Intro-FirstOrderDiffEqSp03(2pp)

Intro-FirstOrderDiffEqSp03(2pp) - FIRST AND SECOND-ORDER...

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Unformatted text preview: FIRST AND SECOND-ORDER TRANSIENT CIRCUITS IN CIRCUITS WITH INDUCTORS AND CAPACITORS VOLTAGES AND CURRENTS CANNOT CHANGE INSTANTANEOUSLY. EVEN THE APPLICATION, OR REMOVAL, OF CONSTANT SOURCES CREATES A TRANSIENT BEHAVIOR LEARNING GOALS FIRST ORDER CIRCUITS Circuits that contain a single energy storing element. Either a capacitor or an inductor SECOND ORDER CIRCUITS Circuits with two energy storing elements in any combination ANALYSIS OF LINEAR CIRCUITS WITH INDUCTORS AND/OR CAPACITORS THE CONVENTIONAL ANALYSIS USING MATHEMATICAL MODELS REQUIRES THE DETERMINATION OF (A SET OF) EQUATIONS THAT REPRESENT THE CIRCUIT. ONCE THE MODEL IS OBTAINED ANALYSIS REQUIRES THE SOLUTION OF THE EQUATIONS FOR THE CASES REQUIRED. FOR EXAMPLE, IN NODE OR LOOP ANALYSIS OF RESISTIVE CIRCUITS ONE REPRESENTS THE CIRCUIT BY A SET OF ALGEBRAIC EQUATIONS THE MODEL WHEN THERE ARE INDUCTORS OR CAPACITORS THE MODELS BECOME LINEAR ORDINARY DIFFERENTIAL EQUATIONS (ODEs). HENCE, IN GENERAL, ONE NEEDS ALL THOSE TOOLS IN ORDER TO BE ABLE TO ANALYZE CIRCUITS WITH ENERGY STORING ELEMENTS. A METHOD BASED ON THEVENIN WILL BE DEVELOPED TO DERIVE MATHEMATICAL MODELS FOR ANY ARBITRARY LINEAR CIRCUIT WITH ONE ENERGY STORING ELEMENT. THE GENERAL APPROACH CAN BE SIMPLIFIED IN SOME SPECIAL CASES WHEN THE FORM OF THE SOLUTION CAN BE KNOWN BEFOREHAND. THE ANALYSIS IN THESE CASES BECOMES A SIMPLE MATTER OF DETERMINING SOME PARAMETERS. TWO SUCH CASES WILL BE DISCUSSED IN DETAIL FOR THE CASE OF CONSTANT SOURCES. ONE THAT ASSUMES THE AVAILABILITY OF THE DIFFERENTIAL EQUATION AND A SECOND THAT IS ENTIRELY BASED ON ELEMENTARY CIRCUIT ANALYSIS… BUT IT IS NORMALLY LONGER WE WILL ALSO DISCUSS THE PERFORMANCE OF LINEAR CIRCUITS TO OTHER SIMPLE INPUTS 1 AN INTRODUCTION INDUCTORS AND CAPACITORS CAN STORE ENERGY. UNDER SUITABLE CONDITIONS THIS ENERGY CAN BE RELEASED. THE RATE AT WHICH IT IS RELEASED WILL DEPEND ON THE PARAMETERS OF THE CIRCUIT CONNECTED TO THE TERMINALS OF THE ENERGY STORING ELEMENT With the switch on the left the capacitor receives charge from the battery. Switch to the right and the capacitor discharges through the lamp GENERAL RESPONSE: FIRST ORDER CIRCUITS Including the initial conditions, the model for the capacitor t t x t t 1 voltage or the inductor current et x ( t ) - e t x ( t0 ) = ò e t fTH ( x )dx * / e - t will be shown to be of the form tt 0 0 dx (t ) + ax (t ) = f (t ); x (0+) = x0 dt t dx + x = fTH ; x (0+) = x0 dt x(t ) = e - t - t0 t t x (t0 ) + 1e t tò 0 t-x t fTH ( x )dx THIS EXPRESSION ALLOWS THE COMPUTATION OF THE RESPONSE FOR ANY FORCING FUNCTION. WE WILL CONCENTRATE ON THE SPECIAL CASE WHEN THE RIGHT HAND SIDE IS CONSTANT Solving the differential equation using integrating factors, one tries to convert the LHS into an exact derivative t is called the " time constant." t It will be shown to provide significan t dx 1 t e t ò t0 dt t t + x = f TH /* e t t t t t t 1 dx 1 + e x = e fTH t dt t t t dæ t ö 1 t ç e x ÷ = e fTH ÷t dt ç ø è information on the reaction speed of the circuit The initial time, to , is arbitrary. The general expression can be used to study sequential switchings . 2 FIRST ORDER CIRCUITS WITH CONSTANT SOURCES dx t + x = fTH ; x (0+) = x0 dt x (t ) = e - t -t0 t 1t x (t 0 ) + ò e tt t-x t fTH ( x )dx 0 x(t ) = e x(t ) = e - - t -t 0 t t -t0 t t f -t æ x ö x(t0 ) + TH e t çte t ÷ t è øt t ö æ tt x(t0 ) + fTH e ç e - e t ÷ ø è - t t x(t ) = e t -t 0 t e t-x t f t -t-x x(t0 ) + TH ò e t dx tt - t t x(t ) = e t -t 0 t t -t0 t The form of the solution is x(t ) = K1 + K 2e - t -t 0 t ; t ³ t0 TIME CONSTANT TRANSIENT x t Any variable in the circuit is of the form =e e Þ - - t ³ t0 0 - 0 x(t ) = fTH + ( x(t0 ) - fTH )e If the RHS is constant - 0 f -t t x x (t0 ) + TH e t ò et dx t t 0 y (t ) = K1 + K 2e - t -t0 t ; t ³ t0 Only the values of the constants K_1, K_2 will change EVOLUTION OF THE TRANSIENT AND INTERPRETATION OF THE TIME CONSTANT Tangent reaches x-axis in one time constant Drops 0.632 of initial value in one time constant With less than 2% error transient is zero beyond this point A QUALITATIVE VIEW: THE SMALLER THE TIME CONSTANT THE FASTER THE TRANSIENT DISAPPEARS 3 THE TIME CONSTANT The following example illustrates the physical meaning of time constant Charging a capacitor v C - v S RS a [email protected] : RS + dv v - v S =0 C c+ C + dt RS vS Cv c _ The model C dvC b dt RTH C Assume v S = VS , vC (0) = 0 t t - t 2t 3t 4t et 0.368 With less than 1% 0.135 error the transient 0.0498 is negligible after 0.0183 five time constants 5t 0.0067 dvC + vC = vTH dt t = RTH C The solution can be shown to be vC (t ) = VS - VS e - t t transient For practical purposes the capacitor is charged when the transient is negligible t = RTH C CIRCUITS WITH ONE ENERGY STORING ELEMENT THE DIFFERENTIAL EQUATION APPROACH CONDITIONS 1. THE CIRCUIT HAS ONLY CONSTANT INDEPENDENT SOURCES 2. THE DIFFERENTIAL EQUATION FOR THE VARIABLE OF INTEREST IS SIMPLE TO OBTAIN. NORMALLY USING BASIC ANALYSIS TOOLS; e.g., KCL, KVL. . . OR THEVENIN 3. THE INITIAL CONDITION FOR THE DIFFERENTIAL EQUATION IS KNOWN, OR CAN BE OBTAINED USING STEADY STATE ANALYSIS FACT : WHEN ALL INDEPENDEN T SOURCES ARE CONSTANT FOR ANY VARIA BLE, y ( t ), IN THE CIRCUIT THE SOLUTION IS OF THE FORM y ( t ) = K1 + K 2e - ( t - tO ) t , t > tO SOLUTION STRATEGY: USE THE DIFFERENTIAL EQUATION AND THE INITIAL CONDITIONS TO FIND THE PARAMETERS K1 , K 2 ,t 4 If the diff eq for y is known in the form dy + a0 y = f We can use this dt info to find the unknowns y (0 + ) = y 0 Use the initial condition to get one more equation y ( 0 + ) = K1 + K 2 a1 Use the diff eq to find two more equations by replacing the form of solution into the differential equation t t t Kdy y (t ) = K1 + K 2e , t > 0 Þ =- 2e t t dt - æ K -t a1ç - 2 e t çt è t ö æ ÷ + a 0 ç K1 + K 2 e t ÷ ç ø è f a 0 K1 = f Þ K1 = a0 ö ÷= f ÷ ø K 2 = y ( 0 + ) - K1 SHORTCUT: WRITE DIFFERENTIAL EQ. IN NORMALIZED FORM WITH COEFFICIENT OF VARIABLE = 1. a1 dy a dy f + a0 y = f Þ 1 +y= dt a0 dt a0 t K1 t a æ a1 ö ç - + a0 ÷ K 2 e t = 0 Þ t = 1 a0 èt ø LEARNING EXAMPLE FIND v (t ), t > 0. ASSUME v (0) = VS 2 5 LEARNING EXAMPLE FIND v (t ), t > 0. ASSUME v (0) = VS x ( t ) = K1 + K 2 e 2 - t t ,t > 0 K1 = x (¥); K1 + K 2 = x (0+ ) ANSWER : v (t ) = VS - (VS / 2)e - t RC , t >0 STEP 2 STEADY STATE ANALYSIS SOLUTION IS v (t ) = K1 + K 2e - t t ,t >0 for t > 0 and t ® ¥, v(t) ® K1 (steady state value) MODEL FOR t > 0. USE KCL @ v (t ) v (t ) - V S dv +C (t ) = 0 dt R */ R initial condition v (0) = VS / 2 (DIFF. EQ. KNOWN, INITIAL CONDITION KNOWN) STEP 1 TIME CONSTANT dy t +y= f dt dv RC (t ) + v (t ) = Vs dt K1 = VS dy + y = f THEN K1 = f dt STEP 3 USE OF INITIAL CONDITION AT t = 0 v (0) = K1 + K 2 Þ K 2 = v (0) - K1 K 2 = v (0) - f v (0) = VS / 2 Þ K 2 = -VS / 2 IF THE MODEL IS t Get time constant as coefficient of derivative LEARNING EXAMPLE IN STEADY STATE THE SOLUTION IS A CONSTANT. HENCE ITS DERIVATIVE IS ZERO. FROM DIFF EQ. dv = 0 Þ v = VS Steady state value from diff. eq. dt \ (equating steady state values) FIND i (t ), t > 0 i (t ) 6 FIND i (t ), t > 0 LEARNING EXAMPLE x ( t ) = K1 + K 2 e i ( t ) = K1 + K 2e + i (t ) VS = v R + v L = Ri (t ) + L t t ,t > 0 K1 = x (¥); K1 + K 2 = x (0+ ) + vR - KVL - t t ,t > 0 MODEL. USE KVL FOR t > 0 vL - - i (t ) di (t ) dt INITIAL CONDITION t < 0 Þ i ( 0- ) = 0 ü ýi ( 0 + ) = 0 inductor Þ i (0-) = i (0+) þ L L di V STEP 1 (t ) + i (t ) = S t = R R dt R STEP 2 STEADY STATE i (¥) = K = VS 1 R t STEP 3 INITIAL CONDITION æ ö L÷ VS ç ANS : i (t ) = 1- e R ÷ i ( 0 + ) = K1 + K 2 Rç ç ÷ è ø LEARNING BY DOING 7 LEARNING BY DOING i ( t ) = K1 + K 2e - t t ,t > 0 v (t ) MODEL. KCL FOR t > 0 v (t ) + i (t ) R L di di v (t ) = L (t ) Þ I S = (t ) + i (t ) INITIAL CONDITION : i (0+ ) = 0 R dt dt IS = L R STEP 2 i (¥) = I S Þ K1 = I S STEP 1 t = STEP 3 i ( 0 + ) = 0 = K1 + K 2 t ö æ L÷ ç ANS : i (t ) = I S ç1 - e R ÷ ÷ ç ø è 8 i (t ) = K1 + K 2e - t t ,t > 0 INITIAL CONDITIONS CIRCUIT IN STEADY STATE FOR t < 0 MODEL FOR t > 0 v C ( 0- ) = i (t ) = v (t ) R2 IT IS SIMPLER TO DETERMINE MODEL FOR CAPACITOR VOLTAGE v (t ) dv v (t ) + C (t ) + = 0; RP = R1 || R2 R1 dt R2 RP = 3k || 6k = 2kW dv v (t ) C (t ) + =0 dt RP 3k (12) = 4V Þ v (0+ ) = 4V 3k + 6k STEP 1 t = RP C = (2 ´ 103 W)(100 ´ 10 -6 F ) = 0.2 s STEP 2 v (¥) = K1 = 0 STEP 3 v (0+ ) = K1 + K 2 = 4V Þ K 2 = 4V t 0.2 [V ], t >0 t 4 - 0.2 e [mA ], t >0 v ( t ) = 4e ANS : i (t ) = 3 - LEARNING EXAMPLE FIND vO (t ), t > 0 9 LEARNING EXAMPLE x ( t ) = K1 + K 2 e FIND vO (t ), t > 0 - t - t t ,t > 0 K1 = x (¥); K1 + K 2 = x (0+ ) vO (t ) = K1 + K 2e t , t > 0 i (t ) STEP 2: FIND K1 USING STEADY STATE ANALYSIS KVL(t>0) 0.5 dvO (t ) + vO (t ) = 6 Þ vO (¥ ) = 6V dt v O ( ¥ ) = K1 \ K1 = 6V MODEL FOR t > 0. USE KVL di - VS1 + R1i ( t ) + L (t ) + R3i (t ) = 0 dt di 2 (t ) + 4i (t ) = 12 vO (t ) = 2i (t )[V ] dt di 0.5 (t ) + i (t ) = 3[ A] dt dv 0.5 O (t ) + vO (t ) = 6V t = 0.5 STEP dt THE NEXT STEP REQUIRES THE INITIAL VALUE OF THE VARIABLE, vO (0+ ) FOR THE INITIAL CONDITION ONE NEEDS THE INDUCTOR CURRENT FOR t<0 AND USES THE CONTINUITY OF THE INDUCTOR CURRENT DURING THE SWITCHING . THE STEADY STATE ASSUMPTION FOR t<0 SIMPLIFIES THE ANALYSIS 1 CIRCUIT IN STEADY STATE (t<0) t <0 10 CIRCUIT IN STEADY STATE (t<0) a i L (t ) RTH = 2 || 2 = 1W I1 t <0 - 12 + 4 I1 - 4 = 0 I1 = 4[ A] KVL KVL VTH = VOC = 2 I1 - 4 = 4[V ] 4 i L (0-) = i (0+ ) = [ A] 3 4 8 i (0+) = Þ vO (0+ ) = [V ] 3 3 8 10 K1 + K 2 = = 6 - K 2 Þ K 2 = 3 3 b MUST FIND i L (t ) FOR EXAMPLE, USE THEVENIN ASSUMING INDUCTOR IN STEADY STATE - t vO (t ) = K1 + K 2e t , t > 0 t 5 i (t ) = 3 - e 3 - t 0.5 , t >0 vO ( t ) = 6 - 10 - 0.5 e [V ], t > 0 3 LEARNING EXTENSION FIND vO (t ), t > 0 R1 R2 C 11 LEARNING EXTENSION FIND vO (t ), t > 0 v C ( t ) = K1 + K 2 e - t t ,t > 0 K1 = vC (¥); K1 + K 2 = i1 (0+) R1 R2 C DETERMINE vc (t ) MODEL FOR t > 0. USE KCL dv dv 2 1 vC vC (t ) = vC (t ) C C (t ) + = 0 Þ ( R1 + R2 )C C ( t ) + vc = 0 vO (t ) = 2+4 3 dt R1 + R2 dt STEP 1 t = ( R1 + R2 )C = (6 ´ 103 W)(100 ´ 10 -6 F ) = 0.6 s STEP 2 t vC (t ) = K1 + K 2e t , t > 0 K1 = 0 t 8vO (t ) = e 0.6 [V ], t > 0 3 INITIAL CONDITIONS. CIRCUIT IN STEADY STATE t<0 STEP 3 vC (0+ ) = 8 = K1 + K 2 Þ K 2 = 8[V ] + 6 vC (0-) = (12)V 9 - vC (t ) = 8e - t 0.6 [V ], t >0 LEARNING EXTENSION FIND i1 ( t ), t > 0 12 LEARNING EXTENSION FIND i1 ( t ), t > 0 i1 (t ) = K1 + K 2e - t t ,t > 0 K1 = i1 (¥); K1 + K 2 = i1 (0+ ) CIRCUIT IN STEADY STATE PRIOR TO SWITCHING i1 (t ) i1 (0-) + vL - L MODEL FOR t > 0. USE KVL di 1 di1 L 1 + 18i1 (t ) = 0 Þ (t ) + i1 (t ) = 0 dt 9 dt 1 STEP 1 t = s 9 STEP 3 i1 (0-) = i1 (0+) = K1 + K 2 Þ K 2 = 1[ A] K1 = 0 STEP 2 12V = 1A 12W i1 (0-) = t FOR INITIAL CONDITIONS ONE NEEDS 1 ANS : i1(t ) = e 9 [ A] = e -9 t [ A], t > 0 INDUCTOR CURRENT FOR t<0 USING THEVENIN TO OBTAIN MODELS Obtain the voltage across the capacitor or the current through the inductor a Circuit with resistances and sources Inductor or Capacitor b RTH a Thevenin Inductor or Capacitor + VTH - b Representation of an arbitrary circuit with one storage element RTH a VTH + - iR C ic vc _ b Case 1.1 Voltage across capacitor C RTH C RTH a [email protected] node a + dvC + vC = vTH dt ic + iR = 0 dv ic = C C dt vC - vTH iR = RTH dvC vC - vTH + =0 dt RTH VTH + Use KVL + vR - - L iL b - Case 1.2 Current through inductor æL ç çR è TH vR + vL = vTH vR = RTH iL di vL = L L dt + vL L diL + RTH iL = vTH dt ö diL vTH ÷ ÷ dt + iL = R = iSC TH ø 13 Thevenin for t>0 at inductor terminals EXAMPLE Find iO (t); t > 0 6W 6W + 24V - t =0 6W iO (t ) 6W 3H 6W + 24V - 6W t >0 6W a 6W b vTH = 0 RTH = 6 + (6 || (6 + 6)) The variable of interest is the inductor current. The model is L diO v + iO = TH RTH dt RTH t= 3H L = = 0.3 s RTH 10W 0.3 And the solution is of the form t t iO (t ) = K1 + K 2 e ; t > 0 diO + iO = 0 ; t > 0 dt t æ K -t ö 0.3ç - 2 e 0.3 ÷ + K1 + K 2 e 0.3 = 0 è 0.3 ø K1 = 0 Þ iO (t ) = K 2e - t 0.3 ;t >0 Next: Initial Condition Determine iO ( 0 + ). Use steady state assumption and continuity of inductor current 6W + 24V - t <0 iC (0+) = i3 6(i3 - i1 ) + 6(i3 - i2 ) = 0 iO (t ) 6W Loop analysis - 24 + 6(i2 - i1 ) + 6(i2 - i3 ) = 0 Circuit for t<0 6W 6i1 + 6(i1 - i3 ) + 6(i1 - i2 ) = 0 v1 v1 v1 - 24 ++ = 0 Þ v1 = 8 Node analysis 66 6 solution : 3H iC (0 +) = 32 mA 6 6W Since K1=0 the solution is i1 6W 6W + 24V - i2 t <0 iO (0-) = iO (0+) 6W v1 6W iO (t ) = K 2e - t 0.3 ;t>0 Evaluating at 0+ K 2 = i3 iO (0+) = 24 v1 + 66 t 32 6 32 iO (t ) = e 0.3 ; t > 0 6 14 EXAMPLE Find iO (t), t > 0 100m F 6k + vC - iO (t ) 12V + - t =0 6k EXAMPLE Find iO (t), t > 0 100m F 6k 12V vTH = 6V 6k t =0 6k For t > 0 iO = 6k RTH = 6k || 6k = 3k + vC - iO (t ) + - 6k vC 6k 6k t = 3 *103 W *100 *10-6 F = 0.3s Model for v C 0.3 Hence, if the capacitor voltage is known the problem is solved dvC + vC = vTH dt 6k 12V a iO (t ) + - 6k + vTH - v C = K1 + K 2 e - t 0.3 t æ K2 - t ö 1.5 ÷ + K + K e 0.3 = 6 ç1.5 e 1 2 ç 1.5 ÷ è ø Model for v_c RTH C dvC + vC = 6 dt b K1 = 6 6k t >0 6k Now we need to determine the initial value v_c(0+) using continuity and the steady state assumption 15 12V 12V circuit in steady state before the switching 6k 6k + vC (0-) iO (t ) + - t <0 6k 6k circuit in steady state before the switching 6k 6k + vC (0-) iO (t ) + - t <0 6k 6k vC (0-) = 6V Continuity of capacitor voltage vC (0+ ) = 6V K1 + K 2 = vC (0+) K1 = 6 Þ K 2 = 0 vC (t ) = 6V ; t > 0 Þ iO (t ) = vC = 1mA ; t > 0 6k Diff Eq Approach 16 ...
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