Applications to Linear Difference Equations

This presentation discusses the applications and properties of linear difference equations. It covers topics such as discrete-time signals, linear independence in the space of signals, and provides examples. Find study material and solved assignments on Desklib.

Applications to
Linear Difference
Equations

Outline
• Introduction
• Applications
• Properties
• Discrete time signals
• Linear Independence in
the space S of Signals
• Examples

Introduction
Difference Equations are often used to produce numerical solutions when
differential equations is used to model a continuous process.
Given scalars a0, …, an, with a0 and an nonzero, and given a signal {zk}, the
equation
a0 yk+n + a1 yk+n-1 + + an-1 yk+1 + an yk = zk for all k (3)⋯
is called a linear difference equation (or linear recurrence relation) of order n. For
simplicity, a0 is often taken equal to 1.

Applications
• Difference Equations are used in a vector space
dealing with time (t) so the application is used
in digital signals that arise in electrical and
control systems: such as stop lights.

Discrete-time signals
• A signal in S is a function defined only on the integers and is
visualized as a sequence of numbers, say, {yk}.
• Figure shows four typical signals whose general terms (.7) k, 1k, (-
1) k, and (-.7) k can be chosen from the pulldown menu.
• Digital signals obviously arise in electrical and control systems
engineering, but discrete-data sequences are also generated in
biology, physics, economics, demography, and many other areas,
wherever a process is measured, or sampled, at discrete time
intervals. When a process begins at a specific time, it is sometimes
convenient to write a signal as a sequence of the form (y0, y1, y2,
…). The terms yk for k < 0 either are assumed to be zero or are
simply omitted.

Linear Independence in the Space S
of Signals
To simplify notation, we consider a set of only three signals in !, say, {uk}, {vk}, and
{wk}. They are linearly independent precisely when the equation
c1 uk + c2 vk + c3 wk = 0 for all k
implies that c1 = c2 = c3 = 0.
The phrase “for all k” means for all integers—positive, negative, and zero.
One could also consider signals that start with k = 0, for example, in which case,
“for all k” would mean for all integers k ≥ 0.

Casorati
Matrix
Suppose , c1, c2, c3 satisfy (1). Then equation (1) holds for any three
consecutive values of k, say, k, k + 1, and k + 2. Thus (1) implies that
c1 uk+1 + c2 vk+1 + c3 wk+1 = 0
and
c1 uk+2 + c2 vk+2 + c3 wk+2 = 0
Hence c1, c2, c3 satisfy
= for all k.
 The coefficient matrix in this system is called the Casorati matrix of the
signals.
 The determinant of the matrix is called the Casoratian of {uk}, {vk}, and {wk}.
 If the Casorati matrix is invertible for at least one value of k, then (2) will
imply that c1 = c2 = c3 = 0, which will prove that the three signals are linearly
independent.

Example 1
Write the following difference equation as a first-order system:
2 – 5 + 6 = 0 for all k.
So for each k,
= .
The difference equation if solved for :
= + 5 + 2 which is
= =
= .
This implies
= A for all k where A = .

Example 2
Find a basis for the solution space of the difference equation. Also prove
that the solution spans the solution set.
Given the equation: – + = 0.
An auxiliary equation can be written as:
– r + = 0 or
9 – 9r +2 =0.
From the quadratic formula, the solution for r:
r = = = and .
With these two solutions: and for the signals implies they are linearly
independent because neither solutions are multiples of the other.
Solution space is two-dimensional by the nth-order
homogeneous linear difference equation (Thm.) with these two signal for a
basis for the solution space by the Basis Theorem.

References

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