5.7 Impulse Functions
In some applications, it is necessary to deal with
phenomena of an impulsive nature—for example, voltages
or forces of large magnitude that act over very short time
intervals. Such problems often lead to differential equations
of the form ay'' + by' + cy = g(t), where g(t) is large during a
short time interval t0 ≤ t < t0+ε and is otherwise zero.
is a measure of the strength of
the input function.
Example
Let I0 be a real number and
let ε be a small positive
constant. Suppose that t0 is
zero and that g(t) is given by
g(t) = I0δε(t), where
Example (continued)
A model for a simple, undamped, massspring system (m = 1, γ = 0, k = 1) in a state
of equilibrium at t = 0, and subjected to the
force I0δε(t) is y'' + y = I0δε(t), y(0) = 0, y'(0) =
0.
 The response, obtained by using Laplace
transforms, is

Example (Continued)
Definition of the Unit Impulse
Function δ, the Dirac delta function.
The unit impulse function δ, that imparts an
impulse of magnitude one at t = t0, but is zero
for all values of t other than t0.
 Properties that define δ
 1.


2.

3.
The Laplace Transform of δ(t − t0).

Since e−st is a continuous function of t for all t
and s,

for all t0 ≥ 0. In the important case when t0 = 0,
Example
Find the solution of the initial value problem
2y'' + y' + 2y = δ(t − 5), y(0) = 0, y' (0) = 0.

δ(t−t0) as the Derivative of u(t−t0).

In view of the definition of δ(t − t0) given
above
or

for all t ≠ t0. Formally differentiating both sides
with respect to t and treating δ(t − t0) as an
ordinary function such that the indefinite
integral on the left-hand side of Eq. actually
satisfies the fundamental theorem of calculus,
we find that δ(t − t0) = u'(t − t0), that is, the
delta function is the derivative of the unit step
function.
5.8 Convolution Integrals and
Their Applications

DEFINITION 5.8.1Let f (t) and g(t) be
piecewise continuous functions on [0,∞). The
convolution of f and g is defined by
The integral in this Eq is known as a
convolution integral.
 The convolution integral can be thought of as
a “generalized product” by writing
h(t) = ( f ∗ g)(t).

THEOREM 5.8.2 – Properties of *

f ∗ g = g ∗ f (commutative law)

f∗(g1+g2) = f∗g1+f∗g2 (distributive law)

( f ∗ g) ∗ h = f ∗ (g ∗ h) (associative
law)

f ∗ 0 = 0 ∗ f = 0.
THEOREM 5.8.3 - Convolution
Theorem.
If F(s) = L{ f (t)} and G(s) = L{g(t)} both
exist for s > a ≥ 0, then
H(s) = F(s)G(s) = L{h(t)}, s > a,
where
Example
Test Theorem 5.8.3 by finding the
convolution of f (t) = t and g(t) = e−2t in
two different ways:
(i) by direct evaluation of the
convolution integral (3) in Definition
5.8.1, and
(ii) by directly computing the inverse
Laplace transform of F(s)G(s), the
product of the Laplace transforms of f
(t) and g(t).

Free and Forced Responses of
Input–Output Problems.
Input–output problem The differential
equation ay'' + by' + cy = g(t), where a, b,
and c are real constants and g is a given
function, together with the initial conditions
y(0) = y0, y'(0) = y1 is often referred to as an
input–output problem.
 The solution of the initial value by setting
g(t)=0 is called the free response of the
system in the t-domain.
 The solution of the initial value problem with
initial values y0=0 and y1=0 is called the
forced response of the system.

TABLE 5.8.1

The total response of ay'' + by' + cy = g(t), y(0)
= y0, y'(0) = y1 as a sum of the free response
and the forced response in both the s-domain
and the t-domain.
Transfer Functions and Impulse
Responses.
DEFINITION 5.8.4 - The transfer function
of the input–output problem is the ratio of
the forced response to the input in the sdomain. Equivalently, the transfer function
is the factor in the equation for Y(s)
multiplying the Laplace transform of the
input, G(s).
 The transfer function for the input–output
problem is H(s) defined as,
H(s) = 1/(as2+bs+c).
 The inverse Laplace transform of the
transfer function, h(t) = L−1{H(s)}, is called
the impulse response of the system.

To obtain the system output in the
t-domain
1. Find the transfer function H(s).
 2. Find the Laplace transform of the
input, G(s).
 3. Construct the output in the sdomain, a simple algebraic operation
Yg(s) = H(s)G(s).
 4. Compute the output in the t-domain,
yg(t) = L−1{Yg(s)}.

Example
Consider the input–output system
y'' + 2y' + 5y = g(t),
where g(t) is any piecewise continuous function of
exponential order.
1. Find the transfer function and the impulse
response.
2. Using a convolution integral to represent the
forced response, find the general solution of Diff. eq.
3. Find the total response if the initial state of the
system is prescribed by y(0) = 1, y(0)=−3.
4. Compute the forced response when g(t) = t.
5.9 Linear Systems and Feedback
Control - open-loop and closed-loop
systems.

An open-loop control system is one in which the
control action is independent of the output. An
example of an open-loop system is a toaster that is
controlled by a timer.
A closed-loop control system is one in which the
control action depends on the output in some manner,
for example, a home heating and cooling system
controlled by a thermostat. The output is specified by
Y(s) = H(s)F(s), where F(s) represents the control
input. The output is used as
an input to an element,
represented by a transfer function
G(s), called a controller.

Example: Block Diagram
Block diagram of a feedback control
system with both positive and negative
feedback loops.
Y1(s) = H1(s) [F(s) − U2(s)]
Y2(s) = H2(s) [Y1(s) + U1(s)]
U1(s) = G1(s)Y2(s)
U2(s) = G2(s)Y2(s).
One can solve this system for H(s) as
H(s) =

Poles, Zeros, and Stability.
DEFINITION 5.9.1 - A function f (t)
defined on 0 ≤ t <∞is said to be
bounded if there is a number M such
that | f (t)| ≤ M for all t ≥ 0.
 DEFINITION 5.9.2 - A system is said
to be bounded-input boundedoutput (BIBO) stable if every
bounded input results in a bounded
output.

THEOREM 5.9.3

An input–output system with strictly proper
transfer function
is BIBO stable if and only if all of the poles
have negative real parts.
* The zeros of H(s) are the solutions of P(s)
while poles of H(s) are the solutions of Q(s).
If transfer functions have more poles than
zeros, that is, n > m then they are called
strictly proper.
Root-Locus Analysis.
Theorem 5.9.3 points out the importance of pole
locations with regard to the linear system stability
problem. For rational transfer functions, the
mathematical problem is to determine whether all of
the roots of the polynomial equation
sn + an−1sn−1 +… +a0 = 0 have negative real parts.
Given numerical values of the polynomial coefficients,
powerful computer programs can then be used to
find the roots. It is often the case that one or more of
the coefficients of the polynomial depend on a
parameter. Then it is of major interest to understand
how the locations of the roots in the complex splane change as the parameter is varied. The graph
of all possible roots of Eq. relative to some particular
parameter is called a root locus. The design
technique based on this graph is called the root
locus method of analysis.

EXAMPLE


Suppose the open-loop
transfer function for a
plant is given by H(s) =
K/(s2 + 2s + 2) and that
it is desired to
synthesize a closedloop system with
negative feedback
using the controller
G(s) = 1/(s + 0.1).
The root locus of the
polynomial equation s3
+ 2.1s2 + 2.2s + K +
0.2 = 0 as K varies
from 1 to 10.
The Routh Stability Criterion.
THEOREM 5.9.4 The Routh Criterion.
 All of the roots of the polynomial
equation sn + an−1sn−1 +… +a0 = 0
have negative real parts if and only if
the elements of the first column of the
Routh table have the same sign.
Otherwise the number of roots with
positive real parts is equal to the
number of changes of sign.
Routh table is defined by

where 1, an−1, an−2,
. . . , a0 are the
coefficients of the
polynomial.
b1, b2, . . . ,
c1, c2, . . . are
defined by
these
quotients
Example
Chapter Summary
Summary (Ctd.)
Summary (Ctd.)
Summary (Ctd.)