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v o [ 1 ] = α size 12{v rSub { size 8{o} } \[ 1 \] =α} {} ,

v o [ 2 ] = ( 1 α ) α + α size 12{v rSub { size 8{o} } \[ 2 \] = \( 1 - α \) α+α} {} ,

v o [ 3 ] = ( 1 α ) 2 α + ( 1 α ) α + α size 12{v rSub { size 8{o} } \[ 3 \] = \( 1 - α \) rSup { size 8{2} } α+ \( 1 - α \) α+α} {} ,

v o [ k ] = α k = 0 n 1 ( 1 α ) k = 1 ( 1 α ) n size 12{v rSub { size 8{o} } \[ k \] =α Sum cSub { size 8{k=0} } cSup { size 8{n - 1} } { \( 1 - α \) rSup { size 8{k} } } =1 - \( 1 - α \) rSup { size 8{n} } } {}

[We have made use of an important formula for the summation of a geometric series which we will prove shortly.] Hence, the solution is

v o [ n ] = ( 1 ( 1 α ) n ) u [ n ] size 12{v rSub { size 8{o} } \[ n \] = \( 1 - \( 1 - α \) rSup { size 8{n} } \) u \[ n \]} {}

The solution

v o [ n ] = ( 1 ( 1 α ) n ) u [ n ] size 12{v rSub { size 8{o} } \[ n \] = \( 1 - \( 1 - α \) rSup { size 8{n} } \) u \[ n \]} {}

is shown plotted below for α size 12{α} {} = 0.25.

4/ Important side issue — the summation of finite geometric series

Suppose

S = n = l k a n = a l + a l + 1 + . . . + a k size 12{S= Sum cSub { size 8{n=l} } cSup { size 8{k} } {a rSup { size 8{n} } } =a rSup { size 8{l} } +a rSup { size 8{l+1} } + "." "." "." +a rSup { size 8{k} } } {}

Then

αS = α n = l k a n = a l + 1 + a l + 2 + . . . + a k + 1 size 12{αS=α Sum cSub { size 8{n=l} } cSup { size 8{k} } {a rSup { size 8{n} } } =a rSup { size 8{l+1} } +a rSup { size 8{l+2} } + "." "." "." +a rSup { size 8{k+1} } } {}

Hence,

( 1 α ) S = ( 1 α ) n = l k a n = a l a k + 1 size 12{ \( 1 - α \) S= \( 1 - α \) Sum cSub { size 8{n=l} } cSup { size 8{k} } {a rSup { size 8{n} } } =a rSup { size 8{l} } - a rSup { size 8{k+1} } } {}

and

S = n = l k a n = a l a k + 1 ( 1 α ) size 12{S= Sum cSub { size 8{n=l} } cSup { size 8{k} } {a rSup { size 8{n} } } = { {a rSup { size 8{l} } - a rSup { size 8{k+1} } } over { \( 1 - α \) } } } {}

Conclusion

  • Difference equations arise in a variety of interesting contexts.
  • Simple difference equations can be solved iteratively.
  • We will develop more systematic and efficient methods to solve arbitrary difference equations.

X. GENERAL LINEAR DIFFERENCE EQUATION

For a DT system, such as the ones shown previously, we can write the relation between an input variable x[n] and an output variable y[n]by a general difference equation of the form

k = 0 K a k y [ n + k ] = l = 0 L b l x [ n + 1 ] size 12{ Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } y \[ n+k \] } = Sum cSub { size 8{l=0} } cSup { size 8{L} } {b rSub { size 8{l} } x \[ n+1 \]} } {}

We seek a solution of this system for an input that is of the form

x [ n ] = Xz n u [ n ] size 12{x \[ n \] = ital "Xz" rSup { size 8{n} } u \[ n \]} {}

where X and z are in general complex quantities and u[n] is the unit step function. We will also assume that the system is at rest for n<0, so that y[n] = 0 for n<0.

We will seek the solution for n>0 by finding the homogeneous (unforced solution) and then a particular solution (forced solution).

1/ Homogeneous solution

a/ Geometric (exponential) solution

Let the homogeneous solution be y h [ n ] size 12{y rSub { size 8{h} } \[ n \] } {} , then the homogeneous equation is

k = 0 K a k y [ n + k ] = 0 size 12{ Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } y \[ n+k} \] =0} {}

To solve this equation we assume a solution of the form

y h [ n ] = n size 12{y rSub { size 8{h} } \[ n \] =Aλ rSup { size 8{n} } } {}

Since

y h [ n + k ] = n + k size 12{y rSub { size 8{h} } \[ n+k \] =Aλ rSup { size 8{n+k} } } {}

we have

k = 0 K a k n + k = 0 size 12{ Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } Aλ rSup { size 8{n+k} } } =0} {}

b/ Characteristic polynomial

The equation can be factored to yield

( k = 0 K a k λ k ) n = 0 size 12{ \( Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } λ rSup { size 8{k} } \) } Aλ rSup { size 8{n} } =0} {}

We are not interested in the trivial solution

y h [ n ] = n = 0 size 12{y rSub { size 8{h} } \[ n \] =Aλ rSup { size 8{n} } =0} {}

Therefore, we can divide by n size 12{Aλ rSup { size 8{n} } } {} to obtain the characteristic polynomial

k = 0 K a k λ k = 0 size 12{ Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } λ rSup { size 8{k} } } =0} {}

This polynomial of order K has K roots which can be exposed by writing the polynomial in factored form

k = 1 K ( λ λ k ) = 0 size 12{ Prod cSub { size 8{k=1} } cSup { size 8{K} } { \( λ - λ rSub { size 8{k} } \) } =0} {}

c/ Natural frequencies

The roots of the characteristic polynomial { λ 1 , λ 2 , . . . λ K } size 12{ lbrace λ rSub { size 8{1} } ,λ rSub { size 8{2} } , "." "." "." λ rSub { size 8{K} } rbrace } {} are called natural frequencies. These are frequencies for which there is an output in the absence of an input.

If the natural frequencies are distinct, i.e., if λ i λ k size 12{λ rSub { size 8{i} }<>λ rSub { size 8{k} } } {} for i k size 12{i<>k} {} the most general homogeneous solution has the form

y h [ n ] = k = 1 K A k λ k n size 12{y rSub { size 8{h} } \[ n \] = Sum cSub { size 8{k=1} } cSup { size 8{K} } {A rSub { size 8{k} } λ rSub { size 8{k} rSup { size 8{n} } } } } {}

Two-minute miniquiz problem

Problem 10-1 — Natural frequencies of the ladder network

The electric ladder network shown below has the following difference equation

v o [ n + 1 ] + ( 2 + rg ) v o [ n ] v o [ n 1 ] = rgv i [ n ] size 12{ - v rSub { size 8{o} } \[ n+1 \] + \( 2+ ital "rg" \) v rSub { size 8{o} } \[ n \]- v rSub { size 8{o} } \[ n - 1 \] = ital "rgv" rSub { size 8{i} } \[ n \]} {}

Find the natural frequencies.

Solution

The natural frequencies are determined by the homogeneous solution

v o [ n + 1 ] + ( 2 + rg ) v o [ n ] v o [ n 1 ] = 0 size 12{ - v rSub { size 8{o} } \[ n+1 \] + \( 2+ ital "rg" \) v rSub { size 8{o} } \[ n \]- v rSub { size 8{o} } \[ n - 1 \] =0} {}

Substituting a solution of the form v 0 [ n ] = n size 12{v rSub { size 8{0} } \[ n \] =Aλ rSup { size 8{n} } } {} results in

{} n + 1 + ( 2 + rg ) n n 1 1 = 0 size 12{ - Aλ rSup { size 8{n+1} } + \( 2+ ital "rg" \) Aλ rSup { size 8{n} } - Aλ rSup { size 8{n - 1} } - 1=0} {}

which yields the characteristic polynomial

λ 2 ( 2 + rg ) λ + 1 = 0 size 12{λ rSup { size 8{2} } - \( 2+ ital "rg" \) λ+1=0} {}

whose roots are the characteristic frequencies

λ 1,2 = ( 1 + rg 2 ) ± ( 1 + rg 2 ) 2 1 size 12{λ rSub { size 8{1,2} } = \( 1+ { { ital "rg"} over {2} } \) +- sqrt { \( 1+ { { ital "rg"} over {2} } \) rSup { size 8{2} } - 1} } {}

d/ Natural frequencies of the ladder network

The electric ladder network shown below has the following difference equation

v o [ n + 1 ] + ( 2 + rg ) v o [ n ] v o [ n 1 ] = rgv i [ n ] size 12{ - v rSub { size 8{o} } \[ n+1 \] + \( 2+ ital "rg" \) v rSub { size 8{o} } \[ n \]- v rSub { size 8{o} } \[ n - 1 \] = ital "rgv" rSub { size 8{i} } \[ n \]} {}

The natural frequencies are:

λ 1,2 = ( 1 + rg 2 ) ± ( 1 + rg 2 ) 2 1 size 12{λ rSub { size 8{1,2} } = \( 1+ { { ital "rg"} over {2} } \) +- sqrt { \( 1+ { { ital "rg"} over {2} } \) rSup { size 8{2} } - 1} } {}

The natural frequencies depend on rg,

Questions & Answers

where we get a research paper on Nano chemistry....?
Maira Reply
what are the products of Nano chemistry?
Maira Reply
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
Google
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
Hafiz Reply
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
Jyoti Reply
I only see partial conversation and what's the question here!
Crow Reply
what about nanotechnology for water purification
RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
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Alexandre
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
what is a peer
LITNING Reply
What is meant by 'nano scale'?
LITNING Reply
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Signals and systems. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10803/1.1
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