1. 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

2. 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.:

3. 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:

4. 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)

5. 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)

6. 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.

7. 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),

8. 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:

9. 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!
Fig. 4-2

10. 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)

11. 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

12. 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

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

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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

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

21. 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:

22. 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

23. 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.