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Chapter 8

Chapter 8. Second-Order Circuit. What is second-order circuit?. A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements.

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Chapter 8

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  1. Chapter 8 Second-Order Circuit SJTU

  2. What is second-order circuit? A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements. Typical examples of second-order circuits: a) series RLC circuit, b) parallel RLC circuit, c) RL circuit, d) RC circuit SJTU

  3. The Series RLC Circuit • The Parallel RLC Circuit • Second-Order Circuit Complete Response SJTU

  4. 1. The Series RLC Circuit FORMULATING SERIES RLC CIRCUIT EQUATIONS Eq.(7-33) SJTU

  5. The initial conditions To solve second-order equation, there must be two initial values. SJTU

  6. ZERO-INPUT RESPONSE OF THE SERIES RLC CIRCUIT With VT=0(zero-input) Eq.(7-33) becomes Eq.(3-37) try a solution of the form then characteristic equation Eq.(7-39) SJTU

  7. Case A: If                              , there are two real, unequal roots Case B: If                              , there are two real, equal roots Case C: If                              , there are two complex conjugate roots In general, a quadratic characteristic equation has two roots: Eq.(7-40) three distinct possibilities: SJTU

  8. A source-free series RLC circuit Special case: Vc(0)=V0, IL(0)=0 V0 V(t) tM I(t) SJTU

  9. t > tM tM>t>0 What happens when R=0? SJTU

  10. Second Order Circuit with no Forcing Function vc(0) = Vo , iL(0) = Io. I. OVER DAMPED: R=2 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A iL(t) = -0.7 e -0.354t +2.7 e -5.646t A vc(t) = 1.318 e -0.354t -0.318 e -5.646t V SJTU

  11. SJTU

  12. SJTU

  13. II. CRITICALLY DAMPED: R=0.943 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A iL(t) = 2e -1.414t -5.83t e -1.414t A vc(t) = e -1.414t+ 2.75 t e -1.414t V SJTU

  14. SJTU

  15. SJTU

  16. III. UNDER DAMPED: R=0.5 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A iL(t) =4.25 e -0.75t cos(1.2t + 1.081) A vc(t) = 2 e -0.75t cos(1.2t - 1.047) V SJTU

  17. SJTU

  18. SJTU

  19. IV. UNDAMPED: R=0 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A iL(t) =2.915 cos(1.414t + 0.815) A vc(t) =1.374 cos(1.414t - 0.756) V SJTU

  20. SJTU

  21. SJTU

  22. EXAMPLE 7-14 A series RLC circuit has C=0.25uF and L=1H. Find the roots of the characteristic equation for RT=8.5kohm, 4kohm and 1kohm SOLUTION: For RT=8.5kohm, the characteristic equation is whose roots are * These roots illustrate case A. The quantity under the radical is positive, and there are two real, unequal roots at S1=-500 and S2=-8000. SJTU

  23. For RT=4kohm, the characteristic equation is whose roots are This is an example of case B. The quantity under the radical is zero, and there are two real, equal roots at S1=S2=-2000. * For RT=1kohm the characteristic equation is whose roots are The quantity under the radical is negative, illustrating case C.                                               In case C the two roots are complex conjugates. * SJTU

  24. In case A the two roots are real and unequal                                     and the zero-input response is the sum of two exponentials of the form Eq.(7-48a) In case B the two roots are real and equal                      and the zero-input response is the sum of an exponential and a damped ramp. Eq.(7-48b) In case C the two roots are complex conjugates                                           and the zero-input response is the sum of a damped cosine and a damped sine. Eq.(7-48c) SJTU

  25. EXAMPLE 7-15 The circuit of Figure 7-31 has C=0.25uF and L=1H. The switch has been open for a long time and is closed at t=0. Find the capacitor voltage for t  0 for (a) R=8.5k ohm, (b) R=4k ohm, and (c) R=1k ohm. The initial conditions are Io=0 and Vo=15V. SOLUTION: • (a) In Example 7-14 the value R=8.5kohm yields case A with roots S1=-500 and S2=-8000. The corresponding zero-input solution takes the form in Eq.(7-48a). Fig. 7-31 SJTU

  26. The initial conditions yield two equations in the constants K1 and K2: Solving these equations yields K1=16 and K2 =-1, so that the zero-input response is SJTU

  27. The initial conditions yield two equations in the constants K1 and K2: • (b) In Example 7-14 the value R=4kohm yields case B with roots S1=S2=-2000. The corresponding zero-input response takes the form in Eq.(7-48b): Solving these equations yields K1=15 and K2= 2000 x 15, so the zero-input response is SJTU

  28. c) In Example 7-14 the value R=1k ohm yields case C with roots                                 . The corresponding zero-input response takes the form in Eq.(7-48c): The initial conditions yield two equations in the constants K1 and K2: Solving these equations yields K1=15 and K2=( ) , so the zero-input response is SJTU

  29. Fig. 7-32 SJTU

  30. In general, a quadratic characteristic equation has two roots: Eq.(7-40) three distinct possibilities: Case A: If                              , there are two real, unequal roots Case B: If                              , there are two real, equal roots Case C: If                              , there are two complex conjugate roots Overdamped situation Ciritically damped situation Underdamped situation SJTU

  31. 2. The Parallel RLC Circuit FORMULATING PARALLEL RLC CIRCUIT EQUATIONS Eq. 7-55 SJTU

  32. Equation(7-55) is second-order linear differential equation of the same form as the series RLC circuit equation in Eq.(7-33). In fact, if we interchange the following quantities: we change one equation into the other. The two circuits are duals, which means that the results developed for the series case apply to the parallel circuit with the preceding duality interchanges. The initial conditions iL(0)=Io and SJTU

  33. set iN=0 in Eq.(7-55) and obtain a homogeneous equation in the inductor current: A trial solution of the form IL=Kest leads to the characteristic equation Eq. 7-56 SJTU

  34. There are three distinct cases: Case A: If (GNL)2-4LC>0, there are two unequal real roots and the zero-input response is the overdamped form Case B: (GNL)2-4LC=0, there are two real equal roots and the zero-input response is the critically damped form Case C:(GNL)2-4LC<0, there are two complex, conjugate roots and the zero-input response is the underdamped form SJTU

  35. EXAMPLE 7-16 In a parallel RLC circuit RT=1/GN=500ohm, C=1uF, L=0.2H. The initial conditions are Io=50 mA and Vo=0. Find the zero-input response of inductor current, resistor current, and capacitor voltage SOLUTION: From Eq.(7-56) the circuit characteristic equation is The roots of the characteristic equation are SJTU

  36. Evaluating this expression at t=0 yields SJTU

  37. SJTU

  38. EXAMPLE 7-17 The switch in Figure 7-34 has been open for a long time and is closed at t=0 (a) Find the initial conditions at t=0 (b) Find the inductor current for t0 (c) Find the capacitor voltage and current through the switch for t 0 Fig. 7-34 SOLUTION: (a) For t<0 the circuit is in the dc steady state                    SJTU

  39. (b) For t 0 the circuit is a zero-input parallel RLC circuit with initial conditions found in (a). The circuit characteristic equation is The roots of this equation are The circuit is overdamped (case A), The general form of the inductor current zero-input response is using the initial conditions SJTU

  40. The initial capacitor voltage establishes an initial condition on the derivative of the inductor current since The derivative of the inductor response at t=0 is The initial conditions on inductor current and capacitor voltage produce two equations in the unknown constants K1 and K2: SJTU

  41. Solving these equations yields K1=30.3 mA and K2=-0.309 ma The zero-input response of the inductor current is (c) Given the inductor current in (b), the capacitor voltage is For t0 the current isw(t) is the current through the 50 ohm resistor plus the current through the 250 ohm resistor SJTU

  42. 3. Second-order Circuit Complete Response The general second-order linear differential equation with a step function input has the form Eq. 7-60 The complete response can be found by partitioning y(t) into forced and natural components: Eq. 7-61 yN(t) --- general solution of the homogeneous equation (input set to zero), yF(t) is a particular solution of the equation ∴ yF=A/ao SJTU

  43. Combining the forced and natural responses Eq. 7-67 EXAMPLE 7-18 The series RLC circuit in Figure 7-35 is driven by a step function and is in the zero state at t=0. Find the capacitor voltage for t  0. SOLUTION: Fig. 7-35 SJTU

  44. By inspection, the forced response is vCF=10V. In standard format the homogeneous equation is the natural response is underdamped (case C) SJTU

  45. The constants K1 and K2 are determined by the initial conditions. These equations yield K1= -10 and K2= -2.58. The complete response of the capacitor voltage step response is SJTU

  46. General second-order circuit • Steps: • Set a second-order differential equation • Find the natural response yN(t) from the homogeneous equation (input set to zero) • Find a particular solution yF(t) of the equation • Determine K1 and K2 by the initial conditions • Yield the required response SJTU

  47. Summary • Circuits containing linear resistors and the equivalent of two energy storage elements are described by second-order differential equations in which the dependent variable is one of the state variables. The initial conditions are the values of the two state variables at t=0. • The zero-input response of a second-order circuit takes different forms depending on the roots of the characteristic equation. Unequal real roots produce the overdamped response, equal real roots produce the critically damped response, and complex conjugate roots produce underdamped responses. • Computer-aided circuit analysis programs can generate numerical solutions for circuit transient responses. Some knowledge of analytical methods and an estimate of the general form of the expected response are necessary to use these analysis tools. SJTU

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