ECE53A RLC Circuits W. Ku 11/29/2007 First Order Linear Time Invariant Circuits Second Order LTI Circuits State Equations KCL, KVL, Branch Equations Generic Second-Order Characteristic Equation Appendix: S Domain and Time Domain Solutions Homework Examples Delay of RC and RLC Lines

For n = 1, the state-variable description is given by First Order Linear Time Invariant (LTI) Systems or Circuits ECE 53A Fall 2007 For n = 1, the state-variable description is given by The first-order differential equation (1) can be solved: Homogeneous Eq.:

Second Order LTI Systems or Circuits ECE 53A Fall 2007 Assume we have a single-input single-output system A is a 2x2 constant matrix, b is a constant 2-vector b = [b1 b2]T, c and d are constant. I.C. needed is x(0) = [x1(0), x2(0)]T Using the second-order RLC network studied previously in example and the 2x2 state matrix is given by:

Second Order LTI Systems or Circuits ECE 53A Fall 2007 Example: Analysis of a second-order RLC network using (a) Loop, (b) Nodal, and (c) state-variable for the second-order RLC network shown in Fig. E-1, set up differential equations in each case For loop analysis, we use KVL for the two loops shown For loop i1 For loop i2 From (1) and (2) Fig. E-1 To solve for the loop currents i1(t) and i2(t)

Second Order RLC Networks ECE 53A Fall 2007 We have studied the first-order RC-networks (and their dual RL networks), we now consider the second-order RLC-networks containing two energy storage elements, L and C. Consider the following RLC network The series RLC network in figure (4-1) has a single loop with loop current i(t) as shown in above figure and the KVL (loop or mesh) equation is given by Since and Fig. (4-1)

Second Order RLC Networks ECE 53A Fall 2007 Substituting these relationships to (1), we have Differentiate (13), we obtain Equation (5) is a second-order differential equation with constant coefficients. This is due to the fact that the circuit in Fig. 4-1 is a linear time-invariant (LTI) system with lumped circuit elements. The homogeneous part of the differential equation is And the characteristic equation is This characteristic equation is quadratic and has two roots.

Second Order RLC Networks ECE 53A Fall 2007 Before we study the nature of the roots we introduce the damping factor ³ and undamped natural frequency (in rad/sec) !n The generic second-order characteristic equation is Relating ³ and !n with the circuit elements R, L, and C by combining (7) and (8), From (10), and coupled with (9),

Second Order RLC Networks ECE 53A Fall 2007 Eq. (11) states that !n is determined by the LC product. In fact, for network analysis and system design, we can have two normalizations; one is for impedance and the other is for frequency. With impedance normalization, the circuit element value can be normalized, e.g., let R=1. With Frequency Normalization, we can normalize !n to 1rad/sec,. These normalizations simplify the analysis and design of circuits. This is especially convenient for the design and synthesis of circuits from specifications. With this in mind let !n = 1 rad/sec, the characteristic equation (8) reduces to The roots of this simplified quadratic equation are given by:

Second Order RLC Networks ECE 53A Fall 2007 The location of these zeros follow a locus, known as root locus, as shown in Fig. 4-2 j! ¾ Fig. 4-2

Appendix: State Variable Formulation for the General Linear Circuits and System ECE 53A Fall 2007 General nth-order LTI Circuits and System described by the state-variable formulation: For LTI systems, A, B, C and D are constant matrices State vector x = [x1, x2, …, xn]T Input vector u = [u1, u2, …, um]T Output vector y = [y1, y2, …, yp]T First-order LTI system, n=1 Second-Order LTI System, n=2 (assume single input single output)

Solution of First Order State Equation in Transform Domain ECE 53A Fall 2007 Laplace Transform of (3): Where From (6) Given the input u(t) and I.C. x(0), x(t) can be obtained by using the inverse Laplace Transform of (7). Let u(t)=δ(t), U(s)=1 and assume x(0)=0, we have

Solution of Second Order State Equation in Transform Domain ECE 53A Fall 2007 For n=2, x=[x1, x2]T and Assume a single-input single-output LTI System, u and y are scalars and The Laplace Transform of (12) and (13) are given by or Where In = n x n identity matrix

Solution of General nth Order State Equation in Transform Domain ECE 53A Fall 2007

ECE 53A Homework Problem P - 1 Fall 2007 (a) Given the resistive network shown in Figure P-1, find voltages and currents of all circuit elements (R’s, voltage and current sources), vk and ik k=1,2,…6. Use KCL and KVL equations based on trees and co-trees shown in Figure P-2. (b) Find the power entering (or leaving) of all circuit elements and show that the conservation of power is satisfied. Is4=1A R1=1 R2=10 Vs1=1V Vs2=1V Is3=2A Figure P-1. A resistive network with 2 independent voltages sources and 2 independent current sources. Figure P-2.Trees and co-trees for resistive network in Figure P-1. Tree: solid lines, Cuts: dotted lines

Homework Problem P - 2 ECE 53A Fall 2007 Is3 R1 R2 I3 ic Vc Vs1 Vs2 i1 For the passive RC network shown below, (a) Write the loop KVL equations. Note that there are three loops, but one of loop current is known. It is constrained by the ideal independent current source Is3. The unknowns are loop current i1 and i2 . (b) Write the KCL nodal equation shown in Figure E2-2, choose tree branches: #1, #2 and #3, loops: #4, #5 and #6.n=4, n-1=3. At node 1 and 2,V1=Vs1, V3=-Vs2 are known so there is only one unknown node voltage: Vc at node 2. Therefore, for this passive circuit, nodal or KCL method is preferred. (c) Let Vs1=1V, Vs2=1V, Is3=1A, R1=1, R2=R, solve for Vc(t), assume Vc(0)=0. #6 Is3 V2 R1 R2 #4 2 #5 V3 I3 1 3 V1 ic Vc Vs1 Vs2 i1 i2 #1 #2 #3 C 4 V4 Figure E2-1. A resistive network with 2 independent voltages sources and 2 independent current sources. Figure P-2.Trees and co-trees for resistive network in Figure P-1. Tree: solid lines, Cuts: dotted lines

ECE 53A Example 2 - 1 Fall 2007 For the passive RC network shown in Figure E2-1, (a) What is the order of this network? (b) Write the loop (KVL) equations. (c) Let Vs1=1V, Vs2=1V, is3=2A, Rs1=Rs2=0, R1=1, R2=R, from the results in part (b), solve for the i1(t) and i2(t). What is the time constant for this network? (d) Write the nodal (KCL) equations and solve for v1(t) and v2(t). Set the parameters same as given in part (c), solve for v1(t), v2(t) and v(t). (e) Write the state equation for this network, and solve for X(t) = vc(t). Since v(t)=vc(t), your result should check that obtained in part (d). Is3 Is3 Rs1 R1 R2 Rs2 V V1 V2 Vs1 Vs2 i1 i2 Figure E2-1.Trees and co-trees for resistive network in Figure P-1. Tree: solid lines, Cuts: dotted lines

Example 2 - 1 ECE 53A Last problem of FINAL (20%) Fall 2007 Last problem of FINAL (20%) (15%)(a) In this RC network, let R1=R2=1, and R3=R, find Vc(t). Note that since there is one energy storage element C in this circuit, it is a first-order RC network and you should get a first-order differential equation. (5%)(b) Assuming that the initial condition is given by vc(0)=0, and the two constant voltage sources vs1 and vs2 are applied at t=0, find vc(t), t¸0. R3 R1 R2 Vc(t) ic + V2 Vs1=1V Vc Vs2=1V - Figure E2-1.Trees and co-trees for resistive network in Figure P-1. Tree: solid lines, Cuts: dotted lines

Example 2 - 1 ECE 53A R1, R2, R3 are links. Fall 2007 R1, R2, R3 are links. 1 3 2 KCL at node 3 (vc)  first order differential equation. 4 Vs1 and Vs2 are constants. Assume Vc(0)=0

Gate Delay and RC Circuits ECE 53A Fall 2007 Delays in Logic Circuits Transition time in CMOS Delays in connected CMOS inverters and CMOS logic gates Even simple models of gate switching will provide reasonable estimate of system limitations

Series RLC Circuit Model of Cascaded Inverter Gates ECE 53A Fall 2007 Pull-Up: Pull-Down: Loop Eq (KVL):

Series RLC Circuit Model of Cascaded Inverter Gates ECE 53A Fall 2007 A b State Eq. (n=2) ==> 2 coupled 1st-order differential Eq. Eigenvalues of A:

Equivalent Circuit For the Cascaded Inverter Pair, Using a Two Lumped Interconnect Model ECE 53A Fall 2007 State Variables: x1=vc1 x2=vc2 Homogeneous Eq.: s-plane s = ¾+j! j! Natural Frequencies δ1, δ2 >0 -¾2 -¾1 ¾

Voltage vC(t) at the Input of the load inverter in Example 6.2: ECE 53A Fall 2007 Voltage vC(t) at the Input of the load inverter in Example 6.2: Note that the coupled inverters undergo a pull-down transition at t=0 and are subsequently pulled up at t=0.3ns. Note also that the voltage vc(t) in the RLC model continues to fall for a period after the second switching event; this is due to the presence of the inductor, which attempts to force the current to flow in the same direction.