Laplace Transform
Solutions of Transient
Circuits
Dr. Holbert
March 5, 2008
Lect13
EEE 202
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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Laplace Circuit Solutions
• In this chapter we will use previously
established techniques (e.g., KCL, KVL,
nodal and loop analyses, superposition,
source transformation, Thevenin) in the
Laplace domain to analyze circuits
• The primary use of Laplace transforms
here is the transient analysis of circuits
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Laplace Circuit Element Models
• Here we develop s-domain models of circuit
elements
• DC voltage and current sources basically remain
unchanged except that we need to remember
that a dc source is really a constant, which is
transformed to a 1/s function in the Laplace
domain
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Resistor
• We start with a simple (and trivial) case, that of the
resistor, R
• Begin with the time domain relation for the element
v(t) = R i(t)
• Now Laplace transform the above expression
V(s) = R I(s)
• Hence a resistor, R, in the time domain is simply that
same resistor, R, in the s-domain
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Capacitor
• Begin with the time domain relation for the element
d v(t)
i(t)  C
dt
• Now Laplace transform the above expression
I(s) = s C V(s) – C v(0)
• Interpretation: a charged capacitor (a capacitor with
non-zero initial conditions at t=0) is equivalent to an
uncharged capacitor at t=0 in parallel with an
impulsive current source with strength C·v(0)
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Capacitor (cont’d.)
• Rearranging the above expression for the capacitor
I(s) v(0)
V(s) 

sC
s
• Interpretation: a charged capacitor can be replaced
by an uncharged capacitor in series with a stepfunction voltage source whose height is v(0)
• Circuit representations of the Laplace transformation
of the capacitor appear on the next page
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Capacitor (cont’d.)
iC(t)
+
Time
Domain
vC(t)
C
–
IC(s)
+
VC(s)
–
IC(s)
+
1/sC
+
–
1/sC
Cv(0)
VC(s)
–
v(0)
s
Laplace (Frequency) Domain Equivalents
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Inductor
• Begin with the time domain relation for the
element
d i(t)
v(t)  L
dt
• Now Laplace transform the above expression
V(s) = s L I(s) – L i(0)
• Interpretation: an energized inductor (an
inductor with non-zero initial conditions) is
equivalent to an unenergized inductor at t=0 in
series with an impulsive voltage source with
strength L·i(0)
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Inductor (cont’d.)
• Rearranging the above expression for the
inductor
V(s) i(0)
I(s) 
sL

s
• Interpretation: an energized inductor at t=0 is
equivalent to an unenergized inductor at t=0 in
parallel with a step-function current source with
height i(0)
• Circuit representations of the Laplace
transformation of the inductor appear on the
next page
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Inductor (cont’d.)
+
Time
Domain
vL(t)
iL(0)
L
–
IL(s)
+
sL
VL(s)
–
IL(s)
+
–
+
VL(s)
–
Li(0)
sL
i(0)
s
Laplace (Frequency) Domain Equivalents
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Analysis Techniques
• In this section we apply our tried and
tested analysis tools and techniques to
perform transient circuit analyses
– KVL, KCL, Ohm’s Law
– Voltage and Current division
– Loop/mesh and Nodal analyses
– Superposition
– Source Transformation
– Thevenin’s and Norton’s Theorems
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Transient Analysis
• Sometimes we not only must Laplace transform
the circuit, but we must also find the initial
conditions
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Element
Capacitor
DC Steady-State
I = 0; open circuit
Inductor
V = 0; short circuit
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Class Examples
• Drill Problems P6-4, P6-5
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