Second Order Circuits

6 Key excerpts on "Second Order Circuits"

• 

  eBook - ePub

  Electrical, Electronics, and Digital Hardware Essentials for Scientists and Engineers

  • Ed Lipiansky(Author)
  • 2012(Publication Date)
  • Wiley-IEEE Press

    (Publisher)

  4 FIRST- AND SECOND-ORDER CIRCUITS UNDER SINUSOIDAL AND STEP EXCITATIONS

  4.1  INTRODUCTION

  First-order circuits are very important in electrical and electronic engineering. Many higher order circuits can be reduced to a first-order circuit. Analyzing the behavior either in the time or in the frequency domains of a first-order circuit is unquestionably simpler than analyzing that of a higher order circuit. Essentially, first-order circuits have a single energy storage device. Such devices can be either a capacitor or an inductor. Examples of circuits that can be reduced to first-order circuits under certain conditions are electronic amplifiers, operational amplifiers, servomechanisms, electric motors, and other control networks.

  Let us present an example of a first-order circuit.

  Example 4.1  RL Series First-Order Circuit

  Given a circuit that contains one resistor in series with an inductor, such as the one shown in Figure 4.1 , we can calculate the output voltage to input voltage ratio as a function of frequency. Such ratio of voltages in the frequency domain is commonly referred to as H (jω ), where H (jω ) is called the circuit transfer function.

  (4.1)  

  Figure 4.1  Circuit for Example 4.1 , a first-order series RL circuit.

  Previously reviewing the material from Chapter 3 on AC analysis we can easily calculate H (jω ) for the circuit in Figure 4.1 .

  The current in the RL circuit is calculated as follows:

  (4.2)  

  where I and

  Vin

  are respectively the current and voltage phasors of the circuit.

  Now the output voltage

  Vout

  is calculated by multiplying the circuit current times the impedance or reactance of inductor L . Thus, we obtain

  (4.3)  

  And finally, our transfer function is

  (4.4)  

  Furthermore, rationalizing the denominator, that is, multiplying numerator and denominator of Equation (4.4) by the complex conjugate of the denominator, (R  − jω L ), we obtain

  (4.5)  

  (4.6)  

  Equation (4.6) is of the form a  + jb where the terms a and jb are frequency dependent. Additionally term jb

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  eBook - PDF

  Engineering Circuit Analysis

  • J. David Irwin, R. Mark Nelms(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  The following examples will demonstrate the analysis techniques. 6.3 Second-Order Circuits 307 EXAMPLE 6.7 Consider the parallel RLC circuit shown in Fig. 6.16. The second-order differential equation that describes the voltage υ(t) is d 2 υ ___ dt 2 + 1 ___ RC dυ ___ dt + υ ___ LC = 0 L υ(t) C + − υ C (0) i L (0) R FIGURE 6.16 Parallel RLC circuit. A comparison of this equation with Eqs. (6.14) and (6.15) indicates that for the parallel RLC circuit the damping term is 1/2 RC and the undamped natural frequency is 1/ √ ___ LC . If the circuit parameters are R = 4 Ω, C = 1/10 F, and L = 10 H, the equation becomes d 2 υ ___ dt 2 + 2.5 dυ ___ dt + υ = 0 Let us assume that the initial conditions on the storage elements are i L (0) = −1/2 A and υ C (0) = 4 V. Let us find the node voltage υ(t) and the inductor current. Solution The characteristic equation for the network is s 2 + 2.5s + 1 = 0 and the roots are s 1 = −0.5 s 2 = −2 Since the roots are real and unequal, the circuit is overdamped, and υ (t) is of the form υ(t) = K 1 e −2t + K 2 e −0.5t The initial conditions are now employed to determine the constants K 1 and K 2 . Since υ (t) = υ C (t), υ C (0) = υ (0) = 4 = K 1 + K 2 The second equation needed to determine K 1 and K 2 is normally obtained from the expression dυ(t) _____ dt = −2K 1 e −2t − 0.5K 2 e −0.5t However, the second initial condition is not dυ (0)/dt. If this were the case, we would simply evaluate the equation at t = 0. This would produce a second equation in the unknowns K 1 and K 2 . We can, however, circumvent this problem by noting that the node equation for the circuit can be written as C dυ (t) _____ dt + υ (t) ____ R + i L (t) = 0 or dυ(t) _____ dt = −1 ___ RC υ(t) − i L (t) ____ C At t = 0, dυ(0) _____ dt = −1 ___ RC υ(0) − 1 __ C i L (0) = −2.5(4) − 10(−1/2) = −5

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  eBook - PDF

  The Analysis and Design of Linear Circuits

  • Roland E. Thomas, Albert J. Rosa, Gregory J. Toussaint(Authors)
  • 2019(Publication Date)
  • Wiley

    (Publisher)

  So it is natural that we investigate the step response of second-order circuits. In Chapter 11, we will develop general techniques for determining the step response of any linear circuit. However, in this introduction we use classical methods, as in Figure 7–1, of solving differential equations to find the step response of second-order circuits. The general second-order linear differential equation with a step function input has the form a 2 d 2 y t dt 2 + a 1 dy t dt + a 0 y t = Au t (7 –60) where y t is a voltage or current response, Au t is the step function input, and a 2 , a 1 , and a 0 are constant coefficients. The step response is the general solution of this differential equation for t ≥ 0. The step response can be found by partitioning y t into forced and natural components: y t = y N t + y F t (7 –61) The natural response y N t is the general solution of the homogeneous equation (input set to zero), while the forced response y F t is a particular solution of the equation a 2 d 2 y F t dt 2 + a 1 dy F t dt + a 0 y F t = A t ≥ 0 Since A is a constant, it follows that dA dt and d 2 A dt 2 are both zero, so it is readily apparent that y F t = A a 0 is a particular solution of this differential equation. So much for the forced response. Turning now to the natural response, we seek a general solution of the homo- geneous equation. The natural response has the same form as the zero-state response studied in the previous section. In a second-order circuit, the zero-state and natural responses take one of the three possible forms: overdamped, criti- cally damped, or underdamped. To describe the three possible forms, we intro- duce two new parameters: ω 0 (omega zero) and ζ (zeta). These parameters are defined in terms of the coefficients of the general second-order equation in Eq. (7–60): ω 2 0 = a 0 a 2 and 2ζω 0 = a 1 a 2 (7 –62) The parameter ω 0 is called the undamped natural frequency and ζ is called the damp- ing ratio.

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  eBook - PDF

  Basic Engineering Circuit Analysis

  • J. David Irwin, R. Mark Nelms(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  The following examples will demonstrate the analysis techniques. 7.3 Second-Order Circuits 241 Consider the parallel RLC circuit shown in Fig. 7.16. The second-order differential equation that describes the voltage υ(t) is d 2 υ ___ dt 2 + 1 ___ RC dυ ___ dt + υ ___ LC = 0 L υ(t) C + − υ C (0) i L (0) R FIGURE 7.16 Parallel RLC circuit. A comparison of this equation with Eqs. (7.14) and (7.15) indicates that for the parallel RLC circuit the damping term is 1/2 RC and the undamped natural frequency is 1/ √ ___ LC . If the circuit param- eters are R = 2 Ω, C = 1/5 F, and L = 5 H, the equation becomes d 2 υ ___ dt 2 + 2.5 dυ ___ dt + υ = 0 Let us assume that the initial conditions on the storage elements are i L (0) = −1 A and υ C (0) = 4 V. Let us find the node voltage υ(t) and the inductor current. Solution The characteristic equation for the network is s 2 + 2.5s + 1 = 0 and the roots are s 1 = −0.5 s 2 = −2 Since the roots are real and unequal, the circuit is overdamped, and υ (t) is of the form υ(t) = K 1 e −2t + K 2 e −0.5t The initial conditions are now employed to determine the constants K 1 and K 2 . Since υ (t) = υ C (t), υ C (0) = υ (0) = 4 = K 1 + K 2 The second equation needed to determine K 1 and K 2 is normally obtained from the expression dυ(t) _____ dt = −2K 1 e −2t − 0.5K 2 e −0.5t However, the second initial condition is not dυ (0)/dt. If this were the case, we would simply eval- uate the equation at t = 0. This would produce a second equation in the unknowns K 1 and K 2 . We can, however, circumvent this problem by noting that the node equation for the circuit can be written as C dυ (t) _____ dt + υ (t) ____ R + i L (t) = 0 or dυ(t) _____ dt = −1 ___ RC υ(t) − i L (t) ____ C At t = 0, dυ(0) _____ dt = −1 ___ RC υ(0) − 1 __ C i L (0) = −2.5(4) − 5(−1) = −5 EXAMPLE 7.7

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  eBook - ePub

  Introduction to Energy, Renewable Energy and Electrical Engineering

  Essentials for Engineering Science (STEM) Professionals and Students

  • Ewald F. Fuchs, Heidi A. Fuchs(Authors)
  • 2020(Publication Date)
  • Wiley

    (Publisher)

  3 DC Circuit Transient Analysis

  In this chapter, we examine the behavior of a DC network as a function of time after a sudden change in the network occurs due to opening or closing of switches. The DC network may contain resistors, inductors, and capacitors. First‐ and second‐order ordinary differential equations will be analytically and numerically solved by well‐known solution techniques and numerical methods such as PSPICE [1 –4 ], MATLAB/Simulink [5 ], and Mathematica [6 ]. Note that there are standard resistor [7 ], capacitor [8 ], and inductor [9 ] values. In electric circuit designs, any (certain) resistor, capacitor, and inductor values might be required based on the desired design performance. These certain (e.g. resistance) values are not necessarily manufactured and sold, simplifying the manufacturing process of resistors: only well‐defined resistance values of resistors are manufactured. For example, a design requires the resistance value of 25 Ω; then a series combination of two resistors with standard resistance values of 10 and 15 Ω must be used because 10 Ω + 15 Ω = 25 Ω. For a 6 Ω resistor, the two resistors with standard values 10 and 15 Ω can be connected in parallel such that .

  3.1 Capacitors

  3.1.1 Energy Stored in a Capacitor

  Ideal lossless (with no losses) capacitors consist of two conducting plates separated by a nonconducting, or dielectric, material. The simplest capacitor consists of two parallel plates as shown in Figure 3.1 . Note the capacitor is the name of a device having the capacitance C. Any conducting or nonconducting material – which is electrically isolated – can be charged with endowed (see Chapter 1 , Appendix 1.B) electrons and has the property of capacitance C

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  eBook - ePub

  The Fundamentals of Electrical Engineering

  for Mechatronics

  • Felix Hüning(Author)
  • 2014(Publication Date)
  • De Gruyter Oldenbourg

    (Publisher)

  C (0)). Due to the open switch the current is zero for t < 0 s and due to the inductor it stays zero at t = 0 s.

  Fig. 7.17 : Series connection of resistor, inductor and capacitor as an easy example for a second order circuit.

  Applying KVL to the circuit yields for t > 0 s:

  In the next step the voltages across the inductor and the resistor are expressed in terms of the capacitor’s voltage uC (t) using Ohm’s law and the relation between the capacitor’s voltage and the current:

  Finally we get this for the capacitor’s voltage: This is a homogeneous second order ODE for the capacitor’s voltage.   Excursus into mechanics: damped harmonic oscillator

  A system which exhibits mathematically identical behavior to that of a similar, but physically different system is analogous to this system. In the case of the RLC-circuit the homogeneous second order ODE has the same structure (same but the constants) like the ODE for a damped harmonic oscillator such as given in Fig. 7.18 . A mass m is connected to a spring and the movement x(t) is damped by friction. Balance of forces yields:

  Here the force for the spring (proportional to the position of the mass) was used:

  Also the damping force (proportional to the velocity v(t) = dx(t)/dt) was used:

  Fig. 7.18 : A mechanical analog to the electrical RLC circuit.

  As the mathematical behavior of both mechanical and electrical systems is identical, these systems are analog. Force causes velocity just as voltage causes current. A damper dissipates mechanical energy into heat just like a resistor dissipates electrical energy into heat. Springs and masses store energy in two different ways (potential energy and kinetic energy respectively) just as capacitors and inductors store energy in two different forms (electric and magnetic field respectively).

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