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First-Order Circuits Cont’d PowerPoint PPT Presentation


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First-Order Circuits Cont’d. Dr. Holbert April 17, 2006. Introduction. In a circuit with energy storage elements, voltages and currents are the solutions to linear, constant coefficient differential equations. Real engineers almost never solve the differential equations directly.

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First order circuits cont d l.jpg

First-Order Circuits Cont’d

Dr. Holbert

April 17, 2006

ECE201 Lect-20


Introduction l.jpg

Introduction

  • In a circuit with energy storage elements, voltages and currents are the solutions to linear, constant coefficient differential equations.

  • Real engineers almost never solve the differential equations directly.

  • It is important to have a qualitative understanding of the solutions.

ECE201 Lect-20


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Important Concepts

  • The differential equation for the circuit

  • Forced (particular) and natural (complementary) solutions

  • Transient and steady-state responses

  • 1st order circuits: the time constant ()

ECE201 Lect-20


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The Differential Equation

  • Every voltage and current is the solution to a differential equation.

  • In a circuit of order n, these differential equations have order n.

  • The number and configuration of the energy storage elements determines the order of the circuit.

    n # of energy storage elements

ECE201 Lect-20


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The Differential Equation

  • Equations are linear, constant coefficient:

  • The variable x(t) could be voltage or current.

  • The coefficients an through a0 depend on the component values of circuit elements.

  • The function f(t) depends on the circuit elements and on the sources in the circuit.

ECE201 Lect-20


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Building Intuition

  • Even though there are an infinite number of differential equations, they all share common characteristics that allow intuition to be developed:

    • Particular and complementary solutions

    • Effects of initial conditions

ECE201 Lect-20


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Differential Equation Solution

  • The total solution to any differential equation consists of two parts:

    x(t) = xp(t) + xc(t)

  • Particular (forced) solution is xp(t)

    • Response particular to a given source

  • Complementary (natural) solution is xc(t)

    • Response common to all sources, that is, due to the “passive” circuit elements

ECE201 Lect-20


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The Forced Solution

  • The forced (particular) solution is the solution to the non-homogeneous equation:

  • The particular solution is usually has the form of a sum of f(t) and its derivatives.

    • If f(t) is constant, then vp(t) is constant

ECE201 Lect-20


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The Natural Solution

  • The natural (or complementary) solution is the solution to the homogeneous equation:

  • Different “look” for 1st and 2nd order ODEs

ECE201 Lect-20


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First-Order Natural Solution

  • The first-order ODE has a form of

  • The natural solution is

  • Tau (t) is the time constant

    • For an RC circuit, t = RC

    • For an RL circuit, t = L/R

ECE201 Lect-20


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Initial Conditions

  • The particular and complementary solutions have constants that cannot be determined without knowledge of the initial conditions.

  • The initial conditions are the initial value of the solution and the initial value of one or more of its derivatives.

  • Initial conditions are determined by initial capacitor voltages, initial inductor currents, and initial source values.

ECE201 Lect-20


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Transients and Steady State

  • The steady-state response of a circuit is the waveform after a long time has passed, and depends on the source(s) in the circuit.

    • Constant sources give DC steady-state responses

      • DC SS if response approaches a constant

    • Sinusoidal sources give AC steady-state responses

      • AC SS if response approaches a sinusoid

  • The transient response is the circuit response minus the steady-state response.

ECE201 Lect-20


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Step-by-Step Approach

  • Assume solution (only dc sources allowed):

    x(t) = K1 + K2 e-t/

  • At t=0–, draw circuit with C as open circuit and L as short circuit; find IL(0–) or VC(0–)

  • At t=0+, redraw circuit and replace C or L with appropriate source of value obtained in step #2, and find x(0)=K1+K2

  • At t=, repeat step #2 to find x()=K1

ECE201 Lect-20


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Step-by-Step Approach

  • Find time constant ()

    Looking across the terminals of the C or L element, form Thevenin equivalent circuit; =RThC or =L/RTh

  • Finish up

    Simply put the answer together.

ECE201 Lect-20


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Class Examples

  • Learning Extension E7.3

  • Learning Extension E7.4

  • Learning Extension E7.5

ECE201 Lect-20


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