Review Capacitors Inductors Voltage current relationship Stored Energy 1st Order Circuits RL RC circuits Steady State Transient response Natural Step response EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu 1 Lecture 5 OUTLINE Chap 4 RC and RL Circuits with General Sources Particular and complementary solutions Time constant Second Order Circuits The differential equation Particular and complementary solutions The natural frequency and the damping ratio Chap 5 Types of Circuit Excitation Why Sinusoidal Excitation Phasors Complex Impedances Reading Chap 4 Chap 5 skip 5 7 EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu 2 1 First Order Circuits vr t R vs t C iL t ic t vc t is t R L vL t KVL around the loop vr t vc t vs t KCL at the node v t 1 R L dv t RC c vc t v s t dt EE40 Summer 2005 Lecture 5 t v x dx i t s L diL t i L t is t R dt Instructor Octavian Florescu 3 Complete Solution Voltages and currents in a 1st order circuit satisfy a differential equation of the form x t dx t f t dt f t is called the forcing function The complete solution is the sum of particular solution forced response and complementary solution natural response x t x p t xc t Particular solution satisfies the forcing function Complementary solution is used to satisfy the initial conditions The initial conditions determine the value of K x p t dx p t dt dxc t 0 dt xc t Ke t xc t f t EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu Homogeneous equation 4 2 The Time Constant The complementary solution for any 1st order circuit is xc t Ke t For an RC circuit RC For an RL circuit L R EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu 5 What Does Xc t Look Like xc t e t 10 4 is the amount of time necessary for an exponential to decay to 36 7 of its initial value 1 is the initial slope of an exponential with an initial value of 1 EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu 6 3 The Particular Solution The particular solution xp t is usually a weighted sum of f t and its first derivative If f t is constant then xp t is constant If f t is sinusoidal then xp t is sinusoidal EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu 7 2nd Order Circuits Any circuit with a single capacitor a single inductor an arbitrary number of sources and an arbitrary number of resistors is a circuit of order 2 Any voltage or current in such a circuit is the solution to a 2nd order differential equation EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu 8 4 A 2nd Order RLC Circuit i t vs t R C L Application Filters A bandpass filter such as the IF amp for the AM radio A lowpass filter with a sharper cutoff than can be obtained with an RC circuit EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu The Differential Equation vs t 9 i t vr t R vc t C vl t KVL around the loop vr t vc t vl t vs t L t 1 di t Ri t i x dx L vs t C dt R di t 1 d 2i t 1 dvs t i t L dt LC dt 2 L dt EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu 10 5 The Differential Equation The voltage and current in a second order circuit is the solution to a differential equation of the following form d 2 x t dx t 2 02 x t f t 2 dt dt x t x p t xc t Xp t is the particular solution forced response and Xc t is the complementary solution natural response EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu 11 The Particular Solution The particular solution xp t is usually a weighted sum of f t and its first and second derivatives If f t is constant then xp t is constant If f t is sinusoidal then xp t is sinusoidal EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu 12 6 The Complementary Solution The complementary solution has the following form x t Ke st c K is a constant determined by initial conditions s is a constant determined by the coefficients of the differential equation st d 2 Ke st dKe 2 02 Ke st 0 2 dt dt s 2 Ke st 2 sKe st 02 Ke st 0 s 2 2 s 02 0 EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu 13 Characteristic Equation To find the complementary solution we need to solve the characteristic equation s 2 2 0 s 02 0 0 The characteristic equation has two rootscall them s1 and s2 xc t K 1e s1t K 2 e s2 t s1 0 0 2 1 s2 0 0 2 1 EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu 14 7 Damping Ratio and Natural Frequency s1 0 0 2 1 0 s2 0 0 2 1 damping ratio The damping ratio determines what type of solution we will get Exponentially decreasing 1 Exponentially decreasing sinusoid 1 The natural frequency is 0 It determines how fast sinusoids wiggle EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu 15 Overdamped Real Unequal Roots If 1 s1 and s2 are real and not equal i t ic t K 1e 2 1 t 0 0 K 2e 1 0 8 0 8 0 6 0 6 0 4 i t 0 4 2 1 t 0 0 0 2 0 2 0 1 00E 06 0 2 0 1 00E 06 t EE40 Summer 2005 Lecture 5 t Instructor Octavian Florescu 16 8 Underdamped Complex Roots If 1 s1 and s2 are complex Define the following constants 0 d 0 1 2 xc t e t A1 cos d t A2 sin d t 1 0 8 0 6 i t 0 4 0 2 0 1 00E 05 0 2 1 00E 05 3 00E 05 0 4 0 6 0 8 1 t EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu 17 Critically damped Real Equal Roots If 1 s1 and s2 are real and equal xc t K1e 0t K 2te 0t EE40 Summer 2005 Lecture 5 Instructor Octavian Florescu 18 9 Example For the example what are and 0 i t d 2i t R di t 1 1 dvs t i t 2 dt L dt LC L dt 10 d 2 xc t dx t 2 0 c 02 xc t 0 2 dt dt 769pF 159 H 02 EE40 Summer 2005 Lecture 5 1 R R 2 0 LC L 2 Instructor Octavian Florescu C L 19 Example 0 011 0 2 455000 Is this system over damped under damped or critically damped What will the current look like 1 0 8 0 6 i t 0 4 0 2 0 1 00E 05 0 2 1 00E 05 3 00E 05 …
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