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caso1

Esto implica que al sumar todas las repeticiones de P(f) cada f b estas deben sumar una constante

caso2

El pulso resultante tiene simetría vestigial

Este filtro puede representarse como la suma de:

Matemáticamente:

P ( f ) = f f b + H 1 ( f ) p ( t ) = 1 t b Sinc t t b + h 1 ( t ) alignl { stack { size 12{P \( f \) = Prod { left ( { {f} over {f rSub { size 8{b} } } } right )} +H rSub { size 8{1} } \( f \) } {} #p \( t \) = { {1} over {t rSub { size 8{b} } } } ital "Sinc" left ( { {t} over {t rSub { size 8{b} } } } right )+h rSub { size 8{1} } \( t \) {} } } {}

H 1 (f) es simétrica y par.

h 1 ( t ) = H 1 ( f ) e jωt df = 2 0 H 1 ( f ) Cos ( ωt ) df h 1 ( t ) = 2 β H 1 ( f ) Cos ( ωt ) df + 2 + β H 1 ( f ) Cos ( ωt ) df c . d . v : { Para la primera Integral : f = x Para la segunda Integral : f = + x Entonces : h 1 ( t ) = 2 0 β H 1 fb 2 x Cos ( ( fb 2 x ) ) tdx + 2 0 β H 1 fb 2 + x Cos ( ( fb 2 + x ) ) tdx Por simetría : H 1 ( fb 2 + x ) = H 1 ( fb 2 x ) h 1 ( t ) = 2 0 β H 1 fb 2 + x Cos 2πt ( fb 2 + x ) Cos 2πt ( fb 2 x ) dx h 1 ( t ) = 4 0 β H 1 fb 2 + x Sen πf b t . Sen ( xt ) dx h 1 ( t ) = 4 Sen πf b t 0 β H 1 fb 2 + x Sen ( xt ) dx alignl { stack { size 12{h rSub { size 8{1} } \( t \) = Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } {H rSub { size 8{1} } \( f \) e rSup { size 8{jωt} } ital "df"=2} Int cSub { size 8{0} } cSup { size 8{ infinity } } {H rSub { size 8{1} } \( f \) ital "Cos" \( ωt \) ital "df"} } {} #h rSub { size 8{1} } \( t \) =2 Int cSub { size 8{ { ital "fb"} wideslash {2} - β} } cSup { size 8{ { ital "fb"} wideslash {2} } } {H rSub { size 8{1} } \( f \) ital "Cos" \( ωt \) ital "df"} +2 Int cSub { size 8{ { ital "fb"} wideslash {2} } } cSup { size 8{ { ital "fb"} wideslash {2} +β} } {H rSub { size 8{1} } \( f \) ital "Cos" \( ωt \) ital "df"} {} # matrix {{} # {} # {} # matrix { {} # {} # c "." d "." v: left lbrace matrix {matrix { ital "Para"` ital "la"` ital "primera"` ital "Integral": {} # f= { ital "fb"} wideslash {2} - x {} # {}} {} ## {} ##matrix { ital "Para"` ital "la"` ital "segunda"` ital "Integral": {} # f= { ital "fb"} wideslash {2} +x {} # {}} {} } right none {}} {} } {} #ital "Entonces": {} # h rSub { size 8{1} } \( t \) =2 Int cSub { size 8{0} } cSup { size 8{β} } {H rSub { size 8{1} } left ( { { ital "fb"} over {2} } - x right ) ital "Cos" \( 2π \( { { ital "fb"} over {2} } - x \) \) ital "tdx"} +2 Int cSub { size 8{0} } cSup { size 8{β} } {H rSub { size 8{1} } left ( { { ital "fb"} over {2} } +x right ) ital "Cos" \( 2π \( { { ital "fb"} over {2} } +x \) \) ital "tdx"} {} #{} # matrix {matrix { matrix {{} # {} } {} # {}} {} # ital "Por"` ital "simetría": {} # {} # H rSub { size 8{1} } \( { { ital "fb"} over {2} } +x \) {} } = - H rSub { size 8{1} } \( { { ital "fb"} over {2} } - x \) {} #{} # h rSub { size 8{1} } \( t \) =2 Int cSub { size 8{0} } cSup { size 8{β} } {H rSub { size 8{1} } left ( { { ital "fb"} over {2} } +x right ) left [ ital "Cos" left (2πt \( { { ital "fb"} over {2} } +x \) right ) - ital "Cos" left (2πt \( { { ital "fb"} over {2} } - x \) right ) right ]ital "dx"} {} # {} #h rSub { size 8{1} } \( t \) = - 4 Int cSub { size 8{0} } cSup { size 8{β} } {H rSub { size 8{1} } } left ( { { ital "fb"} over {2} } +x right ) left [ ital "Sen" left (πf rSub { size 8{b} } t right ) "." ital "Sen" \( 2π ital "xt" \) right ] ital "dx" {} #{} # h rSub { size 8{1} } \( t \) = - 4 ital "Sen" left (πf rSub { size 8{b} } t right ) Int cSub { size 8{0} } cSup { size 8{β} } {H rSub { size 8{1} } } left ( { { ital "fb"} over {2} } +x right ) ital "Sen" \( 2π ital "xt" \) ital "dx" {}} } {}

Para cada ntb el término que se encuentra fuera de la integral se anulará. De esta forma, se evita la interferencia.

A partir de este criterio podemos implementar el filtro de simetría vestigial de tipo Coseno Alzado : Este se caracteriza porque puede reducir la ISI. La parte no nula del espectro es un coseno que, en su forma más simple, está alzado (es decir, se encuentra por encima del eje de frecuencia):

Representación del pulso en los dominios de frecuencia y tiempo.

Segundo criterio de nyquist

En este criterio se busca no sólo eliminar la interferencia, también se presenta como objetivo el disminuir el ancho de banda. Esto se hace definiendo, en el transmisor, una interacción conocida entre pulsos vecinos. El sacrificio, en este caso, es un mayor consumo de potencia.

Entonces, en vez de transmitir a k (secuencia original), se enviará y k =a k +a k-1 . De esta forma se pueden enviar dos bits haciendo uso del mismo ancho de banda. Supongamos el siguiente ejemplo:

Secuencia original de bits: 01010011

Secuencia original 0 1 0 1 0 0 1 1
ak -1 1 -1 1 -1 -1 1 1
yk 0 0 0 0 -2 0 2

El filtro que se coloca en el transmisor pudiera modelarse como:

Teniendo considerada la condición de un sistema con interferencia, ahora se debe tomar en cuenta cuando se introduce ruido AWGN al canal. Supongamos que a la entrada de un sistema de comunicaciones se tiene una secuencia aleatoria, con código de línea NRZ y duración tb. La Densidad Espectral de Potencia sería:

G ( f ) = P ( f ) 2 tb ; P ( f ) Transformada de Fourier de la señal de entrada . alignl { stack { size 12{G \( f \) = { { lline P \( f \) rline rSup { size 8{2} } } over { ital "tb"} } ;} {} #P \( f \) rightarrow ital "Transformada"` ital "de"` ital "Fourier"` ital "de"` ital "la"` ital "señal"` ital "de"` ital "entrada" "." {} } } {}

Asumiendo un sistema como sigue:

La salida del sistema sería una sucesión de pulsos y(t), asociada a un pulso de salida p R (t) y a los de entrada:

A k P R ( f ) = P ( f ) . H T ( f ) . H c ( f ) . H R ( f ) Ecuación ( 1 ) Si la potencia de transmisión es : S T = P ( f ) 2 H T ( f ) 2 tb df Pudiéramos Expresarla en función de la ecuación ( 1 ) : tb . S T = A k 2 P R ( f ) 2 H c ( f ) . H R ( f ) df alignl { stack { size 12{A rSub { size 8{k} } lline P rSub { size 8{R} } \( f \) rline = lline P \( f \) rline "." lline H rSub { size 8{T} } \( f \) rline "." lline H rSub { size 8{c} } \( f \) rline "." lline H rSub { size 8{R} } \( f \) rline ~ rightarrow ` ital "Ecuación"` \( 1 \) } {} #{} # ital "Si"` ital "la"` ital "potencia"` ital "de"` ital "transmisión"` ital "es": {} #S rSub { size 8{T} } = Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } { { { lline P \( f \) rline rSup { size 8{2} } lline H rSub { size 8{T} } \( f \) rline rSup { size 8{2} } } over { ital "tb"} } } ital "df" {} # ital "Pudiéramos"` ital "Expresarla"` ital "en"` ital "función"` ital "de"` ital "la"` ital "ecuación"` \( 1 \) : {} #{} # matrix {{} # {} # {} } ital "tb" "." S rSub { size 8{T} } =A rSub { size 8{k} rSup { size 8{2} } } Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } { { { lline P rSub { size 8{R} } \( f \) rline rSup { size 8{2} } } over { lline H rSub { size 8{c} } \( f \) rline "." lline H rSub { size 8{R} } \( f \) rline } } } ital "df" {}} } {}

Ahora bien, como nuestro objetivo es maximizar la relación señal a ruido, despejamos el valor de Ak (Amplitud del pulso) y definimos σ² (que debe ser minimizado):

A k 2 = tb . S T P R ( f ) 2 H c ( f ) . H R ( f ) df y σ 2 = Gn ( f ) H R ( f ) 2 df Por lo que : A k 2 σ 2 = tb . S T Gn ( f ) H R ( f ) 2 df . P R ( f ) 2 H c ( f ) . H R ( f ) df Minimizar alignl { stack { size 12{A rSub { size 8{k} rSup { size 8{2} } } = { { ital "tb" "." S rSub { size 8{T} } } over { Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } { { { lline P rSub { size 8{R} } \( f \) rline rSup { size 8{2} } } over { lline H rSub { size 8{c} } \( f \) rline "." lline H rSub { size 8{R} } \( f \) rline } } } ital "df"} } } {} #y {} # σ rSup { size 8{2} } = Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } { ital "Gn" \( f \) } lline H rSub { size 8{R} } \( f \) rline rSup { size 8{2} } ital "df" {} #ital "Por"` ital "lo"` ital "que": {} # {} #{ {A rSub { size 8{k} rSup { size 8{2} } } } over {σ rSup { size 8{2} } } } = { { ital "tb" "." S rSub { size 8{T} } } over { Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } { ital "Gn" \( f \) } lline H rSub { size 8{R} } \( f \) rline rSup { size 8{2} } ital "df" "." Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } { { { lline P rSub { size 8{R} } \( f \) rline rSup { size 8{2} } } over { lline H rSub { size 8{c} } \( f \) rline "." lline H rSub { size 8{R} } \( f \) rline } } } ital "df"} } matrix { {} ##{} ## rightarrow ` ital "Minimizar"} {} } } {}

A través de la igualdad de Schwartz podemos cumplir el objetivo:

V ( f ) . W ( f ) df V ( f ) 2 . df . W ( f ) 2 . df Que será igual cuando V ( f ) = k . W ( f ) . Si : W ( f ) = P R ( f ) H C ( f ) H R ( f ) y V ( f ) = H R ( f ) Gn ( f ) Entonces : H R ( f ) Gn ( f ) = k P R ( f ) H C ( f ) H R ( f ) H R ( f ) 2 = k P R ( f ) H C ( f ) Gn ( f ) alignl { stack { size 12{ lline Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } {V \( f \) "." W rSup { size 8{*} } \( f \) ital "df"} rline<= Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } { lline V \( f \) rline rSup { size 8{2} } "." ital "df"} "." Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } { lline W \( f \) rline rSup { size 8{2} } "." ital "df"} } {} # {} #ital "Que"` ital "será"` ital "igual"` ital "cuando"`V \( f \) =k "." W \( f \) "." ` ital "Si": {} # {} #W \( f \) = { { lline P rSub { size 8{R} } \( f \) rline } over { lline H rSub { size 8{C} } \( f \) rline lline H rSub { size 8{R} } \( f \) rline } } ~y~V \( f \) = lline H rSub { size 8{R} } \( f \) rline sqrt { ital "Gn" \( f \) } {} # {} #ital "Entonces": {} # {} #lline H rSub { size 8{R} } \( f \) rline sqrt { ital "Gn" \( f \) } =k left [ { { lline P rSub { size 8{R} } \( f \) rline } over { lline H rSub { size 8{C} } \( f \) rline lline H rSub { size 8{R} } \( f \) rline } } right ] {} #{} # matrix {{} # {} # lline H rSub { size 8{R} } \( f \) rline rSup { size 8{2} } =k left [ { { lline P rSub { size 8{R} } \( f \) rline } over { lline H rSub { size 8{C} } \( f \) rline sqrt { ital "Gn" \( f \) } } } right ]{}} {} } } {}

Finalmente, con la ecuación (1) tenemos que:

H T ( f ) 2 = A k 2 P R ( f ) Gn ( f ) k H C ( f ) P R ( f ) 2 size 12{ lline H rSub { size 8{T} } \( f \) rline rSup { size 8{2} } = { {A rSub { size 8{k} rSup { size 8{2} } } lline P rSub { size 8{R} } \( f \) rline sqrt { ital "Gn" \( f \) } } over {k lline H rSub { size 8{C} } \( f \) rline lline P rSub { size 8{R} } \( f \) rline rSup { size 8{2} } } } } {}

Simulaciones en labview

El VI correspondiente a la teoría de este módulo puede descargarse a través del siguiente enlace:

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
hey
Giriraj
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
nanocopper obvius
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, Laboratorio digital interactivo. OpenStax CNX. Feb 09, 2011 Download for free at http://cnx.org/content/col11274/1.1
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