Transformada de Fourier Discreta

7/21/2019 Transformada de Fourier Discreta

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Procesamiento Digital de Señales

Yahir Hernández Mier

Universidad Politécnica de Victoria

Transformada de Forier Discreta



● !stdiar la definici"n de la Transformada de Forier Discreta

● !stdiar las #ro#iedades de la Transformada de Forier Discreta

● $#render a inter#retar los resltados es#ectrales de la DFT

● $#licar ventanas #ara redcir las fgas de la DFT

● $#licar la DFT a sinsoides reales % com#le&as

'(&etivos



Transformada de Forier contina

X ( f )=∫−∞

∞

x(t )e− j 2 π f t

dt

Im

Re1

j

0

)dentidad de !ler



Transformada de Forier contina

X ( f )=∫−∞

∞

x(t )e− j 2 π f t

dt

Im

Re1

j

cos

sen

0

e

jω

=cos(ω)+ j sen(ω)

)dentidad de !ler



Transformada de Forier discreta

X (m)=∑n=0

N −1

x(n)e

− j 2 π n m

N

*a DFT es n #rocedimiento matemático tilizado #ara determinar el

contenido armónico o frecuencial de na señal discreta

*a DFT es +til #ara analizar secencias discretas sin im#ortar lo ,e

esta secencia re#resente




X (m)=∑n=0

N −1

x (n)e

− j 2 πm n

N




X (m)=∑n=0

N −1

x (n)e

− j 2 πm n

N

=∑n=0

N −1

x (n)e− jω n



Series geométricas

s=1+2

3+

4

9+

8

27+ … -.ál es la sma total/




Series geométricas

s=1+2

3+

4

9+

8


-Por ,é valor se de(e mlti#licar el término anterior #ara o(tener el término sigiente/




Series geométricas

s=1+2

3+

4

9+

8



r =2

30az"n com+n




Series geométricas

s=1+2

3+

4

9+

8



r =2

30az"n com+n

r s=2

3 s=

2

3+

4

9+

8

27+

16

81+ …




Series geométricas

s=1+2

3+

4

9+

8



r =2

30az"n com+n

r s=2

3 s=

2

3+

4

9+

8

27+

16

81+ …

s−r s= s−2

3 s=1+

2

3+

4

9+

8

27+ …−(2

3+

4

9+

8

27+

16

81+ …)




Series geométricas

s=1+2

3+

4

9+

8



r =2

30az"n com+n

r s=2

3 s=

2

3+

4

9+

8

27+

16

81+ …

s−r s= s−2

3 s=1+

2

3+

4

9+

8

27+ …−(2

3+

4

9+

8

27+

16

81+ …)

s− 23 s=1




Series geométricas

s=1+2

3+

4

9+

8



r =2

30az"n com+n

r s=2

3 s=

2

3+

4

9+

8

27+

16

81+ …

s−r s= s−2

3 s=1+

2

3+

4

9+

8

27+ …−(2

3+

4

9+

8

27+

16

81+ …)

s− 23 s=1

s=3




Series geométricas

s=1+2

3+

4

9+

8



r =2

30az"n com+n

r s=2

3 s=

2

3+

4

9+

8

27+

16

81+ …

s−r s= s−2

3 s=1+

2

3+

4

9+

8

27+ …−(2

3+

4

9+

8

27+

16

81+ …)

s− 23 s=1

s=3 1ote ,e2 s=1+2

3+

4

9+

8

27+ …=( 2

3 )0

+ ( 2

3 )1

+ (2

3 )2

+ ( 2

3 )3

…

s=1⋅r 0+ 1⋅r

1+ 1⋅r 2+ 1⋅r

3+ …




Series geométricas

De manera general #ara 2r ≠1

s=a+ a⋅r 1+ a⋅r

2+ a⋅r 3+ …+ a⋅r

n−1




Series geométricas


s=a+ a⋅r 1+ a⋅r

2+ a⋅r 3+ …+ a⋅r

n−1a2 !s el #rimer término de la serie

r 2 !s la raz"n com+n




Series geométricas


s=a+ a⋅r 1+ a⋅r

2+ a⋅r 3+ …+ a⋅r

n−1

s=∑k =0

n−1

a r k

a2 !s el #rimer término de la serie





Series geométricas


s=a+ a⋅r 1+ a⋅r

2+ a⋅r 3+ …+ a⋅r

n−1

s=∑k =0

n−1

a r k

s=a⋅r + a⋅r 2+ a⋅r

3+ a⋅r 4+ …+ a⋅r

n−1+ a⋅r n

r n−1

r 1=r

n−1+ 1=r n






Series geométricas


s=a+ a⋅r 1+ a⋅r

2+ a⋅r 3+ …+ a⋅r

n−1

s=∑k =0

n−1

a r k

s=a⋅r + a⋅r 2+ a⋅r

3+ a⋅r 4+ …+ a⋅r

n−1+ a⋅r n

r n−1

r 1=r

n−1+ 1=r n

s−r s=a+ a⋅r 1+ a⋅r

2+ a⋅r 3+ …+ a⋅r

n−1−a⋅r 1−a⋅r

2−a⋅r 3−a⋅r

4−…−a⋅r n






Series geométricas


s=a+ a⋅r 1+ a⋅r

2+ a⋅r 3+ …+ a⋅r

n−1

s=∑k =0

n−1

a r k

s=a⋅r + a⋅r 2+ a⋅r

3+ a⋅r 4+ …+ a⋅r

n−1+ a⋅r n

r n−1

r 1=r

n−1+ 1=r n


2+ a⋅r 3+ …+ a⋅r

n−1−a⋅r 1−a⋅r

2−a⋅r 3−a⋅r

4−…−a⋅r n

s(1−r )=a−a⋅r n






Series geométricas


s=a+ a⋅r 1+ a⋅r

2+ a⋅r 3+ …+ a⋅r

n−1

s=∑k =0

n−1

a r k

s=a⋅r + a⋅r 2+ a⋅r

3+ a⋅r 4+ …+ a⋅r

n−1+ a⋅r n

r n−1

r 1=r

n−1+ 1=r n


2+ a⋅r 3+ …+ a⋅r

n−1−a⋅r 1−a⋅r

2−a⋅r 3−a⋅r

4−…−a⋅r n


s=a1−r

n

1−r






Series geométricas


s=a+ a⋅r 1+ a⋅r

2+ a⋅r 3+ …+ a⋅r

n−1

s=∑k =0

n−1

a r k

s=a⋅r + a⋅r 2+ a⋅r

3+ a⋅r 4+ …+ a⋅r

n−1+ a⋅r n

r n−1

r 1=r

n−1+ 1=r n


2+ a⋅r 3+ …+ a⋅r

n−1−a⋅r 1−a⋅r

2−a⋅r 3−a⋅r

4−…−a⋅r n


s=a1−r

n

1−r Si % 2∣r ∣< 1 n →∞ s=

a

1−r







X (m)=∑n=0

N −1

x (n)e

− j 2 πm n

N

=∑n=0

N −1

x (n)e

− jω n

Si % 2∣r ∣< 1 N →∞ s=∑n=0

N −1

a r n=

a

1−r s= ∑

n=0

N −1

a r n=a

1−r N

1−r




X (m)=∑n=0

N −1

x (n)e

− j 2 πm n

N

=∑n=0

N −1

x (n)e

− jω n

=∑n=0

N −1

x (n) (e− j ω

)

n

s= ∑n=0

N −1

a r n=a

1−r N

1−r Si % 2∣r ∣< 1 N →∞ s=∑

n=0

N −1

a r n=

a

1−r




X (m)=∑n=0

N −1

x (n)e

− j 2 πm n

N

=∑n=0

N −1

x (n)e

− jω n

=∑n=0

N −1

x (n) (e− j ω

)

n

s= ∑n=0

N −1

a r n=a

1−r N

1−r Si % 2∣r ∣< 1 N →∞ s=∑

n=0

N −1

a r n=

a

1−r

Transformada de Forier discreta de fnciones elementales

Fnci"n im#lso nitario

x (nt s)={1, n=0

0 , n≠0

δ(n)



s= ∑n=0

N −1

a r n=a

1−r N

1−r Si % 2∣r ∣< 1 N →∞ s=∑

n=0

N −1

a r n=

a

1−r


Fnci"n escal"n nitario x (nt s)={1 , n=0,1,2,…0 , n<0

1(nt s)



s= ∑n=0

N −1

a r n=a

1−r N

1−r Si % 2∣r ∣< 1 N →∞ s=∑

n=0

N −1

a r n=

a

1−r


ant s

x (n)={an, n=0,1,2,…

0 , n<0Fnci"n #olinomial



s= ∑n=0

N −1

a r n=a

1−r N

1−r Si % 2∣r ∣< 1 N →∞ s=∑

n=0

N −1

a r n=

a

1−r


e−a nt sFnci"n e3#onencial x (n)={e

−ant s , n=0,1,2,…

0 , n<0



s= ∑n=0

N −1

a r n=a

1−r N

1−r Si % 2∣r ∣< 1 N →∞ s=∑

n=0

N −1

a r n=

a

1−r


Fnci"n senoidal

x(t )={sen(ωt ), t ≥0

0 , t < 0

e jωt =cos(ω t )+ j sen(ωt )e

− jω t =cos(ωt )− j sen(ω t )

sen(ω t )=e jω t −e

− jωt

2 jcos(ω t )=

e jωt + e

− jω t

2

)dentidad de !ler




X (m)=∑n=0

N −1

x(n)e

− j 2 π n m

N

Forma e3#onencial

T f d d F i di t




X (

m)=∑n=0

N −1

x(n

)e

− j 2 π n m

N

De la identidad de !ler2

e− j ω=cos(ω)− j sen(ω)

Forma e3#onencial

T f d d F i di t




X (m)=

∑n=0

N −1

x(n)e

− j 2 π n m

N



Forma e3#onencial

X (m)=∑n=0

N −1

x(n)[cos( 2 π n m

N )− j sen(2 π n m

N )] Forma rectanglar

T f d d F i di t




X (m)=

∑n=0

N −1

x(n)e

− j 2 π n m

N



Forma e3#onencial

X (m)=∑n=0

N −1

x(n)[cos( 2 π n m


N )] Forma rectanglar

Donde2 X(m) es el m4ésimo com#onente de frecencia5 es decir5 X(0), X(1), X(2), ...m es el índice de la salida de la DFT en el dominio de la frecuencia

m= 0, 1, 2, 3, ..., N-1 x(n) es la secencia de mestras de entrada, x(0), x(1), x(2), x(3), ...n es el 6ndice en el dominio del tiem#o de las mestras de entrada , n=0, 1, 2, ..., N-1 j = N es el n+mero de mestras de la secencia de entrada % el n+mero de #ntos de

frecencia en la salida DFT

√ −1

Significado de la Transformada de Fo rier discreta



Significado de la Transformada de Forier discreta

!&em#lo2 X (m)=∑n=0

N −1

x(n)[cos( 2 π n m


N )] N =4

X (m)=∑n=0

3

x (n)[cos( 2π n m

4 )− j sen( 2π n m

4 )]

m=0 X (0)= x(0)cos( 2⋅π⋅0⋅0

4 )− j x(0)sen( 2⋅π⋅0⋅0

4 )+ x (1)cos( 2⋅π⋅1⋅0

4 )− j x(1)sen ( 2⋅π⋅1⋅0

4 )+ x (2)cos( 2⋅π⋅2⋅0

4 )− j x(2)sen( 2⋅π⋅2⋅0

4 )+ x (3)cos( 2⋅π⋅3⋅0

4 )− j x(3)sen (2⋅π⋅3⋅0

4 )m=1

.ada término X(m) de la salida de la DFT es la sma del #rodcto elemento #or

elemento entre na señal de entrada % na sinsoide com#le&a cos(ϕ)− j sen(ϕ)




!&em#lo2 X (m)=∑n=0

N −1

x(n)[cos( 2 π n m


N )] N =4

X (m)=∑n=0

3

x (n)[cos( 2π n m

4 )− j sen( 2π n m

4 )]

m=2

m=3









*a frecuencia e3acta de cada sinsoide de#ende de % de f s N








Frecencia a la cal laseñal original femestreada

1+mero de mestras,e se toman en laDFT










!&em#lo2

f s=500 Hz (muestras/s)

DFT de 16 untos en !os datos muestreados










!&em#lo2



*a frecuencia fundamental de las sinsoides es2

f s

N =

500

16 =31"25 Hz










!&em#lo2




f s

N =

500

16 =31"25 Hz

*as frecuencias de análisis son m+lti#los enteros de la frecencia fndamental

X(0) 1o 0

31.25 = 0 Hz

Término DFT Término de frecencia Frecencia de análisis

X(1) 2o 1 31.25 = 31.25 Hz

X(2) 3o 2 31.25 = 62.5 Hz

X(3) 4o 3 31.25 = 3.!5 Hz 7

7

7

7

7

7

7

7

7

X(15) 16o 15 31.25 = 46".!5 Hz










!&em#lo2




f s

N =

500

16 =31"25 Hz


X(0) 1o 0

31.25 = 0 Hz


X(1) 2o 1 31.25 = 31.25 Hz

X(2) 3o 2 31.25 = 62.5 Hz

X(3) 4o 3 31.25 = 3.!5 Hz 7

7

7

7

7

7

7

7

7

X(15) 16o 15 31.25 = 46".!5 Hz

f an#!$s$s (m)=m f s

N










!&em#lo2



*a frecuencia fundamental de las sinsoides es2 f s

N =

500

16 =31"25 Hz


X(0) 1o 0

31.25 = 0 Hz


X(1) 2o 1 31.25 = 31.25 Hz

X(2) 3o 2 31.25 = 62.5 Hz

X(3) 4o 3 31.25 = 3.!5 Hz 7

7

7

7

7

7

7

7

7

X(15) 16o 15 31.25 = 46".!5 Hz

f an#!$s$s(m)=m f s

N

Magnitdes

de las

com#onentes

de la señal





ϕ

0elaci"n trigonométrica de n valor com#le&o X(m) individal de salida de la DFT


!ste #nto re#resenta el n+mero

com#le&o

%&e $ma'$nar$o j

%&e rea!

X (m)= X rea# (m)+ j X $ma% (m)

X (m)= X rea# (m)+ j X $ma% (m)= X ma% (m)a un #n'u!o X ϕ(m)




ϕ


g


com#le&o

%&e $ma'$nar$o j

%&e rea!

X (m)= X rea# (m)+ j X $ma% (m)





ϕ


g


com#le&o

%&e $ma'$nar$o j

%&e rea!

X (m)= X rea# (m)+ j X $ma% (m)


X ma% (m)=∣ X (m)∣=√ X rea# (m)2+ X $ma% (m)2 Magnitd de X (m) X ma% (m)

X ma% (m)




ϕ


g


com#le&o

%&e $ma'$nar$o j

%&e rea!

X (m)= X rea# (m)+ j X $ma% (m)


X ma% (m)=∣ X (m)∣=√ X rea# (m)2+ X $ma% (m)2

X ϕ(m)=tan−1(

X $ma% (m) X rea# (m) )

Magnitd de

8nglo de fase de X ϕ(m) X (m)

X (m) X ma% (m)

X ma% (m)




ϕ


g


com#le&o

%&e $ma'$nar$o j

%&e rea!

X (m)= X rea# (m)+ j X $ma% (m)


X ma% (m)=∣ X (m)∣=√ X rea# (m)2+ X $ma% (m)2

X ϕ(m)=tan−1

( X $ma% (m) X rea# (m) )

X &' (m)= X ma% (m)2= X rea# (m)2+ X $ma% (m)2

Magnitd de

8nglo de fase de X ϕ(m) X (m)

X (m)

!s#ectro de #otencia de X &' (m) X (m)

X ma% (m)

X ma% (m)




!&em#lo 9

Mestree % o(tenga la DFT de : #ntos de na señal contina conteniendo

com#onentes de frecencia a 9 ;Hz % <;Hz5 e3#resada como2

x$n(t )=sen(2 π 1000 t )+ 0"5sen (2 π 2000 t +3

4π) 3

4 π=135




!&em#lo 9



x$n(t )=sen(2 π 1000 t )+ 0"5sen (2 π 2000 t +3

4π) 3

4 π=135

Mestreando a na frecencia 5 se toman mestras cada segndos2 f s1

f s=t s

x (n)=sen (2π 1000 n t s)+ 0"5sen (2 π 2000n t s+ 34 π)




!&em#lo 9

m f s

N 0 Hz ,1 Hz ,2 Hz ,… ,7 Hz Frecencias de análisis 2



x$n(t )=sen(2 π 1000 t )+ 0"5sen (2 π 2000 t +3

4π) 3

4 π=135


f s=t s


Si f s=8000 muestras/sm=0 m=1 m=2 … m=7




!&em#lo 9

m f s




x$n

(t )=sen(2 π 1000 t )+0"5sen(2 π 2000 t +3

4π) 3

4 π=135


f s=t s



Mestras 2 x (n)=sen (2π 1000 n t s)+ 0"5sen (2 π 2000 n t s+3

4 π) x (0)=sen (2⋅π⋅1000⋅0⋅t s)+ 0"5sen(2⋅π⋅2000⋅n⋅t s+

3

4π)=0"3535534

x (1)=0"3535534

x (2)=0"6464466

x(3)=1"0606602

x (4)=0"3535534

x (5)=−1"0606602

x (6)=−1"3535534

x (7)=−0"3535534




!&em#lo 9

Mestras 2 x (n)=sen (2π 1000 n t s)+ 0"5sen (2 π 2000 n t s+3

4

π)

x (0)=sen (2⋅π⋅1000⋅0⋅t s)+ 0"5sen(2⋅π⋅2000⋅n⋅t s+34

π)=0"3535534

x (1)=0"3535534

x (2)=0"6464466

x(3)=1"0606602

x (4)=0"3535534

x (5)=−1"0606602

x (6)=−1"3535534

x (7)=−0"3535534

x $n(t )sen(2 π 1000 t )0"5sen (2π 2000 t +

3

4 π)




!&em#lo 9

X (m)=∑n=0

7

x (n)

[cos

(2π n m

8 )− j sen

(2π n m

8 )] X (1)=0− j 4"0=4∢−90

X (2)=1"4142+ j 1"4142=2∢45

X (3)=0− j 0=0∢0

X (4)=0− j 0=0∢0

X (5)=0− j 0=0∢0

X (6)=1"4142− j 1"4142=2∢−45

X (7)=0+ j 4"0=4∢90




!&em#lo 9

X (1)=∑n=0

7

x(n)

[cos

(2π 1

8 n

)− j sen

(2π 1

8 n

)]m=1

f = *

f s+ * = f f s

1(t )=cos (2 π 1000 t )

f s =8000muestras/s + * =1

88000=1000 Hz

2(t )=sen (2 π 1000 t )




!&em#lo 9

X (1)=∑n=0

7

x(n)

[cos

(2π 1

8 n

)− j sen

(2π 1

8 n

)]m=1

f = *

f s+ * = f f s

1(t )=cos (2 π 1000 t )


88000=1000 Hz

2(t )=sen (2 π 1000 t )

X (2)=∑n=0

7

x (n)[cos(2 π1

4 n)− j sen (2 π

1

4 n)]

m=2

f = *

f s+ * = f f s

1(t )=cos (2 π 2000 t )

f s=8000 muestras/s + * =1

48000=2000 Hz

2(t )=sen (2 π 2000 t )




!&em#lo 9

X (1)=∑n=0

7

x(n)

[cos

(2π 1

8 n

)− j sen

(2π 1

8 n

)]

m=1

f = *

f s+ * = f f s

1(t )=cos (2 π 1000 t )


88000=1000 Hz

2(t )=sen (2 π 1000 t )

X (2)=∑n=0

7

x (n)[cos(2 π1

4 n)− j sen (2 π

1

4 n)]

m=2

f = *

f s+ * = f f s

1(t )=cos (2 π 2000 t )

f s=8000 muestras/s + * =1

48000=2000 Hz

2(t )=sen (2 π 2000 t )




!&em#lo 9

m=3 1(t )=cos (2π 3000 t ) 2(t )=sen (2π 3000 t )

m=7 1(t )=cos (2 π 2000 t ) 2(t )=sen (2 π 2000 t )

⋮ ⋮

!& l 9

Significado de la Transformada de Forier discreta(4) 0 0 0 0



!&em#lo 9

X (1)=0− j 4"0=4∢−90

X (2)=1"4142+ j 1"4142=2∢45

X (3)=0− j 0=0∢0

X (4)=0− j 0=0∢0

X (5)=0− j 0=0∢0

X (6)=1"4142− j 1"4142=2∢−45

X (7)=0+ j 4"0=4∢90

!& l 9

Significado de la Transformada de Forier discretaX (4) 0 j 0 0∢0



!&em#lo 9

X (1)=0− j 4"0=4∢−90

X (2)=1"4142+ j 1"4142=2∢45

X (3)=0− j 0=0∢0

X (4)=0− j 0=0∢0

X (5)=0− j 0=0∢0

X (6)=1"4142− j 1"4142=2∢−45

X (7)=0+ j 4"0=4∢90

Pro#iedades de la Transformada de Forier discreta



Práctica =79797 '(&etivo2 !stdiar las gráficas de la #arte real5 imaginaria5 magnitd

% ánglo de fase de la DFT5 mediante n #rograma de cálclo nmérico7

97 Tome : mestras de la fnci"n

a na frecencia de mestreo fs = 8000 Hz 7

x (t )=sen(2 π 1000 t )+0"5 sen(2 π 2000 t + 34

π)

<7 .alcle la DFT X(m) de 8 #ntos so(re las mestras ad,iridas x(n)7

>7 !n na #rimera s(gráfica5 grafi,e la #arte real de X(m)7

=7 !n na segnda s(gráfica5 grafi,e la #arte imaginaria de X(m)7

?7 !n na tercera s(gráfica5 grafi,e la magnitd de X(m)

@7 !n na carta s(gráfica5 grafi,e el ánglo de fase de X(m)

A7 '(serve la simetr6a en las gráficas5 -,é #ede conclir/




x (n)=sen (2π 1000 n t s)+ 0"5sen (2 π 2000 n t s+ 34

π)

x (0)=0"3535534

x (1)=0"3535534

x (2)=0"6464466

x(3)=1"0606602

x (4)=0"3535534

x (5)=−1"0606602

x (6)=−1"3535534

x (7)=−0"3535534

f s=8000 muestras/s

N =8

Práctica =7979 0ealizar n #rograma en Scila( ,e calcle la DFT de : #ntos

#ara la señal discreta del e&em#lo 92






x (n)=sen (2π 1000 n t s)+ 0"5sen (2 π 2000 n t s+ 34

π)

X (1)=0− j 4"0=4∢−90 (

X (2)=1"4142+ j 1"4142=2∢45(

X (3)=0− j 0=0∢0(

X (4)=0− j 0=0∢0(

X (5)=0− j 0=0∢0 (

X (6)=1"4142− j 1"4142=2∢−45(

X (7)=0+ j 4"0=4∢90 (

x (0)=0"3535534

x (1)=0"3535534

x (2)=0"6464466

x(3)=1"0606602

x (4)=0"3535534

x (5)=−1"0606602

x (6)=−1"3535534

x (7)=−0"3535534

DFT2

f s=8000 muestras/s

N =8

X (m)=∑n=0

N −1

x(n)[cos( 2 π n m


N )]


P á ti = 9 9 0 li S il ( l l l DFT d : t



DFT2

X (m)=∑n=0

N −1

x(n)[cos( 2 π n m


N )]




P á ti = 9 9 0 li S il ( l l l DFT d : t



DFT2

X (m)=∑n=0

N −1

x(n)[cos( 2 π n m


N )]






Significado de la Transformada de Forier discreta X (4)=0− j 0=0∢0!&em#lo 9

( )



X (1)=0− j 4"0=4∢−90(

X (2)=1"4142+ j 1"4142=2∢45

X (3)=0− j 0=0∢0

( ) j

X (5)=0− j 0=0∢0

X (6)=1"4142− j 1"4142=2∢−45

X (7)=0+ j 4"0=4∢90

X (0)=0

!&em#lo 9Significado de la Transformada de Forier discreta



∣ X (m)∣= -0

N

2 -0=

2∣ X (m)∣ N

0elaci"n entre la am#litd de la señal % la

magnitd de s DFT

x$n (t )=sen(2 π 1000 t )+0"5sen (2 π 2000 t +3

4

π)

N =8

Simetr6a




S e a

Simetr6a




Simetr6a




*a DFT ace#ta entrada com!le"a5 #ero la ma%or6a de las señales f6sicas son reales

Simetr6a





.ando es real las salidas com#le&as de la DFT desde hastason redndantes con valores de salida de frecencia #ara

x(n) m=1 m=

N

2 −1m

N

2

Simetr6a






x(n) m=1 m=

N

2 −1m

N

2*a m-s$ma salida BrealC de la DFT tiene el mismo valor ,e la salida (N-m)-s$ma

Simetr6a






x(n) m=1 m=

N

2 −1m

N

2*a m-s$ma salida BrealC de la DFT tiene el mismo valor ,e la salida (N-m)-s$ma

!l án#ulo de fase de la m-s$ma salida es el negativo del ánglo de la salida

(N-m)-s$ma

Simetr6a






x(n) m=1 m=

N

2 −1m

N

2


(N-m)-s$ma

X (m)=∣ X (m)∣ a X ϕ(m) 'rados

X (m)=∣ X ( N −m)∣ a X ϕ( N −m) 'rados

*a m-s$ma salida BrealC de la DFT tiene el mismo valor ,e la salida (N-m)-s$ma

Simetr6a






x(n) m=1 m=

N

2 −1m

N

2


(N-m)-s$ma

X (m)=∣ X (m)∣ a X ϕ(m) 'rados

X (m)=∣ X ( N −m)∣ a X ϕ( N −m) 'rados

.ando x(n) es real $

X (m)= X ( N −m)

x=a+ j / x=a− j / + x=e

j ϕ x

=e− j ϕ

0ecordar2

*a m-s$ma salida BrealC de la DFT tiene el mismo valor ,e la salida (N-m)-s$ma

Simetr6a




Demostrar la #ro#iedad de simetr6a de la DFT2

X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n( N −m)

N

Simetr6a


D l i d d d i 6 d l DFT



X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n( N −m)

N

X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n N

N e

j 2π n m

N


Simetr6a


D t l i d d d i t 6 d l DFT



X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n( N −m)

N

X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n N

N e

j 2π n m

N

X ( N −m)=∑n=0

N −1

x (n)e− j 2 π n

e

j 2π n m

N


Simetr6a


D t l i d d d i t 6 d l DFT



X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n( N −m)

N

X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n N

N e

j 2π n m

N

X ( N −m)=∑n=0

N −1

x (n)e− j 2 π n

e

j 2π n m

N

e− j ϕ=cos(ϕ)− j sen(ϕ)


Simetr6a





X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n( N −m)

N

X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n N

N e

j 2π n m

N

X ( N −m)=∑n=0

N −1

x (n)e− j 2 π n

e

j 2π n m

N


e− j 2 π n=cos(2π n)− j sen(2π n)


Simetr6a





X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n( N −m)

N

X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n N

N e

j 2π n m

N

X ( N −m)=∑n=0

N −1

x (n)e− j 2 π n

e

j 2π n m

N



e− j 2 π n=1


Simetr6a





X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n( N −m)

N

X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n N

N e

j 2π n m

N

X ( N −m)=∑n=0

N −1

x (n)e− j 2 π n

e

j 2π n m

N



e− j 2 π n=1

X ( N −m)=∑n=0

N −1

x(n)(1)e

j 2 π nm

N


Simetr6a





X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n( N −m)

N

X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n N

N e

j 2π n m

N

X ( N −m)=∑n=0

N −1

x (n)e− j 2 π n

e

j 2π n m

N



e− j 2 π n=1

X ( N −m)=∑n=0

N −1

x(n)(1)e

j 2 π nm

N

X ( N −m)=∑n=0

N −1

x(n)e

j 2 π n m

N


Simetr6a





X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n( N −m)

N

X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n N

N e

j 2π n m

N

X ( N −m)=∑n=0

N −1

x (n)e− j 2 π n

e

j 2π n m

N



e− j 2 π n=1

X ( N −m)=∑n=0

N −1

x(n)(1)e

j 2 π nm

N

X ( N −m)=∑n=0

N −1

x(n)e

j 2 π n m

N


X (m)=∑n=0

N −1

x (n)e− j 2 πn m

N

Simetr6a





X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n( N −m)

N

X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n N

N e

j 2π n m

N

X ( N −m)=∑n=0

N −1

x (n)e− j 2 π n

e

j 2π n m

N



e− j 2 π n=1

X ( N −m)=∑n=0

N −1

x(n)(1

)e

j 2 π nm

N

X ( N −m)=∑n=0

N −1

x(n)e

j 2 π n m

N

X (m)= X (( N −m)


X (m)=∑n=0

N −1

x (n)e− j 2 πn m

N

Simetr6a


2Demostrar la #ro#iedad de simetr6a de la DFT2



X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n( N −m)

N

X ( N −m)=∑n=0

N −1

x (n)e

− j 2π n N

N e

j 2π n m

N

X ( N −m)=∑n=0

N −1

x (n)e− j 2 π n

e

j 2π n m

N



e− j 2 π n=1

X ( N

−m

)=∑n=0

N −1

x(n

)(1

)e

j 2 π nm

N

X ( N −m)=∑n=0

N −1

x(n)e

j 2 π n m

N

.om#le&o con&gado


X (m)= X (( N −m)

X (m)=∑n=0

N −1

x (n)e− j 2 πn m

N

*inealidad


*a DFT de la sma de dos señales es igal a la sma de las DFT de cada señal




x sma (n)= x1(n)+ x 2(n)

X sma (m)= X 1(m)+ X 2(m)

X s0ma(m)=∑n=0

N −1

x s0ma(n)e− j 2 πnm

N

Demostraci"n2

*inealidad






x s0ma(n)= x1(n)+ x2(n)

X s0ma(m)= X 1(m)+ X 2(m)

X s0ma (m)=∑n=0

N −1

x s0ma (n)e− j 2π n m

N

X s0ma(m)=∑n=0

N −1

[ x1(n)+ x2(n)] e− j 2πnm

N

Demostraci"n2

*inealidad






x s0ma(n)= x1(n)+ x2(n)

X s0ma(m)= X 1(m)+ X 2(m)

X s0ma (m)=∑n=0

N −1


N

X s0ma(m)=∑n=0

N −1

[ x1(n)+ x2(n)] e− j 2πnm

N

X s0ma(m)=∑n=0

N −1

x1(n)e− j 2 π nm

N + ∑n=0

N −1

x2(n)e− j 2πnm

N

Demostraci"n2

*inealidad






x s0ma(n)= x1(n)+ x2(n)

X s0ma(m)= X 1(m)+ X 2(m)

X s0ma (m)=∑n=0

N −1


N

X s0ma(m)=∑n=0

N −1

[ x1(n)+ x2(n)] e− j 2πnm

N

X s0ma(m)=∑n=0

N −1

x1(n)e− j 2 π nm

N + ∑n=0

N −1

x2(n)e− j 2πnm

N

X sma (m)= X 1(m)+ X 2(m)

Demostraci"n2

Periodicidad


Si la DFT es evalada #ara todos los valores de n5 % no solo #ara n=0, , %&'5



# 5 % # , , 5

entonces la secencia infinita resltante es na e3tensi"n #eri"dica5 con #eriodo% 5 de la DFT7

Periodicidad





X (m+ N )=∑n=0

N −1

x (n)e

− j 2π n(m+ N ) N

Demostraci"n2

# 5 % # , , 5


Periodicidad





X (m+ N )=∑n=0

N −1

x (n)e

− j 2π n(m+ N ) N

Demostraci"n2

X (m+ N )=∑n=0

N −1

x (n)e− j 2π n m

N e− j 2 πn N

N


Periodicidad





X (m+ N )=∑n=0

N −1

x (n)e

− j 2π n(m+ N ) N

Demostraci"n2

X (m+ N )=∑n=0

N −1

x (n)e− j 2π n m

N e− j 2 πn N

N

X (m+ N )=∑n=0

N −1

x (n)e− j 2π n m

N e− j 2π n


Periodicidad





X (m+ N )=∑n=0

N −1

x (n)e

− j 2π n(m+ N ) N

Demostraci"n2

X (m+ N )=∑n=0

N −1

x (n)e− j 2π n m

N e− j 2 πn N

N

X (m+ N )=∑n=0

N −1

x (n)e− j 2π n m

N e− j 2π n

e− j 2π n=cos(2π n)− j sen(2π n)=1


Periodicidad





X (m+ N )=∑n=0

N −1

x (n)e

− j 2π n(m+ N ) N

Demostraci"n2

X (m+ N )=∑n=0

N −1

x (n)e− j 2π n m

N e− j 2 πn N

N

X (m+ N )=∑n=0

N −1

x (n)e− j 2π n m

N e− j 2π n


X (m+ N )=∑n=0

N −1

x (n)e− j 2π n m

N


Periodicidad





X (m+ N )=∑n=0

N −1

x (n)e

− j 2π n(m+ N ) N

Demostraci"n2

X (m+ N )=∑n=0

N −1

x (n)e− j 2π n m

N e− j 2 πn N

N

X (m+ N )=∑n=0

N −1

x (n)e− j 2π n m

N e− j 2π n


X (m+ N )=∑n=0

N −1

x (n)e− j 2π n m

N

X (m)=∑n=0

N −1

x (n)e− j 2 πn m

N


Periodicidad





X (m+ N )=∑n=0

N −1

x (n)e

− j 2π n(m+ N ) N

Demostraci"n2

X (m+ N )=∑n=0

N −1

x (n)e− j 2π n m

N e− j 2 πn N

N

X (m+ N )=∑n=0

N −1

x (n)e− j 2π n m

N e− j 2π n


X (m+ N )=∑n=0

N −1

x (n)e− j 2π n m

N

X (m)=∑n=0

N −1

x (n)e− j 2 πn m

N

X (m+ N )= X (m)


Me&ora de la resolci"n de la DFT

0ellenadoE con zeros2 * =3"0 Hz 16 muestras



16 ceros





16 ceros

48 ceros





16 ceros

48 ceros 112 ceros


0ellenadoE con zeros en Scila(2

443n G <5 ?5 4@5 A5 <5 9I JJ Señal a rellenar 3n G



3n G

<7 ?7 4 @7 A7 <7 97

44vc G zerosB95 ><C JJ Vector de ceros Bmás elementos ,e 3nC vc G colmn 9 to 9: K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7

colmn 9L to ><K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7

44vcB 92lengthB3nC C G 3n JJ .o#iamos 3n al inicio del vector de ceros vc G colmn 9 to 9: <7 ?7 4 @7 A7 <7 97 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7

colmn 9L to >< K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7 K7

443nrellena G vc JJ 3nrellena toma el valor de vc

1

Práctica =797<7 '(&etivo2 .om#ro(ar la #ro#iedad de simetr6a mediante la

graficaci"n de la magnitd de la DFT en n #rograma de cálclo nmérico7




x(n)=3sen (2 π1

10 n)97 .onsidere la sigiente fnci"n discreta n=[0,10] ! "<7 .on la mestras o(tenidas5 calcle la DFT de ' #ntos7

>7 rafi,e la magnitd de la DFT % ss #artes real e imaginaria Bcon #lot5 no con

sciNstemC7

=7 !sco&a n valor de frecencia m B#or e&em#lo m = C7 Mestre el resltado dela sigiente diferencia2 X(m) * X(%&m+)7 0ecerde ,e los 6ndices en los

#rogramas de cálclo nmérico como Scila( % Matla( comienzan en '7

?7 0e#ita los #asos anteriores #ero ahora con la fnci"n com#le&a

@7 -Oé #ede conclir so(re la #ro#iedad de simetr6a de la DFT/

x (n)=e j π

1

10 n

Práctica =797>7 '(&etivo2 .om#ro(ar la #ro#iedad de linealidad mediante la





x1(n)=3cos(2 π1

7 n)

97 .onsidere la sigientes fnciones discretas % la sma de ellas2

<7 .alcle la DFT de #ntos so(re x '(n), x (n) - x s(n)7

>7 !n na #rimera s(gráfica5 grafi,e la magnitd de X s(m) Bcon #lotC7

=7 !n na segnda s(gráfica5 grafi,e la magnitd de X s(m) = X '(m) + X (m)7

?7 Mestre la sma de la diferencia entre X s(m) % X s(m), mediante la instrcci"n

Ben Scila(C dis!( sum( a.s(Xsm * Xsm) ) )@7 .on fndamento en las gráficas % el resltado de la sma de la diferencia

entre las magnitdes de las DFT5 -,é #ede conclir so(re la #ro#iedad de

linealidad de la DFT/

x2(n)=0"8sen(2 π2

5 n )

x s(n)= x1(n)+ x2(n) n=[0,10 ] ! "

Práctica =797=7 '(&etivo2 .om#ro(ar la #ro#iedad de #eriodicidad mediante la





x(n)=(−0"9)n

97 .onsidere la sigiente fnci"n discreta n=[0,10 ] ! "<7 .on la mestras o(tenidas5 calcle la DFT de ' #ntos7

>7 .on la mestras o(tenidas5 calcle la DFT de ' #ntos en dos ciclos con la

sigiente fnci"n2

=7 rafi,e en na #rimera s(gráfica la magnitd de la DFT de n ciclo7

?7 rafi,e en na segnda s(gráfica la magnitd de la DFT de dos ciclos7


@7 0e#ita los #asos anteriores5 #ero #ara la sigiente fnci"n2

j π3

n



A7 !n fnci"n de los resltados de las dos #artes de la #ráctica5 -Oé #ede

conclir so(re la #ro#iedad de #eriodicidad de la DFT/

x(n)=(0"9e )

Teorema de des#lazamiento #ara la DFT

!sta(lece ,e n des#lazamiento tem#oral de na secencia #eri"dica x(n) se

manifiesta como n des#lazamiento de fase en los ánglos asociados con la DFT



X des1 (m)=∑n=0

N −1

x (n+ k )e− j 2π n m

N


!sta(lece ,e n des#lazamiento tem#oral de na secencia #eri"dica 3BnC se




X des1 (m)=∑n=0

N −1


N

2=n+ k






X des1 (m)=∑n=0

N −1


N

2=n+ k n=2−k






X des1(m)= ∑2=k

N −1+ k

x (2)e

− j 2 π(2−k )m

N

X des1 (m)=∑n=0

N −1


N

2=n+ k n=2−k






X des1 (m)=∑2=0

N −1

x (2)e− j 2π 2 m

N e j 2π k m

N

X des1(m)= ∑2=k

N −1+ k

x (2)e

− j 2 π(2−k )m

N

2=n+ k n=2−k

X des1 (m)=∑n=0

N −1


N






X des1 (m)=e j 2π k m

N ∑2=0

N −1

x (2)e− j 2π 2 m

N

X des1 (m)=∑2=0

N −1

x (2)e− j 2π 2 m

N e j 2π k m

N

X des1 (m)= ∑2=k

N −1+ k

x (2)e

− j 2 π(2−k )m

N

2=n+ k n=2−k

X des1(m)=∑n=0

N −1

x (n+ k )e− j 2π n m N







N ∑2=0

N −1

x (2)e− j 2π 2 m

N


N ∑2=0

N −1

x (2)e− j 2π 2 m

N

X des1 (m)=∑2=0

N −1

x (2)e− j 2π 2 m

N e j 2π k m

N

X des1 (m)= ∑2=k

N −1+ k

x (2)e

− j 2 π(2−k )m

N

2=n+ k n=2−k

X des1(m)=∑n=0

N −1 x (n+ k )e

− j 2π n m N







N X (m)


N ∑2=0

N −1

x (2)e− j 2π 2 m

N


N ∑2=0

N −1

x (2)e− j 2π 2 m

N

X des1 (m)=∑2=0

N −1

x (2)e− j 2π 2 m

N e j 2π k m

N

X des1 (m)= ∑2=k

N −1+ k

x (2)e

− j 2 π(2−k )m

N

2=n+ k n=2−k

X des1(m)=∑n=0

N −1 x (n+ k )e

− j 2π n m N






2π k m

N rad$anes o

360 k m

N 'rados

Des#lazamiento2


N X (m)


N ∑2=0

N −1

x (2)e− j 2π 2 m

N


N ∑2=0

N −1

x (2)e− j 2π 2 m

N

X des1(m)=∑2=0

N −1

x (2)e− j 2π 2 m

N e j 2π k m

N

X des1 (m)= ∑2=k

N −1+ k

x (2)e

− j 2 π(2−k )m

N

2=n+ k n=2−k

X des1(m)=∑n=0

N −1 x (n+ k )e

− j 2π n m N

!&em#lo <

S#onga ,e la señal del e&em#lo 9 es mestreada con n adelanto en el

tiem#o / =2


x (t )=sen(2 π 1000 t )+ 0"5sen (2 π 2000 t +3

π)



$n

( ) ( ) (4

)

x (n+ 3)=sen (2 π 1000(n+ 3)t s)+ 0"5sen(2 π 2000(n+ 3) t s+3

4 π)

!&em#lo <

x (t )=sen(2 π 1000 t )+ 0"5sen (2 π 2000 t +3

π)


S#onga ,e la señal del e&em#lo 9 es mestreada con n adelanto en el

tiem#o / =2



$n 4

Mestras en el e&em#lo 9

Mestras en el e&em#lo <

x (n+ 3)=sen (2 π 1000(n+ 3)t s)+ 0"5sen(2 π 2000(n+ 3) t s+3

4 π)

!&em#lo <


Mestras 2 x (n+ 3)=sen (2 π 1000(n+ 3)t s)+ 0"5sen(2 π 2000(n+ 3) t s+3

4 π)



x (0)=1"0606602

x (1)=0"3535534

x (2)=−1"0606602

x(3)=−1"3535534

x (4)=−0"3535534 x (5)=0"3535534

x (6)=0"3535534

x (7)=0"6464466

!&em#lo <


Dominio de la frecencia 2



!&em#lo <


Dominio de la frecencia 2 X des1(m)=e j 2π k m

N X (m)

1 N 8 k 3



m=1 N =8 k =3

!&em#lo <



N X (m)

1 N 8 k 3



m=1

X des1 (1)=e

j 2 π(3)(1)8 X (1)

N =8 k =3

!&em#lo <



N X (m)

1 N 8 k 3



m=1

X des1 (1)=e

j 2 π(3)(1)8 X (1)

X (1)=4∢−90=4 e− j

π2

N =8 k =3

!&em#lo <



N X (m)

1 N 8 k 3



m=1

X des1 (1)=e

j 2 π(3)(1)8 X (1)

X (1)=4∢−90=4 e− j

π2

N =8 k =3

X des1 (1)=e j 6π8 4 e

− j π2

!&em#lo <



N X (m)

m 1 N 8 k 3



m=1

X des1 (1)=e

j 2 π(3)(1)8 X (1)

X (1)=4∢−90=4 e− j

π2

N =8 k =3

X des1 (1)=e j 6π8 4 e

− j π2

X des1 (1)=e j 3π

4 4 e− j

π2

!&em#lo <



N X (m)

m 1 N 8 k 3



m=1

X des1 (1)=e

j 2 π(3)(1)8 X (1)

X (1)=4∢−90=4 e− j

π2

N =8 k =3

X des1 (1)=e j 6π

8 4 e− j

π2

X des1 (1)=e j 3π

4 4 e− j

π2

X des(1)=4 e j π(3

4−

2

4 )

!&em#lo <



N X (m)

m 1 N 8 k 3



m=1

X des1 (1)=e

j 2 π(3)(1)8 X (1)

X (1)=4∢−90=4 e− j

π2

N =8 k =3

X des1 (1)=e j 6π

8 4 e− j

π2

X des1 (1)=e j 3π

4 4 e− j

π2

X des(1)=4 e j π(3

4−

2

4 )

X des1 (1)=4 e j π

4

!&em#lo <


Dominio de la frecencia 2 X des1 (m)=e j 2π k m

N X (m)

m=1 N =8 k=3



m=1

X des1(1)=e

j 2 π(3)(1)8 X (1)

X (1)=4∢−90(=4 e− j π

2

N =8 k =3

X des1 (1)=e j 6π

8 4 e− j π2

X des1 (1)=e j 3π

4 4 e− j π

2

X des1 (1)=4 e j π( 3

4−

2

4 )

X des1(1)=4 e j π4

X des1(1)=4∢45(

!&em#lo <


Dominio de la frecencia 2 X des1 (1)=4∢45(



!&em#lo 9Dominio de la frecencia 2

X (1)=4∢−90(

Práctica =797?7 '(&etivo2 .om#ro(ar el teorema de des#lazamiento mediante la

graficaci"n de la DFT de na señal des#lazada / mestras7


97 .onsidere la fnci"n del e&em#lo 92 x (t )=sen(2 π1000 t )+0"5sen(2π 2000 t +3

4π)



>7 .alcle la DFT X(m) de 8 #ntos so(re las mestras ad,iridas en n+7

=7 !n na #rimera s(gráfica5 grafi,e la #arte real de X(m)7

?7 !n na segnda s(gráfica5 grafi,e la #arte imaginaria de X(m)7

@7 !n na tercera s(gráfica5 grafi,e la magnitd de X(m)

A7 !n na carta s(gráfica5 grafi,e el ánglo de fase de X(m)

:7 .on fndamento en las gráficas5 -,é #ede conclir so(re el teorema de

des#lazamiento de la DFT/

4<7 Mestree la señal a fs = 8000 Hz % o(tenga 8 mestras n = 10,23 5 #ero #ara n

adelantada tres mestras B n+ C7

Definici"n2

Transformada de Forier Discreta )nversa B)DFTC

x (n)=1

N ∑

m=0

N −1

X (m)e j 2πm n

N



Definici"n2


x (n)=1

N ∑

m=0

N −1

X (m)e j 2πm n

N

( )1 ∑

N −1

( )[ (2πm n

) (2πm n

)]



x (n)=1

N ∑

m=0

X (m)[cos(2πm n

N )+ j sen(

2πm n

N )]

Definici"n2


x (n)=1

N ∑

m=0

N −1

X (m)e j 2πm n

N

( )1 ∑

N −1

( )[ (2πm n

) (2πm n

)]



x (n)=1

N ∑

m=0

X (m)[cos(2πm n

N )+ j sen(

2πm n

N )]

Una señal en tiem#o discreto #ede ser considerada como na sma de

varias frecencias anal6ticas sinsoidales

Definici"n2Transformada de Forier Discreta )nversa B)DFTC

x (n)=1

N ∑

m=0

N −1

X (m)e j 2πm n

N

( )1 ∑

N −1

( )[ (2πm n

) (2πm n

)]



x (n)=1

N ∑

m=0

X (m)[cos(2πm n

N )+ j sen(

2πm n

N )]



*as salidas X(m) de la DFT son n con&nto de N valores com#le&os ,e

indican la magnitd % la fase de cada frecencia de análisis involcrada en

la sma

Definici"n2Transformada de Forier Discreta )nversa B)DFTC

x (n)=1

N ∑

m=0

N −1

X (m)e j 2πm n

N

( )1 ∑

N −1

X ( )[ (2πm n

) j (2πm n

)]



x (n)=1

N ∑

m=0

X (m)[cos(2πm n

N )+ j sen(

2πm n

N )]



*as salidas X(m) de la DFT son n con&nto de N valores com#le&os ,e

indican la magnitd % la fase de cada frecencia de análisis involcrada en

la sma

X (1)=0− j 4"0=4∢−90

X (2)=1"4142+ j 1"4142=2∢45

X (3)=0− j 0=0∢0

X (4)=0− j 0=0∢0

X (5)=0− j 0=0∢0

X (6)=1"4142− j 1"4142=2∢−45

X (7)=0+ j 4"0=4∢90

X (0)=0− j 0=0∢0


X (1)=0− j 4"0=4∢−90

X (2)=1"4142+ j 1"4142=2∢45

X (4)=0− j 0=0∢0

X (5)=0− j 0=0∢0

X (6)=1"4142− j 1"4142=2∢−45

X (0)=0− j 0=0∢0



x (n)=1

N

∑m=0

N −1

X (m)

[cos(

2πm n

N

)+ j sen(2πm n

N

)

]

( ) X (3)=0− j 0=0∢0

( ) X (7)=0+ j 4"0=4∢90(


X (1)=0− j 4"0=4∢−90

X (2)=1"4142+ j 1"4142=2∢45

X (4)=0− j 0=0∢0

X (5)=0− j 0=0∢0

X (6)=1"4142− j 1"4142=2∢−45

X (0)=0− j 0=0∢0



x (n)=1

N

∑m=0

N −1

X (m)

[cos(

2πm n

N

)+ j sen(2πm n

N

)

]

( ) X (3)=0− j 0=0∢0

( ) X (7)=0+ j 4"0=4∢90(

x (0)=0"3535534+ j 0"0

x (1)=0"3535534+ j 0"0

x (2)=0"6464466+ j 0"0

x(3)=1"0606602+ j 0"0

x (4)=0"3535534+ j 0"0

x (5)=−1"0606602+ j 0"0

x (6)=−1"3535534+ j 0"0

x (7)=−0"3535534+ j 0"0

Práctica =797@7 '(&etivo2 .om#render la relaci"n entre la DFT % la )DFT mediante

la #rogramaci"n de esta +ltima en n #rograma de cálclo nmérico7

Transformada de Forier discreta )nversa

97 .onsidere la fnci"n del e&em#lo 92 x (t )=sen(2 π 1000 t )+0"5sen(2π 2000 t +3

4

π)

< Mestree la señal a fs = 8000 Hz % o(tenga 8 mestras n = 10 23



?7 .alcle la )DFT de 8 #ntos so(re X(m) con la fnci"n miN)DFT7sce7

@7 .om#are el resltado de la )DFT con las mestras x(n)7 -,é #ede conclir/

<7 Mestree la señal a fs = 8000 Hz % o(tenga 8 mestras n = 10,23 7

>7 .alcle la DFT de la señal con la fnci"n miNDFT7sce7

=7 Partiendo de la fnci"n miNDFT7sce5 #rograme la ecaci"n de la )DFT2

!&em#lo 9 x$n

(t )=sen(2 π 1000 t )+0"5sen(2 π 2000 t +3

4π)

3

4 π=135


f s=t s

x (n)=sen (2π 1000 n t s)+ 0"5sen (2 π 2000n t s+3

4 π)

Fgas Blea/a#eC en la DFT



m f s



X (1)=0− j 4"0=4∢−90 (

X (2)=1"4142+ j 1"4142=2∢45(

X (3)=0− j 0=0∢0 (

X (4)=0− j 0=0∢0 (

X (5)=0− j 0=0∢0 ( X (6)=1"4142− j 1"4142=2∢−45(

X (7)=0+ j 4"0=4∢90 (

X (0)=0


*as fgas hacen ,e los resltados de la DFT sean solo na a#ro3imaci"n

al verdadero es#ectro de la señal antes del mestreo7

Se #ede redcir mas no eliminar



0ecordar ,e2

f an#$s$s(m)=m f s

N , m=1,2,… , N −1







0ecordar ,e2

f an#$s$s(m)=m f s

N , m=1,2,… , N −1

*a DFT #rodce resltados correctos solo cando la secencia de datos de

entrada contiene energ6a #recisamente en la frecencias de análisis

Fgas Blea/a#eC en la DFT.onsidere na señal de frecencia igal a 3 Hz 5 de la cal se e3traen 64

mestras (n=0... 63) % es mestreada a 7 .alclando la DFT2 f s=64 muestras/s





m f s

N 0 Hz ,1 Hz ,2 Hz ,… ,63 Hz

Frecencias de análisis 2





m f s

N 0 Hz ,1 Hz ,2 Hz ,… ,63 Hz






m f s

N 0 Hz ,1 Hz ,2 Hz ,… ,63 Hz

Frecencias de análisis 2 * =3"4 Hz





m f s

N 0 Hz ,1 Hz ,2 Hz ,… ,63 Hz


0 Hz ,1 Hz ,2 Hz ,… ,63 Hz




Fgas Blea/a#eC en la DFT-Oé casa ese com#ortamiento/



mG= 't#t (in B.ontenedor de salidaC

Fgas Blea/a#eC en la DFT-Oé casa ese com#ortamiento/



mG= 't#t (in B.ontenedor de salidaC

*as fgas hacen ,e cal,ier señal c%a frecencia no se encentre en el

centro del contenedor B(inC se fu#uen hacia los otros contenedores7

Fgas Blea/a#eC en la DFT*a res#esta en am#litd de na DFT de %&!untos so(re na fnci"n sinsoidal

en términos del 6ndice del contenedor m #ede ser a#ro3imada #or la fnci"n

sinc2

X (m)# N 2 ⋅sen(π(k −m))

π(k −m)



( )

4*5 cos=sen( x)

sen( x

N )

, osen( x )

x , o

sen ( N x

2 )

sen ( x

2 )

Fgas Blea/a#eC en la DFT X (m)#

N

2 ⋅

sen(π(k −m))π(k −m)

Demostraci"n2

x r (n)=cos( 2π n k

N )Fnci"n real2

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x

2 )

sen( x

2)




N

2 ⋅


Demostraci"n2


N )Fnci"n real2

1+mero de ciclos enterosen 1 mestras

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x

2 )

sen( x

2)




N

2 ⋅


Demostraci"n2


N )Fnci"n real2


X (m)=∑ N −1

x (n)e− j

2πn m

N

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x

2 )

sen( x

2)



X r (m) ∑n=0

x r (n)e


N

2 ⋅


Demostraci"n2


N )Fnci"n real2


X r (m)=∑ N −1

x r (n)e− j

2πn m

N =∑ N −1

cos( 2π n k )e− j

2π n m

N

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x

2 )

sen( x

2)



X r (m) ∑n=0

x r (n)e ∑n=0

cos( N )e


N

2 ⋅


Demostraci"n2


N )Fnci"n real2


X r (m)=∑ N −1

x r (n)e− j

2πn m

N =∑ N −1

cos( 2π n k

N )e− j

2π n m

N

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x

2 )

sen( x

2)



r ( ) ∑n=0

r ( ) ∑n=0

( N )cos(ω t )=

e jω t +e− j ω t

2)dentidad de !ler


N

2 ⋅


Demostraci"n2


N )Fnci"n real2


X r (m)=∑ N −1

x r (n)e− j

2πn m

N =∑ N −1

cos( 2π n k

N )e− j

2π n m

N

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x

2 )

sen( x

2)



r ( ) ∑n=0

r ( ) ∑n=0

( N )cos(ω t )=


2)dentidad de !ler

X r (m)=1

2∑n=0

N −1 (e j

2π n k

N +e− j

2 πn k

N )e− j

2πn m

N


N

2 ⋅


Demostraci"n2


N )Fnci"n real2


X r (m)=∑ N −1

x r (n)e− j

2πn m

N =∑ N −1

cos( 2π n k

N )e− j

2π n m

N

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x

2 )

sen( x

2)



r ( ) ∑n=0

r ( ) ∑n=0

( N )cos(ω t )=


2)dentidad de !ler

X r (m)=1

2∑n=0

N −1 (e j

2π n k

N +e− j

2 πn k

N )e− j

2πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn k

N e− j

2π n m

N +1

2∑n=0

N −1

e− j

2π n k

N e− j

2 πn m

N


N

2 ⋅


Demostraci"n2


N )Fnci"n real2


X r (m)=∑ N −1

x r (n)e− j

2πn m

N =∑ N −1

cos( 2π n k

N )e− j

2π n m

N

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x

2 )

sen( x

2)



n=0 n=0( N )

cos(ω t )=e jω t +e− j ω t

2)dentidad de !ler

X r (m)=1

2∑n=0

N −1 (e j

2π n k

N +e− j

2 πn k

N )e− j

2πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn k

N e− j

2π n m

N +1

2∑n=0

N −1

e− j

2π n k

N e− j

2 πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn (k −m ) N +

1

2∑n=0

N −1

e− j

2π n(k +m ) N


N

2 ⋅


Demostraci"n2


N )Fnci"n real2


X r (m)=∑ N −1

x r (n)e− j

2πn m

N =∑ N −1

cos( 2π n k

N )e− j

2π n m

N

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x

2 )

sen( x

2)



n=0 n=0( N )


2)dentidad de !ler

X r (m)=1

2∑n=0

N −1 (e j

2π n k

N +e− j

2 πn k

N )e− j

2πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn k

N e− j

2π n m

N +1

2∑n=0

N −1

e− j

2π n k

N e− j

2 πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn (k −m ) N +

1

2∑n=0

N −1

e− j

2π n(k +m ) N


N

2 ⋅


Demostraci"n2


N )Fnci"n real2


X r (m)=∑ N −1

x r (n)e− j

2πn m

N =∑ N −1

cos( 2π n k

N )e− j

2π n m

N

2=−2 π(k −m)

N

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x

2 )

sen( x

2)



n=0 n=0( N )


2)dentidad de !ler

X r (m)=1

2∑n=0

N −1

(e j

2π n k

N +e− j

2 πn k

N )e− j

2πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn k

N e− j

2π n m

N +1

2∑n=0

N −1

e− j

2π n k

N e− j

2 πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn (k −m ) N +

1

2∑n=0

N −1

e− j

2π n(k +m ) N


N

2 ⋅


Demostraci"n2


N )Fnci"n real2


X r (m)=∑0

N −1

x r (n)e− j

2πn m

N =∑0

N −1

cos( 2π n k

N )e− j

2π n m

N

2=−2 π(k −m)

N

∑ N −1

e j

2 πn(k −m ) N =∑

N −1

e− j n 2

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x

2 )

sen( x

2)



n=0 n=0( N )


2)dentidad de !ler

X r (m)=1

2∑n=0

N −1

(e j

2π n k

N +e− j

2 πn k

N )e− j

2πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn k

N e− j

2π n m

N +1

2∑n=0

N −1

e− j

2π n k

N e− j

2 πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn (k −m ) N +

1

2∑n=0

N −1

e− j

2π n(k +m ) N

∑n=0

∑n=0


N

2 ⋅


Demostraci"n2


N )Fnci"n real2


X r (m)=∑0

N −1

x r (n)e− j

2πn m

N =∑0

N −1

cos( 2π n k

N )e− j

2π n m

N

2=−2 π(k −m)

N

∑ N −1

e j

2 πn(k −m ) N =∑

N −1

e− j n 2 =∑

N −1

(e− j 2 )n

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)



n=0 n=0( N )


2)dentidad de !ler

X r (m)=1

2∑n=0

N −1

(e j

2π n k

N +e− j

2 πn k

N )e− j

2πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn k

N e− j

2π n m

N +1

2∑n=0

N −1

e− j

2π n k

N e− j

2 πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn (k −m ) N +

1

2∑n=0

N −1

e− j

2π n(k +m ) N

∑n=0

∑n=0

∑n=0

( )


N

2 ⋅


Demostraci"n2


N )Fnci"n real2


X r (m)=∑n=0

N −1

x r (n)e− j

2πn m

N =∑n=0

N −1

cos( 2π n k

N )e− j

2π n m

N

2=−2 π(k −m)

N

∑ N −1

e j

2 πn(k −m ) N =∑

N −1

e− j n 2 =∑

N −1

(e− j 2 )n

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)



s= ∑n=0

N −1

a r n=a

1−r N

1−r

n=0 n=0( N )


2)dentidad de !ler

X r (m)=1

2∑n=0

N −1

(e j

2π n k

N +e− j

2 πn k

N )e− j

2πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn k

N e− j

2π n m

N +1

2∑n=0

N −1

e− j

2π n k

N e− j

2 πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn (k −m ) N +

1

2∑n=0

N −1

e− j

2π n(k +m ) N

∑n=0

∑n=0

∑n=0


N

2 ⋅


Demostraci"n2


N )Fnci"n real2


X r (m)=∑n=0

N −1

x r (n)e− j

2πn m

N =∑n=0

N −1

cos( 2π n k

N )e− j

2π n m

N

2=−2 π(k −m)

N

∑ N −1

e j

2 πn(k −m ) N =∑

N −1

e− j n 2 =∑

N −1

(e− j 2 )n

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)



s= ∑n=0

N −1

a r n=a

1−r N

1−r

n=0 n=0( )


2)dentidad de !ler

X r (m)=1

2∑n=0

N −1

(e j

2π n k

N +e− j

2 πn k

N )e− j

2πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn k

N e− j

2π n m

N +1

2∑n=0

N −1

e− j

2π n k

N e− j

2 πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn (k −m ) N +

1

2∑n=0

N −1

e− j

2π n(k +m ) N

n=0 n=0 n=0

∑n=0

N −1

( e− j 2 )n

=1−e

− j 2 N

1−e− j 2


N

2 ⋅


Demostraci"n2


N )Fnci"n real2


X r (m)=∑n=0

N −1

x r (n)e− j

2πn m

N =∑n=0

N −1

cos( 2π n k

N )e− j

2π n m

N

2=−2 π(k −m)

N

∑ N −1

e j

2 πn(k −m ) N =∑

N −1

e− j n 2 =∑

N −1

(e− j 2 )n

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)



s= ∑n=0

N −1

a r n=a

1−r N

1−r

n 0 n 0( )


2)dentidad de !ler

X r (m)=1

2∑n=0

N −1

(e j

2π n k

N +e− j

2 πn k

N )e− j

2πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn k

N e− j

2π n m

N +1

2∑n=0

N −1

e− j

2π n k

N e− j

2 πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn (k −m ) N +

1

2∑n=0

N −1

e− j

2π n(k +m ) N

n=0 n=0 n=0

∑n=0

N −1

( e− j 2 )n

=1−e

− j 2 N

1−e− j 2 (

e j 2

N

2

e j 2 N

2

e j

2

2

e j

2

2

)


N

2 ⋅


Demostraci"n2


N )Fnci"n real2


X r (m)=∑n=0

N −1

x r (n)e− j

2πn m

N =∑n=0

N −1

cos( 2π n k

N )e− j

2π n m

N

2=−2 π(k −m)

N

∑0

N −1

e j

2 πn(k −m ) N =∑

0

N −1

e− j n 2 =∑

0

N −1

(e− j 2 )n

4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)



s= ∑n=0

N −1

a r n=a

1−r N

1−r

n 0 n 0( )


2)dentidad de !ler

X r (m)=1

2∑n=0

N −1

(e j

2π n k

N +e− j

2 πn k

N )e− j

2πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn k

N e− j

2π n m

N +1

2∑n=0

N −1

e− j

2π n k

N e− j

2 πn m

N

X r (m)=1

2∑n=0

N −1

e j

2 πn (k −m ) N +

1

2∑n=0

N −1

e− j

2π n(k +m ) N

n=0 n=0 n=0

∑n=0

N −1

( e− j 2 )n

=1−e

− j 2 N

1−e− j 2 (

e j 2

N

2

e j 2 N

2

e j

2

2

e j

2

2

)=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2

e j

2

2

−e− j 2 e

j 2

2

e j

2

2


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2− j 2 e

j 2

2



e

e j

2

2

−e− j 2 e

e j

2

2


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2− j 2 e

j 2

2

=

e j 2

N

2 −e− j 2 N

e j 2

N

2

e j 2

N

2

e j

2

2−e− j 2

e j

2

2



e

e j

2

2

−e j 2 e

e j

2

2

e e e

e j

2

2


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2− j 2 e

j 2

2

=

e j 2

N

2 −e− j 2 N

e j 2

N

2

e j 2

N

2

e j

2

2−e− j 2

e j

2

2



e j

2

2

−e j 2

e j

2

2 e j

2

2

=e

j 2

2

(e j 2

N

2 −e− j 2

N

2

)e

j 2 N

2 (e j

2

2−e− j

2

2 )


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2

e− j 2 e

j 2

2

=

e j 2

N

2 −e− j 2 N

e j 2

N

2

e j 2

N

2

e j

2

2−e− j 2

e j

2

2



e j

2

2

−ej 2

e j

2

2 e j

2

2

=e

j 2

2

(e j 2

N

2 −e− j 2

N

2

)e

j 2 N

2 (e j

2

2−e− j

2

2 ) =

e j

2

2 e− j 2

N

2

(e j 2

N

2 −e− j 2

N

2

)(e

j 2

2−e− j

2

2 )


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2

e− j 2 e

j 2

2

=

e j 2

N

2 −e− j 2 N

e j 2

N

2

e j 2

N

2

e j

2

2−e− j 2

e j

2

2



e j

2

2

−ej 2

e j

2

2 e j

2

2

=e

j 2

2

(e j 2

N

2 −e− j 2

N

2

)e

j 2 N

2 (e j

2

2−e− j

2

2 ) =

e j

2

2 e− j 2

N

2

(e j 2

N

2 −e− j 2

N

2

)(e

j 2

2−e− j

2

2 )


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2

−e− j 2 e

j 2

2

=

e j 2

N

2 −e− j 2 N

e j 2

N

2

e j 2

N

2

e j

2

2−e− j 2

e j

2

2



e j

2

2

−e

e j

2

2 e j

2

2

=e

j 2

2

(e j 2

N

2 −e− j 2

N

2

)e

j 2 N

2 (e j

2

2−e− j

2

2 ) =

e j

2

2 e− j 2

N

2

(e j 2

N

2 −e− j 2

N

2

)(e

j 2

2−e− j

2

2 )

e

j 2

2 −

j 2

2 N


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2

−e− j 2 e

j 2

2

=

e j 2

N

2 −e− j 2 N

e j 2

N

2

e j 2

N

2

e j

2

2−e− j 2

e j

2

2



e j

2

2

e

e j

2

2 e j

2

2

=e

j 2

2

(e j 2

N

2 −e− j 2

N

2

)e

j 2 N

2 (e j

2

2−e− j

2

2 ) =

e j

2

2 e− j 2

N

2

(e j 2

N

2 −e− j 2

N

2

)(e

j 2

2−e− j

2

2 )

e

j 2

2 −

j 2

2 N

=e

j2

2 (1− N )


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2

−e− j 2 e

j 2

2

=

e j 2

N

2 −e− j 2 N

e j 2

N

2

e j 2

N

2

e j

2

2−e− j 2

e j

2

2



e j

2

2

e

e j

2

2 e j

2

2

=e

j 2

2

(e j 2

N

2 −e− j 2

N

2

)e

j 2 N

2 (e j

2

2−e− j

2

2 ) =

e j

2

2 e− j 2

N

2

(e j 2

N

2 −e− j 2

N

2

)(e

j 2

2−e− j

2

2 )

e

j 2

2 −

j 2

2 N

=e

j2

2 (1− N )

=e

j2

2 [−( N −1)]


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2

−e− j 2 e

j 2

2

=

e j 2

N

2 −e− j 2 N

e j 2

N

2

e j 2

N

2

e j

2

2−e− j 2

e j

2

2



e j

2

2

e

e j

2

2 e j

2

2

=e

j 2

2

(e j 2

N

2 −e− j 2

N

2

)e

j 2 N

2 (e j

2

2−e− j

2

2 ) = e

j 2

2 e− j 2

N

2

(e j 2

N

2 −e− j 2

N

2

)(e

j 2

2−e− j

2

2 )

e

j 2

2 −

j 2

2 N

=e

j2

2 (1− N )

=e

j2

2 [−( N −1)]

=e

− j 2 ( N −1)2


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2

−e− j 2 e

j 2

2

=

e j 2

N

2 −e− j 2 N

e j 2

N

2

e j 2

N

2

e j

2

2−e− j 2

e j

2

2



e j

2

2 e j

2

2 e j

2

2

=e

j 2

2

(e j 2

N

2 −e− j 2

N

2

)e

j 2 N

2 (e j

2

2−e− j

2

2 ) = e

j 2

2 e− j 2

N

2

(e j 2

N

2 −e− j 2

N

2

)(e

j 2

2−e− j

2

2 )

e

j 2

2 −

j 2

2 N

=e

j2

2 (1− N )

=e

j2

2 [−( N −1)]

=e

− j 2 ( N −1)2

=e

− j 2 ( N −1)2

(e j 2

N

2 −e− j 2

N

2 )(e

j 2

2−e− j

2

2 )


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2

2−e

− j 2 e j

2

2

2

=

e j 2

N

2 −e− j 2 N

e j 2

N

2

e j 2

N

2

e j

2

2−e− j 2

e j

2

2

2



e j

2

2 e j

2

2 e j

2

2

= e j

2

2

(e j 2

N

2 −e− j 2

N

2

)e

j 2 N

2 (e j

2

2−e− j

2

2 ) = e

j 2

2 e− j 2

N

2

(e j 2

N

2 −e− j 2

N

2

)(e

j 2

2−e− j

2

2 )

e

j 2

2 −

j 2

2 N

=e

j2

2 (1− N )

=e

j2

2 [−( N −1)]

=e

− j 2 ( N −1)2


− jω t

2 j

)dentidad de !ler

=e

− j 2 ( N −1)2

(e j 2

N

2 −e− j 2

N

2 )(e

j 2

2−e− j

2

2 )


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2

2−e

− j 2 e j

2

2

2

=

e j 2

N

2 −e− j 2 N

e j 2

N

2

e j 2

N

2

e j

2

2−e− j 2

e j

2

2

2

=e

− j 2 ( N −1)

2

2 j sen

(2

N

2

)2 j sen ( 2

2 )



e j

2

2 e j

2

2 e j

2

2

= e j

2

2

(e j 2

N

2 −e− j 2

N

2

)e

j 2 N

2 (e j

2

2−e− j

2

2 ) = e

j 2

2 e− j 2

N

2

(e j 2

N

2 −e− j 2

N

2

)(e

j 2

2−e− j

2

2 )

e

j 2

2 −

j 2

2 N

=e

j2

2 (1− N )

=e

j2

2 [−( N −1)]

=e

− j 2 ( N −1)2


− jω t

2 j

)dentidad de !ler

=e

− j 2 ( N −1)2

(e j 2

N

2 −e− j 2

N

2 )(e

j 2

2−e− j

2

2 )


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2

2−e

− j 2 e j

2

2

2

=

e j 2

N

2 −e− j 2 N

e j 2

N

2

e j 2

N

2

e j

2

2−e− j 2

e j

2

2

2

=e

− j 2 ( N −1)

2

2 j sen

(2

N

2

)2 j sen ( 2

2 )2=

2 π (k −m)



e j

2

2 e j

2

2 e j

2

2

= e j

2

2

(e j 2

N

2 −e− j 2

N

2

)e

j 2 N

2 (e j

2

2−e− j

2

2 ) = e

j 2

2 e− j 2

N

2

(e j 2

N

2 −e− j 2

N

2

)(e

j 2

2−e− j

2

2 )

e

j 2

2 −

j 2

2 N

=e

j2

2 (1− N )

=e

j2

2 [−( N −1)]

=e

− j 2 ( N −1)2


− jω t

2 j

)dentidad de !ler

=e

− j 2 ( N −1)2

(e j 2

N

2 −e− j 2

N

2 )(e

j 2

2−e− j

2

2 )

2=−( ) N


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2

j2−e

− j 2 e j

2

2

j2

=

e j 2

N

2 −e− j 2 N

e j 2

N

2

e j 2

N

2

e j

2

2−e− j 2

e j

2

2

j2

=e

− j 2 ( N −1)

2

2 j sen

(2

N

2

)2 j sen ( 2

2 )2=−

2 π (k −m)



e j

2

2 e j

2

2 e j

2

2

= e j

2

2

(e j 2

N

2 −e− j 2

N

2

)e

j 2 N

2 (e j

2

2−e− j

2

2 ) = e

j 2

2 e− j 2

N

2

(e j 2

N

2 −e− j 2

N

2

)(e

j 2

2−e− j

2

2 )

e

j 2

2 −

j 2

2 N

=e

j2

2 (1− N )

=e

j2

2 [−( N −1)]

=e

− j 2 ( N −1)2


− jω t

2 j

)dentidad de !ler

=e

− j 2 ( N −1)2

(e j 2

N

2 −e− j 2

N

2 )(e

j 2

2−e− j

2

2 )

2=− N

=e

− j

(−

2π (k −m ) N

)( N −1)

2

sen

(−

2π(k −m)

N

N

2

)sen(−2π(k −m)

N

1

2 )


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2

j 2−e

− j 2 e j

2

2

j 2

=

e j 2

N

2 −e− j 2 N

e j 2

N

2

e j 2

N

2

e j

2

2−e− j 2

e j

2

2

j 2

=e

− j 2 ( N −1)

2

2 j sen

(2

N

2

)2 j sen ( 2

2 )2=−

2 π (k −m)



e j

2

2 e j

2

2 e j

2

2

= e j

2

2

(e j 2

N

2 −e− j 2

N

2

)e

j 2 N

2 (e j

2

2−e− j

2

2 ) = e

j 2

2 e− j 2

N

2

(e j 2

N

2 −e− j 2

N

2

)(e j

2

2−e− j

2

2 )

e

j 2

2 −

j 2

2 N

=e

j2

2 (1− N )

=e

j2

2 [−( N −1)]

=e

− j 2 ( N −1)2


− jω t

2 j

)dentidad de !ler

=e

− j 2 ( N −1)2

(e j 2

N

2 −e− j 2

N

2 )(e

j 2

2−e− j

2

2 )

2= N

=e

− j

(−

2π (k −m ) N

)( N −1)

2

sen

(−

2π(k −m)

N

N

2

)sen(−2π(k −m)

N

1

2 )=e

j [ π(k −m ) N ]( N −1)sen (π(k −m))

sen

(π(k −m)

N

)


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

∑n=0

N −1

( e− j 2 )n

=

e j 2

N

2

e j 2

N

2

−e− j 2 N e

j 2 N

2

e j 2

N

2

e j

2

2

j 2−e

− j 2 e j

2

2

j 2

=

e j 2

N

2 −e− j 2 N

e j 2

N

2

e j 2

N

2

e j

2

2−e− j 2

e j

2

2

j 2

=e

− j 2 ( N −1)

2

2 j sen

(2

N

2

)2 j sen ( 2

2 )2=−

2 π (k −m)



e j

2

2 e j

2

2 e j

2

2

= e j

2

2

(e j 2

N

2 −e− j 2

N

2

)e

j 2 N

2 (e j

2

2−e− j

2

2 ) = e

j 2

2 e− j 2

N

2

(e j 2

N

2 −e− j 2

N

2

)(e j

2

2−e− j

2

2 )

e

j 2

2 −

j 2

2 N

=e

j2

2 (1− N )

=e

j2

2 [−( N −1)]

=e

− j 2 ( N −1)2


− jω t

2 j

)dentidad de !ler

=e

− j 2 ( N −1)2

(e j 2

N

2 −e− j 2

N

2 )(e

j 2

2−e− j

2

2 )

2 N

=e

− j

(−

2π (k −m ) N

)( N −1)

2

sen

(−

2π(k −m)

N

N

2

)sen(−2π(k −m)

N

1

2 )=e

j [ π(k −m ) N ]( N −1)sen (π(k −m))

sen

(π(k −m)

N

)=e j [π( k −m)−

π( k −m) N ]sen (π(k −m))

sen (π(k −m) N )


N

2 ⋅


4*5 cos=sen( x)

sen( x

N )

, osen ( x)

x , o

sen( N x2 )

sen( x

2)

Demostraci"n2

1 N −1

j2 πn (k −m )

1 N −1

j2π n(k +m )

∑n=0

N −1

( e− j 2 )n

=e j [π(k −m)−

π (k −m ) N ]sen(π (k −m))

sen

(π(k −m)

N ) 2=−

2 π (k −m) N



X r (m)=1

2∑n=0

e j

( ) N +

1

2∑n=0

e− j

( ) N

X r (m)=e j [π( k −m)−

π( k −m ) N ] 1

2

sen (π(k −m))

sen (π(k −m) N )

+e− j [π(k −m)−

π (k −m ) N ] 1

2

sen (π(k −m))

sen ( π(k −m) N )

Fgas Blea/a#eC en la DFTS#onga ,e na sinsoide de frecencia 8 4Hz se mestrea a 4Hz % se

o(tienen mestras




o(tienen mestras




o(tienen mestras

S#onga ,e la señal

mestreada es de

8 4Hz



S#onga ahora ,ela señal mestreada

es de 82 4Hz


o(tienen mestras

S#onga ,e la señal

mestreada es de

8 4Hz



S#onga ahora ,ela señal mestreada

es de 82 4Hz

Fgas Blea/a#eC en la DFT!l es#ectro de frecencia de na señal contina es #eri"dico2

* =3"4 Hz




* =3"4 Hz




* =3"4 Hz * =−3"4 Hz * =28"6 Hz * =16"4 Hz



Ventanas*as ventanas redcen las fgas en la DFT al minimizar la magnitd de los

l"(los laterales de la fnci"n sinc

X (m)# N

2 ⋅




!sto se logra al forzar a ,e la am#litd al inicio % al final de la señal de entrada

alcance savemente n mismo valor de am#litd

!sto es lo ,e se (sca redcir

Ventanas*as ventanas redcen las fgas en la DFT al minimizar la magnitd de los

l"(los laterales de la fnci"n sinc

X (m)# N

2 ⋅




!sto se logra al forzar a ,e la am#litd al inicio % al final de la señal de entrada

alcance savemente n mismo valor de am#litd

!sto es lo ,e se (sca redcir

Ventanas

Rectangular

Triangular



Hanning

Hamming

Ventanas

Rectangular

Triangular

(n)=1 + n=0,1,2,… , N −1

(n)={ n

N $2 + n=0,1,2,… , N $2

2− n

N $2 + n= N $2,… , N



Hanning

Hamming

{ N $2

(n)=0"5−0"5cos(2 π n

N ) + n=0,1,2,… , N −1

(n)=0"54−0"46cos(2 π n

N )+ n=0,1,2,… , N −1

Ventanas

Rectangular

Triangular

(n)=1 + n=0,1,2,… , N −1

n=0,1,2,… , N −1



Hanning

Hamming

(n)=0"5−0"5cos(2 π n

N ) + n=0,1,2,… , N −1

(n)=0"54−0"46cos(2 π n

N )+ n=0,1,2,… , N −1

Ventanas2 res#esta en frecencia

7 (m)=∑n=0

N −1

(n)e

− j 2 π nm

N



7 ma% (m)=∣7 (m)∣

7 (m)=∑n=0

N −1

(n)e

− j 2 π nm

N

7 d8(m)=20⋅!o'10(∣7 (m)∣∣7 (0)∣ )

Magnitud de la respuesta en frecuencia enescala lineal

Magnitud de la respuesta en frecuencia enescala logarítmica

Rectangular Rectangular

H i




Hamming

Triangular

Hanning

Hamming

Triangular

Hanning

f s N

0 2 f s

N

3 f s

N

4 f s

N

5 f s

N

f s N

0 2 f s

N

3 f s

N

4 f s

N

5 f s

N

Práctica =797A7 '(&etivo2 '(servar la res#esta en frecencia de distintas

fnciones ventana


97 .alcle las fnciones ventana 5ectan#ular 5 Trian#ular 5 Hannin# % Hammin# 7

rafi,e s am#litd en s(gráficas7<7 .alcle la DFT de ' #ntos de las fnciones ventana7

>7 rafi,e la magnitd de las DFT de las fnciones ventana

=7 rafi,e la magnitd de las DFT de las fnciones ventana en escala logar6tmica



, g g

Ben deci(elesC7

Ventanas2 minimizaci"n de fgas en la DFT

X (m)=∑n=0

N −1

(n)⋅ x (n)e

− j 2π nm

N




X (m)=∑n=0

N −1

(n)⋅ x (n)e

− j 2π nm

N

Sinsoide de >7=Fnci"n ventana



x (n)=[0"5−0"5cos(

2 π n

64 )]⋅[sen (

2 π3"4 n

64 )]


X (m)=∑n=0

N −1

(n)⋅ x (n)e

− j 2π nm

N

Ventana Hanning




x(n

)=[0"5

−0"5cos

(

2 π n

64 )]⋅[sen

(

2 π3"4 n

64 )]


X (m)=∑n=0

N −1

(n)⋅ x (n)e

− j 2π nm

N

Ventana Hanning

x (n)=[0"5−0"5cos (2 π n

64)]⋅[sen (

2 π3"4 n

64)+ 0"1sen(

2π 7 n

64)]

Sinsoide de >7= % A ciclosFnci"n ventana




x (n)=[0"5−0"5cos(2 π n

64)]⋅[sen (

2 π3"4 n

64)]

64 64 64


X (m)=∑n=0

N −1

(n)⋅ x (n)e

− j 2π nm

N

Ventana Hanning

x (n)=[0"5−0"5cos (2 π n

64)]⋅[sen (

2 π3"4 n

64)+ 0"1sen(

2π 7 n

64)]





x (n)=[0"5−0"5cos(2 π n

64)]⋅[sen (

2 π3"4 n

64)]


X (m)=∑n=0

N −1

(n)⋅ x (n)e

− j 2π nm

N

Ventana Hanning

x (n)=[0"5−0"5cos (2 π n

64)]⋅[sen (

2 π3"4 n

64)+ 0"1sen(

2π 7 n

64)]





x (n)=[0"5−0"5cos(2 π n

64)]⋅[sen (

2 π3"4 n

64)]

DFT de na fnci"n rectanglar general

* ret =sen( x)

sen( x

N ) , o

sen( x) x

, o

sen ( N x

2 )

sen ( x

2)

X (m)= ∑n=− N

2 + 1

N

2

x (n)e− j 2π nm

N x (n)

1 N $2− N $2+ 1



0


* ret =sen( x)

sen( x

N ) , o

sen( x) x

, o

sen ( N x

2 )

sen ( x

2)

X (m)= ∑n=− N

2 + 1

N

2

x (n)e− j 2π nm

N x (n)

1 N $2− N $2+ 1



0

n=−n0+ ( −1)n=−n0 n=n0


* ret =sen( x)

sen( x

N ) , o

sen( x) x

, o

sen ( N x

2 )

sen ( x

2)

X (m)= ∑n=− N

2 + 1

N

2

x (n)e− j 2π nm

N

( ) ∑−n0+ ( −1) − j 2π n m

N

x (n)

1 N $2− N $2+ 1



X (m)= ∑n=−n0

1⋅e N

0

n=−n0+ ( −1)n=−n0 n=n0


4*5 ret =sen ( x )

sen ( x

N )

, osen ( x)

x , o

sen ( N x

2 )

sen ( x

2)

X (m)= ∑n=− N

2 + 1

N

2

x (n)e− j 2π nm

N

X ( ) ∑−n0+ ( −1)

1

− j 2π n m

N

x (n)

1 N $2− N $2+ 1



X (m)= ∑n=−n0

1⋅e N

0

n=−n0+ ( −1)n=−n0

X ()= ∑n=−n0

−n0+ ( −1)

1⋅e− j n

=2π m

N


X ()= ∑n=−n0

−n0+ ( −1)

1⋅e− j n




X ()= ∑n=−n0

−n0+ ( −1)

1⋅e− j n

X ()=

∑n=−n0

−n0+ ( −1)

e− j n




X ()= ∑n=−n0

−n0+ ( −1)

1⋅e− j n

X ()=

∑n=−n0

−n0+ ( −1)

e− j n

=e− j (−n0)+ e

− j (−n0+ 1)+ e− j (−n0+ 2)+ …+ e

− j (−n0+ ( −1))




X ()= ∑n=−n0

−n0+ ( −1)

1⋅e− j n

X ()=

∑n=−n0

−n0+ ( −1)

e− j n

=e− j (−n0)+ e

− j (−n0+ 1)+ e− j (−n0+ 2)+ …+ e

− j (−n0+ ( −1))

=e− j (−n0)+ e

− j (−n0)⋅e− j 1+ e

− j (−n0)⋅e− j 2 + …+ e

− j (−n0)⋅e− j ( −1)



=e− j (−n0)⋅e

− j 0+ e− j (−n0)⋅e

− j 1+ e− j (−n0)⋅e

− j 2 + …+ e− j (−n0)⋅e

− j( −1)


X ()= ∑n=−n0

−n0+ ( −1)

1⋅e− j n

X ()=

∑n=−n0

−n0+ ( −1)

e− j n

=e− j (−n0)+ e

− j (−n0+ 1)+ e− j (−n0+ 2)+ …+ e

− j (−n0+ ( −1))

=e− j (−n0)+ e

− j (−n0)⋅e− j 1+ e

− j (−n0)⋅e− j 2 + …+ e

− j (−n0)⋅e− j ( −1)



=e− j (−n0)⋅e

− j 0+ e− j (−n0)⋅e

− j 1+ e− j (−n0)⋅e

− j 2 + …+ e− j (−n0)⋅e

− j( −1)

=e j n0⋅[e− j 0 + e

− j 1+ e− j 2 + …+ e

− j( −1) ]


X ()= ∑n=−n0

−n0+ ( −1)

1⋅e− j n

X ()=

∑n=−n0

−n0+ ( −1)

e− j n

=e− j (−n0)+ e

− j (−n0+ 1)+ e− j (−n0+ 2)+ …+ e

− j (−n0+ ( −1))

=e− j (−n0)+ e

− j (−n0)⋅e− j 1+ e

− j (−n0)⋅e− j 2 + …+ e

− j (−n0)⋅e− j ( −1)



=e− j (−n0)⋅e

− j 0+ e− j (−n0)⋅e

− j 1+ e− j (−n0)⋅e

− j 2 + …+ e− j (−n0)⋅e

− j( −1)

=e j n0⋅[e− j 0 + e

− j 1+ e− j 2 + …+ e

− j( −1) ]

X ()=e j n0

∑ =0

−1

e− j


X ()= ∑n=−n0

−n0+ ( −1)

1⋅e− j n

X ()=

∑n=−n0

−n0+ ( −1)

e− j n

=e− j (−n0)+ e

− j (−n0+ 1)+ e− j (−n0+ 2)+ …+ e

− j (−n0+ ( −1))

=e− j (−n0)+ e

− j (−n0)⋅e− j 1+ e

− j (−n0)⋅e− j 2 + …+ e

− j (−n0)⋅e− j ( −1)



=e− j (−n0)⋅e

− j 0+ e− j (−n0)⋅e

− j 1+ e− j (−n0)⋅e

− j 2 + …+ e− j (−n0)⋅e

− j( −1)

=e j n0⋅[e− j 0 + e

− j 1+ e− j 2 + …+ e

− j( −1) ]

X ()=e j n0

∑ =0

−1

e− j

∑ =0

−1

e− j =∑ =0

−1

(e− j )


X ()= ∑n=−n0

−n0+ ( −1)

1⋅e− j n

X ()=

∑n=−n0

−n0+ ( −1)

e− j n

=e− j (−n0)+ e

− j (−n0+ 1)+ e− j (−n0+ 2)+ …+ e

− j (−n0+ ( −1))

=e− j (−n0)+ e

− j (−n0)⋅e− j 1+ e

− j (−n0)⋅e− j 2 + …+ e

− j (−n0)⋅e− j ( −1)



=e− j (−n0)⋅e

− j 0+ e− j (−n0)⋅e

− j 1+ e− j (−n0)⋅e

− j 2 + …+ e− j (−n0)⋅e

− j( −1)

=e j n0⋅[e− j 0 + e

− j 1+ e− j 2 + …+ e

− j( −1) ]

X ()=e j n0

∑ =0

−1

e− j

∑ =0

−1

e− j =∑ =0

−1

(e− j )

Serie geométrica


Series geométricas


s=a+ a⋅r 1+ a⋅r

2+ a⋅r 3+ …+ a⋅r

n−1

s=∑k =0

n−1

a r k

s=a⋅r + a⋅r 2+ a⋅r

3+ a⋅r 4+ …+ a⋅r

n−1+ a⋅r n

r n−1

r 1=r

n−1+ 1=r n





s−r s=a+ a⋅r 1

+ a⋅r 2

+ a⋅r 3

+ …+ a⋅r n−1

−a⋅r 1

−a⋅r 2

−a⋅r 3

−a⋅r 4

−…−a⋅r n


s=a1−r

n

1−r

Si % 2∣r ∣< 1 n →∞ s= a

1−r


∑ =0

−1

(e− j )




∑ =0

−1

(e− j ) s=∑k =0

n−1

a r k =a

1−r n

1−r




∑ =0

−1

(e− j ) s=∑k =0

n−1

a r k =a

1−r n

1−r

∑ =0

−1

(e− j ) =1−e

− j

1−e

− j




∑ =0

−1

(e− j ) s=∑k =0

n−1

a r k =a

1−r n

1−r

∑ =0

−1

(e− j ) =1−e

− j

1−e

− j

Mlti#licando #or 9E2




∑ =0

−1

(e− j ) s=∑k =0

n−1

a r k =a

1−r n

1−r

∑ =0

−1

(e− j ) =1−e

− j

1−e

− j


=1−e

− j

⋅

e

j

2

e

j

2



=1−e− j

⋅

e j 2

e

j

2


∑ =0

−1

(e− j ) s=∑k =0

n−1

a r k =a

1−r n

1−r

∑ =0

−1

(e− j ) =1−e

− j

1−e

− j


=1−e

− j

⋅

e

j

2

e

j

2



=1−e− j

e j 2

e

j

2

=

e

j

2 −e− j

e

j

2

e

j

2

e

j

2 −e− j

e

j

2

e

j

2


∑ =0

−1

(e− j ) s=∑k =0

n−1

a r k =a

1−r n

1−r

∑ =0

−1

(e− j ) =1−e

− j

1−e

− j


=1−e

− j

⋅

e

j

2

e

j

2

=

e

j

2 (e j

2 −e

− j

2 )e

j 2 (e

j 2 −e

− j 2 )



1−e− j

e j 2

e

j

2

=

e

j

2 −e− j

e

j

2

e

j

2

e

j

2 −e− j

e

j

2

e

j

2


∑ =0

−1

(e− j ) s=∑k =0

n−1

a r k =a

1−r n

1−r

∑ =0

−1

(e− j ) =1−e

− j

1−e

− j


=1−e

− j

⋅

e

j

2

e

j

2

=

e

j

2 (e j

2 −e

− j

2 )e

j 2 (e

j 2 −e

− j 2 )

=e

− j

2 (e j

2 −e

− j

2 )− j j − j



1−e− j

e j 2

e

j

2

=

e

j

2 −e− j

e

j

2

e

j

2

e

j

2 −e− j

e

j

2

e

j

2

e 2

(e2 −e 2

)


∑ =0

−1

(e− j ) s=∑k =0

n−1

a r k =a

1−r n

1−r

∑ =0

−1

(e− j ) =1−e

− j

1−e

− j


=1−e

− j

⋅

e

j

2

e

j

2

=

e

j

2 (e j

2 −e

− j

2 )e

j 2 (e

j 2 −e

− j 2 )

=e

− j

2 (e j

2 −e

− j

2 )− j j − j



1−e− je

j 2

e

j

2

=

e

j

2 −e− j

e

j

2

e

j

2

e

j

2 −e− j

e

j

2

e

j

2

e 2

(e2 −e 2

)

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )


∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )




∑ 1=0

) −1

e− j 2 1=e

− j2 ( ) −1)2 ⋅

(e j 2 )

2 −e

− j 2 ) 2 )

(e j2

2 −e

− j 22 )

sen(%)=e j%−e

− j%

2 j




∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

sen(%)=e j%−e

− j%

2 j



=e

− j ( −1)2 ⋅

2 j sen (

2 )2 j sen (

2 )


∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

sen(%)=e j%−e

− j%

2 j



=e

− j ( −1)2 ⋅

2 j sen (

2 )2 j sen (

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

sen

(

2 )sen (

2 )


∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

( )

X ()=e j n0 ∑

=0

−1

e− j

sen(%)=e j%−e

− j%

2 j



=e

− j ( −1)2 ⋅

2 j sen (

2 )2 j sen (

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

sen

(

2 )sen (

2 )


∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

( )

X ()=e j n0 ∑

=0

−1

e− j

X ()=e j n0⋅e

− j ( −1)2 ⋅

sen (

2 )sen

(2 )

sen(%)=e j%−e

− j%

2 j



=e

− j ( −1)2 ⋅

2 j sen (

2 )2 j sen (

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

sen

(

2 )sen (

2 )


∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

( )

X ()=e j n0 ∑

=0

−1

e− j

X ()=e j n0⋅e

− j ( −1)2 ⋅

sen (

2 )sen

(2 )

X ()=e j (n0−

−1

2 )⋅

sen (

2 )sen

(

2

)

sen(%)=e j%−e

− j%

2 j



=e

− j ( −1)2 ⋅

2 j sen (

2 )2 j sen (

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

sen

(

2 )sen (

2 )

(2

)


∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

( )

X ()=e j n0 ∑

=0

−1

e− j

X ()=e j n0⋅e

− j ( −1)2 ⋅

sen (

2 )sen

(2 )

X ()=e j (n0−

−1

2 )⋅

sen (

2 )sen

(

2

)2 m

sen(%)=e j%−e

− j%

2 j



=e

− j ( −1)2 ⋅

2 j sen (

2 )2 j sen (

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

sen

(

2

)sen (

2 )

( )= 2π m N


∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

(

)

X ()=e j n0 ∑

=0

−1

e− j

X ()=e j n0⋅e

− j ( −1)2 ⋅

sen (

2 )sen

(2 )

X ()=e j (n0−

−1

2 )⋅

sen (

2 )sen

(

2

)2 m

sen(%)=e j%−e

− j%

2 j



=e

− j ( −1)2 ⋅

2 j sen (

2 )2 j sen (

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

sen

(

2

)sen (

2 )

( )= 2π m N

X (m)=e j( 2 π m

N )(n0− −1

2 )⋅

sen (2 π m

2 N )sen (2 π m

2 N )


∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

(

)

X ()=e j n0 ∑

=0

−1

e− j

X ()=e j n0⋅e

− j ( −1)2 ⋅

sen (

2 )sen

(2 )

X ()=e j (n0−

−1

2 )⋅

sen (

2 )sen

(

2

)2 m

sen(%)=e j%−e

− j%

2 j



=e

− j ( −1)2 ⋅

2 j sen (

2 )2 j sen (

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

sen

(

2

)sen (

2 )

( )= 2π m N

X (m)=e j( 2 π m

N )(n0− −1

2 )⋅

sen (2 π m

2 N )sen (2 π m

2 N )

X (m)=e j( 2 π m

N )(n0− −1

2 )⋅sen

( π m N )

sen ( π m

N )


∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

(e j

2 −e

− j

2 )(e

j

2 −e

− j

2 )

(

)

X ()=e j n0 ∑

=0

−1

e− j

X ()=e j n0⋅e

− j ( −1)2 ⋅

sen (

2 )sen

(2 )

X ()=e j (n0−

−1

2 )⋅

sen (

2 )sen

(

2

)= 2π m

sen(%)=e j%−e

− j%

2 j



=e

− j ( −1)2 ⋅

2 j sen (

2 )2 j sen (

2 )

∑ =0

−1

e− j =e

− j ( −1)2 ⋅

sen

(

2

)sen (

2 )

( )= 2π m N

X (m)=e j( 2 π m

N )(n0− −1

2 )⋅

sen (2 π m

2 N )sen (2 π m

2 N )

X (m)=e j( 2 π m

N )(n0− −1

2 )⋅sen

( π m N )

sen ( π m

N )

Forma General del

Kernel de Dirichlet

Tarea =797 '(&etivo2 '(tener la DFT de na fnci"n rectanglar simétrica % de

na fnci"n rectanglar donde todos ss elementos son 97

Tarea

97 asándose en la secci"n >79>7< % >79>7> del li(ro Understanding Digital Signal

Processing de 07 *%ons5 o(tenga la DFT de na fnci"n rectanglar simétrica % de

na fnci"n rectanglar donde todos ss mestras tienen am#litd 97

<7 0ealice el desarrollo en s caderno7 !scanee o fotograf6e el resltado5

fa(ri,e n archivo en formato PDF % s(a al sitio del crso7



DFT )nversa de na fnci"n rectanglar general

X (m) → x (n) X (m)

1

0

m=−m0+ ( −1)m=−m0

N $2− N $2+ 1

m




x (n)=1

N ∑m=− N

2 + 1

N

2

X (m)e j 2π mn

N

X (m) → x (n) X (m)

1

0

m=−m0+ ( −1)m=−m0

N $2− N $2+ 1

m




x (n)=1

N ∑m=− N

2 + 1

N

2

X (m)e j 2π mn

N

X (m) → x (n) X (m)

1

0

m=−m0+ ( −1)m=−m0

N $2− N $2+ 1

m

( 2 )( 1)sen( π n

)



x (n)=e− j( 2π n

N )(m0− −1

2 )⋅

1

N ⋅

sen( π n N )

sen( π n

N )

DFT )nversa de na fnci"n rectanglar simétrica H (m)

1

0m=( −1)$ 2m=−m0=−( −1)$ 2

N $2− N $2+ 1

m



DFT )nversa de na fnci"n rectanglar simétrica

H (m)

1

0m=( −1)$ 2m=−m0=−( −1)$ 2

N $2

m

x (n)=e− j( 2 π n

N )(m0− −1

2 )⋅

1

N ⋅

sen( π n

N )sen( π n

N ))DFT de na fnci"n rectanglar general

− N $2+ 1




H (m)

1

0m=( −1)$ 2m=−m0=−( −1)$ 2

N $2

m

x (n)=e− j( 2 π n

N )(m0− −1

2 )⋅

1

N ⋅

sen( π n

N )sen( π n


Sstit%endo el valor de m0= −1

2

− N $2+ 1




H (m)

1

0m=( −1)$ 2m=−m0=−( −1)$ 2

N $2

m

x (n)=e− j( 2 π n

N )(m0− −1

2 )⋅

1

N ⋅

sen( π n

N )sen( π n



2

9(n)=e

− j (2π n

N )( ) −1

2− ) −1

2 )⋅

1

N⋅

sen (πn )

N )(πn )

− N $2+ 1



( ) N sen (πn

N )


H (m)

1

0m=( −1)$ 2m=−m0=−( −1)$ 2

N $2

m

x (n)=e− j( 2 π n

N )(m0− −1

2 )⋅

1

N ⋅

sen( π n

N )sen( π n



2

9(n)=e

− j (2π n

N )( ) −1

2− ) −1

2 )⋅

1

N ⋅

sen (πn )

N )(πn )

− N $2+ 1



( )sen (πn

N )

9(n)=1

N ⋅

sen ( π n

N )sen

(π n

N

)

Tarea =7<7 '(&etivo2 .om#render el algoritmo de la transformada rá#ida de

Forier BFFTC7

Tarea

97 Haga n resmen del ca#6tlo = del li(ro Understanding Digital Signal

Processing de 07 *%ons7 Ponga es#ecial cidado en inclir los diagramas % las

ecaciones más im#ortantes7

<7 0ealice el resmen en s caderno7 !scanee o fotograf6e el resltado5

fa(ri,e n archivo en formato PDF % s(a al sitio del crso7



Práctica =7 '(&etivo2 .om#render la tilidad de la DFT en el análisis de señales

mediante la o(tenci"n del contenido frecencial de na señal contina en el

tiem#o formada #or dos sinsoides de diferente frecencia

Práctica integradora de la nidad

97 Tome 0' #ntos de na señal ,e está formada #or la sma de dos sinsoides

de frecencias 0 % 00 Hz 5 la cal es mestreada a 000 Hz

<7 .alcle la DFT de 607 #ntos de la señal discreta7

=7 rafi,e la señal contina en na #rimera s(gráfica7 rafi,e la magnitd de

>7 .alcle el vector de frecencias de análisis corres#ondiente a la mitad de las

frecencias de la DFT7



, # g , gla mitad de las frecencias de la DFT de la señal en otra s(gráfica7

?7 $#li,e na fnci"n ventana so(re la señal % velva a calclar la DFT7

@7 rafi,e de neva centa la DFT de la señal7 -Oé #ede conclir/

Transformada de Fourier Discreta

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