2
Chapter 15- Circuit Analysis in the sDomain
Key Concepts
 Introduction
 15.1 Z(s) and Y(s)
 15.2 Nodal and Mesh Analysis in the sDomain
 15.4 Poles, Zeros, and Transfer Functions
Chapter 15- Circuit Analysis in the s-Domain
3
Introduction
You have studied the Laplace transform, included
are the transforms of the impulse function, the step
function, the exponential function, the ramp function,
the sine and cosine functions. In addition, the
consequences in the s-domain of the time-domain
operations of addition, multiplication by a constant,
differentiation, and integration. These results are
collected in Tables 14.1 and 14.2; several others
which are derived in Appendix 7 are also included
Chapter 15- Circuit Analysis in the s-Domain
4
Chapter 15- Circuit Analysis in the s-Domain
5
Chapter 15- Circuit Analysis in the s-Domain
6
 Having been introduced to the concept of complex frequency
and to the Laplace transform technique, we now are ready to
see the details of how circuit analysis in the s-domain
actually works.
 As the Chap. 10 has already been studied, in fact several
shortcuts are routinely applied. The first of these is to create
a new way of viewing capacitors and inductors so that sdomain nodal and mesh equations can be written directly.
 As part of this method, we will learn how to take care to
account for initial conditions. Another “shortcut” is the
concept of a circuit transfer function. This general function
can be exploited to predict the response of a circuit to
various inputs, its stability, and even its frequency-selective
response.
Chapter 15- Circuit Analysis in the s-Domain
7
15.1 Z(s) AND Y(s)
 The key concept that makes phasors so useful in the
analysis of sinusoidal steady-state circuits is the
transformation of resistors, capacitors, and inductors into
impedances.
 Circuit analysis then proceeds using the basic techniques of
nodal
or
mesh
analysis,
superposition,
source
transformation, as well as Thévenin or Norton equivalents.
This concept can be extended to the s-domain, since the
sinusoidal steady state is included in s-domain analysis as a
special case (where σ = 0).
Resistors in the Frequency Domain
Let’s begin with the simplest situation: a resistor connected to
a voltage source v(t). Ohm’s law specifies that
v(t) = R i(t)
Taking the Laplace transform of both sides,
V(s) = R I(s)
Chapter 15- Circuit Analysis in the s-Domain
8
Resistors in the Frequency Domain
Thus, the ratio of the frequency-domain representation of the
voltage to the frequency-domain representation of the current
is simply the resistance, R. Since we are working in the
frequency domain, we refer to this quantity as an impedance
for the sake of clarity, but still assign it the unit ohms (Ω):
Just as we found in working with phasors in the sinusoidal
steady state, the impedance of a resistor does not depend on
frequency. The admittance Y(s) of a resistor, defined as the
ratio of I(s) to V(s), is simply 1/R; the unit of admittance is the
siemen (S).
11
Inductors in the Frequency Domain
Equation [3] may be further simplified if we are only interested
in the sinusoidal steady-state response. It is permissible to
neglect the initial conditions in such instances as they only
affect the nature of the transient response. Thus, we substitute
s = jω and find
Z( jω) = jωL
as was obtained previously in Chap. 10.
V(s) = sLI(s) − Li(0−)
12
Note that the inductor symbol labeled with an admittance
Y(s) = 1/sL,
it can also be viewed as an impedance
Z(s) = sL;
.
13
Example 15.1
14
PRACTICE
Determine the current i (t) in the circuit of Fig.
15
Modeling Capacitors in the s-Domain
16
The same concepts apply to capacitors in the s-domain as well.
Following the passive sign convention as illustrated in Fig. 15.5a, the
governing equation for capacitors is
Taking the Laplace transform of both sides results in
Modeling Capacitors in the s-Domain
17
which can be modeled as an admittance sC in parallel with a current source
Cv(0−) as shown in Fig. 15.5b. Performing a source transformation on this
circuit (taking care to follow the passive sign convention) results in an
equivalent model for the capacitor consisting of an impedance 1/sC in series
with a voltage source v(0−)/s, as shown in Fig. 15.5c.
In working with these s-domain equivalents, we should be careful not to be
confused with the independent sources being used to include initial
conditions. The initial condition for an inductor is given as i (0−); this term
may appear as part of either a voltage source or a current source, depending
on which model is chosen. The initial condition for a capacitor is given as
v(0−); this term may thus appear as part of either a voltage source or a
current source. A very common mistake working with s-domain analysis for
the first time is to always use v(0−) for the voltage source component of the
model, even when dealing with an inductor.
EXAMPLE 15.2 Determine vC (t) in the circuit of Fig.
15.6a, given an initial voltage vC (0−) = −2 V.
18
19
20
Consider the circuit in Fig. Find the value of the voltage
across the capacitor assuming that the value of vs(t) = 10 u(t) V and
assume that at t = 0, -1 A flows through the inductor and +5 V is across the
capacitor.
Example:
21
15.2 NODAL AND MESH ANALYSIS IN THE s-DOMAIN
In Chap. 10, we learned how to transform time-domain
circuits driven by sinusoidal sources into their
frequency-domain equivalents. The benefits of this
transformation were immediately evident, as we were
no longer required to solve integrodifferential
equations. Nodal and mesh analysis of such circuits
(restricted to determining only the steady-state
response) resulted in algebraic expressions in terms of
jω, ω being the frequency of the sources.
22
Example:
23
24
Example:
25
26
27
15.4 POLES, ZEROS, AND TRANSFER FUNCTIONS
28
Consider the simple circuit in Fig. 15.19a. The s-domain
equivalent is given in Fig. 15.19b, and nodal analysis yields
15.4 POLES, ZEROS, AND TRANSFER FUNCTIONS
29
where H(s) is the transfer function of the circuit, defined as
the ratio of the output to the input.
The concept of a transfer function is very important, There
are two reasons for this.
 First, once we know the transfer function of a particular
circuit, we can easily find the output that results from any
input. All we need to do is multiply H(s) by the input
quantity, and take the inverse transform of the resulting
expression.
 Second, the form of the transfer function contains a great
deal of information about the behavior we might expect from
a particular circuit (or system).
15.4 POLES, ZEROS, AND TRANSFER FUNCTIONS
30
 In order to evaluate the stability of a system it is necessary
to determine the poles and zeros of the transfer function
H(s); we will explore this issue in detail shortly. Writing Eq.
[7] as
 we see that the magnitude of this function approaches zero
as s→∞. Thus, we say that H(s) has a zero at s=∞.
 The function approaches infinity at s = −1/RC; we therefore
say that H(s) has a pole at s = −1/RC.
 These frequencies are termed critical frequencies, and
their early identification simplifies the construction of the
response curves.
Example: For the RC circuit in Fig (a), obtain the transfer function
Vo / Vs and its frequency response. Let vs = Vm cosωt.
Solution:
31
The frequency-domain
equivalent of the circuit is in
Fig (b). By voltage division,
the transfer function is given
by

32

33
Thanks
