SlideShare ist ein Scribd-Unternehmen logo
REPÚBLICA BOLIVARIANA DE VENEZUELA
MINISTERIO DEL PODER POPULAR
PARA LA EDUCACIÓN UNIVERSITARIA
INSTITUTO UNIVERSITARIO POLITÉCNICO
“SANTIAGO MARIÑO”
EXTENSIÓN San Cristóbal.
Convolución y su
transformada de Fourier
Autor:
Fernando A. Mora M.
C.I:
19.777.425
San Cristóbal. Septiembre 2018
DEMOSTRACIONES DE CONVOLUCION Y TRANSFORMADA DE FOURIER
I. CONVOLUCION
Sean dos señales u(t) y v(t) totalmente integrales en R, su convolucion esta dada por la siguiente
formula:
(u ∗ v)(t) = ∫ u(s)v(t − s)ds
∞
−∞
Una de las señales debe ser acotadas en R y la otra debe ser totalmente integrable en R.
Teorema 1: La convolución es conmutativa
Demostración. Hacemos el siguiente cambio de variable t − s = λ, de modo que:
u ∗ v̂(t) = ∫ u(s)
∞
−∞
v(t − s)ds = − ∫ u(t − λ)v(λ)dλ
−∞
∞
…. (Cambio de variable)
u ∗ v̂(t) = ∫ v(λ)u(t − λ)dλ =
∞
−∞
v ∗ û(t)
u ∗ v̂(t) = v ∗ û(t)
Teorema 2: La convolucion es asociativa.
Sean tres señales u(t), v(t)y w(t) , su convolucion viene dada por:
u(t) ∗ v(t) ∗ w(t) = (u(t) ∗ v(t)) ∗ w(t) = u(t) ∗ (v(t) ∗ w(t))
Demostracion:
Usando la definición de convolucion:
u(t) ∗ h(t) = ∫ u(s)h(t − s)ds
∞
−∞
Tenemos:
C1 = [u(t) ∗ v(t)]∗ w(t) = [∫ u(suv)v(t − suv)dsuv] ∗ w(t)
∞
−∞
La expresión global quedaría:
C1 = [u(t) ∗ v(t)] ∗ w(t) = ∫ [∫ u(suv)v(svw − suv)dsuv]w(t − svw)
∞
−∞
∞
−∞
dsvw]
Para:
C2 = u(t) ∗ [v(t) ∗ w(t)] = u(t) ∗ [∫ v(svw)w(t − svw)dsvw]
∞
−∞
C2 = u(t) ∗ [v(t) ∗ w(t)] = ∫ u(suv)[
∞
−∞
∫ v(svw)w(t − suv − svw)dsvw]dsuv
∞
−∞
Lo que yo quiero probar es:
C1 = C2
∫ ∫ u(suv)v(svw − suv)w(t − svw)
∞
−∞
∞
−∞
dsuvdsvw
= ∫ ∫ u(suv)v(svw)w(t − suv − svw)dsvwdsuv
∞
−∞
∞
−∞
Hacemos el siguiente cambio de variable: λ = suv + svw y dλ = dsuv entonces:
C2 = ∫ ∫ u(λ − svw)v(
∞
−∞
∞
−∞
svw)w(t − λ)dλdsvw
Luego hacemos el siguiente cambio de variable: θ = λ − svw y dθ = −dsvw
C2 = −∫ ∫ u(θ)v(λ−
−∞
∞
∞
−∞
θ)w(t − λ)dλdθ
C2 = ∫ ∫ u(θ)v(λ−
−∞
−∞
∞
−∞
θ)w(t − λ)dλdθ
Entonces, nos queda:
∫ ∫ u(suv)v(svw − suv)w(t − svw)
∞
−∞
∞
−∞
dsuvdsvw = ∫ ∫ u(θ)v(λ −
−∞
−∞
∞
−∞
θ)w(t − λ)dλdθ
Teorema 3: La convolucion es distributiva
Sean tres señales u(t), v(t)y w(t) , se cumple que:
u(t) ∗ [v(t) + w(t)] = u(t) ∗ v(t) + u(t) ∗ w(t)
Demostración:
u(t) ∗ [v(t) + w(t)] = ∫ u(t)[v(t− s) + w(t − s)]ds
∞
−∞
u(t) ∗ [v(t) + w(t)] = ∫ u(t)v(t − s)ds
∞
−∞
+ ∫ u(t)w(t − s)ds
∞
−∞
u(t) ∗ [v(t) + w(t)] = u(t) ∗ v(t) + u(t) ∗ w(t)
Teorema 4: La convolucion es bilineal
Sean tres señales u(t), v(t)y w(t) , se cumple que:
(αu(t) + βv(t)) ∗ w(t) = α(u(t) ∗ w(t)) + β(v(t) ∗ w(t))
Demostración:
(αu(t) + βv(t)) ∗ w(t) = ∫ (αu(s)+ βv(s))w(t − s)ds
∞
−∞
(αu(t) + βv(t)) = ∫ αu(s)w(t − s)ds+ ∫ βv(s)w(t − s)ds
∞
−∞
∞
−∞
(αu(t) + βv(t)) = α(u(t) ∗ w(t) + β(v(t) ∗ w(t))
Teorema 5: Propiedad de escalabilidad
Sea:
y(t) = u(t) ∗ h(t) y z(t) = u(at) ∗ h(at), a > 0
Entonces:
z(t) = ∫ u(as)h(a(t − s))ds
∞
−∞
Demostración:
Haciendo el siguiente cambio de variable: λ = as, ds =
dλ
a
, para a > 0
z(t) =
1
a
∫ u(λ)h(at − λ)dλ
∞
−∞
Donde:
y(t) = u(t) ∗ h(t) = ∫ u(s)h(t − s)ds
∞
−∞
Entonces podemos definir, las siguientes relaciones:
z(t) = (
1
a
) y(at)
(
1
a
) y(at) = u(at)h(at)
Entonces quedaría:
y(at) = |a|u(at)h(at)
Teorema 6: Propiedad de desplazamiento en el tiempo
u(t + a) ∗ v(t + b) = z(t + a + b)
Donde:
z(t) = ∫ u(s)v(t − s)ds
∞
−∞
Demostración:
u(t + a) ∗ v(t + b) = ∫ u(s + a)v(t + b − s)ds
∞
−∞
Aplicando el cambio de variable: s + a = r
u(t + a) ∗ v(t + b) = ∫ u(r)v(t + a + b − r)dr
∞
−∞
u(t + a) ∗ v(t + b) = z(t + a + b)
Teorema 7: Si 𝒖( 𝒕) 𝒚 𝒗(𝒕) son señales continuas y totalmente integrables en R, entonces su
convolucion 𝒖( 𝒕) ∗ 𝒗(𝒕) es tambien una señal totalmente integrable. Se verifica la
desigualdad.
‖ 𝑢( 𝑡) ∗ 𝑣(𝑡)‖ 𝐿1
(𝑅) ≤ ‖ 𝑢(𝑡)‖ 𝐿1
(𝑅)‖ 𝑣(𝑡)‖ 𝐿1
(𝑅)
Antes de demostrar dicho teorema, definimos el teorema de fubini.
Teorema de Fubini:
Afirma que si :
𝑓( 𝑥, 𝑦) = 𝑓( 𝑥) 𝑔(𝑦)d
Entonces:
∫ 𝑓( 𝑥) 𝑑𝑥 ∫ 𝑔( 𝑦) 𝑑𝑦 = ∫ 𝑓( 𝑥) 𝑔( 𝑦) 𝑑(𝑥, 𝑦)
𝐴𝑋𝐵𝐵𝐴
Demostración:
Ahora retomemos la demostración:
‖ 𝑢( 𝑡) ∗ 𝑣( 𝑡)‖ 𝐿1( 𝑅) = ∫ | 𝑢( 𝑡) ∗ 𝑣( 𝑡)| 𝑑𝑡
∞
−∞
‖ 𝑢( 𝑡) ∗ 𝑣( 𝑡)‖ 𝐿1 ( 𝑅) = ∫ |∫ 𝑢( 𝑠) 𝑣( 𝑡 − 𝑠) 𝑑𝑠
∞
−∞
| 𝑑𝑡
∞
−∞
‖ 𝑢( 𝑡) ∗ 𝑣( 𝑡)‖ 𝐿1( 𝑅) ≤ ∫ ∫ | 𝑢( 𝑠)|| 𝑣( 𝑡 − 𝑠)| 𝑑𝑠𝑑𝑡
∞
−∞
∞
−∞
‖ 𝑢( 𝑡) ∗ 𝑣(𝑡)‖ 𝐿1
(𝑅) ≤ ∫ {∫ | 𝑢(𝑠)| 𝑑𝑠
∞
−∞
} | 𝑣(𝑡 − 𝑠)| 𝑑𝑡 (𝐴𝑝𝑙𝑖𝑐𝑎𝑛𝑑𝑜 𝐹𝑢𝑏𝑖𝑛𝑖)
∞
−∞
‖ 𝑢( 𝑡) ∗ 𝑣(𝑡)‖ 𝐿1
(𝑅) ≤ ‖ 𝑢(𝑡)‖ 𝐿1
(𝑅) ∫ | 𝑣(𝑡 − 𝑠)| 𝑑𝑡 =
∞
−∞
‖ 𝑢(𝑡)‖ 𝐿1
(𝑅)‖ 𝑣(𝑡)‖ 𝐿1
(𝑅)
II. TRANSFORMADA DE FOURIER
Sean las siguientes funciones: 𝑓1( 𝑡) 𝑦 𝑓2(𝑡) y sus transformadas de Fourier son:
𝐹1 = ∫ 𝑓1(𝑡)𝑒−𝑗𝜔𝑡
𝑑𝑡
∞
−∞
𝐹2 = ∫ 𝑓2(𝑡)𝑒−𝑗𝜔𝑡
𝑑𝑡
∞
−∞
Sus transformadas inversas son:
𝑓1( 𝑡) =
1
2𝜋
∫ 𝐹1 (𝜔)𝑒 𝑗𝜔𝑡
𝑑𝜔
∞
−∞
𝑓2( 𝑡) =
1
2𝜋
∫ 𝐹2 ( 𝜔) 𝑒 𝑗𝜔𝑡
𝑑𝜔
∞
−∞
Entonces:
Teorema 1:
𝑓1 ∗ 𝑓2 = 𝐹−1
[𝐹1 ( 𝜔) 𝐹2( 𝜔)]
Demostración:
𝑓1 ∗ 𝑓2 = ∫ 𝑓1( 𝑠) 𝑓2( 𝑡 − 𝑠) 𝑑𝑠
∞
−∞
𝑓1 ∗ 𝑓2 = 2𝜋 ∫ 𝑓1( 𝑠)[
1
2𝜋
∫ 𝐹2(𝜔)𝑒 𝑗𝜔(𝑡−𝑠)
∞
−∞
∞
−∞
𝑑𝜔]𝑑𝑠
𝑓1 ∗ 𝑓2 = 2𝜋 ∫ 𝐹2( 𝜔)[
1
2𝜋
∫ 𝑓1(𝑠)𝑒−𝑗𝜔𝑠
𝑑𝑠]𝑒 𝑗𝜔𝑡
𝑑𝜔
∞
−∞
∞
−∞
𝑓1 ∗ 𝑓2 = 2𝜋 ∫
1
2𝜋
𝐹2( 𝜔) 𝐹1(𝜔)𝑒 𝑗𝜔𝑡
𝑑𝜔
∞
−∞
𝒇 𝟏 ∗ 𝒇 𝟐 = 𝟐𝝅𝑭−𝟏
[𝑭 𝟏( 𝝎) 𝑭 𝟐( 𝝎)]
Teorema 2:
𝐹−1[ 𝐹1( 𝜔) ∗ 𝐹2( 𝜔)] = 2𝜋𝑓1 𝑓2
Demostración:
𝐹−1[ 𝐹1( 𝜔) ∗ 𝐹2 ( 𝜔)] = 𝐹−1
[∫ 𝐹1( 𝑠) 𝐹2( 𝜔 − 𝑠) 𝑑𝑠]
∞
−∞
𝐹−1[ 𝐹1( 𝜔) ∗ 𝐹2( 𝜔)] =
1
2𝜋
∫ [
∞
−∞
∫ 𝐹1( 𝑠) 𝐹2 ( 𝜔 − 𝑠) 𝑑𝑠]𝑒 𝑗𝜔𝑡
𝑑𝜔
∞
−∞
Hacemos el siguiente cambio de variable: 𝜔 − 𝑠 = 𝑥, 𝜔 = 𝑥 + 𝑠, 𝑑𝜔 = 𝑑𝑥
𝐹−1[ 𝐹1( 𝜔) ∗ 𝐹2( 𝜔)] =
1
2𝜋
∫ [
∞
−∞
∫ 𝐹1 ( 𝑠) 𝐹2( 𝑥) 𝑑𝑠]𝑒 𝑗(𝑥+𝑠)𝑡
𝑑𝑥
∞
−∞
𝐹−1[ 𝐹1 ( 𝜔)∗ 𝐹2 ( 𝜔)] =
1
2𝜋
∫ 𝐹1 ( 𝑠)
∞
−∞
∫ 𝐹2(𝑥)
∞
−∞
𝑒 𝑗( 𝑥+𝑠) 𝑡
𝑑𝑥𝑑𝑠
𝐹−1[ 𝐹1( 𝜔) ∗ 𝐹2 ( 𝜔)] =
1
2𝜋
∫ 𝐹1 ( 𝑠) 𝑒 𝑗𝑠𝑡
[∫ 𝐹2(𝑥)𝑒 𝑗𝑥𝑡
𝑑𝑥]𝑑𝑠
∞
−∞
∞
−∞
Multiplicamos y dividimos por 2𝝅
𝐹−1[ 𝐹1( 𝜔) ∗ 𝐹2( 𝜔)] = 2𝜋[
1
2𝜋
∫ 𝐹1( 𝜔) 𝑒 𝑗𝜔𝑡
𝑑𝜔][
1
2𝜋
∫ 𝐹2 (𝜔)𝑒 𝑗𝜔𝑡
𝑑𝜔]
∞
−∞
∞
−∞
𝑭−𝟏[ 𝑭 𝟏( 𝝎)∗ 𝑭 𝟐( 𝝎)] = 𝟐𝝅𝒇 𝟏( 𝒕) 𝒇 𝟐(𝒕)
Teorema 3:
La transformada de Fourier de la convolucion de dos funciones 𝒇 𝟏( 𝒕) 𝒚 𝒇 𝟐(𝒕) es la
multiplicación de las transformadas de Fourier de ambas funciones.
𝐹(𝑓1( 𝑡) ∗ 𝑓2( 𝑡)) = 𝐹1( 𝜔) 𝐹2 (𝜔)
Demostración:
𝐹(𝑓1( 𝑡) ∗ 𝑓2( 𝑡)) = ∫ [∫ 𝑓1( 𝑠) 𝑓2( 𝑡 − 𝑠) 𝑑𝑠]𝑒−𝑗𝜔𝑡
𝑑𝑡
∞
−∞
∞
−∞
𝐹(𝑓1( 𝑡) ∗ 𝑓2( 𝑡)) = ∫ 𝑓1( 𝑠)[∫ 𝑓2(𝑡 − 𝑠)𝑒−𝑗𝜔𝑡
𝑑𝑡]𝑑𝑠
∞
−∞
∞
−∞
Pero:
∫ 𝑓2(𝑡 − 𝑠)𝑒−𝑗𝜔𝑡
𝑑𝑡
∞
−∞
= 𝐹[ 𝑓2( 𝑡 − 𝑠)] = 𝐹2 (𝜔)𝑒−𝑗𝜔𝑠
Finalmente reemplazamos en la expresión:
𝐹(𝑓1( 𝑡) ∗ 𝑓2 ( 𝑡)) = ∫ 𝑓1( 𝑠)
∞
−∞
𝐹2( 𝜔) 𝑒−𝑗𝜔𝑠
𝑑𝑠
𝐹(𝑓1( 𝑡) ∗ 𝑓2 ( 𝑡)) = ∫ 𝑓1 ( 𝑠)
∞
−∞
𝑒−𝑗𝜔𝑠
𝑑𝑠𝐹2(𝜔)
𝑭(𝒇 𝟏( 𝒕)∗ 𝒇 𝟐( 𝒕)) = 𝑭 𝟏( 𝝎) 𝑭 𝟐(𝝎)

Weitere ähnliche Inhalte

Empfohlen

2024 State of Marketing Report – by Hubspot
2024 State of Marketing Report – by Hubspot2024 State of Marketing Report – by Hubspot
2024 State of Marketing Report – by Hubspot
Marius Sescu
 
Everything You Need To Know About ChatGPT
Everything You Need To Know About ChatGPTEverything You Need To Know About ChatGPT
Everything You Need To Know About ChatGPT
Expeed Software
 
Product Design Trends in 2024 | Teenage Engineerings
Product Design Trends in 2024 | Teenage EngineeringsProduct Design Trends in 2024 | Teenage Engineerings
Product Design Trends in 2024 | Teenage Engineerings
Pixeldarts
 
How Race, Age and Gender Shape Attitudes Towards Mental Health
How Race, Age and Gender Shape Attitudes Towards Mental HealthHow Race, Age and Gender Shape Attitudes Towards Mental Health
How Race, Age and Gender Shape Attitudes Towards Mental Health
ThinkNow
 
AI Trends in Creative Operations 2024 by Artwork Flow.pdf
AI Trends in Creative Operations 2024 by Artwork Flow.pdfAI Trends in Creative Operations 2024 by Artwork Flow.pdf
AI Trends in Creative Operations 2024 by Artwork Flow.pdf
marketingartwork
 
Skeleton Culture Code
Skeleton Culture CodeSkeleton Culture Code
Skeleton Culture Code
Skeleton Technologies
 
PEPSICO Presentation to CAGNY Conference Feb 2024
PEPSICO Presentation to CAGNY Conference Feb 2024PEPSICO Presentation to CAGNY Conference Feb 2024
PEPSICO Presentation to CAGNY Conference Feb 2024
Neil Kimberley
 
Content Methodology: A Best Practices Report (Webinar)
Content Methodology: A Best Practices Report (Webinar)Content Methodology: A Best Practices Report (Webinar)
Content Methodology: A Best Practices Report (Webinar)
contently
 
How to Prepare For a Successful Job Search for 2024
How to Prepare For a Successful Job Search for 2024How to Prepare For a Successful Job Search for 2024
How to Prepare For a Successful Job Search for 2024
Albert Qian
 
Social Media Marketing Trends 2024 // The Global Indie Insights
Social Media Marketing Trends 2024 // The Global Indie InsightsSocial Media Marketing Trends 2024 // The Global Indie Insights
Social Media Marketing Trends 2024 // The Global Indie Insights
Kurio // The Social Media Age(ncy)
 
Trends In Paid Search: Navigating The Digital Landscape In 2024
Trends In Paid Search: Navigating The Digital Landscape In 2024Trends In Paid Search: Navigating The Digital Landscape In 2024
Trends In Paid Search: Navigating The Digital Landscape In 2024
Search Engine Journal
 
5 Public speaking tips from TED - Visualized summary
5 Public speaking tips from TED - Visualized summary5 Public speaking tips from TED - Visualized summary
5 Public speaking tips from TED - Visualized summary
SpeakerHub
 
ChatGPT and the Future of Work - Clark Boyd
ChatGPT and the Future of Work - Clark Boyd ChatGPT and the Future of Work - Clark Boyd
ChatGPT and the Future of Work - Clark Boyd
Clark Boyd
 
Getting into the tech field. what next
Getting into the tech field. what next Getting into the tech field. what next
Getting into the tech field. what next
Tessa Mero
 
Google's Just Not That Into You: Understanding Core Updates & Search Intent
Google's Just Not That Into You: Understanding Core Updates & Search IntentGoogle's Just Not That Into You: Understanding Core Updates & Search Intent
Google's Just Not That Into You: Understanding Core Updates & Search Intent
Lily Ray
 
How to have difficult conversations
How to have difficult conversations How to have difficult conversations
How to have difficult conversations
Rajiv Jayarajah, MAppComm, ACC
 
Introduction to Data Science
Introduction to Data ScienceIntroduction to Data Science
Introduction to Data Science
Christy Abraham Joy
 
Time Management & Productivity - Best Practices
Time Management & Productivity -  Best PracticesTime Management & Productivity -  Best Practices
Time Management & Productivity - Best Practices
Vit Horky
 
The six step guide to practical project management
The six step guide to practical project managementThe six step guide to practical project management
The six step guide to practical project management
MindGenius
 
Beginners Guide to TikTok for Search - Rachel Pearson - We are Tilt __ Bright...
Beginners Guide to TikTok for Search - Rachel Pearson - We are Tilt __ Bright...Beginners Guide to TikTok for Search - Rachel Pearson - We are Tilt __ Bright...
Beginners Guide to TikTok for Search - Rachel Pearson - We are Tilt __ Bright...
RachelPearson36
 

Empfohlen (20)

2024 State of Marketing Report – by Hubspot
2024 State of Marketing Report – by Hubspot2024 State of Marketing Report – by Hubspot
2024 State of Marketing Report – by Hubspot
 
Everything You Need To Know About ChatGPT
Everything You Need To Know About ChatGPTEverything You Need To Know About ChatGPT
Everything You Need To Know About ChatGPT
 
Product Design Trends in 2024 | Teenage Engineerings
Product Design Trends in 2024 | Teenage EngineeringsProduct Design Trends in 2024 | Teenage Engineerings
Product Design Trends in 2024 | Teenage Engineerings
 
How Race, Age and Gender Shape Attitudes Towards Mental Health
How Race, Age and Gender Shape Attitudes Towards Mental HealthHow Race, Age and Gender Shape Attitudes Towards Mental Health
How Race, Age and Gender Shape Attitudes Towards Mental Health
 
AI Trends in Creative Operations 2024 by Artwork Flow.pdf
AI Trends in Creative Operations 2024 by Artwork Flow.pdfAI Trends in Creative Operations 2024 by Artwork Flow.pdf
AI Trends in Creative Operations 2024 by Artwork Flow.pdf
 
Skeleton Culture Code
Skeleton Culture CodeSkeleton Culture Code
Skeleton Culture Code
 
PEPSICO Presentation to CAGNY Conference Feb 2024
PEPSICO Presentation to CAGNY Conference Feb 2024PEPSICO Presentation to CAGNY Conference Feb 2024
PEPSICO Presentation to CAGNY Conference Feb 2024
 
Content Methodology: A Best Practices Report (Webinar)
Content Methodology: A Best Practices Report (Webinar)Content Methodology: A Best Practices Report (Webinar)
Content Methodology: A Best Practices Report (Webinar)
 
How to Prepare For a Successful Job Search for 2024
How to Prepare For a Successful Job Search for 2024How to Prepare For a Successful Job Search for 2024
How to Prepare For a Successful Job Search for 2024
 
Social Media Marketing Trends 2024 // The Global Indie Insights
Social Media Marketing Trends 2024 // The Global Indie InsightsSocial Media Marketing Trends 2024 // The Global Indie Insights
Social Media Marketing Trends 2024 // The Global Indie Insights
 
Trends In Paid Search: Navigating The Digital Landscape In 2024
Trends In Paid Search: Navigating The Digital Landscape In 2024Trends In Paid Search: Navigating The Digital Landscape In 2024
Trends In Paid Search: Navigating The Digital Landscape In 2024
 
5 Public speaking tips from TED - Visualized summary
5 Public speaking tips from TED - Visualized summary5 Public speaking tips from TED - Visualized summary
5 Public speaking tips from TED - Visualized summary
 
ChatGPT and the Future of Work - Clark Boyd
ChatGPT and the Future of Work - Clark Boyd ChatGPT and the Future of Work - Clark Boyd
ChatGPT and the Future of Work - Clark Boyd
 
Getting into the tech field. what next
Getting into the tech field. what next Getting into the tech field. what next
Getting into the tech field. what next
 
Google's Just Not That Into You: Understanding Core Updates & Search Intent
Google's Just Not That Into You: Understanding Core Updates & Search IntentGoogle's Just Not That Into You: Understanding Core Updates & Search Intent
Google's Just Not That Into You: Understanding Core Updates & Search Intent
 
How to have difficult conversations
How to have difficult conversations How to have difficult conversations
How to have difficult conversations
 
Introduction to Data Science
Introduction to Data ScienceIntroduction to Data Science
Introduction to Data Science
 
Time Management & Productivity - Best Practices
Time Management & Productivity -  Best PracticesTime Management & Productivity -  Best Practices
Time Management & Productivity - Best Practices
 
The six step guide to practical project management
The six step guide to practical project managementThe six step guide to practical project management
The six step guide to practical project management
 
Beginners Guide to TikTok for Search - Rachel Pearson - We are Tilt __ Bright...
Beginners Guide to TikTok for Search - Rachel Pearson - We are Tilt __ Bright...Beginners Guide to TikTok for Search - Rachel Pearson - We are Tilt __ Bright...
Beginners Guide to TikTok for Search - Rachel Pearson - We are Tilt __ Bright...
 

Convolucion y transformada_de_fourier fmm

  • 1. REPÚBLICA BOLIVARIANA DE VENEZUELA MINISTERIO DEL PODER POPULAR PARA LA EDUCACIÓN UNIVERSITARIA INSTITUTO UNIVERSITARIO POLITÉCNICO “SANTIAGO MARIÑO” EXTENSIÓN San Cristóbal. Convolución y su transformada de Fourier Autor: Fernando A. Mora M. C.I: 19.777.425 San Cristóbal. Septiembre 2018
  • 2. DEMOSTRACIONES DE CONVOLUCION Y TRANSFORMADA DE FOURIER I. CONVOLUCION Sean dos señales u(t) y v(t) totalmente integrales en R, su convolucion esta dada por la siguiente formula: (u ∗ v)(t) = ∫ u(s)v(t − s)ds ∞ −∞ Una de las señales debe ser acotadas en R y la otra debe ser totalmente integrable en R. Teorema 1: La convolución es conmutativa Demostración. Hacemos el siguiente cambio de variable t − s = λ, de modo que: u ∗ v̂(t) = ∫ u(s) ∞ −∞ v(t − s)ds = − ∫ u(t − λ)v(λ)dλ −∞ ∞ …. (Cambio de variable) u ∗ v̂(t) = ∫ v(λ)u(t − λ)dλ = ∞ −∞ v ∗ û(t) u ∗ v̂(t) = v ∗ û(t) Teorema 2: La convolucion es asociativa. Sean tres señales u(t), v(t)y w(t) , su convolucion viene dada por: u(t) ∗ v(t) ∗ w(t) = (u(t) ∗ v(t)) ∗ w(t) = u(t) ∗ (v(t) ∗ w(t)) Demostracion: Usando la definición de convolucion: u(t) ∗ h(t) = ∫ u(s)h(t − s)ds ∞ −∞ Tenemos: C1 = [u(t) ∗ v(t)]∗ w(t) = [∫ u(suv)v(t − suv)dsuv] ∗ w(t) ∞ −∞ La expresión global quedaría:
  • 3. C1 = [u(t) ∗ v(t)] ∗ w(t) = ∫ [∫ u(suv)v(svw − suv)dsuv]w(t − svw) ∞ −∞ ∞ −∞ dsvw] Para: C2 = u(t) ∗ [v(t) ∗ w(t)] = u(t) ∗ [∫ v(svw)w(t − svw)dsvw] ∞ −∞ C2 = u(t) ∗ [v(t) ∗ w(t)] = ∫ u(suv)[ ∞ −∞ ∫ v(svw)w(t − suv − svw)dsvw]dsuv ∞ −∞ Lo que yo quiero probar es: C1 = C2 ∫ ∫ u(suv)v(svw − suv)w(t − svw) ∞ −∞ ∞ −∞ dsuvdsvw = ∫ ∫ u(suv)v(svw)w(t − suv − svw)dsvwdsuv ∞ −∞ ∞ −∞ Hacemos el siguiente cambio de variable: λ = suv + svw y dλ = dsuv entonces: C2 = ∫ ∫ u(λ − svw)v( ∞ −∞ ∞ −∞ svw)w(t − λ)dλdsvw Luego hacemos el siguiente cambio de variable: θ = λ − svw y dθ = −dsvw C2 = −∫ ∫ u(θ)v(λ− −∞ ∞ ∞ −∞ θ)w(t − λ)dλdθ C2 = ∫ ∫ u(θ)v(λ− −∞ −∞ ∞ −∞ θ)w(t − λ)dλdθ Entonces, nos queda: ∫ ∫ u(suv)v(svw − suv)w(t − svw) ∞ −∞ ∞ −∞ dsuvdsvw = ∫ ∫ u(θ)v(λ − −∞ −∞ ∞ −∞ θ)w(t − λ)dλdθ Teorema 3: La convolucion es distributiva Sean tres señales u(t), v(t)y w(t) , se cumple que: u(t) ∗ [v(t) + w(t)] = u(t) ∗ v(t) + u(t) ∗ w(t)
  • 4. Demostración: u(t) ∗ [v(t) + w(t)] = ∫ u(t)[v(t− s) + w(t − s)]ds ∞ −∞ u(t) ∗ [v(t) + w(t)] = ∫ u(t)v(t − s)ds ∞ −∞ + ∫ u(t)w(t − s)ds ∞ −∞ u(t) ∗ [v(t) + w(t)] = u(t) ∗ v(t) + u(t) ∗ w(t) Teorema 4: La convolucion es bilineal Sean tres señales u(t), v(t)y w(t) , se cumple que: (αu(t) + βv(t)) ∗ w(t) = α(u(t) ∗ w(t)) + β(v(t) ∗ w(t)) Demostración: (αu(t) + βv(t)) ∗ w(t) = ∫ (αu(s)+ βv(s))w(t − s)ds ∞ −∞ (αu(t) + βv(t)) = ∫ αu(s)w(t − s)ds+ ∫ βv(s)w(t − s)ds ∞ −∞ ∞ −∞ (αu(t) + βv(t)) = α(u(t) ∗ w(t) + β(v(t) ∗ w(t)) Teorema 5: Propiedad de escalabilidad Sea: y(t) = u(t) ∗ h(t) y z(t) = u(at) ∗ h(at), a > 0 Entonces: z(t) = ∫ u(as)h(a(t − s))ds ∞ −∞ Demostración: Haciendo el siguiente cambio de variable: λ = as, ds = dλ a , para a > 0 z(t) = 1 a ∫ u(λ)h(at − λ)dλ ∞ −∞ Donde:
  • 5. y(t) = u(t) ∗ h(t) = ∫ u(s)h(t − s)ds ∞ −∞ Entonces podemos definir, las siguientes relaciones: z(t) = ( 1 a ) y(at) ( 1 a ) y(at) = u(at)h(at) Entonces quedaría: y(at) = |a|u(at)h(at) Teorema 6: Propiedad de desplazamiento en el tiempo u(t + a) ∗ v(t + b) = z(t + a + b) Donde: z(t) = ∫ u(s)v(t − s)ds ∞ −∞ Demostración: u(t + a) ∗ v(t + b) = ∫ u(s + a)v(t + b − s)ds ∞ −∞ Aplicando el cambio de variable: s + a = r u(t + a) ∗ v(t + b) = ∫ u(r)v(t + a + b − r)dr ∞ −∞ u(t + a) ∗ v(t + b) = z(t + a + b) Teorema 7: Si 𝒖( 𝒕) 𝒚 𝒗(𝒕) son señales continuas y totalmente integrables en R, entonces su convolucion 𝒖( 𝒕) ∗ 𝒗(𝒕) es tambien una señal totalmente integrable. Se verifica la desigualdad. ‖ 𝑢( 𝑡) ∗ 𝑣(𝑡)‖ 𝐿1 (𝑅) ≤ ‖ 𝑢(𝑡)‖ 𝐿1 (𝑅)‖ 𝑣(𝑡)‖ 𝐿1 (𝑅) Antes de demostrar dicho teorema, definimos el teorema de fubini.
  • 6. Teorema de Fubini: Afirma que si : 𝑓( 𝑥, 𝑦) = 𝑓( 𝑥) 𝑔(𝑦)d Entonces: ∫ 𝑓( 𝑥) 𝑑𝑥 ∫ 𝑔( 𝑦) 𝑑𝑦 = ∫ 𝑓( 𝑥) 𝑔( 𝑦) 𝑑(𝑥, 𝑦) 𝐴𝑋𝐵𝐵𝐴 Demostración: Ahora retomemos la demostración: ‖ 𝑢( 𝑡) ∗ 𝑣( 𝑡)‖ 𝐿1( 𝑅) = ∫ | 𝑢( 𝑡) ∗ 𝑣( 𝑡)| 𝑑𝑡 ∞ −∞ ‖ 𝑢( 𝑡) ∗ 𝑣( 𝑡)‖ 𝐿1 ( 𝑅) = ∫ |∫ 𝑢( 𝑠) 𝑣( 𝑡 − 𝑠) 𝑑𝑠 ∞ −∞ | 𝑑𝑡 ∞ −∞ ‖ 𝑢( 𝑡) ∗ 𝑣( 𝑡)‖ 𝐿1( 𝑅) ≤ ∫ ∫ | 𝑢( 𝑠)|| 𝑣( 𝑡 − 𝑠)| 𝑑𝑠𝑑𝑡 ∞ −∞ ∞ −∞ ‖ 𝑢( 𝑡) ∗ 𝑣(𝑡)‖ 𝐿1 (𝑅) ≤ ∫ {∫ | 𝑢(𝑠)| 𝑑𝑠 ∞ −∞ } | 𝑣(𝑡 − 𝑠)| 𝑑𝑡 (𝐴𝑝𝑙𝑖𝑐𝑎𝑛𝑑𝑜 𝐹𝑢𝑏𝑖𝑛𝑖) ∞ −∞ ‖ 𝑢( 𝑡) ∗ 𝑣(𝑡)‖ 𝐿1 (𝑅) ≤ ‖ 𝑢(𝑡)‖ 𝐿1 (𝑅) ∫ | 𝑣(𝑡 − 𝑠)| 𝑑𝑡 = ∞ −∞ ‖ 𝑢(𝑡)‖ 𝐿1 (𝑅)‖ 𝑣(𝑡)‖ 𝐿1 (𝑅)
  • 7. II. TRANSFORMADA DE FOURIER Sean las siguientes funciones: 𝑓1( 𝑡) 𝑦 𝑓2(𝑡) y sus transformadas de Fourier son: 𝐹1 = ∫ 𝑓1(𝑡)𝑒−𝑗𝜔𝑡 𝑑𝑡 ∞ −∞ 𝐹2 = ∫ 𝑓2(𝑡)𝑒−𝑗𝜔𝑡 𝑑𝑡 ∞ −∞ Sus transformadas inversas son: 𝑓1( 𝑡) = 1 2𝜋 ∫ 𝐹1 (𝜔)𝑒 𝑗𝜔𝑡 𝑑𝜔 ∞ −∞ 𝑓2( 𝑡) = 1 2𝜋 ∫ 𝐹2 ( 𝜔) 𝑒 𝑗𝜔𝑡 𝑑𝜔 ∞ −∞ Entonces: Teorema 1: 𝑓1 ∗ 𝑓2 = 𝐹−1 [𝐹1 ( 𝜔) 𝐹2( 𝜔)] Demostración: 𝑓1 ∗ 𝑓2 = ∫ 𝑓1( 𝑠) 𝑓2( 𝑡 − 𝑠) 𝑑𝑠 ∞ −∞ 𝑓1 ∗ 𝑓2 = 2𝜋 ∫ 𝑓1( 𝑠)[ 1 2𝜋 ∫ 𝐹2(𝜔)𝑒 𝑗𝜔(𝑡−𝑠) ∞ −∞ ∞ −∞ 𝑑𝜔]𝑑𝑠 𝑓1 ∗ 𝑓2 = 2𝜋 ∫ 𝐹2( 𝜔)[ 1 2𝜋 ∫ 𝑓1(𝑠)𝑒−𝑗𝜔𝑠 𝑑𝑠]𝑒 𝑗𝜔𝑡 𝑑𝜔 ∞ −∞ ∞ −∞ 𝑓1 ∗ 𝑓2 = 2𝜋 ∫ 1 2𝜋 𝐹2( 𝜔) 𝐹1(𝜔)𝑒 𝑗𝜔𝑡 𝑑𝜔 ∞ −∞ 𝒇 𝟏 ∗ 𝒇 𝟐 = 𝟐𝝅𝑭−𝟏 [𝑭 𝟏( 𝝎) 𝑭 𝟐( 𝝎)] Teorema 2: 𝐹−1[ 𝐹1( 𝜔) ∗ 𝐹2( 𝜔)] = 2𝜋𝑓1 𝑓2
  • 8. Demostración: 𝐹−1[ 𝐹1( 𝜔) ∗ 𝐹2 ( 𝜔)] = 𝐹−1 [∫ 𝐹1( 𝑠) 𝐹2( 𝜔 − 𝑠) 𝑑𝑠] ∞ −∞ 𝐹−1[ 𝐹1( 𝜔) ∗ 𝐹2( 𝜔)] = 1 2𝜋 ∫ [ ∞ −∞ ∫ 𝐹1( 𝑠) 𝐹2 ( 𝜔 − 𝑠) 𝑑𝑠]𝑒 𝑗𝜔𝑡 𝑑𝜔 ∞ −∞ Hacemos el siguiente cambio de variable: 𝜔 − 𝑠 = 𝑥, 𝜔 = 𝑥 + 𝑠, 𝑑𝜔 = 𝑑𝑥 𝐹−1[ 𝐹1( 𝜔) ∗ 𝐹2( 𝜔)] = 1 2𝜋 ∫ [ ∞ −∞ ∫ 𝐹1 ( 𝑠) 𝐹2( 𝑥) 𝑑𝑠]𝑒 𝑗(𝑥+𝑠)𝑡 𝑑𝑥 ∞ −∞ 𝐹−1[ 𝐹1 ( 𝜔)∗ 𝐹2 ( 𝜔)] = 1 2𝜋 ∫ 𝐹1 ( 𝑠) ∞ −∞ ∫ 𝐹2(𝑥) ∞ −∞ 𝑒 𝑗( 𝑥+𝑠) 𝑡 𝑑𝑥𝑑𝑠 𝐹−1[ 𝐹1( 𝜔) ∗ 𝐹2 ( 𝜔)] = 1 2𝜋 ∫ 𝐹1 ( 𝑠) 𝑒 𝑗𝑠𝑡 [∫ 𝐹2(𝑥)𝑒 𝑗𝑥𝑡 𝑑𝑥]𝑑𝑠 ∞ −∞ ∞ −∞ Multiplicamos y dividimos por 2𝝅 𝐹−1[ 𝐹1( 𝜔) ∗ 𝐹2( 𝜔)] = 2𝜋[ 1 2𝜋 ∫ 𝐹1( 𝜔) 𝑒 𝑗𝜔𝑡 𝑑𝜔][ 1 2𝜋 ∫ 𝐹2 (𝜔)𝑒 𝑗𝜔𝑡 𝑑𝜔] ∞ −∞ ∞ −∞ 𝑭−𝟏[ 𝑭 𝟏( 𝝎)∗ 𝑭 𝟐( 𝝎)] = 𝟐𝝅𝒇 𝟏( 𝒕) 𝒇 𝟐(𝒕) Teorema 3: La transformada de Fourier de la convolucion de dos funciones 𝒇 𝟏( 𝒕) 𝒚 𝒇 𝟐(𝒕) es la multiplicación de las transformadas de Fourier de ambas funciones. 𝐹(𝑓1( 𝑡) ∗ 𝑓2( 𝑡)) = 𝐹1( 𝜔) 𝐹2 (𝜔) Demostración: 𝐹(𝑓1( 𝑡) ∗ 𝑓2( 𝑡)) = ∫ [∫ 𝑓1( 𝑠) 𝑓2( 𝑡 − 𝑠) 𝑑𝑠]𝑒−𝑗𝜔𝑡 𝑑𝑡 ∞ −∞ ∞ −∞ 𝐹(𝑓1( 𝑡) ∗ 𝑓2( 𝑡)) = ∫ 𝑓1( 𝑠)[∫ 𝑓2(𝑡 − 𝑠)𝑒−𝑗𝜔𝑡 𝑑𝑡]𝑑𝑠 ∞ −∞ ∞ −∞ Pero:
  • 9. ∫ 𝑓2(𝑡 − 𝑠)𝑒−𝑗𝜔𝑡 𝑑𝑡 ∞ −∞ = 𝐹[ 𝑓2( 𝑡 − 𝑠)] = 𝐹2 (𝜔)𝑒−𝑗𝜔𝑠 Finalmente reemplazamos en la expresión: 𝐹(𝑓1( 𝑡) ∗ 𝑓2 ( 𝑡)) = ∫ 𝑓1( 𝑠) ∞ −∞ 𝐹2( 𝜔) 𝑒−𝑗𝜔𝑠 𝑑𝑠 𝐹(𝑓1( 𝑡) ∗ 𝑓2 ( 𝑡)) = ∫ 𝑓1 ( 𝑠) ∞ −∞ 𝑒−𝑗𝜔𝑠 𝑑𝑠𝐹2(𝜔) 𝑭(𝒇 𝟏( 𝒕)∗ 𝒇 𝟐( 𝒕)) = 𝑭 𝟏( 𝝎) 𝑭 𝟐(𝝎)