Presentation is loading. Please wait.

Presentation is loading. Please wait.

SJTU1 Chapter 8 Second-Order Circuit. SJTU2 A second-order circuit is characterized by a second- order differential equation. It consists of resistors.

Similar presentations


Presentation on theme: "SJTU1 Chapter 8 Second-Order Circuit. SJTU2 A second-order circuit is characterized by a second- order differential equation. It consists of resistors."— Presentation transcript:

1 SJTU1 Chapter 8 Second-Order Circuit

2 SJTU2 A second-order circuit is characterized by a second- order differential equation. It consists of resistors and the equivalent of two energy storage elements. What is second-order circuit? Typical examples of second-order circuits: a) series RLC circuit, b) parallel RLC circuit, c) RL circuit, d) RC circuit

3 SJTU3 1.The Series RLC Circuit 2.The Parallel RLC Circuit 3.Second-Order Circuit Complete Response

4 SJTU4 1. The Series RLC Circuit FORMULATING SERIES RLC CIRCUIT EQUATIONS Eq.(7-33)

5 SJTU5 The initial conditions To solve second-order equation, there must be two initial values.

6 SJTU6 ZERO-INPUT RESPONSE OF THE SERIES RLC CIRCUIT With V T =0(zero-input) Eq.(7-33) becomes Eq.(3-37) try a solution of the form then Eq.(7-39) characteristic equation

7 SJTU7 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

8 SJTU8 A source-free series RLC circuit Special case: Vc(0)=V 0, I L (0)=0 V(t) V0V0 I(t) tMtM

9 SJTU9 t M >t>0 t > t M What happens when R=0?

10 SJTU10 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 t +2.7 e t A vc(t) = e t e t V

11 SJTU11

12 SJTU12

13 SJTU13 II. CRITICALLY DAMPED: R=0.943, L= 1/3 H, C=1.5F, Vo=1V, Io=2A iL(t) = 2e t -5.83t e t A vc(t) = e t t e t V

14 SJTU14

15 SJTU15

16 SJTU16 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 ) A vc(t) = 2 e -0.75t cos(1.2t ) V

17 SJTU17

18 SJTU18

19 SJTU19 IV. UNDAMPED: R=0, L= 1/3 H, C=1.5F, Vo=1V, Io=2A iL(t) =2.915 cos(1.414t ) A vc(t) =1.374 cos(1.414t ) V

20 SJTU20

21 SJTU21

22 SJTU22 EXAMPLE 7-14 A series RLC circuit has C=0.25uF and L=1H. Find the roots of the characteristic equation for R T =8.5kohm, 4kohm and 1kohm SOLUTION: For R T =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= *

23 SJTU23 For R T =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= For R T =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. * *

24 SJTU24 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)

25 SJTU25 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 I o =0 and V o =15V. SOLUTION: Fig (a) In Example 7-14 the value R=8.5kohm yields case A with roots S 1 =-500 and S 2 = The corresponding zero-input solution takes the form in Eq.(7-48a).

26 SJTU26 The initial conditions yield two equations in the constants K1 and K2: Solving these equations yields K 1 =16 and K 2 =-1, so that the zero-input response is

27 SJTU27 The initial conditions yield two equations in the constants K 1 and K 2 : Solving these equations yields K 1 =15 and K 2 = 2000 x 15, so the zero-input response is (b) In Example 7-14 the value R=4kohm yields case B with roots S 1 =S 2 = The corresponding zero-input response takes the form in Eq.(7-48b):

28 SJTU28 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 K 1 and K 2 : Solving these equations yields K 1 =15 and K 2 =( ), so the zero-input response is

29 SJTU29 Fig. 7-32

30 SJTU30 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

31 SJTU31 2. The Parallel RLC Circuit FORMULATING PARALLEL RLC CIRCUIT EQUATIONS Eq. 7-55

32 SJTU32 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 i L (0)=I o and

33 SJTU33 set i N =0 in Eq.(7-55) and obtain a homogeneous equation in the inductor current: A trial solution of the form I L =Ke st leads to the characteristic equation Eq. 7-56

34 SJTU34 There are three distinct cases: Case A: If (G N L) 2 -4LC>0, there are two unequal real roots and the zero-input response is the overdamped form Case B: (G N L) 2 -4LC=0, there are two real equal roots and the zero-input response is the critically damped form Case C:(G N L) 2 -4LC<0, there are two complex, conjugate roots and the zero-input response is the underdamped form

35 SJTU35 EXAMPLE 7-16 In a parallel RLC circuit R T =1/G N =500ohm, C=1uF, L=0.2H. The initial conditions are I o =50 mA and V o =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

36 SJTU36 Evaluating this expression at t=0 yields

37 SJTU37

38 SJTU38 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 SOLUTION: (a) For t<0 the circuit is in the dc steady state

39 SJTU39 (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

40 SJTU40 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 K 1 and K 2 :

41 SJTU41 Solving these equations yields K 1 =30.3 mA and K 2 = ma The zero-input response of the inductor current is (c) Given the inductor current in (b), the capacitor voltage is For t 0 the current i sw (t) is the current through the 50 ohm resistor plus the current through the 250 ohm resistor

42 SJTU42 3. Second-order Circuit Complete Response The general second-order linear differential equation with a step function input has the form Eq The complete response can be found by partitioning y(t) into forced and natural components: Eq y N (t) --- general solution of the homogeneous equation (input set to zero), y F (t) is a particular solution of the equation y F =A/a o

43 SJTU43 Combining the forced and natural responses Eq 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. Fig SOLUTION:

44 SJTU44 By inspection, the forced response is v CF =10V. In standard format the homogeneous equation is the natural response is underdamped (case C)

45 SJTU45 The constants K 1 and K 2 are determined by the initial conditions. These equations yield K 1 = -10 and K 2 = The complete response of the capacitor voltage step response is

46 SJTU46 General second-order circuit Steps: 1.Set a second-order differential equation 2.Find the natural response y N (t) from the homogeneous equation (input set to zero) 3.Find a particular solution y F (t) of the equation 4.Determine K 1 and K 2 by the initial conditions 5.Yield the required response

47 SJTU47 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.


Download ppt "SJTU1 Chapter 8 Second-Order Circuit. SJTU2 A second-order circuit is characterized by a second- order differential equation. It consists of resistors."

Similar presentations


Ads by Google