1. 1 Lecture #1 EGR 272 – Circuit Theory II Reading Assignment: Chapter 8 in Electric Circuits, 6th Edition by Nilsson Welcome to EGR 272 Circuit Theory II Syllabus Homework Web page Office hours EGR 271: Chapters 1 - 7 in Electric Circuits, 6th Edition by Nilsson EGR 272: Chapters 8 - 15 in Electric Circuits, 6th Edition by Nilsson

2. 2 Lecture #1 EGR 272 – Circuit Theory II Review - Key relationships for capacitors and inductors CapacitorsInductors Current Voltage Powerp(t) = v(t)i(t)p(t) = v(t)i(t) Energy Cannot changev C (0 + ) = v C (0 - )i L (0 + ) = i L (0 - ) instantaneously Steady-state behavioropen-circuitshort-circuit

3. 3 Lecture #1 EGR 272 – Circuit Theory II Chapter 8 – Second-Order Circuits Order of a circuit Order of the differential equation (DE) required to describe the circuit The number of independent* energy storage elements (C’s and L’s) == * C’s and L’s are independent if they cannot be combined with other C’s and L’s (in series or parallel, for example) Example: Draw several circuits with C’s and L’s and identify the order of each circuit.

4. 4 Lecture #1 EGR 272 – Circuit Theory II 2 nd -order circuits have 2 independent energy storage elements (inductors and/or capacitors) Analysis of a 2 nd -order circuit yields a 2 nd -order differential equation (DE) A 2nd-order differential equation has the form: Solution of a 2 nd -order differential equation requires two initial conditions: x(0) and x’(0) All higher order circuits (3 rd, 4 th, etc) have the same types of responses as seen in 1 st - order and 2 nd -order circuits Since 2nd-order circuits have two energy-storage types, the circuits can have the following forms: 1) Two capacitors 2) Two inductors 3) One capacitor and one inductor A) Series RLC circuit * B) Parallel RLC circuit * C) Others * The textbook focuses on these two types of 2nd-order circuits

5. 5 Lecture #1 EGR 272 – Circuit Theory II Form of the solution to differential equations As seen with 1 st -order circuits in Chapter 7, the general solution to a differential equation has two parts: x(t) = x h + x p = homogeneous solution + particular solution or x(t) = x n + x f = natural solution + forced solution where x h or x n is due to the initial conditions in the circuit and x p or x f is due to the forcing functions (independent voltage and current sources for t > 0). The forced response The forced response is due to the independent sources in the circuit for t > 0. Since the natural response will die out once the circuit reaches steady-state, the forced response can be found by analyzing the circuit at t = . In particular, x f = x(  )

6. 6 Lecture #1 EGR 272 – Circuit Theory II The natural response A 2 nd -order differential equation has the form: where x(t) is a voltage v(t) or a current i(t). To find the natural response, set the forcing function f(t) (the right-hand side of the DE) to zero. Substituting the general form of the solution Ae st yields the characteristic equation: s 2 + a 1 s + a o = 0 Finding the roots of this quadratic (called the characteristic roots or natural frequencies) yields: The roots of the quadratic equation above may be real and distinct, repeated, or complex. Thus, the natural response to a 2 nd -order circuit has 3 possible forms:

7. 7 Lecture #1 EGR 272 – Circuit Theory II 1) Overdamped response Roots are real and distinct [ (a 1 ) 2 > 4a o ] Solution has the form: Sketch the form of the solution. Discuss the concept of the dominant root.

8. 8 Lecture #1 EGR 272 – Circuit Theory II 2) Critically damped response Roots are repeated [ (a 1 ) 2 = 4a o ] so s 1 = s 2 = s = -a 1 /2 Solution has the form: Sketch the form of the solution.

9. 9 Lecture #1 EGR 272 – Circuit Theory II 3) Underdamped response Roots are complex [ (a 1 ) 2 < 4a o ] so s 1, s 2 =   j  Show that the solution has the form: Sketch the form of the solution. Discuss the concept of the exponential envelope. Sketch x n if A 1 = 0, A 2 = 10,  =-1, and  = . Sketch x n if A 1 = 0, A 2 = 10,  =-10, and  = 100 .

10. 10 Lecture #1 EGR 272 – Circuit Theory II Illustration: The transient response to a 2 nd -order circuit must follow one of the forms indicated above (overdamped, critically damped, or underdamped). Consider the circuit shown below. v(t) is 0V for t < 0 and the steady-state value of v(t) is 10V. How does it get from 0 to 10V? Discuss the possible responses for v(t) Define the terms damping, rise time, ringing, and % overshoot

11. 11 Lecture #1 EGR 272 – Circuit Theory II Examples: When is each of the 3 types of responses desired? Discuss the following cases: An elevator A cruise-control circuit The output of a logic gate The start up voltage waveform for a DC power supply

12. 12 Lecture #1 EGR 272 – Circuit Theory II Procedure for analyzing 2 nd -order circuits 1.Write the differential equation for x(t). It is generally easiest to solve for inductor currents or capacitor voltages and then to use the result to find other variables. Note: If the circuit is a series RLC circuit or a parallel RLC circuit, standard formulas for  and w o can be used to find the characteristic equation: s 2 + 2  s + w o 2. 2.Find the initial conditions, x(0) and x’(0). A)Find x(0) by analyzing the circuit at t = 0 - (find all capacitor voltages and inductor currents) B)Find x’(0) by analyzing the circuit at t = 0 + (use the values for v C and i L found at t = 0 - in the circuit). 3.Solve the differential equation. A)Find the forced response, x f, by analyzing the circuit at t = . x f = x(  ). B)Find the natural response, x n 1)set the right side of the DE to zero 2)find the characteristic equation and the roots of the equation 3)assume that x n has the standard form of the solution for an overdamped, underdamped, or critically damped circuit. There will be two unknowns in this solution. C)Find the total response, x(t) = x n + x f. Use the two initial conditions to solve for the two unknowns in the total response. Next class: detailed circuit examples