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Transient Circuit Analysis
Cont’d.
Dr. Holbert
November 27, 2001
ECE201 Lect-24
1
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.
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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 ()
• 2nd order circuits: natural frequency (ω0)
and the damping ratio (ζ)
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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
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The Differential Equation
• Equations are linear, constant coefficient:
d n x(t )
d n 1 x(t )
an
 an 1
 ...  a0 x(t )  f (t )
n
n 1
dt
dt
• 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.
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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
– Roots of the characteristic equation
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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
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The Forced Solution
• The forced (particular) solution is the
solution to the non-homogeneous equation:
d n x(t )
d n 1 x(t )
an
 an 1
 ...  a0 x(t )  f (t )
n
n 1
dt
dt
• 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
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The Natural Solution
• The natural (or complementary) solution is
the solution to the homogeneous equation:
d n x(t )
d n 1 x(t )
an
 an 1
 ...  a0 x(t )  0
n
n 1
dt
dt
• Different “look” for 1st and 2nd order ODEs
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First-Order Natural Solution
• The first-order ODE has a form of
dxc (t ) 1
 xc (t )  0
dt

• The natural solution is
xc (t )  Ke
 t /
• Tau () is the time constant
• For an RC circuit,  = RC
• For an RL circuit,  = L/R
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Second-Order Natural Solution
• The second-order ODE has a form of
2
d x(t )
dx(t )
2

2



0
0 x (t )  0
2
dt
dt
• To find the natural solution, we solve the
characteristic equation:
s  2 0 s    0
2
2
0
• Which has two roots: s1 and s2.
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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.
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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.
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Step-by-Step Approach
1. Assume solution (only dc sources allowed):
i.
ii.
x(t) = K1 + K2 e-t/
x(t) = K1 + K2 es t + K3 es t
1
2
2. At t=0–, draw circuit with C as open circuit and
L as short circuit; find IL(0–) and/or VC(0–)
3. At t=0+, redraw circuit and replace C and/or L
with appropriate source of value obtained in
step #2, and find x(0)=K1+K2 (+K3)
4. At t=, repeat step #2 to find x()=K1
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Step-by-Step Approach
5.
Find time constant (), or characteristic roots (s)
i.
ii.
6.
Looking across the terminals of the C or L element,
form Thevenin equivalent circuit; =RThC or =L/RTh
Write ODE at t>0; find s from characteristic equation
Finish up
i.
ii.
Simply put the answer together.
Typically have to use dx(t)/dt│t=0 to generate another
algebraic equation to solve for K2 & K3 (try repeating
the circuit analysis of step #5 at t=0+, which basically
means using the values obtained in step #3)
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Class Examples
• Learning Extension E6.3
• Learning Extension E6.4
• Learning Extension E6.11
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