Bài giảng Electric circuit theory - Chapter VIII: Second Order Circuits - Nguyễn Công Phương

Nguy ễn Công Ph ươ ng Electric Circuit Theory Second-Order Circuits Contents I. Basic Elements Of Electrical Circuits II. Basic Laws III. Electrical Circuit Analysis IV. Circuit Theorems V. Active Circuits VI. Capacitor And Inductor VII. First Order Circuits VIII.Second Order Circuits IX. Sinusoidal Steady State Analysis X. AC Power Analysis XI. Three-phase Circuits XII. Magnetically Coupled Circuits XIII.Frequency Response XIV.The Laplace Transform XV. Two-port Networks Second-Order Circuits - sites.google.com/site/ncpdhbkhn 2 Second-Order Circuits 1. The Source-Free Series RLC Circuit 2. Initial Conditions 3. The Characteristic Equation 4. The Classical Method 5. Second-Order Op Amp Circuits Second-Order Circuits - sites.google.com/site/ncpdhbkhn 3 v The Source-Free Series RLC Circuit (1) + – R E vVE(0)==0 ;(0) iI == 0+ 0 + R L C RR0 RR 0 i R 2 E 0 di 1 t di Rdi i A Ri+ L + idt = 0 → + + = 0 + – ∫−∞ 2 dt C dt Ldt LC t = 0 st i= Ae B R A st 2 R 1  →Ase2 st + A se st + e st = 0 →Ae s ++ s  = 0 R 1 L LC L LC  →+s2 s + = 0 L LC i= Ae st ≠ 0  2 R R  1 s =−+  − =−+α α2 − ω 2  1 2L 2 L  LC 0 → → =st1 + st 2  it( ) Ae1 Ae 2    2 =−−R R −1 =−−α α2 − ω 2 s2   0  2L 2 L  LC Second-Order Circuits - sites.google.com/site/ncpdhbkhn 4 v The Source-Free Series RLC Circuit (2) + – R E vVE(0)==0 ;(0) iI == 0+ 0 + R L C RR0 RR 0 i R E 0 di 1 t A RiL+ + idt = 0 + – ∫−∞ dt C t = 0 B R 1 →+s2 s + = 0 L LC → =st1 + st 2 it( ) Ae1 Ae 2 i(0) = Ae0 + Ae 0 = I 1 2 0 A →  1 A =st1 + st 2 =+  2 i'(0) ( Ase11 Ase 22) As 11 As 22 t=0 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 5 v The Source-Free Series RLC Circuit (3) + – it( ) = Aest1 + Ae st 2 1 2 L C R i 2 R R R  1 2 2 E 0 s =−±  − =−±α α − ω A 1,2 2L 2 L  LC 0 + – R 1 t = 0 α=, ω = B 2L 0 LC α> ω α= ω αω< =−± αω 0 : 0 : 0,s 1,2 j d : −α =st1 + st 2 = + st =ω + ω t it( ) Ae1 Ae 2 it()( A1 Ate 2 ) it() ( A1 cosd tA 2 sin d te ) i( t ) i( t ) i( t ) t t t 0 0 0 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 6 v + – R E vVE(0)==0 ;(0) iI == 0+ 0 + R L C RR0 RR 0 i R E 0 di 1 t A RiL+ + idt = 0 + – ∫−∞ dt C t = 0 B R 1 →+s2 s + = 0 L LC st st Initial Conditions → =1 + 2 it( ) Ae1 Ae 2 i(0) = Ae0 + Ae 0 = I 1 2 0 A →  1 A =st1 + st 2 =+  2 i'(0) ( Ase11 Ase 22) As 11 As 22 t=0 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 7 Second-Order Circuits 1. The Source-Free Series RLC Circuit 2. Initial Conditions 3. The Characteristic Equation 4. The Classical Method 5. Second-Order Op Amp Circuits Second-Order Circuits - sites.google.com/site/ncpdhbkhn 8 Initial Conditions (1) v Ex. 1 + – di(0) dv (0) 25 Ω 5H Find i(0), v (0), , ? i 50mF dt dt 30V A 5Ω − 30 i(0)= i (0) = = 1A + – 25+ 5 t = 0 B − 5 v(0)= v (0 ) = 30 = 5V 25+ 5 di di (0) 255i+ + v = 0 →25(0)i + 5 += v (0) 0 dt dt di (0) 25× 1 + 5 → =− =− 6 A/s dt 5 − dv − dv (0) dv (0) 1 i =50 × 10 3 →i(0) = 50 × 10 3 → = = 20 V/s dt dt dt 50× 10 −3 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 9 Initial Conditions (2) t = 0 Ex. 2 A di(0) dv (0) 4Ω Find i(0), v (0), , ? B dt dt + + – 6Ω 0.1F − 12 i(0)= i (0) = = 3A v 12V 4 i 0.5H – v(0)= v (0− ) = 12V di di (0) di (0) 12 0.5=v → 0.5 = v (0) → = = 24 A/s dt dt dt 0.5 12 3+ v dv v(0) dv (0) dv (0) +0.1 +i = 0 →+0.1 +=i (0) 0 → =−6 =− 50V/s 6 dt 6 dt dt 0.1 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 10 Initial Conditions (3) Ex. 3 t = 0 di(0) dv (0) Find i(0), v (0), , ? 4Ω 2Ω dt dt + + − 6Ω 1mF v – i(0)= i (0 ) = 4A 0.01H i – − 6+ 4 4A 24V v(0)= v (0) = 24 = 20V + + 6 4 2 i4 i2 i= i + i 4 6 i 6 4Ω 2Ω − dv + + i= i + 10 3 2 4 dt 0.01H i 6Ω 1mF v – – di 0.01= 6 i 24V dt 6 + − = 4i4 6 i 6 v 0 + = 2i2 v 24 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 11 Initial Conditions (4) Ex. 3 i4 i2 di(0) dv (0) Find i(0), v (0), , ? i6 4Ω 2Ω dt dt + + − 0.01H i 6Ω 1mF v – i(0)= i (0 ) = 4A – − 6+ 4 24V v(0)= v (0) = 24 = 20V 6+ 4 + 2 i= i + i = + 4 6 i4(0) i (0) i 6 (0) −3 dv = + −3 dv (0) i2 i 4 10 i(0)= i (0) + 10 dt 2 4 dt di = di (0) 0.01 6 i6 = dt 0.01 6i6 (0) dt 4i+ 6 i − v = 0 + − = 4 6 4i4 (0) 6 i 6 (0) v (0) 0 + = 2i (0)+ v (0) = 24 2i2 v 24 2 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 12 Initial Conditions (5) Ex. 3 i4 i2 di(0) dv (0) Find i(0), v (0), , ? i6 4Ω 2Ω dt dt + + − 0.01H i 6Ω 1mF v – i(0)= i (0 ) = 4A – − 6+ 4 24V v(0)= v (0) = 24 = 20V 6+ 4 + 2 = + = + i4(0) i (0) i 6 (0) i4(0) 4 i 6 (0) = + −3 dv (0) = + −3 dv (0) i2(0) i 4 (0) 10 i2(0) i 4 (0) 10 dt dt dv (0) = −  2400 V/s di (0) di (0)  dt 0.01= 6i (0) 0.01= 6i (0) →  6 6 di (0) dt dt  = 240 A/s + − = + − =  4i4 (0) 6 i 6 (0) v (0) 0 4i4 (0) 6 i 6 (0) 20 0 dt + = + = 2i2 (0) v (0) 24 2i2 (0) 20 24 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 13 Initial Conditions (6) Ex. 3 t = 0 i(0) 0.01H 4Ω 2Ω 4Ω 2Ω + + + + – 6Ω – 6Ω v 1mF v(0) i 1mF – 0.01H – 4A 24V 4A 24V 1. Consider the old state’s circuit i(0)= i (0− ) = 4A (switch is on the old position) 2. Find iL(0) & vC(0) − 6+ 4 3. Consider the new state’s circuit v(0)= v (0) = 24 = 20V 6+ 4 + 2 (switch moved to the new position) 4. Write the branch current equations, di dv with vL=& iC = Ldt C dt 5. Consider the moment t = 0 6. Find di (0)/ dt & dv (0)/ dt Second-Order Circuits - sites.google.com/site/ncpdhbkhn 14 Initial Conditions (7) i i Ex. 3 4 2 t = 0 i i6 4Ω 2Ω 4Ω 2Ω + + + + – 6Ω 1mF v – 6Ω 1mF v – 0.01H i – 0.01H 24V 4A 24V = + 1. Consider the old state’s circuit i4 i i 6 (switch is on the old position) − dv 2. Find i (0) & v (0) i= i + 10 3 L C 2 4 dt dv (0) = − 3. Consider the new state’s circuit  2400 V/s di  dt (switch moved to the new position) 0.01= 6 i →  6 di (0) 4. Write the branch current equations, dt  = 240 A/s di dv  with vL=& iC = + − =  dt L C 4i4 6 i 6 v 0 dt dt 5. Consider the moment t = 0 + = 2i2 v 24 6. Find di (0)/ dt & dv (0)/ dt Second-Order Circuits - sites.google.com/site/ncpdhbkhn 15 Initial Conditions (8) Ex. 4 1H diL(0) dv C (0) 30 Ω Find iL(0), v C (0), , ? 1mF dt dt 10 Ω + 20 Ω – + 120V − 120 – i (0 )= = 3A 2 1 L 10+ 30 40V → =− = iL(0) i L (0 ) 3A 1H 30 Ω 1mF 10 Ω + − = = × = 20 Ω – vC( 0) v R 1 10 3 30V 120V → =− = vC(0) v C (0 ) 30V 1 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 16 Initial Conditions (9) Ex. 4 1H diL(0) dv C (0) 30 Ω Find iL(0), v C (0), , ? 1mF dt dt 10 Ω + − − = 20 Ω – iL i2 i 3 0  + 120V  – iL + + += L Ri1L Ri 22 v C E 2 2 1  dt 40V −− + =−  vC Ri22 Ri 33 E 1 E 2 i3 dv i= C c 2 dt 1H 30 Ω  dv −c − = 1mF iL C i 3 0  dt 10 Ω +  20 Ω – iL i2  diL dv c → ++ += + L RiRC1L 2 v C E 2 120V dt dt –  2  dv 40V −−v RCc + Ri =− E E  C 2dt 3312 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 17 Initial Conditions (10) Ex. 4 i(0)= 3A; v (0) = 30V di(0) dv (0) L C i3 Find i(0), v (0),L , C ? L C dt dt 1H 30 Ω  −dv c − = 1mF iL C i 3 0  dt +  10 Ω  i 20 Ω – diL++ dv c += L i2 L RiRC1L 2 v C E 2  dt dt + 120V  dv – −−v RCc + Ri =− E E 2  C 2dt 3312 40V  dv (0)  − dv (0) i(0)− Cc − i (0)0 = 310−3 C −i (0)0 =  1dt 3  dt 3     →diL(0) ++ dv c (0) += →×diL (0) +×+×−3 dv C (0) −= L Ri1L(0) RC 2 v C (0) E 2 1 1032010 3040  dt dt  dt dt  dv (0)  − dv (0) −−v(0) RCc + Ri (0) =− EE 30−× 20 103 C + 30(0)i =− 120 40  C 2dt 33 12  dt 3 di(0) dv (0) →L =24A/s; C = 800V/s dt dt Second-Order Circuits - sites.google.com/site/ncpdhbkhn 18 v + – R E vVE(0)==0 ;(0) iI == 0+ 0 + R L C RR0 RR 0 i R E 0 di 1 t A RiL+ + idt = 0 + – ∫−∞ dt C t = 0 B R 1 →+s2 s + = 0 L LC st st Characteristic Equation → =1 + 2 it( ) Ae1 Ae 2 i(0) = Ae0 + Ae 0 = I 1 2 0 A →  1 A =st1 + st 2 =+  2 i'(0) ( Ase11 Ase 22) As 11 As 22 t=0 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 19 Second-Order Circuits 1. The Source-Free Series RLC Circuit 2. Initial Conditions 3. The Characteristic Equation 4. The Classical Method 5. Second-Order Op Amp Circuits Second-Order Circuits - sites.google.com/site/ncpdhbkhn 20 The Characteristic Equation (1) R L + C i v – R Ls di 1 Ri+ L +∫ idt = 0 + a dt C I( s ) 1 V( s ) i= Ae st ab Cs − b d( Ae st ) 1 →+RAest L +∫ Ae st = 0 dt C =++1  = Vsab () RLs  Is () ZsIsab ()() Cs  st 1  →Ae R ++ Ls  = 0 = Cs  Vab ( s ) 0 st i= Ae ≠ 0 I( s )≠ 0 1 1 →+R Ls + = 0 →Zs()0 =→++ RLs = 0 Cs ab Cs Second-Order Circuits - sites.google.com/site/ncpdhbkhn 21 Ex. 1 The Characteristic Equation (2) t = 0 A Find the characteristic equation? 4Ω B 1 + 0.5 s 2 0.3s+ 0.5 s + 6 + – Z =6 +0.1 s = 6Ω 0.1F ab 1 2 + 12V 0.5 s + 0.05s 1 v 0.1 s i 0.5H – =→2 + += Zab 0 0.3 s 0.5 s 6 0 1 60.5×s 0.3 s2 + 0.5 s + 6 Z = + = cd 0.1s 6+ 0.5 s 0.1(0.5 ss + 6) a c e =→2 + += Zcd 0 0.3 s 0.5 s 6 0 1 b d f 6 2 0.3s+ 0.5 s + 6 Z=0.5 s +0.1 s = 6 ef 1 + 1 6 + 0.6s 1 0.5 s 0.1 s 0.1 s =→2 + += Zef 0 0.3 s 0.5 s 6 0 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 22 Ex. 2 The Characteristic Equation (3) t = 0 Find the characteristic equation? 4Ω 2Ω + + 4 2 6Ω 1mF v – 0.01H i 0.01 s 6 – 1 4A 24V 10 −3 s 1. Consider the new state’s circuit a b (switch moved to the new position) 1 2 2. Deactivate all sources (if any) 6× 0.01 s −3 Z = +4 + 10 s 3. Transform elements to s-domain ab + 1 6 0.01 s 2 + 1  −3 R→ RL; → LsC ; →  10 s Cs  2× 10−4s 2 + 0.168 s + 36 4. Select any 2 neighbouring points = a & b on a wire segment (0.01s+ 6)(0.02 s + 1) 5. Calculate Zab (s) =→×−4 2 + += 6. Let Z (s) = 0 Zab 0 2 10 s 0.168 s 36 0 ab Second-Order Circuits - sites.google.com/site/ncpdhbkhn 23 The Characteristic Equation (4) 2 R 1 R R  1 s2 + s + = 0 →=−±s   − =−±α α2 − ω 2 L LC 1,2 2L 2 L  LC 0 R 1 α=, ω = 2L 0 LC α> ω =st1 + st 2 0 : xt( ) Ae1 Ae 2 α= ω = + st 0 : xt()( A1 Ate 2 ) α< ω =−± α ω =ω + ω −αt 0,s 1,2 j d : xt() ( A1 cosd tA 2 sin d te ) Second-Order Circuits - sites.google.com/site/ncpdhbkhn 24 Second-Order Circuits 1. The Source-Free Series RLC Circuit 2. Initial Conditions 3. The Characteristic Equation 4. The Classical Method 5. Second-Order Op Amp Circuits Second-Order Circuits - sites.google.com/site/ncpdhbkhn 25 The Classical Method (1) v + – RL RL L C i i R i f + C n + C v v R0 f n E0 A – – + – E E t = 0 + – E B + – if, v f in, v n = + v(0) i if i n + – = + R L C v vf v n i(0) R0 E0 A + – Second-Order Circuits - sites.google.com/site/ncpdhbkhn 26 The Classical Method (2) Ex. 1 = + i if i n Find i(t)? v 1. Write the general form + – 2. Find the initial conditions i(0)= 1A; v (0) = 5V 25 Ω 5H 3. Find the forced response i 5Ω 50mF 4. Find the natural response 30V A a. find the characteristic equation + – b. solve the cha. equ. t = 0 c. write the natural response 24V B + – (depends on the cha. roots) 5. Find the integration constants v 6. Write the complete response + – 5H 30 25 Ω i i(0)= = 1A 5Ω 50mF 25+ 5 30V A 5 + – v(0)= 30 = 5V 25+ 5 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 27 The Classical Method (3) Ex. 1 = + 1. Write the general form i if i n Find i(t)? v di (0) + – 2. Find the initial conditions ii(0)(0)== 1A; 1A; v (0) = − = 1.2 5V A/s dt 3. Find the forced response i = 0 25 Ω 5H f i 5Ω 50mF 4. Find the natural response 30V A a. find the characteristic equation + – b. solve the cha. equ. t = 0 c. write the natural response 24V B + – (depends on the cha. roots) 5. Find the integration constants 5H 6. Write the complete response di di (0) 25 Ω i + 50mF 25i+ 5 + v = 24 →25i (0) + 5 += v (0) 24 v dt dt di (0) 24V – → = − 1.2A/s dt + – = i f 0 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 28 The Classical Method (4) Ex. 1 = + 1. Write the general form i if i n Find i(t)? v di (0) + – 2. Find the initial conditions i(0)= 1A; = − 1.2 A/s dt 3. Find the forced response i = 0 25 Ω 5H f i 5Ω 50mF 4. Find the natural response 30V A a. find the characteristic equation + – b. solve the cha. equ. t = 0 c. write the natural response =−4t + − t 24V B in Ae1 Ae 2 + – (depends on the cha. roots) 5. Find the integration constants 25 5s 6. Write the complete response 1 1 =++ = →2 + += Zab 25 5 s −3 0 0.25s 1.25 s 1 0 −3 50× 10 s 50× 10 s → =− =− → =−4t + − t s14; s 2 1 in Ae1 Ae 2 a b Second-Order Circuits - sites.google.com/site/ncpdhbkhn 29 The Classical Method (5) Ex. 1 = + 1. Write the general form i if i n Find i(t)? v di (0) + – 2. Find the initial conditions i(0)= 1A; = − 1.2 A/s dt 3. Find the forced response i = 0 25 Ω 5H f i 5Ω 50mF 4. Find the natural response 30V A a. find the characteristic equation + – b. solve the cha. equ. t = 0 c. write the natural response =−4t + − t 24V B in Ae1 Ae 2 + – (depends on the cha. roots) = = 5. Find the integration constants A10.067; A 2 0.933 6. Write the complete response =0 + 0 = i(0) Ae1 Ae 2 1 A = 0.067 →  1 di (0) 0 0 = =−4Ae − Ae =− 1.2 A2 0.933 dt 1 2 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 30 The Classical Method (6) Ex. 1 = + 1. Write the general form i if i n Find i(t)? v di (0) + – 2. Find the initial conditions i(0)= 1A; = − 1.2 A/s dt 3. Find the forced response i = 0 25 Ω 5H f i 5Ω 50mF 4. Find the natural response 30V A a. find the characteristic equation + – b. solve the cha. equ. t = 0 c. write the natural response =−4t + − t 24V B in Ae1 Ae 2 + – (depends on the cha. roots) = = 5. Find the integration constants A10.067; A 2 0.933 6. Write the complete response it()=+ 0 0.067 e−4t + 0.933 e − t = 0.067 e − 4 t + 0.933 e − t A Second-Order Circuits - sites.google.com/site/ncpdhbkhn 31 v Ex. 1 The Classical Method (7) + – 25 Ω 5H i 5Ω 50mF it( )= 0.067 e−4t + 0.933 e − t A 30V A + – i t = 0 24V B + – 1A 0 0A t 25 Ω 5H i(0) 5Ω 25 Ω 25 Ω 5H in 50mF if 30V A + – B 24V B + – = =−4t + − t = i(0) 1A in Ae1 Ae 2 i f 0 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 32 Ex. 1 The Classical Method (8) it( )= 0.067 e−4t + 0.933 e − t A 1 0.9 0.8 0.7 0.6 0.5 0.4 Current(A) 0.3 0.2 0.1 0 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 Time (ms) Second-Order Circuits - sites.google.com/site/ncpdhbkhn 33 Ex. 2 The Classical Method (9) t = 0 A Find v(t)? = + v vf v n 4Ω B =− =12 = =− = + i(0) i (0 ) 3A; v(0) v (0 ) 12 V + – 4 2Ω 0.1F v 12V v+ dv + = →+v(0) dv (0) += 0.1i 0 0.1i (0) 0 i 1.6H 2 dt 2 dt – + 12 dv (0) 3 → =−2 =− 90V/s PPP 1. Write the general form dt 0.1 PPP 2. Find the initial conditions = PPP 3. Find the forced response v f 0 4. Find the natural response 1 2× 1.6s 0.32 s2 + 1.6 s + 2 PPP a. find the characteristic equation Z = + = + + PPP b. solve the cha. equ. 0.1s 2 1.6 s 0.1 ss (1.6 2) c. write the natural response PPP Z=→0 0.32 s2 + 1.6 s += 2 0 (depends on the cha. roots) 5. Find the integration constants → = =− → = + −2.5 t 6. Write the complete response s1 s 2 2.5 vn ( A1 Ate 2 ) Second-Order Circuits - sites.google.com/site/ncpdhbkhn 34 Ex. 2 The Classical Method (10) t = 0 A Find v(t)? = + v vf v n 4Ω B dv (0) + v(0)= 12V; = − 90V/s + – 2Ω 0.1F dt 12V − v v=0; v = ( AAte + ) 2.5 t f n 1 2 i 1.6H = + =+ + −2.5 t – vvf v n 0 ( AAte1 2 ) =+0 == v(0)( AAe1 2 0) A 1 12 PPP 1. Write the general form dv d − PPP 2. Find the initial conditions =(12 + A t ) e 2.5 t dt dt 2 PPP 3. Find the forced response 4. Find the natural response =−−2.5t + − 2.5 t − − 2.5 t 30e A2 e 2.5 A 2 te PPP a. find the characteristic equation PPP b. solve the cha. equ. dv (0) =−0 + 0 − ××=− 0 30eAe2 2.5 A 2 0 e 90 c. write the natural response dt PPP (depends on the cha. roots) → = − A2 60 PPP 5. Find the integration constants 6. Write the complete response →vt( ) = (12 − 60 te )−2.5 t V PPP Second-Order Circuits - sites.google.com/site/ncpdhbkhn 35 Ex. 2 The Classical Method (11) vt( )= (12 − 60 te )−2.5 t V 12 10 8 6 4 2 Voltage (V) Voltage 0 -2 -4 -6 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 Time (ms) Second-Order Circuits - sites.google.com/site/ncpdhbkhn 36 Ex. 3 The Classical Method (12) =+ = + t = 0 Find i(t) & v(t)? iifn ivv; fn v 4Ω 2Ω − + + i(0)= i (0 ) = 4A 6Ω 1mF v – 0.01H i − 6+ 4 – v(0)= v (0 ) = 24 = 20V 4A 24V 6+ 4 + 2 = + i4 i i 6 − dv i= i +1 × 10 3 2 4 dt dv (0) = −  2400 V/s di →  dt 0.01= 6 i  i4 i2 6 di (0) dt  = 240 A/s i  dt 6 4Ω 2Ω + − = + 4i4 6 i 6 v 0 + 0.01H i 6Ω 1mF v – + = 2i2 v 24 – 24V Second-Order Circuits - sites.google.com/site/ncpdhbkhn 37 Ex. 3 The Classical Method (13) =+ = + t = 0 Find i(t) & v(t)? iifn ivv; fn v = = 4Ω 2Ω i(0) 4A; v (0) 20V + + di(0) dv (0) 6Ω 1mF v – =240 A/s; = − 2400V/s 0.01H i dt dt – 24 4 4A 24V i = = 4A; v =24 = 16V f 4+ 2 f 4+ 2 1 2 4Ω 2Ω × −3 + + =6 0.01 s ++10 s = Z 4 0 – ab 0.01H if 6Ω 1mF vf 6+ 0.01 s + 1 2 − – 10 3 s 24V s= −4.2 + j 0.6 → 1  = − − s2 4.2 j 0.6 4 2  = + −4.2 t 0.01 s 6 in ( A1 cos 0.6 tA 2 sin 0.6 te ) 1 →  − = + −4.2 t 10 3 s vBn (1 cos0.6 tB 2 sin 0.6 te ) Second-Order Circuits - sites.google.com/site/ncpdhbkhna b 38 Ex. 3 The Classical Method (14) =+ = + t = 0 Find i(t) & v(t)? iifn ivv; fn v = = 4Ω 2Ω i(0) 4A; v (0) 20V + + di(0) dv (0) 6Ω 1mF v – =240 A/s; = − 2400V/s 0.01H i dt dt – 24 4 4A 24V i = = 4A; v =24 = 16V f 4+ 2 f 4+ 2 =+−4.2t =+ − 4.2 t iAn (12 cos0.6 tA sin0.6) te ; vBn ( 12 cos0.6 tB sin0.6) te =+=+ + −4.2 t iiif n 4 ( A1 cos0.6 tA 2 sin 0.6 te ) =+ +0 = i(0) 4 ( A1 cos 0 Ae 2 sin 0) 4 = di (0) A1 0 = + −+0 →  (A1 cos0 A 2 sin0)( 4.2 e ) = dt A2 400 +− +0 = ( 0.6A1 sin0 0.6 A 2 cos0) e 240 →i =4 + 400 e−4.2 t sin0.6 t A Second-Order Circuits - sites.google.com/site/ncpdhbkhn 39 Ex. 3 The Classical Method (15) =+ = + t = 0 Find i(t) & v(t)? iifn ivv; fn v = = 4Ω 2Ω i(0) 4A; v (0) 20V + + di(0) dv (0) 6Ω 1mF v – =240 A/s; = − 2400V/s 0.01H i dt dt – 24 4 4A 24V i = = 4A; v =24 = 16V f 4+ 2 f 4+ 2 =+−4.2t =+ − 4.2 t iAn (12 cos0.6 tA sin0.6) te ; vBn ( 12 cos0.6 tB sin0.6) te =+=+ + −4.2 t vvvf n 16 ( B1 cos0.6 tB 2 sin0.6) te =+ +0 = v(0) 16 ( B1 cos0 Be 2 sin 0) 20 = dv (0) B1 4 = + −+0 →  (B1 cos0 B 2 sin0)( 4.2 e ) = − dt B2 3944 +− +0 =− ( 0.6B1 sin0 0.6 B 2 cos0) e 2400 →=+v16 (4cos 0.6 t − 3944sin 0.6 te )−4.2 t V Second-Order Circuits - sites.google.com/site/ncpdhbkhn 40 Ex. 3 The Classical Method (16) v=16 + (4cos0.6 t − 3944sin 0.6 te )−4.2 t V 50 0 -50 -100 Voltage (V) Voltage -150 -200 0 500 1000 1500 2000 2500 3000 Time (ms) Second-Order Circuits - sites.google.com/site/ncpdhbkhn 41 Second-Order Circuits 1. The Source-Free Series RLC Circuit 2. Initial Conditions 3. The Characteristic Equation 4. The Classical Method 5. Second-Order Op Amp Circuits Second-Order Circuits - sites.google.com/site/ncpdhbkhn 42 Second-Order Op Amp Circuits (1) Ex. 10k Ω a t = 0 + 10k Ω Find v(t)? = + b v vf v n – + = = + va (0) 0; v (0) 0 – 4V 20 µF v v− v − b = × 6 dv 100 10 100 µF – 10× 10 3 dt = va v b v− v − dv v(0)− v (0) − dv (0) dv (0) →a =100 × 10 6 →a =100 × 10 6 → = 0 10× 10 3 dt 10× 10 3 dt dt = = = dv vf v b v a 4 V v= v + 2 a dv dv dt →0.2 + 1.2 +=v 4 4 − v− dv 2 a=20 × 10 6 a dt dt 10× 10 3 dt 2 dv+ dv + = 0.2 1.2v 0 2 st →2 + += dt2 dt →(0.2s ++ 1.2 s 1) Ae = 0 0.2s 1.2 s 1 0 v= Ae st → =− =− → =−t + − 5 t s11; s 2 5 vn Ae1 Ae 2 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 43 Second-Order Op Amp Circuits (2) Ex. 10k Ω a t = 0 + 10k Ω Find v(t)? = + b v vf v n – + dv (0) + v(0)= 0; = 0 – 4V 20 µF v dt 100 µF – = v f 4 =−t + − 5 t vn Ae1 Ae 2 = +−t + − 5 t v4 Ae1 Ae 2 =+ + = v(0) 4 A1 A 2 0 A = − 5 −t − 5 t →  1 →=−v4 5 e + e V dv (0) A = 1 =−A −5 A = 0  2 dt 1 2 Second-Order Circuits - sites.google.com/site/ncpdhbkhn 44 Second-Order Op Amp Circuits (3) Ex. v=4 − 5 e−t + e − 5 t V 4 3.5 3 2.5 2 Voltage (V) Voltage 1.5 1 0.5 0 0 1000 2000 3000 4000 5000 6000 7000 8000 Time (ms) Second-Order Circuits - sites.google.com/site/ncpdhbkhn 45