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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

First-Order Circuits Cont’d

Dr. Holbert

April 17, 2006

ECE201 Lect-20

  • 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

important concepts
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

the differential equation
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

the differential equation5
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

building intuition
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

differential equation solution
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

the forced solution
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

the natural solution
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

first order natural solution
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

initial conditions
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

transients and steady state
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

step by step approach
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

step by step approach14
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

Class Examples
  • Learning Extension E7.3
  • Learning Extension E7.4
  • Learning Extension E7.5

ECE201 Lect-20