denominator, and to write
the transfer function in terms of those factors:
H(s) = N (s)
D(s) = K
(s z
1
)(s z
2
) . . . (s z
m1
)(s z
m
)
(s p
1
)(s p
2
) . . . (s p
n1
)(s p
n
) ,
(2)
where the numerator and denominator polynomials, N (s) and D(s), have real coecients dened
by the systems dierential equation and K = b
m
/a
n
. As written in Eq. (2) the z
i
s are the roots
of the equation
N (s) = 0,
(3)
and are dened to be the system zeros, and the
p
i
s are the roots of the equation
D(s) = 0,
(4)
and are dened to be the system poles. In Eq. (2) the factors in the numerator and denominator
are written so that when s = z
i
the numerator N (s) = 0 and the transfer function vanishes, that is
lim
sz
i
H(s) = 0.
and similarly when s = p
i
the denominator polynomial D(s) = 0 and the value of the transfer
function becomes unbounded,
lim
sp
i
H(s) = .
All of the coecients of polynomials N (s) and D(s) are real, therefore the poles and zeros must
be either purely real, or appear in complex conjugate pairs. In general for the poles, either p
i
=
i
,
or else p
i
, p
i+1
=
i
�j
i
. The existence of a single complex pole without a corresponding conjugate
pole would generate complex coecients in the polynomial D(s). Similarly, the system zeros are
either real or appear in complex conjugate pairs.
1 Example
A linear system is described by the dierential equation
d
2
y
dt
2
+ 5 dy
dt + 6y = 2
du
dt + 1.
Find the system poles and zeros.
Solution: From the dierential equation the transfer function is
H(s) =
2s + 1
s
2
+ 5s + 6 .
(5)
which may be written in factored form
H(s) = 12 s + 1/2
(s + 3)(s + 2)
=
1
2
s (1/2)
(s (3))(s (2)) .
(6)
The system therefore has a single real zero at s = 1/2, and a pair of real poles at
s = 3 and s = 2.
The poles and zeros are properties of the transfer function, and therefore of the dierential
equation describing the input-output system dynamics. Together with the gain constant K they
completely characterize the dierential equation, and provide a complete description of the system.
Example
A system has a pair of complex conjugate poles p
1
, p
2
=
1 � j2, a single real zero
z
1
=
4, and a gain factor K = 3. Find the dierential equation representing the
system.
Solution: The transfer function is
H(s) = K
s
z
(s p
1
)(s p
2
)
= 3
s (4)
(s (1 + j2))(s (1 j2))
= 3
(s + 4)
s
2
+ 2s + 5
(7)
and the dierential equation is
d
2
y
dt
2
+ 2 dy
dt + 5y = 3
du
dt + 12u
(8)
2 Figure 1: The pole-zero plot for a typical third-order system with one real pole and a complex
conjugate pole pair, and a single real zero.
1.1
The Pole-Zero Plot
A system is characterized by its poles and zeros in the sense that they allow reconstruction of the
input/output dierential equation. In general, the poles and zeros of a transfer function may be
complex, and the system dynamics may be represented graphically by plotting their locations on
the complex s-plane, whose axes represent the real and imaginary parts of the complex variable s.
Such plots are known as pole-zero plots. It is usual to mark a zero location by a circle (
) and a
pole location a cross (
�). The location of the poles and zeros provide qualitative insights into the
response characteristics of a system. Many computer programs are available to determine the poles
and zeros of a system from either the transfer function or the system state equations [8]. Figure 1
is an example of a pole-zero plot for a third-order system with a single real zero, a real pole and a
complex conjugate pole pair, that is;
H(s) =
(3s + 6)
(s
3
+ 3s
2
+ 7s + 5) = 3
(s (2))
(s (1))(s (1 2j))(s (1 + 2j))
1.2
System Poles and the Homogeneous Response
Because the transfer function completely represents a system dierential equation, its poles and
zeros eectively dene the system response. In particular the system poles directly dene the
components in the homogeneous response. The unforced response of a linear SISO system to a set
of initial conditions is
y
h
(t) =
n
i=1
C
i
e i
t
(9)
where the constants C
i
are determined from the given set of initial conditions and the exponents
i
are the roots of the characteristic equation or the system eigenvalues. The characteristic equation
is
D(s) = s
n
+ a
n1
s
n1
+ . . . + a
0
= 0,
(10)
and its roots are the system poles, that is
i
= p
i
, leading to the following important relationship:
3 Figure 2: The specication of the form of components of the homogeneous response from the system
pole locations on the pole-zero plot.
The transfer function poles are the roots of the characteristic equation, and also the
eigenvalues of the system
A matrix.
The homogeneous response may therefore be written
y
h
(t) =
n
i=1
C
i
e
p
i
t
.
(11)
The location of the poles in the s-plane therefore dene the n components in the homogeneous
response as described below:
1. A real pole p
i
=
in the left-half of the s-plane denes an exponentially decaying component
, Ce
t
, in the homogeneous response. The rate of the decay is determined by the pole
location; poles far from the origin in the left-half plane correspond to components that decay
rapidly, while poles near the origin correspond to slowly decaying components.
2. A pole at the origin p
i
= 0 denes a component that is constant in amplitude and dened by
the initial conditions.
3. A real pole in the right-half plane corresponds to an exponentially increasing component Ce
t
in the homogeneous response; thus dening the system to be unstable.
4. A complex conjugate pole pair � j in the left-half of the s-plane combine to generate a
response component that is a decaying sinusoid of the form Ae
t
sin (t + ) where A and
are determined by the initial conditions. The rate of decay is specied by ; the frequency
of oscillation is determined by .
5. An imaginary pole pair, that is a pole pair lying on the imaginary axis,
�j generates an
oscillatory component with a constant amplitude determined by the initial conditions.
4 6. A complex pole pair in the right half plane generates an exponentially increasing component.
These results are summarized in Fig. 2.
Example
Comment on the expected form of the response of a system with a pole-zero plot shown
in Fig. 3 to an arbitrary set of initial conditions.
Figure 3: Pole-zero plot of a fourth-order system with two real and two complex conjugate poles.
Solution: The system has four poles and no zeros. The two real poles correspond to
decaying exponential terms C
1
e
3t
and C
2
e
0.1t
, and the complex conjugate pole pair
introduce an oscillatory component Ae
t
sin (2t + ), so that the total homogeneous
response is
y
h
(t) = C
1
e
3t
+ C
2
e
0.1t
+ Ae
t
sin (2t + )
(12)
Although the relative strengths of these components in any given situation is determined
by the set of initial conditions, the following general observations may be made:
1. The term e
3t
, with a time-constant of 0.33 seconds, decays rapidly and is
signicant only for approximately 4 or 1.33seconds.
2. The response has an oscillatory component Ae
t
sin(2t + ) dened by the com-
plex conjugate pair, and exhibits some overshoot. The oscillation will decay in
approximately four seconds because of the e
t
damping term.
3. The term e
0.1t
, with a time-constant = 10 seconds, persists for approximately
40 seconds. It is therefore the dominant long term response component in the
overall homogeneous response.
5 Figure 4: Denition of the parameters
n
and for an underdamped, second-order system from
the complex conjugate pole locations.
The pole locations of the classical second-order homogeneous system
d
2
y
dt
2
+ 2
n
dy
dt +
2n
y = 0,
(13)
described in Section 9.3 are given by
p
1
, p
2
= n
�
n 2
1.
(14)
If 1, corresponding to an overdamped system, the two poles are real and lie in the left-half
plane. For an underdamped system, 0
< 1, the poles form a complex conjugate pair,
p
1
, p
2
= n
� j
n
1

2
(15)
and are located in the left-half plane, as shown in Fig. 4. From this gure it can be seen that the
poles lie at a distance
n
from the origin, and at an angle
� cos
1
() from the negative real axis.
The poles for an underdamped second-order system therefore lie on a semi-circle with a radius
dened by
n
, at an angle dened by the value of the damping ratio .
1.3
System Stability
The stability of a linear system may be determined directly from its transfer function. An nth order
linear system is asymptotically stable only if all of the components in the homogeneous response
from a nite set of initial conditions decay to zero as time increases, or
lim
t
n
i=1
C
i
e
p
i
t
= 0.
(16)
where the p
i
are the system poles. In a stable system all components of the homogeneous response
must decay to zero as time increases. If any pole has a positive real part there is a component in
the output that increases without bound, causing the system to be unstable.
6 In order for a linear system to be stable, all of its poles must have negative real parts,
that is they must all lie within the left-half of the s-plane. An unstable pole, lying in
the right half of the s-plane, generates a component in the system homogeneous response
that increases without bound from any nite initial conditions. A system having one
or more poles lying on the imaginary axis of the s-plane has non-decaying oscillatory
components in its homogeneous response, and is dened to be marginally stable.
2
Geometric Evaluation of the Transfer Function
The transfer function may be evaluated for any value of s = + j, and in general, when s is
complex the function H(s) itself is complex. It is common to express the complex value of the
transfer function in polar form as a magnitude and an