1. CSE 245: Computer Aided Circuit Simulation and Verification Instructor: Prof. Chung-Kuan Cheng Winter 2003 Lecture 2: Closed Form Solutions (Linear System)

2. Feb. 22 2003Cheng & Peng @ UCSDLecture2.2 Outline  Time Domain Analysis State Equations and Order of RLC network RLC Network Analysis Response in time domain  Frequency Domain Analysis From time domain to Frequency domain Correspondence between time domain and frequency domain Serial expansion of (sI-A) -1

3. Feb. 22 2003Cheng & Peng @ UCSDLecture2.3 State of a system  The state of a system is a set of data, the value of which at any time t, together with the input to the system at time t, determine uniquely the value of any network variable at time t.  We can express the state in vector form x = Where x i (t) is the state variables of the system

4. Feb. 22 2003Cheng & Peng @ UCSDLecture2.4 State Variable  How to Choose State Variable? The knowledge of the instantaneous values of all branch currents and voltages determines this instantaneous state But NOT ALL these values are required in order to determine the instantaneous state, some can be derived from others. choose capacitor voltages and inductor currents as the state variables! But not all of them are chose

5. Feb. 22 2003Cheng & Peng @ UCSDLecture2.5 Degenerate Network  A network that has a cut-set composed only of inductors and/or current sources or a loop that contains only of capacitors and/or voltage sources is called a degenerate network  Example: The following network is a degenerate network since C 1, C 2 and C 5 form a degenerate capacitor loop

6. Feb. 22 2003Cheng & Peng @ UCSDLecture2.6 Degenerate Network  In a degenerated network, not all the capacitors and inductors can be chose as state variables since there are some redundancy  On the other hand, we choose all the capacitor voltages and inductors currents as state variable in a nondegenerate network  We will give an example of how to choose state variable in the following section

7. Feb. 22 2003Cheng & Peng @ UCSDLecture2.7 Order of Circuit  n = b LC – n C - n L  n the order of circuit, total number of independent state variables  b LC total number of capacitors and inductors in the network  n C number of degenerate loops (C-E loops)  n L number of degenerate cut-sets (L-J cut-sets)  n = 4 – 1 = 3  In a nondegenerate network, n equals to the total number of energy storage elements

8. Feb. 22 2003Cheng & Peng @ UCSDLecture2.8 State Equations  State Equation  Output Equation  State Equation together with Output Equation are called the state equations of the network = Qx(t) + Du(t) = Ax(t) + Bu(t)

9. Feb. 22 2003Cheng & Peng @ UCSDLecture2.9 Outline  Time Domain Analysis State Equations and Order of RLC network RLC Network Analysis Response in time domain  Frequency Domain Analysis From time domain to Frequency domain Correspondence between time domain and frequency domain Serial expansion of (sI-A) -1

10. Feb. 22 2003Cheng & Peng @ UCSDLecture2.10 RLC Network Analysis  A given RLC network  Degenerate Network, Choose only voltages of C 1 and C 5, current of L 6 as our state variable

11. Feb. 22 2003Cheng & Peng @ UCSDLecture2.11 Tree Structure  Take into tree as many capacitors as possible and,  as less inductors as possible  Resistors can be chose as either tree branches or co-tree branches

12. Feb. 22 2003Cheng & Peng @ UCSDLecture2.12 Linear State Equation  By a mixed cut-set and mesh analysis, consider capacitor cut-sets and inductor loops only. we can write the linear state equation as follows M = Gx(t) + Pu(t) Cut-set KCL Loop KVL Cut-set KCL =- +Vs

13. Feb. 22 2003Cheng & Peng @ UCSDLecture2.13 General Form of the State Equation  The state equation is of the form  Or  v t : voltage in the trunk, capacitor voltage  i l : current in the loop, inductor current.  Y and R are the admittance matrix and impedance matrix of cut-set and mesh  E covers the co-tree branches in the cut-set  –E T covers the tree trunks in the mesh analysis = -+ Pu = Gx(t) + Pu(t) M

14. Feb. 22 2003Cheng & Peng @ UCSDLecture2.14 State Equations  If we shift the matrix M to the right hand side, we have  Let A = M -1 G and B = M -1 P, we have the state equation  Together with the output equation  are called the State Equations of the linear system = Gx(t) + Pu(t) M = M -1 Gx(t) + M -1 Pu(t) = Ax(t) + Bu(t) = Qx(t) + Du(t)

15. Feb. 22 2003Cheng & Peng @ UCSDLecture2.15 Outline  Time Domain Analysis State Equations and Order of RLC network RLC Network Analysis Response in time domain  Frequency Domain Analysis From time domain to Frequency domain Correspondence between time domain and frequency domain Serial expansion of (sI-A) -1

16. Feb. 22 2003Cheng & Peng @ UCSDLecture2.16 Response in time domain  We can solve the state equation and get the closed form expression  The output equation can be expressed as Note: * denotes convolution

17. Feb. 22 2003Cheng & Peng @ UCSDLecture2.17 Impulse Response  The Impulse Response of a system is defined as the Zero State Response resulting from an impulse excitation  Thus, in the output equation, replace u(t) by the impulse function  (t), and let x(t 0 )=0 we have h(t) = y(t) = Qe At B

18. Feb. 22 2003Cheng & Peng @ UCSDLecture2.18 Outline  Time Domain Analysis State Equations and Order of RLC network RLC Network Analysis Response in time domain  Frequency Domain Analysis From time domain to Frequency domain Correspondence between time domain and frequency domain Serial expansion of (sI-A) -1

19. Feb. 22 2003Cheng & Peng @ UCSDLecture2.19 From time domain to frequency domain  Laplace Transformation = Ax(t) + Bu(t) = Qx(t) + Du(t) Laplace Transform sx(s) – x(t 0 )= Ax(s) +Bu(s) y(s) = Qx(s) +Du(s) State Equations in S domain State Equations in time Domain

20. Feb. 22 2003Cheng & Peng @ UCSDLecture2.20 Solutions in S domain  By solving the state equation in s domain, we have  Suppose the network has zero state and the output vector depends only on the state vector x, that is, x(t 0 ) = 0 and D = 0, we can derive the transfer function of the network H(s) = = Q(sI-A) -1 B x(s) = (sI-A) -1 x(t 0 )+ (sI-A) -1 Bu(s) y(s) = Qx(s) +Du(s) = Q(sI-A) -1 (x(t 0 ) + Bu(s)) +Du(s)

21. Feb. 22 2003Cheng & Peng @ UCSDLecture2.21 Outline  Time Domain Analysis State Equations and Order of RLC network RLC Network Analysis Response in time domain  Frequency Domain Analysis From time domain to Frequency domain Correspondence between time domain and frequency domain Serial expansion of (sI-A) -1

22. Feb. 22 2003Cheng & Peng @ UCSDLecture2.22 Correspondence between time domain and frequency domain  We can derive the time domain solutions of the network from the s domain solutions by inverse Laplace Transformation of the s domain solutions. State Equations in S domain State Equations in time Domain Inverse Laplace Transform sx(s) – x(t 0 )= Ax(s) +Bu(s) y(s) = Qx(s) +Du(s) x(t) = L -1 [(sI-A) -1 x(t 0 ) + (sI-A) -1 Bu(s)] = L -1 [(sI-A) -1 ]x(t 0 ) + L -1 [(sI-A) -1 ]B*u(t) y(t) = L -1 [Q(sI-A) -1 (x(t 0 ) + Bu(s)) +Du(s)] = Q L -1 [(sI-A) -1 ] x(t 0 ) + {QL -1 [(sI-A) -1 ]B +D  (t)}* u(s)

23. Feb. 22 2003Cheng & Peng @ UCSDLecture2.23 Correspondence between time domain and frequency domain  (sI-A) -1 e At  multiplication of u(s) in s domain corresponds to the convolution in time domain Solution from time domain analysis Solution by inverse Laplace transform x(t) = L -1 [(sI-A) -1 x(t 0 ) + (sI-A) -1 Bu(s)] = L -1 [(sI-A) -1 ]x(t 0 ) + L -1 [(sI-A) -1 ]B*u(t) y(t) = L -1 [Q(sI-A) -1 (x(t 0 ) + Bu(s)) +Du(s)] = Q L -1 [(sI-A) -1 ] x(t 0 ) + {QL -1 [(sI-A) -1 ]B +D  (t)}* u(s)

24. Feb. 22 2003Cheng & Peng @ UCSDLecture2.24 Outline  Time Domain Analysis State Equations and Order of RLC network RLC Network Analysis Response in time domain  Frequency Domain Analysis From time domain to Frequency domain Correspondence between time domain and frequency domain Serial expansion of (sI-A) -1

25. Feb. 22 2003Cheng & Peng @ UCSDLecture2.25 Serial expansion of (sI-A) -1  When s  0 we can write (sI-A) -1 as  Thus, the transfer function can be wrote as  When s   we can write (sI-A) -1 as  The transfer function can be wrote as (sI-A) -1 = -A -1 (I – SA -1 ) = -A -1 (I + SA -1 + S 2 A -2 + … + S k A -k + … ) H(s) = Q(sI-A) -1 B = -QA -1 (I + SA -1 + S 2 A -2 + … + S k A -k + … )B (sI-A) -1 = S -1 (I – S -1 A) -1 = S -1 (I + S -1 A + S -2 A 2 + … + S -k A k + … ) H(s) = Q(sI-A) -1 B = S -1 (I + S -1 A + S -2 A 2 + … + S -k A k + … )B

26. Feb. 22 2003Cheng & Peng @ UCSDLecture2.26  Assume A has non-degenerate eigenvalues and corresponding linearly independent eigenvectors, then A can be decomposed as where and Matrix Decomposition

27. Feb. 22 2003Cheng & Peng @ UCSDLecture2.27 Matrix Decomposition  Then we can write (sI-A) -1 in the following form  (sI-A) -1 in s domain corresponds to the exponential function e At in time domain, we can write e At as (sI-A) -1 = (SI – X -1  X) ‑ 1 = X -1 (SI –  ) -1 X = X -1 X e At = X -1 X