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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

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
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Sanket Reply
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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