ v+ v= Vs 2 dt L dt LC LC ⇒ d2 d v + 40 v + 144 v = 2304 2 dt dt The forced response will be a constant, vf = B so d2 d B + ( 40 ) B + (144 ) B = 2304 ⇒ B = 16 V 2 dt dt The characteristic equation is s 2 + 40 s + 144 = 0 ⇒ s1,2 = −4, −36 . v ( t ) = A1 e− 4 t + A2 e−36 t + 16 . Then Using the initial conditions: 10 = v ( 0+ ) = A1 e0 + A2 e0 + 16 ⇒ 0= d v ( 0+ ) = −4 A1 e0 − 36 A2 e0 dt ⇒ ⇒ − 4 A1 − 36 A2 = 0 A1 + A2 = −6 A1 = −6.75 and A2 = 0.75 So v ( t ) = 0.75 e −36 t − 6.75 e−4 t + 16 V for t > 0 (checked using LNAP on 7/22/03) 9-44 Section 9-10: State Variable Approach to Circuit Analysis P9.10-1 At t = 0− the circuit is source free ∴ iL (0) = 0 and v(0) = 0. Apply KCL at the top node to get After t = 0 iL ( t ) + 1 dv( t ) =4 5 dt (1) Apply KVL to the right mesh to get v( t ) −(1) Solving for i1 ( t ) in (1) and plugging into (2) ⇒ diL ( t ) − 6 iL ( t ) = 0 dt (2) d 2 v( t ) dv ( t ) +6 + 5v ( t ) = 120 . 2 dt dt The characteristic equation is s 2 + 6 s + 5= 0 . The natural frequencies are s1,2 =−1, −5 . The natural response has the form vn (t ) = A1 e−t + A2 e−5t . Try v f ( t ) = B as the forced response. Substituting into the differential equation gives B = 24 so v f ( t ) = 24 V. The complete response has the form v (t ) = A1 e−t + A2 e−5t + 24 . Now consider the initial conditions. From (1) dv(0) = 20 − 5 iL (0) = 20 V . Then s dt v(0) = 0 = A1 + A2 + 24 ⇒ dv(0) = 20 = − A1 −5 A2 dt A1= −25, A2 = 1 Finally v (t ) = − 25e −t + e −5t + 24 V . 9-45 P9.10-2 Before t = 0 there are no sources in the circuit so iL (0) = 0 and v(0) = 0 . After t = 0 we have: Apply KCL at the top node to get iL ( t ) = 4 − 1 dv ( t ) 10 dt (1) Apply KVL to the left mesh to get v( t ) − diL ( t ) − 6iL ( t ) = 0 dt (2) Substituting iL ( t ) from (1) into (2) gives d 2 v( t ) dv ( t ) +6 + 10v ( t ) = 2 40 2 dt dt The characteristic equation is s 2 + 6 s + 10 = 0 . The natural frequencies are s1,2 = −3 ± j . The natural response has the form vn (t ) = e −3t A1 cos t + A2 sin t . Try v f ( t ) = B as the forced response. Substituting into the differential equation gives B = 24 so v f ( t ) = 24 V. The complete response has the form v(t ) = e −3t A1 cos t + A2 sin t + 24 . Now consider the initial conditions. From (1) dv(0) = 40 − 10 iL (0) = 40 V . Then s dt v (0) = 0 = A1 + 24 ⇒ A1 = −24 dv(0) = 40 = −3 A1 + A2 = 72 + A2 ⇒ A2 = −32 dt Finally, v (t ) = e −3t [ −24 cos t −32 sin t ] + 24 V 9-46 P9.10-3 Assume that the circuit is at steady state before t = 0 so iL (0) = −3 A and v(0) = 0 V . After t = 0 we have KCL: i ( t ) + C KVL: v( t ) = L i (t ) + C d 2i ( t ) −6 1 di ( t ) 1 + + i (t ) = 2 dt R C dt LC LC ⇒ dv( t ) v( t ) + + 6=0 dt R di ( t ) dt d di( t ) 1 di ( t ) L + L + 6=0 dt dt R dt d 2i ( t ) di ( t ) + 100 + 250i ( t ) = −1500 2 dt dt The characteristic equation is s 2 + 100 s + 250 = 0 . The natural frequencies are s1,2 = −2.57, − 97.4 . The natural response has the form in (t ) = A1 e−2.57 t + A2 e−97.4t . Try ( t ) = B as the forced response. Substituting into the differential equation gives B = −6 so i f ( t ) = −6 A . The complete response has the form i(t ) = A1 e−2.57 t + A2 e−97.4t − 6 . i f Now consider the initial conditions: i (0) = A1 + A2 − 6 = −3 A1 = 3 .081 di (0) = 0 = − 2.57 A1 −97.4 A2 A2 =−0.081 dt Finally: i (t ) = 3. 081 e v(t ) = .2 −2.57 t −.081e di ( t ) = −1.58e dt −97.4 t −2.57 t −6 A +1.58e−97.4t V 9-47 P9.10-4 Apply KCL to the supernode corresponding to the dependent voltage source to get ix ( t ) − 2ix ( t ) − 0.01 dv ( t ) vx ( t ) + =0 2 dt Apply KCL at node 1 to get i ( t ) − 2ix ( t ) + (Encircled numbers are node numbers.) vx ( t ) =0 2 Apply KVL to the top-right mesh to get vx ( t ) + v ( t ) − 0.1 di ( t ) =0 dt Apply KVL to the outside loop to get ix ( t ) = −2 vx ( t ) − v ( t ) . Eliminate ix ( t ) to get Then eliminate vx ( t ) to get dv ( t ) 5 =0 vx ( t ) + v ( t ) − 0.01 2 dt 9 i ( t ) + vx ( t ) + 2 v ( t ) = 0 2 d i (t ) vx ( t ) = −v ( t ) + 0.01 dt dv ( t ) di ( t ) + 0.25 =0 dt dt di ( t ) −2.5 v ( t ) + i ( t ) + 0.45 =0 dt −1.5 v ( t ) − 0.01 Using the operator s = d we have dt (−1.5 − .01s )v ( t ) + (.25s ) i ( t ) = 0 (−2.5)v( t ) + (1+.45s ) i ( y ) = 0 The characteristic equation is s 2 +13.33 s + 333.33 = 0 . The natural frequencies are s1 , s2 = −6.67 ± j 17 . The natural response has the form vn (t ) = [ A cos 17 t + B sin 17 t ] e−6.67 t . ( t ) = 0 . The complete response has the form + B sin 17 t ] e−6.67 t . The forced response is v v(t ) = [ A cos17 t f 9-48 The given initial conditions are i (0) = 0 and v (0) = 10 V. Then v(0) =10 = A and dv(0) =−111= − 6.67 A +17 B ⇒ B =−2.6 dt Finally i (t ) = [3.27 sin 17 t ] e−6.67 t A . (Checked using LNAP on 7/22/03) P9.10-5 Assume that the circuit is at steady state before t = 0 so v(0) = 10 V and iL (0) = The switch is open when 0 < t < 0.5 s v ( 0) 10 = A. 3 3 For this series RLC circuit we have: α= R 1 = 3 and ω 02 = = 12 2L LC −α ± α 2 −ω02 = s1,2 = −3 ± j 3 The natural response has the form vn (t ) = e −3t ( A cos 1.73 t + B sin 1.72 t ) . There is no source so v f ( t ) =0 . The complete response has the form v (t ) = e −3t ( A cos 1.73 t + B sin 1.72 t ) . Next v(0) = 10 = A A= 10 dv(0) i ( 0 ) 10 3 ⇒ =− =− = − 20 = −3 A +1.73 B B =5.77 dt C 16 so v(t ) = e−3t (10 cos1.73 t + 5.77 sin 1.73 t ) V i(t ) = e−3t ( 3.33 cos1.73 t − 5.77 sin 1.73 t ) A In particular, 1.73 1.73 v (0.5) = e −1.5 10 cos + 5.77 sin = 0.2231× ( 6.4864 + 4.3915 ) = 2.43 V 2 2 and 1.73 1.73 i (0.5) = e −1.5 3.33 cos −5.77 sin = 0.2231× ( 2.1600 − 4.3915 ) = −0.50 A 2 2 9-49 The switch is closed when t > 0.5 s Apply KCL at the top node: v( t ) −30 1 dv ( t ) + iL ( t ) + =0 6 6 dt dv( t ) 1 ⇒ iL ( t ) = 5 − v( t ) + 6 dt ⇒ 2 d iL ( t ) 1 dv ( t ) d v ( t ) =− + dt 6 dt dt 2 Apply KVL to the right mesh: 1 diL ( t ) v( t ) = 3 iL ( t ) + 2 dt The circuit is represented by the differential equation d 2v ( t ) dv ( t ) +7 + 1 8 v ( t ) = 18 0 2 dt dt The characteristic equation is 0 = s 2 + 7 s + 18 . The natural frequencies are s1,2 = −3.7 ± j 2.4 . The natural response has the form vn (t ) = e −3.5 t ( A cos 2.4 t + B sin 2.4 t ) . The forced response is v f ( t ) = 10 V . The complete response has the form v (t ) = e −3.5 t ( A cos 2.4 t + B sin 2.4 t ) + 10 . Next v (0.5) = e −3.5×0.5 ( A cos 1.2 + B sin 1.2 ) = 0.063 A + 0.162 B dv ( 0.5 ) = e−3.5×0.5 ( −3.5 A + 2.4 B ) cos1.2 − ( 3.5B + 2.4 A ) sin 1.2 dt = e−3.5×0.5 ( −3.5 cos 1.2 − 2.4 sin 1.2 ) A + e−3.5×0.5 ( 2.4 cos1.2 − 3.5 sin 1.2 ) B = −0.6091A − 0.4158B Using the initial conditions yields 2.43=v(0.5) = 0.063 A+ 0.162 B A=−20.65 dv(0.5) i ( 0.5 ) −1 2 ⇒ B = 23.03 =− =− =3= −0.6091A− 0.4158B dt C 16 Finally In summary v(t ) = e −3.5 t ( −20.65 cos 2.4 t + 23.03 sin 2.4 t ) + 10 e −3t (10 cos1.73 t + 5.77 sin 1.73 t ) V 0<t <0.5 v(t ) = −3.5 t e ( −20.65 cos 2.4 t + 23.03 sin 2.4 t )+10 V 0.5<t 9-50 Section 9-11: Roots in the Complex Plane P9.11-1 After t = 0 i1 + i 2 + 2 ×10−3 d i1 dt 2 ×10−3 d i1 dt 2000 −6 =0 = 3000 i 2 + 2 ×10−3 d i2 dt d yields dt 2000 + 2 × 10−3 s i1 6 2000 = −3 −3 2 × 10 s 3000 + 2 × 10 s i 2 0 Using the operator s = s 2 + 3.5 × 106 s + 1.5 ×1012 = 0 ⇒ s1,2 = −5 × 105 , −3 ×106 P9.11-2 From P9.7-1 s2 + 1 1 1 1 = 0 ⇒ s2 + =0 s+ s+ −6 RC LC ( 250 ) ( 5 ×10 ) ( 0.8) ( 5 ×10−6 ) ⇒ s 2 + 800 s + 250000 = 0 s1,2 = 400 ± j 300 P9.11-3 dv( t ) v( t ) 1 KCL: i L ( t ) = × 10−6 + 4 dt 4000 di ( t ) KVL: vs ( t ) = 4 L + v( t ) dt d 2v ( t ) dv ( t ) dv( t ) v( t ) d 1 v ( t ) = 4 ×10−6 + + v ( t ) = 10−6 + 10−3 + v (t ) 2 s dt 4 dt 4000 dt dt 9-51 Characteristic equation: s 2 + 103 s + 106 = 0 Characteristic roots: s1, 2 = −500 ± j 866 P9.11-4 Before t = 0 the voltage source voltage is 0 V so vb (0+) = vb (0−) = 0 V and i (0+) = i (0−) = 0 A . Apply KCL at node a to get va (0+ ) −36 v (0+ ) − vb (0+ ) − i (0 + ) + a = 0 ⇒ va (0+ ) + 2 va (0+ ) = 36 ⇒ va (0) = 12 V 12 6 After t = 0 the node equations are: − va ( t ) − vs ( t ) 1 t v ( t ) − va ( t ) + ∫ ( vb (τ ) − va (τ ) ) dτ + b =0 0 12 6 L C Using the operator s = d vb ( t ) vb ( t ) − va ( t ) 1 t + + ∫ ( vb (τ ) − va (τ ) ) dτ = 0 6 dt L0 d we have dt v (t ) 1 1 1 1 1 + + va ( t ) + − − vb ( t ) = s 6 12 s 12 6 s 1 1 1 1 1 − − va ( t ) + s + + vb ( t ) = 0 6 s 18 6 s Using Cramer’s rule 9-52 ( s 2 +5s + 6) vb ( t ) = ( s + 6) vs ( t ) =( s + 6) ( 36 ) The characteristic equation is s 2 + 5s + 6 = 0 . The natural frequencies are s1,2 = −2, −3 . The natural response has the form vn (t ) = A1 e −2 t + A2 e −3 t . Try v f ( t ) = B as the forced response. Substituting into the differential equation gives B = 36 so v f ( t ) = 36 V. The complete response has the form vb (t ) = A1 e −2 t + A2 e −3t + 36 . Next vb (0+ ) = 36 + A1 + A2 dvb + (0 ) = −2 A1 − 3 A2 dt Apply KCL at node a to get At t = 0+ 1 dvb ( t ) vb ( t ) − va ( t ) + + i (t ) = 0 18 dt 6 1 1 d vb ( 0 (−2 A1 − 3 A2 ) = 18 18 dt + ) = v ( 0 )−v ( 0 ) − i + + a b 6 (0+ ) = 126−0 − 0 = 2 So 0 = vb (0+ ) =36+ A1 + A 2 ⇒ 1 ( −2 A 1 − 3 A 2 ) = 2 18 Finally vb ( t ) = 36 − 72e −2 t A1 = −72, A2 = 36 + 36e −3t V for t ≥ 0 9-53 PSpice Problems SP 9-1 Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure to edit the labels of the parts so, for example, there is only one R1.) 9-54 V(C1:2), V(C2:2) and V(C3:2) are the capacitor voltages, listed from top to bottom. 9-55 SP 9-2 Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure to edit the labels of the parts so, for example, there is only one R1.) 9-56 V(R2:2), V(R4:2) and V(R6:2) are the output voltages, listed from top to bottom. 9-57 SP 9-3 9-58 SP 9-4 9-59 Verification Problems VP 9-1 This problem is similar to the verification example in this chapter. First, check the steady-state inductor current v ( t ) 25 i (t ) = s = = 250 mA 100 100 This agrees with the value of 250.035 mA shown on the plot. Next, the plot shows an underdamped response. That requires 12 ×10−3 = L < 4 R 2C = 4(100) 2 (2 × 10−6 ) = 8 × 10−2 This inequality is satisfied, which also agrees with the plot. The damped resonant frequency is given by ω= d 1 LC 1 − 2 RC 2 2 1 1 = − = 5.95 × 103 −6 −6 −3 2(100)(2 × 10 ) ( 2 ×10 ) (12 ×10 ) The plot indicates a maxima at 550.6µs and a minima at 1078.7µs. The period of the damped oscillation is T d = 2 (1078.7 µ s − 550.6 µ s) = 1056.2 µ s Finally, check that 5.95 ×103 = ω d = 2π 2π = = 5.949 × 103 −6 1056.2 ×10 Td The value of ω d determined from the plot agrees with the value obtained from the circuit. The plot is correct. VP 9-2 This problem is similar to the verification example in this chapter. First, check the steady-state inductor current. v ( t ) 15 i (t ) = s = = 150 mA 100 100 9-60 This agrees with the value of 149.952 mA shown on the plot. Next, the plot shows an underdamped response. This requires 8 ×10−3 = L < 4 R 2C = 4 (100)2 (0.2 × 10−6 ) = 8 × 10−3 This inequality is not satisfied. The values in the circuit would produce a critically damped, not underdamped, response. This plot is not correct. Design Problems DP 9-1 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = The specifications require that vC ( ∞ ) = 1 so 2 R2 1 = 2 R1 + R 2 R2 R1 + R 2 1 ⇒ R1 = R 2 Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives: vC ( t ) R2 vs ( t ) = L +C d vC ( t ) = iL ( t ) dt d iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get 9-61 vs ( t ) = L v (t ) d vC ( t ) d d + C vC ( t ) + R1 C + C vC ( t ) + vC ( t ) R2 dt R 2 dt dt = LC L d R1 d2 + R1 C vC ( t ) + 1 + vC ( t ) + vt R2 dt R2 C ( ) dt 2 The characteristic equation is R1 1+ 1 R1 R2 + s+ s2 + R 2 C L LC = s 2 + 6s + 8 = ( s + 2 ) ( s + 4 ) = 0 Equating coefficients of like powers of s: 1 R2 C + R1 L 1+ = 6 and R1 R2 =8 LC Using R1 = R 2 = R gives 1 R + =6 ⇒ RC L 1 =4 LC These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 4 1 3 1 + 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω R 2 4 Pick R = 1.309 Ω. Then vc ( t ) = iL ( t ) = At t = 0+ vC ( t ) 1.309 + 1 + A1 e −2 t + A2 e−4 t V 2 d vC ( t ) = −1.236 A1 e−2 t − 3.236 A2 e−4 t + 0.3819 dt () 0 = vc 0+ = A1 + A2 + 0.5 ( ) = −1.236 A − 3.236 A 0 = iL 0 + 1 2 + 0.3819 Solving these equations gives A1 = -1 and A2 = 0.5, so vc ( t ) = 1 −2 t 1 −4 t −e + e V 2 2 9-62 DP 9-2 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = The specifications require that vC ( ∞ ) = 1 so 4 R2 1 = 4 R1 + R 2 R2 R1 + R 2 1 ⇒ 3 R 2 = R1 Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives: vC ( t ) R2 vs ( t ) = L +C d vC ( t ) = iL ( t ) dt d iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get vs ( t ) = L v (t ) d vC ( t ) d d + C vC ( t ) + R1 C + C vC ( t ) + vC ( t ) R2 dt R 2 dt dt L d R1 d2 v t+ + R1 C vC ( t ) + 1 + vt 2 C( ) R2 dt R2 C ( ) dt The characteristic equation is R1 1+ 1 R1 R2 2 2 2 + s+ = s + 4s + 4 = ( s + 2 ) = 0 s + R 2 C L LC Equating coefficients of like powers of s: R 1+ 1 R1 R2 1 + = 4 and =4 R2 C L LC = LC 9-63 Using R 2 = R and R1 = 3R gives 1 3R + =4 ⇒ RC L 1 =1 LC These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 1 4 1 1 + 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω R 3 3 3 Pick R = 1 Ω. Then R1 = 3 Ω and R 2 = 1 Ω . vc ( t ) = iL ( t ) = vC ( t ) + At t = 0+ 1 + ( A1 + A2 t ) e −2 t V 4 d 1 vC ( t ) = + dt 4 () (( A 0 = vc 0+ = A1 + () 0 = iL 0+ = 2 ) − A1 ) − A2 t e −2 t 1 4 1 + A2 − A1 4 Solving these equations gives A1 = -0.25 and A2 = -0.5, so vc ( t ) = 1 1 1 −2 t − + t e V 4 4 2 DP 9-3 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions R2 vC ( ∞ ) = 1 R1 + R 2 4 The specifications require that vC ( ∞ ) = so 5 9-64 R2 4 = 5 R1 + R 2 ⇒ 4 R1 = R 2 Next, represent the circuit by a 2nd order differential equation: vC ( t ) KCL at the top node of R2 gives: R2 vs ( t ) = L KVL around the outside loop gives: +C d vC ( t ) = iL ( t ) dt d iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get vs ( t ) = L v (t ) d vC ( t ) d d + C vC ( t ) + R1 C + C vC ( t ) + vC ( t ) R2 dt R 2 dt dt = LC L d R1 d2 v t+ + R1 C vC ( t ) + 1 + vt 2 C( ) R2 dt R2 C ( ) dt The characteristic equation is R1 1+ 1 R1 R2 + s+ s2 + R 2 C L LC = s 2 + 4 s + 20 = ( s + 2 − j 4 ) ( s + 2 + j 4 ) = 0 Equating coefficients of like powers of s: R1 1 + = 4 and R2 C L 1+ R1 R2 LC = 20 Using R1 = R and R 2 = 4 R gives 1 R + = 4 and 4R C L 1 = 16 LC 1 1 These equations do not have a unique solution. Try C = F . Then L = H and 8 2 2 + 2 R = 4 ⇒ R2 − 2R + 1 = 0 ⇒ R = 1 Ω R Then R1 = 1 Ω and R 2 = 4 Ω . Next vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V 9-65 iL ( t ) = + vC ( t ) 4 + A 2 −2 t A1 1d vC ( t ) = 0.2 + e cos 4 t − e−2 t sin 4 t 8 dt 2 2 At t = 0 () 0 = vc 0+ = 0.8 + A1 () 0 = iL 0+ = 0.2 + A2 2 Solving these equations gives A1 = −0.8 and A2 = −0.4, so vc ( t ) = 0.8 − e −2 t ( 0.8 cos 4 t + 0.4 sin 4 t ) V DP 9-4 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = The specifications require that vC ( ∞ ) = 1 so 2 R2 1 = 2 R1 + R 2 R2 R1 + R 2 1 ⇒ R1 = R 2 Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives: vC ( t ) R2 vs ( t ) = L +C d vC ( t ) = iL ( t ) dt d iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get 9-66 v (t ) d vC ( t ) d d + C vC ( t ) + R1 C + C vC ( t ) + vC ( t ) R2 dt R 2 dt dt vs ( t ) = L = LC L d R1 d2 vC ( t ) + + R1 C vC ( t ) + 1 + vt R2 dt R2 C ( ) dt 2 The characteristic equation is R1 1+ 1 R1 R2 + s+ s2 + R 2 C L LC = s 2 + 4 s + 20 = ( s + 2 − j 4 ) ( s + 2 + j 4 ) = 0 Equating coefficients of like powers of s: R1 1 + = 4 and R2 C L 1+ R1 R2 LC = 20 Using R1 = R 2 = R gives 1 R + = 4 and RC L Substituting L = 1 = 10 LC 1 into the first equation gives 10 C ( RC ) 2 0.4 ± 0.42 − 4 ( 0.1) 4 1 − ( RC ) + = 0 ⇒ RC = 10 10 2 Since RC cannot have a complex value, the specification cannot be satisfied. DP 9-5 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions 9-67 vC ( ∞ ) = R2 R1 + R 2 1, iL ( ∞ ) = The specifications require that vo ( ∞ ) = 1 so 2 R2 1 R1 + R 2 1 = 2 R1 + R 2 and vo ( ∞ ) = R2 R1 + R 2 1 ⇒ R1 = R 2 Next, represent the circuit by a 2nd order differential equation: KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives: d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L Use the substitution method to get vs ( t ) = R1 C = R1 LC Using iL ( t ) = vo ( t ) gives R2 vs ( t ) = d d d L iL ( t ) + R 2 iL ( t ) + L iL ( t ) + R 2 iL ( t ) + R1 iL ( t ) dt dt dt d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt R1 R2 LC L d R1 + R 2 d2 + R1 C vo ( t ) + v t+ vt 2 o( ) R2 dt R2 o ( ) dt The characteristic equation is R2 1+ 1 R2 R1 + s2 + s+ R1 C L LC Equating coefficients of like powers of s: = s 2 + 6s + 8 = ( s + 2 ) ( s + 4 ) = 0 R2 1 + = 6 and R1 C L 1+ R2 R1 LC =8 Using R1 = R 2 = R gives 1 R + =6 ⇒ RC L 1 =4 LC These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 4 9-68 1 3 1 + 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω R 2 4 Pick R = 1.309 Ω. Then 1 + A1 e −2 t + A2 e −4 t V 2 A1 −2 t A2 −4 t v (t ) 1 iL ( t ) = o e+ eV = + 1.309 2.618 1.309 1.309 vo ( t ) = vC ( t ) = 1.309 iL ( t ) + At t = 0+ 1d 1 iL ( t ) = + 0.6167 A1 e−2 t + 0.2361 A2 e−4 t 4 dt 2 () 0 = iL 0+ = () 0 = vC 0+ = A1 A2 1 + + 2.618 1.309 1.309 1 + 0.6167 A1 + 0.2361 A2 2 Solving these equations gives A1 = -1 and A2 = 0.5, so vo ( t ) = 1 −2 t 1 −4 t −e + e V 2 2 DP 9-6 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = R2 R1 + R 2 1, iL ( ∞ ) = The specifications require that vo ( ∞ ) = 1 R1 + R 2 and vo ( ∞ ) = R2 R1 + R 2 1 3 so 4 R2 3 = 4 R1 + R 2 ⇒ 3R1 = R 2 9-69 Next, represent the circuit by a 2nd order differential equation: d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives: Use the substitution method to get vs ( t ) = R1 C = R1 LC Using iL ( t ) = vo ( t ) gives R2 vs ( t ) = d d d L iL ( t ) + R 2 iL ( t ) + L iL ( t ) + R 2 iL ( t ) + R1 iL ( t ) dt dt dt d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt R1 R2 LC L d R1 + R 2 d2 + R1 C vo ( t ) + vo ( t ) + vt R2 dt R2 o ( ) dt 2 The characteristic equation is R2 1+ 1 R2 R1 + s2 + s+ R1 C L LC Equating coefficients of like powers of s: = s 2 + 4 s + 4 = ( s + 2 )2 = 0 R2 1 + = 4 and R1 C L 1+ R2 R1 LC =4 Using R1 = R and R 2 = 3R gives 1 3R + = 4 and RC L 1 =1 LC These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 1 4 1 1 + 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω R 3 3 3 Pick R = 1 Ω. Then R1 = 1 Ω and R 2 = 3 Ω . vo ( t ) = 3 + ( A1 + A2 t ) e −2 t V 4 9-70 iL ( t ) = vo ( t ) vC ( t ) = 3 iL ( t ) + = 3 1 A1 A2 −2 t t e V + + 4 3 3 3 A1 A2 A2 −2 t d iL ( t ) = + + t e + 4 3 3 3 dt At t = 0+ 0 = iL ( 0 + ) = A1 + 1 4 3 3 A1 A2 0 = vC ( 0 + ) = + + 43 3 Solving these equations gives A1 = −0.75 and A2 = −1.5, so vo ( t ) = 3 3 3 −2 t − + t e V 4 4 2 DP 9-7 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = R2 R1 + R 2 1, iL ( ∞ ) = 1 The specifications require that vo ( ∞ ) = so 5 R2 1 = 5 R1 + R 2 1 R1 + R 2 and vo ( ∞ ) = R2 R1 + R 2 1 ⇒ R1 = 4 R 2 Next, represent the circuit by a 2nd order differential equation: KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives: d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L 9-71 Use the substitution method to get vs ( t ) = R1 C = R1 LC Using iL ( t ) = vo ( t ) gives R2 vs ( t ) = d d d L iL ( t ) + R 2 iL ( t ) + L iL ( t ) + R 2 iL ( t ) + R1 iL ( t ) dt dt dt d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt R1 R2 LC L d R1 + R 2 d2 + R1 C vo ( t ) + v t+ vt 2 o( ) R2 dt R2 o ( ) dt The characteristic equation is R2 1+ 1 R2 R1 + s2 + s+ R1 C L LC = s 2 + 4 s + 20 = ( s + 2 − j 4 ) ( s + 2 + j 4 ) = 0 Equating coefficients of like powers of s: R2 1 + = 4 and R1 C L 1+ R2 R1 LC = 20 Using R 2 = R and R1 = 4 R gives 1 R + = 4 and 4R C L These equations do not have a unique solution. Try C = 1 = 16 LC 1 1 F . Then L = H and 8 2 2 + 2 R = 4 ⇒ R2 − 2R + 2 = 0 ⇒ R = 1 Ω R Then R1 = 4 Ω and R 2 = 1 Ω . Next vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V iL ( t ) = vo ( t ) vC ( t ) = iL ( t ) + At t = 0+ 1 = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V 1d iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e−2 t sin 4 t 2 dt 9-72 () ( 0 ) = 0.2 + 2 A 0 = iL 0+ = 0.2 + A1 0 = vC + 2 Solving these equations gives A1 = −0.8 and A2 = −0.4, so vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) V DP 9-8 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = R2 R1 + R 2 1, iL ( ∞ ) = The specifications require that vC ( ∞ ) = 1 R1 + R 2 and vo ( ∞ ) = R2 R1 + R 2 1 1 so 2 R2 1 = ⇒ R1 = R 2 2 R1 + R 2 Next, represent the circuit by a 2nd order differential equation: KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives: d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L Use the substitution method to get vs ( t ) = R1 C = R1 LC Using iL ( t ) = vo ( t ) gives R2 d d d L iL ( t ) + R 2 iL ( t ) + L iL ( t ) + R 2 iL ( t ) + R1 iL ( t ) dt dt dt d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt 9-73 vs ( t ) = R1 R2 LC L d R1 + R 2 d2 + R1 C vo ( t ) + v t+ vt 2 o( ) R2 dt R2 o ( ) dt The characteristic equation is R2 1+ 1 R2 R1 + s2 + s+ R1 C L LC = s 2 + 4 s + 20 = ( s + 2 − j 4 ) ( s + 2 + j 4 ) = 0 Equating coefficients of like powers of s: R2 1 + = 4 and R1 C L 1+ R2 R1 LC = 20 Using R1 = R 2 = R gives 1 R + =4 ⇒ RC L Substituting L = 1 = 10 LC 1 into the first equation gives 10 C ( RC ) 2 0.4 ± 0.42 − 4 ( 0.1) 4 1 − ( RC ) + = 0 ⇒ RC = 10 10 2 Since RC cannot have a complex value, the specification cannot be satisfied. 9-74 DP 9-9 Let’s simulate the three copies of the circuit simultaneously. Each copy uses a different value of the inductance. The PSpice transient response shows that when L = 1 H the inductor current has its maximum at approximately t=0.5 s. Consequently, we choose L = 1 H. 9-75 ...
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This note was uploaded on 06/10/2011 for the course PHYSICS physics taught by Professor Physics during the Spring '11 term at HKU.

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