1. Initial Conditions & Passive Network Synthesis

2. Sarvajanik College of Engineering & Technology Made by: Dhruvita Shah 130420117051 Khushbu Shah 130420117052 Salman Mister 140423117003 Raj Patel 140423117010

3. INITIAL CONDITIONS IN ELEMENTS: RESISTOR: In an ideal resistor, current and voltage are related by Ohm’s law, v=iRR. On application of a step voltage across the resistor, the resulting current iR has the same shape as the applied voltage but altered by the scale factor 1/R.

4. At t=0-

5. At t=0+

6. Thus, as the current through a resistor will change instantaneously if the voltage changes instantaneously. Similarly, voltage will change instantaneously if the current through it changes instantaneously.

7. Inductor: The current cannot change instantaneously in a system of constant inductance. Consequently, closing a switch to connect an inductor to a source of energy will not cause current to flow at the initial instant. Thus inductor will act as if it were an open circuit independent of the voltage at the terminals of the inductor. If a current of value I0 flows in the inductor at the instant(t=0-) switching takes place, current ( i0 )Will continue to flow just after switching (t=0+). Thus for the initial instant, the inductor can be thought of as a current source of amp. i.e. iL(0-)=il(0+)=I0

8. Capacitor: The voltage cannot change instantaneously in a system of fixed capacitance. If an uncharged capacitor is connected to an energy source, a current will flow instantaneously, the capacitor being equivalent to a short circuit. This follows because voltage and charge are proportional in a capacitive system, Vc= Q/C, so that zero charge corresponds to zero voltage across the terminal of the capacitor. Uncharged capacitor is equivalent to a short circuit i.e. Vc(0-)=0as q(0)=0. Vc(0+)= q(0+/c),where q(0+) is the initial charge. thus capacitor Vc(0+)=Vc(0-).in fig.

9. PROCEDURE FOR EVALUATING INITIAL CONDITION : There is no unique procedure that must be followed for evaluating the initial conditions. The procedure generally followed consists of the following two steps in sequence: 1) Solving for the initial values of the variables namely voltages and currents t= 0+ and 2) Solving for the derivatives of the variables at t= 0+.

10. (1)Solving for initial values of the variables at t= 0+ : Initial values of current or voltage may be found directly from a study of the equivalent network schematic at t= 0+. For each element in the network, we must determine just what will happen when the switching action takes place. From this analysis, anew schematic of an equivalent network for t= 0+ may be constructed according to the following rules : i) Resistors are left in the network without any change. ii) Replace all inductors with open circuits or with current generators having the value of current flowing at t= 0+. iii) Replace all capacitors with short circuit or with a voltage source of the value v(0+)= q(0+)/C if there is an initial charge, q(0+)

11. (2) Solving for the derivatives at t = 0+ : The details and order of multiplication will be different for each different network. A successful approach will not be obvious at all, a fact that adds interest and offers a challenge in the solution of initial-value problem. It can be best understood from the method followed in the different networks studied hereafter.

12. In the network of the fig. the switch k is closed at t=0 with the capacitor uncharged and with zero current in the inductor. find the values of I, di/dt and d²i/dt² at t=0+. If v=100v, L=1H, R=10 ohm, C=10micro farad. Example

13. At t=0+

14. Since there is no initial voltage on the capacitor, it may be replaced by a sort circuit. Similarly inductor may be replaced by an open circuit, being no initial current. The resulting equivalent network at t=0+ is shown in fig. In this particular case, there is no need to write equation for the network. By inspection. i(0+)=0 because of open circuit due to L

15. Ri+L di/dt+1/C ………1 Voltage across capacitor is zero at t=0+ So, Ri(0+)+L di(0+)/dt+0=V………..2 Substituting the values, di/dt (0+) = 100amp/sec……….3 Differentiating eqn 1, R di/dt + L d²i/dt² + i/c = 0……..4 Writing above eqn at t = 0+ R di(0+)/dt + L d²i(0+)/dt² + i(0+)/c = 0 Substituting values, d²i/dt² (0+) = -1000 amp/sec²

16. Passive Network Synthesis

17. Hurwitz polynomial In mathematics, a Hurwitz polynomial, named after Adolf Hurwitz, is a polynomial whose coefficients are positive real numbers and whose roots (zeros) are located in the left half-plane of the complex plane or on the j ω axis, that is, the real part of every root is zero or negative. The term is sometimes restricted to polynomials whose roots have real parts that are strictly negative, excluding the axis (i.e., a Hurwitz stable polynomial)

18. A polynomial function P(s) of a complex variables is said to be Hurwitz if the following conditions are satisfied: 1. P(s) is real when s is real. 2. The roots of P(s) have real parts which are zero or negative. Hurwitz polynomials are important in control systems theory because they represent the characteristic equations of stable linear systems. Whether a polynomial is Hurwitz can be determined by solving the equation to find the roots, or from the coefficients without solving the equation by the Routh-Hurwitz stability criterion.

19. Properties of Hurwitz Polynomial Hurwitz polynomial P(s) is given as P(s) = 1. All the coefficients of the polynomial must be real and positive. 2. The roots of odd and even parts of the polynomial P(s) lie on the imaginary axis. 3. If P(s) is either even or odd, all its roots are on the imaginary axis.

20. 4. The continued fraction expansion of the ratio of odd to even parts or even to add parts of P(s) yields all positive quotients. 5. If the polynomial P(s) is completely even or completely odd, then differentiate P(s) is with respect to s and obtain the polynomial P’(s) = (d P(s)/ds). The continued fraction expansion of P(s)/P’(s) gives all positive quotients for a polynomial P(s) to be Hurwitz.

21. Example: A simple example of a Hurwitz polynomial is the following: x^2 + 2x + 1. The only real solution is −1, as it factors to (x+1)^2. All the coefficients of the polynomial are real & positive. As the real part of the root is negative, it is a hurwitz polynomial.

22. Positive Real Function Definition The term positive-real function was originally defined by Otto Brune to describe any function Z(s) which is rational (the quotient of two polynomials), is real when s is real has positive real part when s has a positive real part

23. Many authors strictly adhere to this definition by explicitly requiring rationality, or by restricting attention to rational functions, at least in the first instance. However, a similar more general condition, not restricted to rational functions had earlier been considered by Cauer, and some authors ascribe the term positive-real to this type of condition, while other consider it to be a generalization of the basic definition.

24. Properties of Positive Real Functions The sum of two PR functions is PR. The composition of two PR functions is PR. In particular, if Z(s) is PR, then so are 1/Z(s) and Z(1/s). All the poles and zeros of a PR function are in the left half plane or on its boundary the imaginary axis. Any poles and zeroes on the imaginary axis are simple (have a multiplicity of one). Any poles on the imaginary axis have real strictly positive residues, and similarly at any zeroes on the imaginary axis, the function has a real strictly positive derivative.

25. Over the right half plane, the minimum value of the real part of a PR function occurs on the imaginary axis (because the real part of an analytic function constitutes a harmonic function over the plane, and therefore satisfies the maximum principle). For a rational PR function, the number of poles and number of zeroes differ by at most one.