DISCRETE-SIGNAL ANALYSIS AND DESIGN- P35 potx

156 DISCRETE-SIGNAL ANALYSIS AND DESIGN N := 8 n := 0,1 N V C := 1.5 IL := 1.0 R := 3 L := 1 C := 0.5 x(n) := V C IL if n = 0 T . + 0 1 L −1 C R L 1 0 0 1 x(n − 1) if n > 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0 0.3 0.6 0.9 1.2 1.5 V(C) I(L) x(n)0 x(n)1 n.T Solution to matrix differential equation for initial conditions of V C = 1.5, I L = 1. 0 C L R I L V C V L + + + − − − u(t) V OUT T := 0.1 Figure A-2 LCR Circuit differential equation solution for initial values of V C and I L , I gen =0. i C = C dv C dt = u − i L v L = L di L dt = v C − Ri L (A-3) v OUT = Ri L ADDITIONAL DISCRETE-SIGNAL ANALYSIS AND DESIGN INFORMATION 157 Rewrite Eq. (A-3) in state-variable format: v · C = 0v C − 1 C i L + 1 C u õ · L = 1 L v C − R L i L + 0u (A-4) v O = Ri L A nodal circuit analysis conÞrms these facts for this example. R, L, and C are constant values, but they can easily be time-varying and/or nonlinear functions of voltage and current. The discrete analysis method deals with all of this very nicely. We now add in the initial conditions at time zero, V C0 and I L0 : v · C = 0(v C + V C0 ) − 1 C (i L + I L0 ) + 1 C u õ · L = 1 L (v C + V C0 ) − R L (i L + I L0 ) +0u (A-5) v O = Ri L The two derivatives appear on the left side. Note that if (v C +V C0 )is multiplied by zero, the rate of change of v C does not depend on that term, and the rate of change of i L does not depend on u if the u is multiplied by zero. The options of Eqs. (A-4) and (A-5) can easily be imagined. Description of ßow-graph methods in [Dorf and Bishop, 2004, Chaps. 2 and 3] and in numerous other references are excellent tools that are commonly used for these problems. We will not be able to get deeply into that subject in this book, but Fig. A-4 is an example. The next step is to rewrite Eq. (A-5) in matrix format. Also, v C is now called X 1 ,andI L is now called X 2 .  ˙ X 1 ˙ X 2  =  0 −1 C 1 L −R L  X 1 X 2  +  1 C 0  (u) (A-6) V O = RX 2 158 DISCRETE-SIGNAL ANALYSIS AND DESIGN Now write the (A-6) equations as follows: ˙ X 1 = 0X 1 − 1 C X 2 + 1 C u ˙ X 2 = 1 L X 1 − R L X 2 + 0u (A-7) V O = RX 2 Next, we will solve Eq. (A-6) [same as Eq. (A-7) for X 1 ( =v c ), X 2 ( =i L ), and V O ]. Component u is the input signal generator. This general idea applies to a wide variety of practical problems (see [Dorf and Bishop, 2004, Chap. 3] and many other references). The meth- ods of matrix algebra and matrix calculus operations are found in many handbooks (e.g., [Zwillinger, 1996]). The general format for state-variable equations, similar to Eqs. (A-4) and (A-5), is ˙ x = Ax + Bu y = Cx + Du (A-8) in which A, B, C,andD are coefÞcient matrices whose numbers may be complex and varying in some manner with time, voltage, or current, u pertains to complex signal sources, and y is the complex output for a complex input signal u. The boldface letters denote matrices speciÞcally. Comparing Eq. (A-6) with Eq. (A-5), the general idea is clear. At any time t , x is the collection (vector) of voltages v and currents i in the circuit, and ˙ x is the collection (vector) of their time derivatives. Matrix algebra or matrix calculus, using Mathcad, Þnds all the values of x and y at each value of time t with great rapidity, and plots graphs of the results. Starting at t = 0, from a set of initial values of V and I ,we can trace the history of the network through the transient period and into the steady state. We will look more closely at the general method of matrix construc- tion for our speciÞc example. Equations (A-6), (A-7), and (A-9) can be Laplace-transformed. In this process the unit matrix [I] =  10 01  ,and this work must be in accordance with the rules of matrix algebra and the ADDITIONAL DISCRETE-SIGNAL ANALYSIS AND DESIGN INFORMATION 159 Laplace transform: 1.sX(s)− x(0) = AX(s) +BU(s) 2.sX(s)− AX(s) = x(0) +BU(s) 3.X(s)[sI − A] = x(0) +BU(s) 4.X(s)= [sI −A] −1 x(0) + [sI − A] −1 BU(s) (A-9) This can be inverse-Laplace-transformed to get x (t), a function of time for each X (s) [Dorf and Bishop, 2004, Sec. 3], but we want to use the dis- crete derivative of Eq. (A-2) as an alternative for discrete-signal analysis and design [Dorf and Bishop, 2004, Sec. 3]. USING THE DISCRETE DERIVATIVE We now replace the ˙x matrix in Eq. (A-6) by incorporating Eq. (A-2). Using the intermediate steps in Eq. (A-10) and using the sequence index x(n) method that we are already very familiar with produces the following Mathcad program, expressed in Word for Windows format. 1.(if n = 0)  x 0 (n) x 1 (n)  =  V C0 I L0  2.(if n>0)  x 0 (n) x 1 (n)  =  T  0 −1 C 1 L −R L  +  10 01  (A-10) ×  x 0 (n −1) x 1 (n −1)  + T  b 0 b 1  u(n) Figure A-3 shows this equation in Mathcad program form. We can imme- diately use three options: 1. Initial conditions x 0 (0) and x 1 (0) can be zero, u(n) can have a dc value or a time function such as a step or sine wave, and in this example [Eqs. (A-6) and (A-7)] b 0 =1, b 1 =0, T =0.1, u(n) =1.0. 2. u(n) can be zero, and the transient response is driven by x (0), an initial value of capacitor voltage or inductor current or both. 160 DISCRETE-SIGNAL ANALYSIS AND DESIGN N := 1000 n := 0,1 N b := u(n) := 100 sin x(n) : = if n = 0 T + 0 1 L −1 C R L 1 0 0 1 .x(n − 1) + T b u(n) if n > 0 T := 0.01 R := 0.01 L := 0.5 C := 0.5 1 0 2 π 4mA n N 0 0 0 200 400 600 800 1000 −400 −200 0 200 400 V(C) I(L) n Figure A-3 Time response of the LCR network with zero initial condi- tions and sine-wave current excitation. 3. x(0) and u(n) can both be operational at t =0 or later than zero. There are a lot of options for this problem. The Mathcad worksheet Fig. A-2 shows option 2 for initial condition values of V C and I L ,andu(n) =0. The values of x (n) 0 and x (n) 1 are plotted. Figure A-3 uses zero initial conditions and u(n) =100 mA sine wave, frequency =4, b 0 =1.0, b 1 =0. The buildups from zero of x 1 (capacitor volts) and x 2 (inductor amperes) are plotted. The lag of I L (peaks at a later time) can be noticed. . circuit analysis conÞrms these facts for this example. R, L, and C are constant values, but they can easily be time-varying and/ or nonlinear functions of voltage and current. The discrete analysis. time for each X (s) [Dorf and Bishop, 2004, Sec. 3], but we want to use the dis- crete derivative of Eq. (A-2) as an alternative for discrete-signal analysis and design [Dorf and Bishop, 2004, Sec Also, v C is now called X 1 ,andI L is now called X 2 .  ˙ X 1 ˙ X 2  =  0 −1 C 1 L −R L  X 1 X 2  +  1 C 0  (u) (A-6) V O = RX 2 158 DISCRETE-SIGNAL ANALYSIS AND DESIGN Now write the (A-6)