Natural and Step Response of Series & Parallel RLC Circuits (Second
Total Page:16
File Type:pdf, Size:1020Kb
Natural and Step Response of Series & Parallel RLC Circuits (Second-order Circuits) Objectives: Determine the response form of the circuit Natural response parallel RLC circuits Natural response series RLC circuits Step response of parallel and series RLC circuits Natural Response of Parallel RLC Circuits The problem – given initial energy stored in the inductor and/or capacitor, find v(t) for t ≥ 0. It is convenient to calculate v(t) for this circuit because A. The voltage must be continuous for all time B. The voltage is the same for all three components C. Once we have the voltage, it is pretty easy to calculate the branch current D. All of the above Natural Response of Parallel RLC Circuits The problem – given initial energy stored in the inductor and/or capacitor, find v(t) for t ≥ 0. dv(t) 1 t v(t) C v(x)dx I 0 KCL : 0 dt L 0 R d 2v(t) 1 1 dv(t) Differentiate both sides to remove the integral : C v(t) 0 dt2 L R dt d 2v(t) 1 1 dv(t) Divide both sides by C to place in standard form: v(t) 0 dt2 LC RC dt Natural Response of Parallel RLC Circuits The problem – given initial energy stored in the inductor and/or capacitor, find v(t) for t ≥ 0. d 2v(t) 1 1 dv(t) Describing equation: v(t) 0 dt2 LC RC dt This equation is Second order Homogeneous Ordinary differential equation With constant coefficients Once again we want to pick a possible solution to this differential equation. This must be a function whose first AND second derivatives have the same form as the original function, so a possible candidate is A. Ksin t B. Keat C. Kt2 Natural Response of Parallel RLC Circuits The problem – given initial energy stored in the inductor and/or capacitor, find v(t) for t ≥ 0. d 2v(t) 1 1 dv(t) Describing equation: v(t) 0 dt2 LC RC dt The circuit has two initial conditions that must be satisfied, so the solution for v(t) must have two constants. Use s1t s2t v(t) A1e A2e V; Substitute : 1 1 (s2 A es1t s2 A es2t ) (s A es1t s A es2t ) (A es1t A es2t ) 0 1 1 2 2 RC 1 1 2 2 LC 1 2 2 s1t 2 s2t [s1 (1 RC)s1 (1 LC)]A1e [s2 (1 RC)s2 (1 LC)]A2e 0 Natural Response of Parallel RLC Circuits The problem – given initial energy stored in the inductor and/or capacitor, find v(t) for t ≥ 0. d 2v(t) 1 1 dv(t) Describing equation: v(t) 0 dt2 LC RC dt s1t s2t Solution: v(t) A1e A2e Where s1 and s2 are solutions for the CHARACTERISTIC EQUATION: s2 (1 RC)s (1 LC) 0 s2 (1 RC)s (1 LC) 0 is called the “characteristic equation” because it characterizes the circuit. A. True B. False Natural Response of Parallel RLC Circuits The problem – given initial energy stored in the inductor and/or capacitor, find v(t) for t ≥ 0. The two solutions to the characteristic equation can be calculated using the quadratic formula: (1 RC) (1 RC)2 4(1 LC) s2 (1 RC)s (1 LC) 0; s 1,2 2 2 2 2 s1,2 (1 2RC) (1 2RC) (1 LC) 0 1 where (the neper frequency in rad/s) 2RC 1 and (the resonant radian frequency in rad/s) 0 LC So far, we know that the parallel RLC natural response is given by s1t s2t v(t) A1e A2e 1 1 s 2 2 where and 1,2 0 2RC 0 LC There are three different forms for s1 and s2. For a parallel RLC circuit with specific values of R, L and C, the form for s1 and s2 depends on A. The value of B. The value of 0 2 2 C. The value of ( - 0 ) Natural Response – Overdamped Example Given V0 = 12 V and I0 = 30 mA, find v(t) for t ≥ 0. 1 1 12,500 rad/s 2RC 2(200)(0.2) 1 1 10,000 rad/s 0 LC (0.05)(0.2) 2 2 o so this is the overdamped case! 2 2 2 2 s1,2 0 12,500 (12,500) (10,000) 12,000 7500 s1 5000 rad/s, s2 20,000 rad/s Natural Response – Overdamped Example Given V0 = 12 V and I0 = 30 mA, find v(t) for t ≥ 0. 5000t 20,000t v(t) A1e A2e V, for t 0 Now we must use the coefficients in the equation to satisfy the initial conditions in the circuit : v(t) v(t) t0 in the equation t0 in the circuit dv(t) dv(t) in the equation in the circuit dt t0 dt t0 Natural Response – Overdamped Example Given V0 = 12 V and I0 = 30 mA, find v(t) for t ≥ 0. 5000(0) 20,000(0) Equation: v(0) A1e A2e A1 A2 Circuit : v(0) V0 12 V A1 A2 12 dv(0) Equation: (5000)A e5000(0) (20,000)A e20,000(0) dt 1 2 5000A1 20,000A2 Now we need the initial value of the first derivative of the voltage from the circuit. The describing equation of which circuit component involves dv(t)/dt? A. The resistor B. The inductor C. The capacitor Natural Response – Overdamped Example Given V0 = 12 V and I0 = 30 mA, find v(t) for t ≥ 0. dv(0) Equation: (5000)Ae5000(0) (20,000)A e20,000(0) dt 1 2 5000A1 20,000A2 dv(t) dv(0) 1 1 Circuit : i (t) C i (0) (i (0) i (0)) C dt dt C C C L R dv(0) 1 v(0) 1 12 iL (0) 0.03 450,000 V/s dt C R 0.2 200 5000A1 20,000A2 450,000 Natural Response – Overdamped Example Given V0 = 12 V and I0 = 30 mA, find v(t) for t ≥ 0. 5000t 20,000t v(t) A1e A2e V, for t 0 A1 A2 12; 5000A1 20,000A2 450,000 Solving simultaneously, A1 14, A2 26 Thus, v(t) 14e5000t 26e20,000t V, for t 0 Checks: v(0) 14 26 12 V (OK) v() 0 (OK) Natural Response – Overdamped Example Given V0 = 12 V and I0 = 30 mA, find v(t) for t ≥ 0. You can solve this problem using the Second-Order Circuits table: 1. Make sure you are on the Natural Response side. 2. Find the parallel RLC column. 3. Use the equations in Row 4 to calculate and 0. 4. Compare the values of and 0 to determine the response form (given in one of the last 3 rows). 5. Use the equations to solve for the unknown coefficients. 6. Write the equation for v(t), t ≥ 0. 7. Solve for any other quantities requested in the problem. Natural Response – Overdamped Example Given V0 = 12 V and I0 = 30 mA, find v(t) for t ≥ 0. You can solve this problem using the Second-Order Circuits table: 1. Make sure you are on the Natural Response side. 2. Find the parallel RLC column. 3. Use the equations in Row 4 to calculate and 0. 4. Compare the values of and 0 to determine the response form (given in one of the last 3 rows). 5. Use the equations to solve for the unknown coefficients. 6. Write the equation for v(t), t ≥ 0. 7. Solve for any other quantities requested in the problem. The values of the __________ determine whether the response is overdamped, underdamped, or critically damped A. Initial conditions B. R, L, and C components C. Independent sources Natural Response of Parallel RLC Circuits The problem – given initial energy stored in the inductor and/or capacitor, find v(t) for t ≥ 0. Recap: 1 1 s2 (1 RC)s (1 LC) 0; s 2 2 ; ; 1,2 0 2RC 0 LC 2 2 s1t s2t 0 : overdamped, so v(t) A1e A2e 2 2 2 2 0 : underdamped, so s1,2 jd where d 0 ( jd )t ( jd )t t jd t t jd t v(t) A1e A2e A1e e A2e e Note Euler's identity : e jx cos x jsin x; e jx cos x jsin x t t v(t) A1e (cosd t jsin d t) A2e (cosd t jsin d t) t t e cosd t(A1 A2 ) e sin d t( jA1 jA2 ) t t v(t) B1e cosd t B2e sin d t When the response is underdamped, the voltage is given by the equation t t v(t) B1e cosd t B2e sindt In this equation, the coefficients B1 and B2 are A. Real numbers B. Imaginary numbers C. Complex conjugate numbers Natural Response – Underdamped Example Given V0 = 0 V and I0 = 12.25 mA, find v(t) for t ≥ 0. 1 1 200 rad/s 2RC 2(20,000)(0.125) 1 1 1000 rad/s 0 LC (8)(0.125) 2 2 o so this is the underdamped case! 2 2 2 2 d 0 (1000) (200) 979.8 rad/s t t v(t) B1e cosd t B2e sin d t 200t 200t B1e cos979.8t B2e sin 979.8t V, t 0 Now we evaluate v(0) and dv(0)/dt from the equation for v(t), and set those values equal to v(0) and dv(0)/dt from the circuit, solving for B1 and B2. The values for v(0) and dv(0)/dt from the circuit do not depend on whether the response is overdamped, underdamped, or critically damped.