Impedence.pdf

336 Figure 6.2.17 Series RLC circuit. CHAPTER 6 R 1 vs Electrical and Electromechanical Systems L i C v1 2 ■ Solution In this circuit the energy is stored in the capacitor and in the inductor. The energy stored in the capacitor is Cv12 /2 and the energy stored in the inductor is Li 2 /2. Th...

336 Figure 6.2.17 Series RLC circuit. CHAPTER 6 R 1 vs Electrical and Electromechanical Systems L i C v1 2 ■ Solution In this circuit the energy is stored in the capacitor and in the inductor. The energy stored in the capacitor is Cv12 /2 and the energy stored in the inductor is Li 2 /2. Thus a suitable choice of state variables is v1 and i. From Kirchhoff’s voltage law, di vs − Ri − L − v1 = 0 dt Solve this for di/dt: di 1 1 R = v s − v1 − i dt L L L This is the first state equation. Now find an equation for dv1 /dt. From the capacitor relation,  1 i dt v1 = C Differentiating gives the second state equation. 1 dv1 = i dt C The two state equations can be expressed in matrix form as follows. ⎡ di ⎤ ⎡ ⎤ ⎡ ⎤ 1 R 1 − − ⎢ dt ⎥ ⎢ L ⎥ i L ⎣ ⎣ ⎦=⎣ 1 ⎦ v + L ⎦ vs dv1 1 0 0 C dt 6.3 TRANSFER FUNCTIONS AND IMPEDANCE The Laplace transform and the transfer function concept enable us to deal with algebraic equations rather than differential equations, and thus ease the task of analyzing circuit models. This section illustrates why this is so, and introduces a related concept, impedance, which is simply the transfer function between voltage and current. 6.3.1 USE OF TRANSFORMED EQUATIONS Applying the Laplace transform to the circuit equations often helps in eliminating intermediate variables, because the transformed equations are algebraic and therefore are easier to solve. This is especially true when the circuit contains dynamic elements (capacitances or inductances). E X A M P L E 6.3.1 An RLC Circuit with Two Inputs ■ Problem Consider the RLC circuit treated in Example 6.2.14 and shown in Figure 6.2.17. Use the Laplace transform to obtain the differential equation model for the current i 3 .