Transcript Applications to Difference Equations

Let S is the space of
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}.
…
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.
The crystal clear sounds from a
compact disc player are
produced from music that has
been sampled at the rate of
44,100 times per second. At each
measurement, the amplitude of
the music signal is recorded as a
number, say, yk.
…
The original music is composed
of many different sounds of
varying frequencies, yet the
sequence {yk} contains enough
information to reproduce all the
frequencies in the sound up to
about 20,000 cycles per second,
higher than the human ear can
sense.
k
1,
k
(-2)
k
3
Verify that
and
are linearly independent
signals.
1
 k+1
Casorati Matrix : 1
1k+ 2

For k = 0 :
k
1 1 1 
1 -2 3


1 4 9 
1 1 1 
0 -3 2 


0 3 8 
(-2)
k
(-2)
k+1
(-2)
k+ 2
3 
k+1 
3 
k+ 2 
3 
k
1 1 1 
0 -3 2 


0 0 10  …
The Casorati matrix
is invertible for k = 0.
k
k
k
So 1 , (-2) and 3 are
linearly independent.
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 ,  k
is called a linear difference
equation (or linear recurrence
relation) of order n.
…
For simplicity, a0 is often
taken equal to 1. If {zk} is the
zero sequence, the equation
is homogeneous; otherwise,
the equation is nonhomogeneous.
In general, a nonzero signal rk
satisfies the homogeneous
difference equation
yk+n  a1 yk+n-1  ...  an1 yk+1  an yk  0, k
if and only if r is a root of the
auxiliary equation
…
r  a1r
n
n-1
 ...  an1r  an .1  0
When the auxiliary equation
has a complex root, the
difference equation has
k
solutions of the form s cos kw
and sk sin kw, for constants s
and w.
Given a1, … , an, consider the
mapping T: S  S that
transforms a signal {yk}into a
signal {wk} given by
wk  yk+n  a1 yk+n-1  ...  an1 yk+1  an yk
It is readily checked that T is a
linear transformation.
…
This implies that the solution
set of the homogeneous
equation
yk+n  a1 yk+n-1  ...  an1 yk+1  an yk  0, k
is the kernel of T (the set of
signals that T maps into the
zero signal) and hence the
solution set is a subspace of
S.
Any linear
combination of
solutions is again
a solution.
If an  0 and if {zk} is given,
the equation
yk+n+a1yk+n-1+…+an1yk+1+anyk=zk,
for all k has a unique
solution whenever y0,…, yn-1
are specified.
The set H of all solutions of
the nth-order homogeneous
linear difference equation
yk+n+a1yk+n-1+…+an-1yk+1+anyk=0,
for all k is an n-dimensional
vector space.
Find a basis for the set of
all solutions to the
difference equation
yk+3 – 2yk+2 – 5yk+1 + 6yk = 0
for all k
Verify that the signal yk = k2
satisfies the difference
equation
yk+2 – 4yk+1 + 3yk = -4k for all k
Then find a description of all
solutions of this equation.
A modern way to study a
homogeneous nth-order
linear difference equation is
to replace it by an equivalent
system of first order
difference equations,
…
written in the form
xk+1 = Axk for k = 0, 1, 2, …
Where the vectors xk are in
Rn and A is an n x n matrix.
Write the following
difference equation as a
first order system:
yk+3 – 2yk+2 – 5yk+1 + 6yk = 0
for all k
It can be shown that the
Sin
k
π
/
2
k
k
signals 2 , 3
, and
Cos
k
π
/
2
k
3
are solutions of
yk+3 – 2yk+2 + 9yk+1 – 18yk = 0
…
Show that these
signals form a basis
for the set of all
solutions of the
difference equation.