Intro-FirstOrderDiffEqSp03(1pp)

Intro-FirstOrderDiffEqSp03(1pp) - 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 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 1t voltage or the inductor current et x (t ) - e t x (t 0 ) = ò e 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 dx t + x = fTH ; x (0+) = x0 dt x(t ) = e - t -t 0 t t 1x(t0 ) + ò e t t0 t-x t f TH ( 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 t is called the " time constant." exact derivative t It will be shown to provide significan t dx 1t t + x = fTH /* e information on the reaction speed of the t dt t t t circuit 1t dx 1 t + e x = e fTH et The initial time, to , is arbitrary. The dt t ò t0 t t ö1 dæ ç e x ÷ = e fTH ÷t dt ç ø è t t t t general expression can be used to study sequential switchings . FIRST ORDER CIRCUITS WITH CONSTANT SOURCES dx t + x = fTH ; x (0+ ) = x0 dt x (t ) = t -t0 et t -t- x et 1 x (t0 ) + ò tt fTH ( x )dx 0 x(t ) = e x(t ) = e - - t -t 0 t t -t 0 t x(t0 ) + t t x(t ) = e fTH x(t0 ) + òe tt t e t-x t - t t t-x t t t dx x(t ) = fTH + ( x(t0 ) - fTH )e x(t ) = e - t -t 0 t x(t ) = K1 + K 2e - t -t0 t ; t ³ t0 TIME CONSTANT TRANSIENT x t fTH x(t0 ) + e t 0 The form of the solution is Any variable in the circuit is of the form =e e Þ t -t0 t 0 t ³ t0 0 - x t t æ tt tö x(t0 ) + fTH e ç e - e ÷ ø è - If the RHS is constant t -t 0 t t fTH æö e çte ÷ t è øt - tt t x t ò e dx t0 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 THE TIME CONSTANT The following example illustrates the physical meaning of time constant Charging a capacitor v C - v S RS a KCL@a : RS dv v - v S + C c+ C =0 + dt RS vS Cv c _ The model t t 2t 3t 4t 5t e - t t 0.368 With less than 1% 0.135 error the transient 0.0498 is negligible after 0.0183 five time constants 0.0067 dvC RTH C + vC = vTH dt dv b CC dt Assume t = RTH C v S = V S , v C ( 0) = 0 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 If the diff eq for y is known in the form dy a1 + a0 y = f We can use this dt info to find the unknowns y (0 + ) = y 0 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 2 e , t > 0 Þ =- 2e t t dt - t K2 -t e æ ö ÷ + a0 ç K1 + K 2 ç ÷ è ø f a 0 K 1 = f Þ K1 = a0 æ a1 ç çt è t t et ö ÷= f ÷ ø a æ a1 ö - + a0 ÷ K 2 e t = 0 Þ t = 1 ç a0 èt ø Use the initial condition to get one more equation y (0 + ) = K1 + K 2 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 LEARNING EXAMPLE FIND v (t ), t > 0. ASSUME v (0) = VS 2 LEARNING EXAMPLE FIND v (t ), t > 0. ASSUME v (0) = VS ANSWER : v (t ) = VS - (VS 2 x ( t ) = K1 + K 2 e K1 = x (¥); t / 2)e RC , t > 0 - t t ,t > 0 K1 + K 2 = x ( 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 ) dv v (t ) - V S +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 Get time constant as coefficient of derivative 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) K1 = V S dy IF THE MODEL IS t + y = f THEN K1 = f dt STEP 3 USE OF INITIAL CONDITION AT t = 0 v ( 0) = K 1 + K 2 Þ K 2 = v ( 0 ) - K1 K 2 = v (0) - f v (0) = VS / 2 Þ K 2 = -VS / 2 LEARNING EXAMPLE FIND i (t ), t > 0 i (t ) LEARNING EXAMPLE FIND i (t ), t > 0 x ( t ) = K1 + K 2 e i (t ) VS = v R + v L = Ri (t ) + L t t ,t > 0 K1 = x (¥); K1 + K 2 = x (0+ ) + vR - KVL - i (t ) = K1 + K 2e + vL - - t t ,t > 0 MODEL. USE KVL FOR t > 0 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 æ ö STEP 3 INITIAL CONDITION L÷ VS ç R ANS : i (t ) = i ( 0 + ) = K1 + K 2 ç1 - e ÷ Rç ÷ è ø LEARNING BY DOING LEARNING BY DOING i (t ) = K1 + K 2e - t t ,t > 0 v (t ) MODEL. KCL FOR t > 0 v (t ) + i (t ) R di L 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 æ ö L÷ ç ANS : i ( t ) = I S ç1 - e R ÷ ç ÷ è ø i ( t ) = K1 + K 2 e - t t ,t > 0 INITIAL CONDITIONS CIRCUIT IN STEADY STATE FOR t < 0 MODEL FOR t > 0 vC ( 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 ) =0 C (t ) + 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 v ( t ) = 4e - t t 0.2 [V ], t >0 4 - 0.2 ANS : i (t ) = e [mA], t > 0 3 LEARNING EXAMPLE FIND vO (t ), t > 0 LEARNING EXAMPLE x ( t ) = K1 + K 2 e FIND vO (t ), t > 0 v O ( t ) = K1 + K 2 i (t ) t et - t t ,t > 0 K1 = x (¥); K1 + K 2 = x (0+ ) ,t > 0 STEP 2: FIND K1 USING STEADY STATE ANALYSIS KVL(t>0) 0.5 dvO ( t ) + vO (t ) = 6 Þ vO (¥) = 6V dt vO ( ¥ ) = 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 CIRCUIT IN STEADY STATE (t<0) a i L (t ) RTH = 2 || 2 = 1W I1 t <0 KVL - 12 + 4 I1 - 4 = 0 I1 = 4[ A] 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 v O ( t ) = K1 + K 2 e - t t ,t > 0 t t 5i (t ) = 3 - e 0.5 , t > 0 3 10 - 0.5 vO ( t ) = 6 - e [V ], t > 0 3 LEARNING EXTENSION FIND vO (t ), t > 0 R1 C R2 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 C R2 DETERMINE vc ( t ) MODEL FOR t > 0. USE KCL vC dv 2 1 dv C C (t ) + = 0 Þ ( R1 + R2 )C C (t ) + vc = 0 vO ( t ) = vC ( t ) = vC ( t ) dt R1 + R2 dt 2+4 3 STEP 1 t = ( R1 + R2 )C = (6 ´ 103 W)(100 ´ 10 -6 F ) = 0.6 s STEP 2 v C ( t ) = K1 + K 2 e - t 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 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 MODEL FOR t > 0. USE KVL di 1 di1 L 1 + 18i1 (t ) = 0 Þ (t ) + i1 (t ) = 0 dt 9 dt 1 t= s STEP 1 9 STEP 2 K1 = 0 L i1 (0-) = 12V = 1A 12W STEP 3 i1 (0-) = i1 (0+ ) = K1 + K 2 Þ K 2 = 1[ A] FOR INITIAL CONDITIONS ONE NEEDS ANS : i1(t ) = e INDUCTOR CURRENT FOR t<0 - t 1 9 [ A] = e -9 t [ A], 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 VTH - b Representation of an arbitrary circuit with one storage element RTH a VTH + iR C ic + RTH a KCL@ node a ic + iR = 0 Inductor or Capacitor + + + vR - Use KVL + vL vR + vL = vTH vR = RTH iL diL vL = L dt L VTH dvC ic = C iL dt b b vC - vTH Case 1.1 Case 1.2 iR = Voltage across capacitor Current through inductor RTH diL L + RTH iL = vTH dvC vC - vTH dt C + =0 dt RTH æ L ö diL v dvC ÷ ç + iL = TH = iSC RTH C + vC = vTH ç R ÷ dt RTH dt è TH ø vc _ Thevenin for t>0 at inductor terminals EXAMPLE Find iO (t); t > 0 6W 6W + 24V - t =0 iO (t ) 6W 3H 6W 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 vTH + iO = RTH dt RTH L 3H t= = = 0. 3 s RTH 10W 0.3 And the solution is of the form - t t iO (t ) = K1 + K 2e ; t > 0 diO + iO = 0 ; t > 0 dt t æ K 2 - 0t.3 ö 0.3ç e ÷ + K1 + K 2e 0.3 = 0 ø è 0.3 K1 = 0 Þ iO (t ) = K 2e Next: Initial Condition - t 0.3 ;t >0 Determine iO ( 0 + ). Use steady state assumption and continuity of 6W + 24V - t <0 iC (0+ ) = i3 6(i3 - i1 ) + 6(i3 - i2 ) = 0 Circuit for t<0 iO (t ) 6W Loop analysis - 24 + 6(i2 - i1 ) + 6(i2 - i3 ) = 0 inductor current 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 32 iO (t ) = e 6 - t 0.3 ; 32 6 t >0 EXAMPLE 12V Find iO (t), t > 0 100m F 6k iO (t ) + - 6k 6k + vC t =0 6k EXAMPLE Find iO (t), t > 0 100m F 6k + - t =0 6k For t > 0 iO = 6k RTH = 6k || 6k = 3k + vC - iO (t ) 12V vTH = 6V vC 6k 6k Hence, if the capacitor voltage is known the problem is solved dvC RTH C + vC = vTH dt 12V a iO (t ) + - 6k + vTH - Model for v C dvC 0.3 + vC = 6 dt vC = K 1 + K 2 e - t 0 .3 t æ K2 - t ö 1.5ç e 1.5 ÷ + K1 + K 2e 0.3 = 6 ç 1.5 ÷ è ø Model for v_c 6k t = 3 *103 W *100 *10 -6 F = 0.3s 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 12V circuit in steady state before the switching 6k 6k + vC (0-) iO (t ) + - 6k t <0 6k 12V 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 ...
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