La Transformada de Fourier

Dr. José Enrique Alvarez Estrada
http://www.software.org.mx/~jalvarez/
Las ideas de
Jean Baptiste Fourier
cualquier señal puede
formarse sumando

funciones seno
de diferentes frecuencias
a diferentes amplitudes
1
0.5
0
-0.5
-1
1
0.5
0
-0.5
-1

1
0.5
0
-0.5
-1
1
0.5
0
-0.5
-1

1
0.5
0
-0.5
-1

1
0.5
0
-0.5
-1
1
0.5
0
-0.5
-1

1
0.5
0
-0.5
-1

1
0.5
0
-0.5
-1

1
0.5
0
-0.5
-1
1
0.5
0
-0.5
-1

1

*1

0.5
0
-0.5

*2

-1

1
0.5

*0

0
-0.5
-1

1
0.5
0
-0.5
-1

*1

+
1
0.5
0
-0.5
-1

1

*1

3.5

0.5
0
-0.5

2.5

*2

-1

1
0.5

*0

0
-0.5
-1

1
0.5
0
-0.5
-1

1.5

+

0.5
-0.5
-1.5
-2.5

*...
1
0.5
0
-0.5
-1
1
0.5
0

+

-0.5
-1
2
1
0
-1
-2
1
0.5
0

+

-0.5
-1
2
1
0
-1

+

-2
1
0.5
0
-0.5
-1
1
0.5
0

+

-0.5
-1
2
1
0
-1

+

-2
1
0.5
0
-0.5
-1
1
0.5
0

+

-0.5
-1
2
1
0
-1

+

-2
1
0.5
0
-0.5
-1
3.5
2.5
1.5
0.5
-0.5
-1.5
-2.5
-3.5

3
Y a la inversa...
Dada una señal compuesta,

¿cuánto tengo que agregar
de cada señal fundamental
para recrearla?
3.5
2.5
1.5
0.5
-0.5
-1.5
-2.5
-3.5

3
1
0.5
0
-0.5
-1

¿Cuánto sen(3t)?
3.5
2.5
1.5
0.5
-0.5
-1.5
-2.5
-3.5

3
1
0.5
0
-0.5
-1
1

¿Cuánto sen(3t)?

0.5
0

3.5

¿Cuánto sen(4t)?

2.5
1.5
0.5
-0.5
-1.5
-2.5
-3.5

-0.5
-1

3
1
0.5
0
-0.5
-1
1

¿Cuánto sen(3t)?

0.5
0

3.5

¿Cuánto sen(4t)?

2.5
1.5
0.5

-1
3

1

-0.5
-1.5
-2.5
-3.5

-0.5

¿Cuánt...
1
0.5
0
-0.5
-1
1

¿Cuánto sen(3t)?

0.5
0

3.5

¿Cuánto sen(4t)?

2.5
1.5
0.5

-1
3

1

-0.5
-1.5
-2.5

-0.5

¿Cuánto sen...
Así que la palabra transformar
significa cambiar el dominio de la señal,
pasando del dominio del tiempo al
dominio de la f...
Fourier
3.5
2.5
1.5
0.5

3

-0.5
-1.5
-2.5
-3.5

tiempo

Fourier
2

3.5

1.8
2.5

1.6

1.5

1.4

0.5

3

-0.5

Fourier

1.2
1
0.8
0.6

-1.5

0.4

-2.5

0.2
0

-3.5

0

tiempo

1

2

3

4
...
Por simplicidad, trabajaremos con
una versión discreta de la señal
5.00
4.00
3.00
2.00
1.00
0.00
-1.00
-2.00
-3.00
-4.00
-5.00

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17
5.00
4.00
3.00
2.00
1.00
0.00
-1.00
-2.00
-3.00
-4.00
-5.00

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17
...
5.00
4.00
3.00
2.00
1.00
0.00
-1.00

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

-2.00
-3.00
-4.00
-5.00

Conver...
5.00
4.00
3.00
2.00
1.00
0.00
-1.00

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

-2.00
-3.00
-4.00
-5.00...
5.00
4.00
3.00
2.00
1.00
0.00
-1.00

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

-2.00
-3.00
-4.00
-5.00...
5.00
4.00
3.00
2.00
1.00
0.00
-1.00

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

-2.00
-3.00
-4.00
-5.00...
5.00
4.00
3.00
2.00
1.00
0.00
-1.00

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

-2.00
-3.00
-4.00
-5.00...
-2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00
5.00
4.00
3.00
2.00
1.00
0.00
-1.00
-2.00
-3.00
-4.00
-5.00

0

1

2

3

4

5

6

7

8

9

10 11 12 13 14 15 16 17

-2.63 ...
Conversor
D/A
5.00
4.00
3.00
2.00
1.00
0.00
-1.00
-2.00
-3.00
-4.00
-5.00

0

1

2

3

4

5

6

7

8

9

10 11 12 13 14 15...
5.00
4.00
3.00
2.00
1.00
0.00
-1.00

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

-2.00
-3.00
-4.00
-5.00...
Hasta aquí, tenemos grabado
un archivo WAV...
Pero un WAV ocupa mucho espacio,
porque guarda todas las muestras.
¿Y si sólo almacenamos
algunas de sus características?
periodo de
muestreo
“T”

5.00
4.00
3.00
2.00
1.00
0.00
-1.00

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17...
5.00
4.00
3.00
2.00
1.00
0.00
-1.00

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

-2.00
-3.00
-4.00
-5.00...
5.00
4.00
3.00
2.00

Transformada
de Fourier

1.00
0.00
-1.00

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

1...
5.00
4.00
3.00
2.00
1.00
0.00
-1.00

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

-2.00
-3.00
-4.00
-5.00...
Hagamos un ejemplo
5
4
3

Intensidad

2
1
0
0

1

2

3

4

5

6

7

8

9

-1
-2
-3
-4
-5

# de muestra

10

11

12

13

14

15

16

17
5
4
3

Intensidad

2
1
0
0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

-1
-2
-3
-4
-5

# de muestra

Esta ...
5
4
3

Intensidad

2
1
0
0

1

2

3

4

5

6

7

8

9

-1
-2
-3
-4
-5

# de muestra

10

11

12

13

14

15

16

17
5
4
3

Intensidad

2
1
0
0

1

2

3

4

5

6

7

8

9

-1
-2
-3
-4
-5

# de muestra

10

11

12

13

14

15

16

17

0
1
2...
¿Qué frecuencias básicas
debemos analizar para
reconstruir la señal
en la ventana?
La frecuencia de muestreo

el doble

debe ser al menos
de la frecuencia máxima
que se requiere analizar.
La frecuencia de muestreo

el doble

debe ser al menos
de la frecuencia máxima
que se requiere analizar.

F s ≥2 F máx
la frecuencia máxima analizable

no puede ser mayor que la mitad
de la frecuencia de muestreo
la frecuencia máxima analizable

no puede ser mayor que la mitad
de la frecuencia de muestreo

Fs
F máx ≤
2
Así que, para
nuestro ejemplo...
20 Hz
F máx ⩽
⩽10 Hz
2
Por tanto,
los valores de “n”
y las frecuencias
a analizar serán:
n=0⇒ F 0=

0
20=0 Hz
10
0
20=0 Hz
10
1
n=1⇒ F 1= 20=2 Hz
10

n=0 ⇒ F 0 =
0
20=0 Hz
10
1
n=1⇒ F 1= 20=2 Hz
10
2
n=2⇒ F 2 = 20=4 Hz
10
n=0 ⇒ F 0 =
0
20=0 Hz
10
1
n=1⇒ F 1= 20=2 Hz
10
2
n=2⇒ F 2 = 20=4 Hz
10
3
n=3 ⇒ F 3 = 20=6 Hz
10
n=0 ⇒ F 0 =
0
20=0 Hz
10
1
n=1⇒ F 1= 20=2 Hz
10
2
n=2⇒ F 2 = 20=4 Hz
10
3
n=3⇒ F 3 = 20=6 Hz
10
4
n=4 ⇒ F 4 = 20=8 Hz
10
n=0 ⇒ F 0 =
0
20=0 Hz
10
1
n=1⇒ F 1 = 20=2 Hz
10
2
n=2⇒ F 2= 20=4 Hz
10
3
n=3 ⇒ F 3 = 20=6 Hz
10
4
n=4 ⇒ F 4 = 20=8 Hz
10
5
n=5⇒ F 5 =...
¿Cuánto aporta n=0, sen(0t),
para formar la señal?
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(0t)
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0....
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(0t)
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0....
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(0t)
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0....
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(0t)
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0....
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(0t)
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0....
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(0t)
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0....
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(0t)
m[t] Producto
0.0000
0.00
0.000
0.0000
4.25
0.000
0.0000
2...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(0t)
m[t] Producto
0.0000
0.00
0.000
0.0000
4.25
0.000
0.0000
2...
¿Cuánto aporta n=1, sen(2t),
para formar la señal?
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(2t)
0.0000
0.5878
0.9511
0.9511
0.5878
0.0000
-0.5878
-0.9511
...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(2t)
0.0000
0.5878
0.9511
0.9511
0.5878
0.0000
-0.5878
-0.9511
...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(2t)
0.0000
0.5878
0.9511
0.9511
0.5878
0.0000
-0.5878
-0.9511
...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(2t)
m[t] Producto
0.0000
0.00
0.000
0.5878
4.25
2.498
0.9511
2...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(2t)
m[t] Producto
0.0000
0.00
0.000
0.5878
4.25
2.498
0.9511
2...
¿Cuánto aporta n=2, sen(4t),
para formar la señal?
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(4t)
0.0000
0.9511
0.5878
-0.5878
-0.9511
0.0000
0.9511
0.5878
...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(4t)
0.0000
0.9511
0.5878
-0.5878
-0.9511
0.0000
0.9511
0.5878
...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(4t)
0.0000
0.9511
0.5878
-0.5878
-0.9511
0.0000
0.9511
0.5878
...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(4t)
m[t] Producto
0.0000
0.00
0.000
0.9511
4.25
4.042
0.5878
2...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(4t)
m[t] Producto
0.0000
0.00
0.000
0.9511
4.25
4.042
0.5878
2...
¿Cuánto aporta n=3, sen(6t),
para formar la señal?
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(6t)
0.0000
0.9511
-0.5878
-0.5878
0.9511
0.0000
-0.9511
0.5878...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(6t)
0.0000
0.9511
-0.5878
-0.5878
0.9511
0.0000
-0.9511
0.5878...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(6t)
0.0000
0.9511
-0.5878
-0.5878
0.9511
0.0000
-0.9511
0.5878...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(6t)
m[t] Producto
0.0000
0.00
0.000
0.9511
4.25
4.042
-0.5878
...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(6t)
m[t] Producto
0.0000
0.00
0.000
0.9511
4.25
4.042
-0.5878
...
¿Cuánto aporta n=4, sen(8t),
para formar la señal?
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(8t)
0.0000
0.5878
-0.9511
0.9511
-0.5878
0.0000
0.5878
-0.9511...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(8t)
0.0000
0.5878
-0.9511
0.9511
-0.5878
0.0000
0.5878
-0.9511...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(8t)
0.0000
0.5878
-0.9511
0.9511
-0.5878
0.0000
0.5878
-0.9511...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(8t)
m[t] Producto
0.0000
0.00
0.000
0.5878
4.25
2.498
-0.9511
...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(8t)
m[t] Producto
0.0000
0.00
0.000
0.5878
4.25
2.498
-0.9511
...
¿Cuánto aporta n=5, sen(10t),
para formar la señal?
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(10t)
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(10t)
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(10t)
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(10t)
m[t] Producto
0.0000
0.00
0.000
0.0000
4.25
0.000
0.0000
...
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

sen(10t)
m[t] Producto
0.0000
0.00
0.000
0.0000
4.25
0.000
0.0000
...
¡Son demasiados cálculos!
Se necesita crear un
método abreviado
que requiera
menos esfuerzo.
1
Fourier (m , n , N , T )=
N

N −1

∑ m [kT ]e
k=0

−2 π n k
j
N
1
Fourier (m , n , N , T )=
N
La Transformada
de Fourier...

N −1

∑ m [kT ]e
k=0

−2 π n k
j
N
1
Fourier (m , n , N , T )=
N
-2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00

N −1

∑ m [kT ]e
k=0

...de una venta...
1
Fourier (m , n , N , T )=
N
n=1
2 Hz

N −1

∑ m [kT ]e
k=0

...para la “n” de la
frecuencia cuyo aporte
se quiere conoce...
1
Fourier (m , n , N , T )=
N
10 muestras

-2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00

N −1

∑ m [kT ]e

−2 π n...
1
Fourier (m , n , N , T )=
N
...y cuyo periodo de
muestreo fue “T”...

1/Fs

N −1

∑ m [kT ]e
k=0

−2 π n k
j
N
1
Fourier (m , n , N , T )=
N

...se calcula como...

N −1

∑ m [kT ]e
k=0

−2 π n k
j
N
1
Fourier (m , n , N , T )=
N

N −1

∑ m [kT ]e
k=0

−2 π n k
j
N
1
Fourier (m , n , N , T )=
N

N −1

∑ m [kT ]e
k=0

−2 π n k
j
N
1
Fourier (m , n , N , T )=
N

N −1

∑ m [kT ]e
k=0

−2 π n k
j
N
1
Fourier (m , n , N , T )=
N

N −1

∑ m [kT ]e
k=0

−2 π n k
j
N
1
Fourier (m , n , N , T )=
N

N −1

∑ m [kT ]e
k=0

−2 π n k
j
N
1
Fourier (m , n , N , T )=
N

N −1

∑ m [kT ]e
k=0

−2 π n k
j
N
¡Difícil de entender!
?
1
Fourier (m , n , N , T )=
N

N −1

∑ m [kT ]e
k=0

−2 π n k
j
N
?
1
Fourier (m , n , N , T )=
N

N −1

∑ m [kT ]e
k=0

−2 π n k
j
N
Para facilitar el cálculo...
aplicando mi
célebre fórmula...
e xj =cos( x)+ sen( x) j
e xj =cos( x)+ sen( x) j
e xj =cos( x)+ sen( x) j
e xj =cos( x)+ sen( x) j

mi Transformada
queda como...
1
N

N −1

∑
k =0

k
k
m[kT ] cos(−2 π n )+ sen(−2 π n ) j
N
N

(

)
1
N

N −1

∑
k =0

k
k
m[kT ] cos(−2 π n )+ sen(−2 π n ) j
N
N

(

)
1
N

N −1

∑
k =0

k
k
m[kT ] cos(−2 π n )+ sen(−2 π n ) j
N
N

(

)
1
N

N −1

∑
k =0

k
k
m[kT ] cos(−2 π n )+ sen(−2 π n ) j
N
N

(

)
1
N

N −1

∑
k =0

k
k
m[kT ] cos(−2 π n )+ sen(−2 π n ) j
N
N

(

)
1
N

N −1

∑
k =0

k
k
m[kT ] cos(−2 π n )+ sen(−2 π n ) j
N
N

(

)
1
N

N −1

∑
k =0

k
k
m[kT ] cos(−2 π n )+ sen(−2 π n ) j
N
N

(

)
Valor
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63
-2.62
-4.25
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63

5
4
3
2
1
Intensida...
Ventana
de análisis
N = 10 muestras
Valor
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63
-2.62
-4.25
0.00
4.25
2.63
2.62
4.25
0...
Valor
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63
-2.62
-4.25
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63

5
4
3
2
1
Intensida...
Valor
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63
-2.62
-4.25
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63

5
4
3
2
1
Intensida...
Valor
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63
-2.62
-4.25
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63

5
4
3
2
1
Intensida...
Valor
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63
-2.62
-4.25
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63

5
4
3
2
1
Intensida...
Valor
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63
-2.62
-4.25
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63

5
4
3
2
1
Intensida...
Valor
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63
-2.62
-4.25
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63

5
4
3
2
1
Intensida...
Valor
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63
-2.62
-4.25
0.00
4.25
2.63
2.62
4.25
0.00
-4.25
-2.63

5
4
3
2
1
Intensida...
¿Cuánto aporta n=0, sen(0t),
para formar la señal en la ventana?
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

n=0
N = 10
T = 0.05

(

)
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=0
N = 10
T = 0.05

)
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=0
N = 10
T = 0.05

sen(-2πnk/N)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=0
N = 10
T = 0.05

cos(-2πnk/N)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=0
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=0
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=0
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
¿Cuánto aporta n=1, sen(2t),
para formar la señal en la ventana?
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=1
N = 10
T = 0.05

)
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=1
N = 10
T = 0.05

sen(-2πnk/N)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=1
N = 10
T = 0.05

cos(-2πnk/N)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=1
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=1
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=1
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
¿Cuánto aporta n=2, sen(4t),
para formar la señal en la ventana?
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=2
N = 10
T = 0.05

)
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=2
N = 10
T = 0.05

sen(-2πnk/N)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=2
N = 10
T = 0.05

cos(-2πnk/N)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=2
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=2
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=2
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
¿Cuánto aporta n=3, sen(6t),
para formar la señal en la ventana?
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=3
N = 10
T = 0.05

)
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=3
N = 10
T = 0.05

sen(-2πnk/N)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=3
N = 10
T = 0.05

cos(-2πnk/N)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=3
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=3
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=3
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
¿Cuánto aporta n=4, sen(8t),
para formar la señal en la ventana?
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=4
N = 10
T = 0.05

)
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=4
N = 10
T = 0.05

sen(-2πnk/N)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=4
N = 10
T = 0.05

cos(-2πnk/N)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=4
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=4
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=4
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
¿Cuánto aporta n=5, sen(10t),
para formar la señal en la ventana?
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=5
N = 10
T = 0.05

)
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=5
N = 10
T = 0.05

sen(-2πnk/N)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=5
N = 10
T = 0.05

cos(-2πnk/N)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=5
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=5
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

n=5
N = 10
T = 0.05

m[kT] cos(-2...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
1
N

N −1

k
k
∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j
k =0

(

k
0
1
2
3
4
5
6
7
8
9

m[kT] cos(-2πnk/N) m[kT]*cos(...)...
Reuniendo todas las componentes
podemos expresar analíticamente que...
5.00
4.00
3.00
2.00
1.00
0.00
-1.00

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

-2.00
-3.00
-4.00
-5.00...
Si representamos los resultados
en un Ecualizador Gráfico...
y = 2*sen(2t)
y = 1*sen(6t)
Pero, ¿cómo y por qué
funciona la Transformada de Fourier?
1
Fourier (m , n , N , T )=
N

N −1

∑ m [kT ]e
k=0

−2 π n k
j
N
Cómo funciona la Transformada de Fourier
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Cómo funciona la Transformada de Fourier

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En el 2012 preparé esta presentación, basada en el libro "Reconocimiento de Voz y Fonética Acústica" de Bernal Bermúdez Et Al, para explicar cómo funciona la Transformada de Fourier.

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Cómo funciona la Transformada de Fourier

  1. 1. La Transformada de Fourier Dr. José Enrique Alvarez Estrada http://www.software.org.mx/~jalvarez/
  2. 2. Las ideas de Jean Baptiste Fourier
  3. 3. cualquier señal puede formarse sumando funciones seno de diferentes frecuencias a diferentes amplitudes
  4. 4. 1 0.5 0 -0.5 -1
  5. 5. 1 0.5 0 -0.5 -1 1 0.5 0 -0.5 -1
  6. 6. 1 0.5 0 -0.5 -1 1 0.5 0 -0.5 -1 1 0.5 0 -0.5 -1
  7. 7. 1 0.5 0 -0.5 -1 1 0.5 0 -0.5 -1 1 0.5 0 -0.5 -1 1 0.5 0 -0.5 -1
  8. 8. 1 0.5 0 -0.5 -1 1 *1 0.5 0 -0.5 *2 -1 1 0.5 *0 0 -0.5 -1 1 0.5 0 -0.5 -1 *1 +
  9. 9. 1 0.5 0 -0.5 -1 1 *1 3.5 0.5 0 -0.5 2.5 *2 -1 1 0.5 *0 0 -0.5 -1 1 0.5 0 -0.5 -1 1.5 + 0.5 -0.5 -1.5 -2.5 *1 -3.5 3
  10. 10. 1 0.5 0 -0.5 -1
  11. 11. 1 0.5 0 + -0.5 -1 2 1 0 -1 -2
  12. 12. 1 0.5 0 + -0.5 -1 2 1 0 -1 + -2 1 0.5 0 -0.5 -1
  13. 13. 1 0.5 0 + -0.5 -1 2 1 0 -1 + -2 1 0.5 0 -0.5 -1
  14. 14. 1 0.5 0 + -0.5 -1 2 1 0 -1 + -2 1 0.5 0 -0.5 -1 3.5 2.5 1.5 0.5 -0.5 -1.5 -2.5 -3.5 3
  15. 15. Y a la inversa...
  16. 16. Dada una señal compuesta, ¿cuánto tengo que agregar de cada señal fundamental para recrearla?
  17. 17. 3.5 2.5 1.5 0.5 -0.5 -1.5 -2.5 -3.5 3
  18. 18. 1 0.5 0 -0.5 -1 ¿Cuánto sen(3t)? 3.5 2.5 1.5 0.5 -0.5 -1.5 -2.5 -3.5 3
  19. 19. 1 0.5 0 -0.5 -1 1 ¿Cuánto sen(3t)? 0.5 0 3.5 ¿Cuánto sen(4t)? 2.5 1.5 0.5 -0.5 -1.5 -2.5 -3.5 -0.5 -1 3
  20. 20. 1 0.5 0 -0.5 -1 1 ¿Cuánto sen(3t)? 0.5 0 3.5 ¿Cuánto sen(4t)? 2.5 1.5 0.5 -1 3 1 -0.5 -1.5 -2.5 -3.5 -0.5 ¿Cuánto sen(6t)? 0.5 0 -0.5 -1
  21. 21. 1 0.5 0 -0.5 -1 1 ¿Cuánto sen(3t)? 0.5 0 3.5 ¿Cuánto sen(4t)? 2.5 1.5 0.5 -1 3 1 -0.5 -1.5 -2.5 -0.5 ¿Cuánto sen(6t)? 0.5 0 -3.5 ¿Cuánto sen(12t)? -0.5 -1 1 0.5 0 -0.5 -1
  22. 22. Así que la palabra transformar significa cambiar el dominio de la señal, pasando del dominio del tiempo al dominio de la frecuencia.
  23. 23. Fourier
  24. 24. 3.5 2.5 1.5 0.5 3 -0.5 -1.5 -2.5 -3.5 tiempo Fourier
  25. 25. 2 3.5 1.8 2.5 1.6 1.5 1.4 0.5 3 -0.5 Fourier 1.2 1 0.8 0.6 -1.5 0.4 -2.5 0.2 0 -3.5 0 tiempo 1 2 3 4 5 6 7 8 frecuencia 9 10 11 12
  26. 26. Por simplicidad, trabajaremos con una versión discreta de la señal
  27. 27. 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 -2.00 -3.00 -4.00 -5.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
  28. 28. 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 -2.00 -3.00 -4.00 -5.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 fenómeno analógico a estudiar
  29. 29. 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 -2.00 -3.00 -4.00 -5.00 Conversor A/D 16 17
  30. 30. 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.00 -3.00 -4.00 -5.00 Conversor A/D 20 Hz Frecuencia de muestreo (Fs)
  31. 31. 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.00 -3.00 -4.00 -5.00 Conversor A/D 20 Hz 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.00 -3.00 -4.00 -5.00 versión discreta (muestras)
  32. 32. 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.00 -3.00 -4.00 -5.00 Conversor A/D 20 Hz 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.00 -3.00 -4.00 -5.00 10 muestras tamaño de ventana “N”
  33. 33. 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.00 -3.00 -4.00 -5.00 Conversor A/D 20 Hz ventana “m” 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -2.00 -3.00 -4.00 -5.00 10 muestras
  34. 34. -2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00
  35. 35. 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 -2.00 -3.00 -4.00 -5.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00
  36. 36. Conversor D/A 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 -2.00 -3.00 -4.00 -5.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00
  37. 37. 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.00 -3.00 -4.00 -5.00 Conversor D/A 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 -2.00 -3.00 -4.00 -5.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00
  38. 38. Hasta aquí, tenemos grabado un archivo WAV...
  39. 39. Pero un WAV ocupa mucho espacio, porque guarda todas las muestras. ¿Y si sólo almacenamos algunas de sus características?
  40. 40. periodo de muestreo “T” 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.00 -3.00 -4.00 -5.00 Conversor A/D 1/Fs 20 Hz 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -2.00 -3.00 -4.00 -5.00 10 muestras
  41. 41. 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.00 -3.00 -4.00 -5.00 Conversor A/D 1/Fs 20 Hz 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -2.00 -3.00 -4.00 -5.00 10 muestras “n” de la frecuencia cuyo aporte se quiere conocer n=5 10Hz
  42. 42. 5.00 4.00 3.00 2.00 Transformada de Fourier 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.00 -3.00 -4.00 -5.00 Conversor A/D 1/Fs 20 Hz 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -2.00 -3.00 -4.00 Fourier -5.00 10 muestras n=5 10Hz
  43. 43. 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.00 -3.00 -4.00 -5.00 Conversor A/D 1/Fs 20 Hz Aporte de la frecuencia que se quiere conocer 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -2.00 -3.00 -4.00 Fourier -5.00 10 muestras 10Hz AFn
  44. 44. Hagamos un ejemplo
  45. 45. 5 4 3 Intensidad 2 1 0 0 1 2 3 4 5 6 7 8 9 -1 -2 -3 -4 -5 # de muestra 10 11 12 13 14 15 16 17
  46. 46. 5 4 3 Intensidad 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -1 -2 -3 -4 -5 # de muestra Esta señal se muestreará a una frecuencia Fs = 20Hz
  47. 47. 5 4 3 Intensidad 2 1 0 0 1 2 3 4 5 6 7 8 9 -1 -2 -3 -4 -5 # de muestra 10 11 12 13 14 15 16 17
  48. 48. 5 4 3 Intensidad 2 1 0 0 1 2 3 4 5 6 7 8 9 -1 -2 -3 -4 -5 # de muestra 10 11 12 13 14 15 16 17 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00
  49. 49. ¿Qué frecuencias básicas debemos analizar para reconstruir la señal en la ventana?
  50. 50. La frecuencia de muestreo el doble debe ser al menos de la frecuencia máxima que se requiere analizar.
  51. 51. La frecuencia de muestreo el doble debe ser al menos de la frecuencia máxima que se requiere analizar. F s ≥2 F máx
  52. 52. la frecuencia máxima analizable no puede ser mayor que la mitad de la frecuencia de muestreo
  53. 53. la frecuencia máxima analizable no puede ser mayor que la mitad de la frecuencia de muestreo Fs F máx ≤ 2
  54. 54. Así que, para nuestro ejemplo...
  55. 55. 20 Hz F máx ⩽ ⩽10 Hz 2
  56. 56. Por tanto, los valores de “n” y las frecuencias a analizar serán:
  57. 57. n=0⇒ F 0= 0 20=0 Hz 10
  58. 58. 0 20=0 Hz 10 1 n=1⇒ F 1= 20=2 Hz 10 n=0 ⇒ F 0 =
  59. 59. 0 20=0 Hz 10 1 n=1⇒ F 1= 20=2 Hz 10 2 n=2⇒ F 2 = 20=4 Hz 10 n=0 ⇒ F 0 =
  60. 60. 0 20=0 Hz 10 1 n=1⇒ F 1= 20=2 Hz 10 2 n=2⇒ F 2 = 20=4 Hz 10 3 n=3 ⇒ F 3 = 20=6 Hz 10 n=0 ⇒ F 0 =
  61. 61. 0 20=0 Hz 10 1 n=1⇒ F 1= 20=2 Hz 10 2 n=2⇒ F 2 = 20=4 Hz 10 3 n=3⇒ F 3 = 20=6 Hz 10 4 n=4 ⇒ F 4 = 20=8 Hz 10 n=0 ⇒ F 0 =
  62. 62. 0 20=0 Hz 10 1 n=1⇒ F 1 = 20=2 Hz 10 2 n=2⇒ F 2= 20=4 Hz 10 3 n=3 ⇒ F 3 = 20=6 Hz 10 4 n=4 ⇒ F 4 = 20=8 Hz 10 5 n=5⇒ F 5 = 20=10 Hz 10 n=0 ⇒ F 0=
  63. 63. ¿Cuánto aporta n=0, sen(0t), para formar la señal?
  64. 64. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(0t) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 5 4 3 2 1 0 0 1 2 3 4 5 6 -1 -2 -3 -4 -5 El valor de sen(0t) en los 21 instantes de muestreo 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  65. 65. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(0t) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 m[t] 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 5 4 3 2 1 0 0 1 2 3 4 5 6 7 -1 -2 -3 -4 -5 El valor de las 21 muestras 8 9 10 11 12 13 14 15 16 17 18 19 20
  66. 66. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(0t) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 m[t] 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -1 -2 -3 -4 -5 ¿Cómo podemos comparar una señal con la otra?
  67. 67. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(0t) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 m[t] Producto 0.00 0.000 4.25 0.000 2.63 0.000 2.62 0.000 4.25 0.000 0.00 0.000 -4.25 0.000 -2.63 0.000 -2.62 0.000 -4.25 0.000 0.00 0.000 4.25 0.000 2.63 0.000 2.62 0.000 4.25 0.000 0.00 0.000 -4.25 0.000 -2.63 0.000 -2.62 0.000 -4.25 0.000 0.00 0.000 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -1 -2 -3 -4 -5 ¿Y si multiplicamos ambas? El resultado nos dará el área de un rectángulo por cada muestra
  68. 68. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(0t) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 m[t] Producto 0.00 0.000 4.25 0.000 2.63 0.000 2.62 0.000 4.25 0.000 0.00 0.000 -4.25 0.000 -2.63 0.000 -2.62 0.000 -4.25 0.000 0.00 0.000 4.25 0.000 2.63 0.000 2.62 0.000 4.25 0.000 0.00 0.000 -4.25 0.000 -2.63 0.000 -2.62 0.000 -4.25 0.000 0.00 0.000 1 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -0.2 -0.4 -0.6 -0.8 -1 Si ambas se parecen, las áreas de los rectángulos serán grandes. Si no se parecen, serán pequeñas.
  69. 69. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(0t) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 m[t] Producto 0.00 0.000 4.25 0.000 2.63 0.000 2.62 0.000 4.25 0.000 0.00 0.000 -4.25 0.000 -2.63 0.000 -2.62 0.000 -4.25 0.000 0.00 0.000 4.25 0.000 2.63 0.000 2.62 0.000 4.25 0.000 0.00 0.000 -4.25 0.000 -2.63 0.000 -2.62 0.000 -4.25 0.000 0.00 0.000 1 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -0.2 -0.4 -0.6 -0.8 -1 Como las señales no se parecen, las áreas de los rectángulos son nulas.
  70. 70. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(0t) m[t] Producto 0.0000 0.00 0.000 0.0000 4.25 0.000 0.0000 2.63 0.000 0.0000 2.62 0.000 0.0000 4.25 0.000 0.0000 0.00 0.000 0.0000 -4.25 0.000 0.0000 -2.63 0.000 0.0000 -2.62 0.000 0.0000 -4.25 0.000 0.0000 0.00 0.000 0.0000 4.25 0.000 0.0000 2.63 0.000 0.0000 2.62 0.000 0.0000 4.25 0.000 0.0000 0.00 0.000 0.0000 -4.25 0.000 0.0000 -2.63 0.000 0.0000 -2.62 0.000 0.0000 -4.25 0.000 0.0000 0.00 0.000 Sumatoria: 0.000 Normalizada: 0.000 1 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -0.2 -0.4 -0.6 -0.8 -1 La sumatoria de las áreas de los rectángulos equivale a 0 unidades.
  71. 71. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(0t) m[t] Producto 0.0000 0.00 0.000 0.0000 4.25 0.000 0.0000 2.63 0.000 0.0000 2.62 0.000 0.0000 4.25 0.000 0.0000 0.00 0.000 0.0000 -4.25 0.000 0.0000 -2.63 0.000 0.0000 -2.62 0.000 0.0000 -4.25 0.000 0.0000 0.00 0.000 0.0000 4.25 0.000 0.0000 2.63 0.000 0.0000 2.62 0.000 0.0000 4.25 0.000 0.0000 0.00 0.000 0.0000 -4.25 0.000 0.0000 -2.63 0.000 0.0000 -2.62 0.000 0.0000 -4.25 0.000 0.0000 0.00 0.000 Sumatoria: 0.000 Normalizada: 0.000 1 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -0.2 -0.4 -0.6 -0.8 -1 Luego sen(0t) no aporta nada a la señal original.
  72. 72. ¿Cuánto aporta n=1, sen(2t), para formar la señal?
  73. 73. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(2t) 0.0000 0.5878 0.9511 0.9511 0.5878 0.0000 -0.5878 -0.9511 -0.9511 -0.5878 0.0000 0.5878 0.9511 0.9511 0.5878 0.0000 -0.5878 -0.9511 -0.9511 -0.5878 0.0000 5 4 3 2 1 0 0 -1 -2 -3 -4 -5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  74. 74. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(2t) 0.0000 0.5878 0.9511 0.9511 0.5878 0.0000 -0.5878 -0.9511 -0.9511 -0.5878 0.0000 0.5878 0.9511 0.9511 0.5878 0.0000 -0.5878 -0.9511 -0.9511 -0.5878 0.0000 m[t] 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 5 4 3 2 1 0 0 -1 -2 -3 -4 -5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  75. 75. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(2t) 0.0000 0.5878 0.9511 0.9511 0.5878 0.0000 -0.5878 -0.9511 -0.9511 -0.5878 0.0000 0.5878 0.9511 0.9511 0.5878 0.0000 -0.5878 -0.9511 -0.9511 -0.5878 0.0000 m[t] Producto 0.00 0.000 4.25 2.498 2.63 2.501 2.62 2.492 4.25 2.498 0.00 0.000 -4.25 2.498 -2.63 2.501 -2.62 2.492 -4.25 2.498 0.00 0.000 4.25 2.498 2.63 2.501 2.62 2.492 4.25 2.498 0.00 0.000 -4.25 2.498 -2.63 2.501 -2.62 2.492 -4.25 2.498 0.00 0.000 5 4 3 2 1 0 0 -1 -2 -3 -4 -5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  76. 76. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(2t) m[t] Producto 0.0000 0.00 0.000 0.5878 4.25 2.498 0.9511 2.63 2.501 0.9511 2.62 2.492 0.5878 4.25 2.498 0.0000 0.00 0.000 -0.5878 -4.25 2.498 -0.9511 -2.63 2.501 -0.9511 -2.62 2.492 -0.5878 -4.25 2.498 0.0000 0.00 0.000 0.5878 4.25 2.498 0.9511 2.63 2.501 0.9511 2.62 2.492 0.5878 4.25 2.498 0.0000 0.00 0.000 -0.5878 -4.25 2.498 -0.9511 -2.63 2.501 -0.9511 -2.62 2.492 -0.5878 -4.25 2.498 0.0000 0.00 0.000 Sumatoria: 39.957 Normalizada: 1.998 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -1 -2 -3 -4 -5 La sumatoria de las áreas de los rectángulos equivale a casi 40 unidades. Normalizada representa casi 2.
  77. 77. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(2t) m[t] Producto 0.0000 0.00 0.000 0.5878 4.25 2.498 0.9511 2.63 2.501 0.9511 2.62 2.492 0.5878 4.25 2.498 0.0000 0.00 0.000 -0.5878 -4.25 2.498 -0.9511 -2.63 2.501 -0.9511 -2.62 2.492 -0.5878 -4.25 2.498 0.0000 0.00 0.000 0.5878 4.25 2.498 0.9511 2.63 2.501 0.9511 2.62 2.492 0.5878 4.25 2.498 0.0000 0.00 0.000 -0.5878 -4.25 2.498 -0.9511 -2.63 2.501 -0.9511 -2.62 2.492 -0.5878 -4.25 2.498 0.0000 0.00 0.000 Sumatoria: 39.957 Normalizada: 1.998 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -1 -2 -3 -4 -5 Luego sen(2t) aporta 2 a la señal original.
  78. 78. ¿Cuánto aporta n=2, sen(4t), para formar la señal?
  79. 79. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(4t) 0.0000 0.9511 0.5878 -0.5878 -0.9511 0.0000 0.9511 0.5878 -0.5878 -0.9511 0.0000 0.9511 0.5878 -0.5878 -0.9511 0.0000 0.9511 0.5878 -0.5878 -0.9511 0.0000 5 4 3 2 1 0 0 -1 -2 -3 -4 -5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  80. 80. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(4t) 0.0000 0.9511 0.5878 -0.5878 -0.9511 0.0000 0.9511 0.5878 -0.5878 -0.9511 0.0000 0.9511 0.5878 -0.5878 -0.9511 0.0000 0.9511 0.5878 -0.5878 -0.9511 0.0000 m[t] 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 5 4 3 2 1 0 0 -1 -2 -3 -4 -5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  81. 81. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(4t) 0.0000 0.9511 0.5878 -0.5878 -0.9511 0.0000 0.9511 0.5878 -0.5878 -0.9511 0.0000 0.9511 0.5878 -0.5878 -0.9511 0.0000 0.9511 0.5878 -0.5878 -0.9511 0.0000 m[t] Producto 0.00 0.000 4.25 4.042 2.63 1.546 2.62 -1.540 4.25 -4.042 0.00 0.000 -4.25 -4.042 -2.63 -1.546 -2.62 1.540 -4.25 4.042 0.00 0.000 4.25 4.042 2.63 1.546 2.62 -1.540 4.25 -4.042 0.00 0.000 -4.25 -4.042 -2.63 -1.546 -2.62 1.540 -4.25 4.042 0.00 0.000 5 4 3 2 1 0 0 -1 -2 -3 -4 -5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  82. 82. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(4t) m[t] Producto 0.0000 0.00 0.000 0.9511 4.25 4.042 0.5878 2.63 1.546 -0.5878 2.62 -1.540 -0.9511 4.25 -4.042 0.0000 0.00 0.000 0.9511 -4.25 -4.042 0.5878 -2.63 -1.546 -0.5878 -2.62 1.540 -0.9511 -4.25 4.042 0.0000 0.00 0.000 0.9511 4.25 4.042 0.5878 2.63 1.546 -0.5878 2.62 -1.540 -0.9511 4.25 -4.042 0.0000 0.00 0.000 0.9511 -4.25 -4.042 0.5878 -2.63 -1.546 -0.5878 -2.62 1.540 -0.9511 -4.25 4.042 0.0000 0.00 0.000 Sumatoria: 0.000 Normalizada: 0.000 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -1 -2 -3 -4 -5 La sumatoria de las áreas de los rectángulos equivale a 0 unidades.
  83. 83. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(4t) m[t] Producto 0.0000 0.00 0.000 0.9511 4.25 4.042 0.5878 2.63 1.546 -0.5878 2.62 -1.540 -0.9511 4.25 -4.042 0.0000 0.00 0.000 0.9511 -4.25 -4.042 0.5878 -2.63 -1.546 -0.5878 -2.62 1.540 -0.9511 -4.25 4.042 0.0000 0.00 0.000 0.9511 4.25 4.042 0.5878 2.63 1.546 -0.5878 2.62 -1.540 -0.9511 4.25 -4.042 0.0000 0.00 0.000 0.9511 -4.25 -4.042 0.5878 -2.63 -1.546 -0.5878 -2.62 1.540 -0.9511 -4.25 4.042 0.0000 0.00 0.000 Sumatoria: 0.000 Normalizada: 0.000 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -1 -2 -3 -4 -5 Luego sen(4t) no aporta nada a la señal original.
  84. 84. ¿Cuánto aporta n=3, sen(6t), para formar la señal?
  85. 85. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(6t) 0.0000 0.9511 -0.5878 -0.5878 0.9511 0.0000 -0.9511 0.5878 0.5878 -0.9511 0.0000 0.9511 -0.5878 -0.5878 0.9511 0.0000 -0.9511 0.5878 0.5878 -0.9511 0.0000 5 4 3 2 1 0 0 -1 -2 -3 -4 -5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  86. 86. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(6t) 0.0000 0.9511 -0.5878 -0.5878 0.9511 0.0000 -0.9511 0.5878 0.5878 -0.9511 0.0000 0.9511 -0.5878 -0.5878 0.9511 0.0000 -0.9511 0.5878 0.5878 -0.9511 0.0000 m[t] 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 5 4 3 2 1 0 0 -1 -2 -3 -4 -5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  87. 87. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(6t) 0.0000 0.9511 -0.5878 -0.5878 0.9511 0.0000 -0.9511 0.5878 0.5878 -0.9511 0.0000 0.9511 -0.5878 -0.5878 0.9511 0.0000 -0.9511 0.5878 0.5878 -0.9511 0.0000 m[t] Producto 0.00 0.000 4.25 4.042 2.63 -1.546 2.62 -1.540 4.25 4.042 0.00 0.000 -4.25 4.042 -2.63 -1.546 -2.62 -1.540 -4.25 4.042 0.00 0.000 4.25 4.042 2.63 -1.546 2.62 -1.540 4.25 4.042 0.00 0.000 -4.25 4.042 -2.63 -1.546 -2.62 -1.540 -4.25 4.042 0.00 0.000 5 4 3 2 1 0 0 -1 -2 -3 -4 -5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  88. 88. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(6t) m[t] Producto 0.0000 0.00 0.000 0.9511 4.25 4.042 -0.5878 2.63 -1.546 -0.5878 2.62 -1.540 0.9511 4.25 4.042 0.0000 0.00 0.000 -0.9511 -4.25 4.042 0.5878 -2.63 -1.546 0.5878 -2.62 -1.540 -0.9511 -4.25 4.042 0.0000 0.00 0.000 0.9511 4.25 4.042 -0.5878 2.63 -1.546 -0.5878 2.62 -1.540 0.9511 4.25 4.042 0.0000 0.00 0.000 -0.9511 -4.25 4.042 0.5878 -2.63 -1.546 0.5878 -2.62 -1.540 -0.9511 -4.25 4.042 0.0000 0.00 0.000 Sumatoria: 19.992 Normalizada: 1.000 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -1 -2 -3 -4 -5 La sumatoria de las áreas de los rectángulos equivale a casi 20 unidades. Normalizada representa casi 1.
  89. 89. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(6t) m[t] Producto 0.0000 0.00 0.000 0.9511 4.25 4.042 -0.5878 2.63 -1.546 -0.5878 2.62 -1.540 0.9511 4.25 4.042 0.0000 0.00 0.000 -0.9511 -4.25 4.042 0.5878 -2.63 -1.546 0.5878 -2.62 -1.540 -0.9511 -4.25 4.042 0.0000 0.00 0.000 0.9511 4.25 4.042 -0.5878 2.63 -1.546 -0.5878 2.62 -1.540 0.9511 4.25 4.042 0.0000 0.00 0.000 -0.9511 -4.25 4.042 0.5878 -2.63 -1.546 0.5878 -2.62 -1.540 -0.9511 -4.25 4.042 0.0000 0.00 0.000 Sumatoria: 19.992 Normalizada: 1.000 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -1 -2 -3 -4 -5 Luego sen(6t) aporta 1 a la señal original.
  90. 90. ¿Cuánto aporta n=4, sen(8t), para formar la señal?
  91. 91. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(8t) 0.0000 0.5878 -0.9511 0.9511 -0.5878 0.0000 0.5878 -0.9511 0.9511 -0.5878 0.0000 0.5878 -0.9511 0.9511 -0.5878 0.0000 0.5878 -0.9511 0.9511 -0.5878 0.0000 5 4 3 2 1 0 0 -1 -2 -3 -4 -5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  92. 92. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(8t) 0.0000 0.5878 -0.9511 0.9511 -0.5878 0.0000 0.5878 -0.9511 0.9511 -0.5878 0.0000 0.5878 -0.9511 0.9511 -0.5878 0.0000 0.5878 -0.9511 0.9511 -0.5878 0.0000 m[t] 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 5 4 3 2 1 0 0 -1 -2 -3 -4 -5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  93. 93. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(8t) 0.0000 0.5878 -0.9511 0.9511 -0.5878 0.0000 0.5878 -0.9511 0.9511 -0.5878 0.0000 0.5878 -0.9511 0.9511 -0.5878 0.0000 0.5878 -0.9511 0.9511 -0.5878 0.0000 m[t] Producto 0.00 0.000 4.25 2.498 2.63 -2.501 2.62 2.492 4.25 -2.498 0.00 0.000 -4.25 -2.498 -2.63 2.501 -2.62 -2.492 -4.25 2.498 0.00 0.000 4.25 2.498 2.63 -2.501 2.62 2.492 4.25 -2.498 0.00 0.000 -4.25 -2.498 -2.63 2.501 -2.62 -2.492 -4.25 2.498 0.00 0.000 5 4 3 2 1 0 0 -1 -2 -3 -4 -5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  94. 94. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(8t) m[t] Producto 0.0000 0.00 0.000 0.5878 4.25 2.498 -0.9511 2.63 -2.501 0.9511 2.62 2.492 -0.5878 4.25 -2.498 0.0000 0.00 0.000 0.5878 -4.25 -2.498 -0.9511 -2.63 2.501 0.9511 -2.62 -2.492 -0.5878 -4.25 2.498 0.0000 0.00 0.000 0.5878 4.25 2.498 -0.9511 2.63 -2.501 0.9511 2.62 2.492 -0.5878 4.25 -2.498 0.0000 0.00 0.000 0.5878 -4.25 -2.498 -0.9511 -2.63 2.501 0.9511 -2.62 -2.492 -0.5878 -4.25 2.498 0.0000 0.00 0.000 Sumatoria: 0.000 Normalizada: 0.000 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -1 -2 -3 -4 -5 La sumatoria de las áreas de los rectángulos equivale a 0 unidades.
  95. 95. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(8t) m[t] Producto 0.0000 0.00 0.000 0.5878 4.25 2.498 -0.9511 2.63 -2.501 0.9511 2.62 2.492 -0.5878 4.25 -2.498 0.0000 0.00 0.000 0.5878 -4.25 -2.498 -0.9511 -2.63 2.501 0.9511 -2.62 -2.492 -0.5878 -4.25 2.498 0.0000 0.00 0.000 0.5878 4.25 2.498 -0.9511 2.63 -2.501 0.9511 2.62 2.492 -0.5878 4.25 -2.498 0.0000 0.00 0.000 0.5878 -4.25 -2.498 -0.9511 -2.63 2.501 0.9511 -2.62 -2.492 -0.5878 -4.25 2.498 0.0000 0.00 0.000 Sumatoria: 0.000 Normalizada: 0.000 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -1 -2 -3 -4 -5 Luego sen(8t) no aporta nada a la señal original.
  96. 96. ¿Cuánto aporta n=5, sen(10t), para formar la señal?
  97. 97. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(10t) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 5 4 3 2 1 0 0 -1 -2 -3 -4 -5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  98. 98. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(10t) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 m[t] 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 5 4 3 2 1 0 0 -1 -2 -3 -4 -5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  99. 99. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(10t) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 m[t] Producto 0.00 0.000 4.25 0.000 2.63 0.000 2.62 0.000 4.25 0.000 0.00 0.000 -4.25 0.000 -2.63 0.000 -2.62 0.000 -4.25 0.000 0.00 0.000 4.25 0.000 2.63 0.000 2.62 0.000 4.25 0.000 0.00 0.000 -4.25 0.000 -2.63 0.000 -2.62 0.000 -4.25 0.000 0.00 0.000 5 4 3 2 1 0 0 -1 -2 -3 -4 -5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
  100. 100. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(10t) m[t] Producto 0.0000 0.00 0.000 0.0000 4.25 0.000 0.0000 2.63 0.000 0.0000 2.62 0.000 0.0000 4.25 0.000 0.0000 0.00 0.000 0.0000 -4.25 0.000 0.0000 -2.63 0.000 0.0000 -2.62 0.000 0.0000 -4.25 0.000 0.0000 0.00 0.000 0.0000 4.25 0.000 0.0000 2.63 0.000 0.0000 2.62 0.000 0.0000 4.25 0.000 0.0000 0.00 0.000 0.0000 -4.25 0.000 0.0000 -2.63 0.000 0.0000 -2.62 0.000 0.0000 -4.25 0.000 0.0000 0.00 0.000 Sumatoria: 0.000 Normalizada: 0.000 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -1 -2 -3 -4 -5 La sumatoria de las áreas de los rectángulos equivale a 0 unidades.
  101. 101. t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sen(10t) m[t] Producto 0.0000 0.00 0.000 0.0000 4.25 0.000 0.0000 2.63 0.000 0.0000 2.62 0.000 0.0000 4.25 0.000 0.0000 0.00 0.000 0.0000 -4.25 0.000 0.0000 -2.63 0.000 0.0000 -2.62 0.000 0.0000 -4.25 0.000 0.0000 0.00 0.000 0.0000 4.25 0.000 0.0000 2.63 0.000 0.0000 2.62 0.000 0.0000 4.25 0.000 0.0000 0.00 0.000 0.0000 -4.25 0.000 0.0000 -2.63 0.000 0.0000 -2.62 0.000 0.0000 -4.25 0.000 0.0000 0.00 0.000 Sumatoria: 0.000 Normalizada: 0.000 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -1 -2 -3 -4 -5 Luego sen(10t) no aporta nada a la señal original.
  102. 102. ¡Son demasiados cálculos!
  103. 103. Se necesita crear un método abreviado que requiera menos esfuerzo.
  104. 104. 1 Fourier (m , n , N , T )= N N −1 ∑ m [kT ]e k=0 −2 π n k j N
  105. 105. 1 Fourier (m , n , N , T )= N La Transformada de Fourier... N −1 ∑ m [kT ]e k=0 −2 π n k j N
  106. 106. 1 Fourier (m , n , N , T )= N -2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 N −1 ∑ m [kT ]e k=0 ...de una ventana “m” de muestras... −2 π n k j N
  107. 107. 1 Fourier (m , n , N , T )= N n=1 2 Hz N −1 ∑ m [kT ]e k=0 ...para la “n” de la frecuencia cuyo aporte se quiere conocer... −2 π n k j N
  108. 108. 1 Fourier (m , n , N , T )= N 10 muestras -2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 N −1 ∑ m [kT ]e −2 π n k j N k=0 ...con un tamaño de ventana de “N” muestras...
  109. 109. 1 Fourier (m , n , N , T )= N ...y cuyo periodo de muestreo fue “T”... 1/Fs N −1 ∑ m [kT ]e k=0 −2 π n k j N
  110. 110. 1 Fourier (m , n , N , T )= N ...se calcula como... N −1 ∑ m [kT ]e k=0 −2 π n k j N
  111. 111. 1 Fourier (m , n , N , T )= N N −1 ∑ m [kT ]e k=0 −2 π n k j N
  112. 112. 1 Fourier (m , n , N , T )= N N −1 ∑ m [kT ]e k=0 −2 π n k j N
  113. 113. 1 Fourier (m , n , N , T )= N N −1 ∑ m [kT ]e k=0 −2 π n k j N
  114. 114. 1 Fourier (m , n , N , T )= N N −1 ∑ m [kT ]e k=0 −2 π n k j N
  115. 115. 1 Fourier (m , n , N , T )= N N −1 ∑ m [kT ]e k=0 −2 π n k j N
  116. 116. 1 Fourier (m , n , N , T )= N N −1 ∑ m [kT ]e k=0 −2 π n k j N
  117. 117. ¡Difícil de entender!
  118. 118. ? 1 Fourier (m , n , N , T )= N N −1 ∑ m [kT ]e k=0 −2 π n k j N
  119. 119. ? 1 Fourier (m , n , N , T )= N N −1 ∑ m [kT ]e k=0 −2 π n k j N
  120. 120. Para facilitar el cálculo...
  121. 121. aplicando mi célebre fórmula...
  122. 122. e xj =cos( x)+ sen( x) j
  123. 123. e xj =cos( x)+ sen( x) j
  124. 124. e xj =cos( x)+ sen( x) j
  125. 125. e xj =cos( x)+ sen( x) j mi Transformada queda como...
  126. 126. 1 N N −1 ∑ k =0 k k m[kT ] cos(−2 π n )+ sen(−2 π n ) j N N ( )
  127. 127. 1 N N −1 ∑ k =0 k k m[kT ] cos(−2 π n )+ sen(−2 π n ) j N N ( )
  128. 128. 1 N N −1 ∑ k =0 k k m[kT ] cos(−2 π n )+ sen(−2 π n ) j N N ( )
  129. 129. 1 N N −1 ∑ k =0 k k m[kT ] cos(−2 π n )+ sen(−2 π n ) j N N ( )
  130. 130. 1 N N −1 ∑ k =0 k k m[kT ] cos(−2 π n )+ sen(−2 π n ) j N N ( )
  131. 131. 1 N N −1 ∑ k =0 k k m[kT ] cos(−2 π n )+ sen(−2 π n ) j N N ( )
  132. 132. 1 N N −1 ∑ k =0 k k m[kT ] cos(−2 π n )+ sen(−2 π n ) j N N ( )
  133. 133. Valor 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 5 4 3 2 1 Intensidad Muestra 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 0 1 2 3 4 5 6 7 8 9 10 -1 -2 -3 -4 -5 # de muestra de operar sobre todo el conjunto de datos, elegiremos una ventana de tamaño “N” En vez “m” 11 12 13 14 15 16 17
  134. 134. Ventana de análisis N = 10 muestras Valor 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 m[0..9] 5 4 3 2 1 Intensidad Muestra 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 0 1 2 3 4 5 6 7 8 9 -1 -2 -3 -4 -5 # de muestra 10 11 12 13 14 15 16 17
  135. 135. Valor 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 5 4 3 2 1 Intensidad Muestra 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 0 1 2 3 4 5 6 7 8 9 -1 -2 -3 -4 -5 # de muestra la ventana se recorre a m[1..10] 10 11 12 13 14 15 16 17
  136. 136. Valor 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 5 4 3 2 1 Intensidad Muestra 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 0 1 2 3 4 5 6 7 8 9 -1 -2 -3 -4 -5 # de muestra m[2..11] 10 11 12 13 14 15 16 17
  137. 137. Valor 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 5 4 3 2 1 Intensidad Muestra 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 0 1 2 3 4 5 6 7 8 9 -1 -2 -3 -4 -5 # de muestra m[3..12] 10 11 12 13 14 15 16 17
  138. 138. Valor 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 5 4 3 2 1 Intensidad Muestra 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 0 1 2 3 4 5 6 7 8 9 -1 -2 -3 -4 -5 # de muestra m[4..13] 10 11 12 13 14 15 16 17
  139. 139. Valor 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 5 4 3 2 1 Intensidad Muestra 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 0 1 2 3 4 5 6 7 8 9 -1 -2 -3 -4 -5 # de muestra m[5..14] 10 11 12 13 14 15 16 17
  140. 140. Valor 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 5 4 3 2 1 Intensidad Muestra 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 0 1 2 3 4 5 6 7 8 9 10 -1 -2 -3 -4 -5 # de muestra m[6..15] 11 12 13 14 15 16 17
  141. 141. Valor 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 -2.62 -4.25 0.00 4.25 2.63 2.62 4.25 0.00 -4.25 -2.63 5 4 3 2 1 Intensidad Muestra 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -1 -2 -3 -4 -5 # de muestra usaremos ésta como ejemplo: m[7..16]
  142. 142. ¿Cuánto aporta n=0, sen(0t), para formar la señal en la ventana?
  143. 143. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 n=0 N = 10 T = 0.05 ( )
  144. 144. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=0 N = 10 T = 0.05 )
  145. 145. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=0 N = 10 T = 0.05 sen(-2πnk/N) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 )
  146. 146. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=0 N = 10 T = 0.05 cos(-2πnk/N) 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 sen(-2πnk/N) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 )
  147. 147. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=0 N = 10 T = 0.05 m[kT] cos(-2πnk/N) -2.63 1.0000 -2.62 1.0000 -4.25 1.0000 0.00 1.0000 4.25 1.0000 2.63 1.0000 2.62 1.0000 4.25 1.0000 0.00 1.0000 -4.25 1.0000 sen(-2πnk/N) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 )
  148. 148. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=0 N = 10 T = 0.05 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 1.0000 -2.6200 0.0000 0.0000 -4.25 1.0000 -4.2500 0.0000 0.0000 0.00 1.0000 0.0000 0.0000 0.0000 4.25 1.0000 4.2500 0.0000 0.0000 2.63 1.0000 2.6300 0.0000 0.0000 2.62 1.0000 2.6200 0.0000 0.0000 4.25 1.0000 4.2500 0.0000 0.0000 0.00 1.0000 0.0000 0.0000 0.0000 -4.25 1.0000 -4.2500 0.0000 0.0000 )
  149. 149. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=0 N = 10 T = 0.05 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 1.0000 -2.6200 0.0000 0.0000 -4.25 1.0000 -4.2500 0.0000 0.0000 0.00 1.0000 0.0000 0.0000 0.0000 4.25 1.0000 4.2500 0.0000 0.0000 2.63 1.0000 2.6300 0.0000 0.0000 2.62 1.0000 2.6200 0.0000 0.0000 4.25 1.0000 4.2500 0.0000 0.0000 0.00 1.0000 0.0000 0.0000 0.0000 -4.25 1.0000 -4.2500 0.0000 0.0000 Suma: 0.0000 0.0000 )
  150. 150. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 1.0000 -2.6200 0.0000 0.0000 -4.25 1.0000 -4.2500 0.0000 0.0000 0.00 1.0000 0.0000 0.0000 0.0000 4.25 1.0000 4.2500 0.0000 0.0000 2.63 1.0000 2.6300 0.0000 0.0000 2.62 1.0000 2.6200 0.0000 0.0000 4.25 1.0000 4.2500 0.0000 0.0000 0.00 1.0000 0.0000 0.0000 0.0000 -4.25 1.0000 -4.2500 0.0000 0.0000 Suma: 0.0000 0.0000 √ 0 + 0 =0 2 n=0 N = 10 T = 0.05 2 )
  151. 151. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 1.0000 -2.6200 0.0000 0.0000 -4.25 1.0000 -4.2500 0.0000 0.0000 0.00 1.0000 0.0000 0.0000 0.0000 4.25 1.0000 4.2500 0.0000 0.0000 2.63 1.0000 2.6300 0.0000 0.0000 2.62 1.0000 2.6200 0.0000 0.0000 4.25 1.0000 4.2500 0.0000 0.0000 0.00 1.0000 0.0000 0.0000 0.0000 -4.25 1.0000 -4.2500 0.0000 0.0000 Suma: 0.0000 0.0000 √ 0 + 0 =0 2 n=0 N = 10 T = 0.05 2 1 . 0=0 10 )
  152. 152. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 1.0000 -2.6200 0.0000 0.0000 -4.25 1.0000 -4.2500 0.0000 0.0000 0.00 1.0000 0.0000 0.0000 0.0000 4.25 1.0000 4.2500 0.0000 0.0000 2.63 1.0000 2.6300 0.0000 0.0000 2.62 1.0000 2.6200 0.0000 0.0000 4.25 1.0000 4.2500 0.0000 0.0000 0.00 1.0000 0.0000 0.0000 0.0000 -4.25 1.0000 -4.2500 0.0000 0.0000 Suma: 0.0000 0.0000 √ 0 + 0 =0 2 n=0 N = 10 T = 0.05 ) 2 1 . 0=0 10 la señal sen(0t) aportó 0 de amplitud a la señal analizada
  153. 153. ¿Cuánto aporta n=1, sen(2t), para formar la señal en la ventana?
  154. 154. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=1 N = 10 T = 0.05 )
  155. 155. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=1 N = 10 T = 0.05 sen(-2πnk/N) 0.0000 -0.5878 -0.9511 -0.9511 -0.5878 0.0000 0.5878 0.9511 0.9511 0.5878 )
  156. 156. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=1 N = 10 T = 0.05 cos(-2πnk/N) 1.0000 0.8090 0.3090 -0.3090 -0.8090 -1.0000 -0.8090 -0.3090 0.3090 0.8090 sen(-2πnk/N) 0.0000 -0.5878 -0.9511 -0.9511 -0.5878 0.0000 0.5878 0.9511 0.9511 0.5878 )
  157. 157. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=1 N = 10 T = 0.05 m[kT] cos(-2πnk/N) -2.63 1.0000 -2.62 0.8090 -4.25 0.3090 0.00 -0.3090 4.25 -0.8090 2.63 -1.0000 2.62 -0.8090 4.25 -0.3090 0.00 0.3090 -4.25 0.8090 sen(-2πnk/N) 0.0000 -0.5878 -0.9511 -0.9511 -0.5878 0.0000 0.5878 0.9511 0.9511 0.5878 )
  158. 158. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=1 N = 10 T = 0.05 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 0.8090 -2.1196 -0.5878 1.5400 -4.25 0.3090 -1.3133 -0.9511 4.0420 0.00 -0.3090 0.0000 -0.9511 0.0000 4.25 -0.8090 -3.4383 -0.5878 -2.4981 2.63 -1.0000 -2.6300 0.0000 0.0000 2.62 -0.8090 -2.1196 0.5878 1.5400 4.25 -0.3090 -1.3133 0.9511 4.0420 0.00 0.3090 0.0000 0.9511 0.0000 -4.25 0.8090 -3.4383 0.5878 -2.4981 )
  159. 159. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=1 N = 10 T = 0.05 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 0.8090 -2.1196 -0.5878 1.5400 -4.25 0.3090 -1.3133 -0.9511 4.0420 0.00 -0.3090 0.0000 -0.9511 0.0000 4.25 -0.8090 -3.4383 -0.5878 -2.4981 2.63 -1.0000 -2.6300 0.0000 0.0000 2.62 -0.8090 -2.1196 0.5878 1.5400 4.25 -0.3090 -1.3133 0.9511 4.0420 0.00 0.3090 0.0000 0.9511 0.0000 -4.25 0.8090 -3.4383 0.5878 -2.4981 Suma: -19.0025 6.1678 )
  160. 160. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 0.8090 -2.1196 -0.5878 1.5400 -4.25 0.3090 -1.3133 -0.9511 4.0420 0.00 -0.3090 0.0000 -0.9511 0.0000 4.25 -0.8090 -3.4383 -0.5878 -2.4981 2.63 -1.0000 -2.6300 0.0000 0.0000 2.62 -0.8090 -2.1196 0.5878 1.5400 4.25 -0.3090 -1.3133 0.9511 4.0420 0.00 0.3090 0.0000 0.9511 0.0000 -4.25 0.8090 -3.4383 0.5878 -2.4981 Suma: -19.0025 6.1678 √−19.0025 + 6.1678 ≈20 2 n=1 N = 10 T = 0.05 2 )
  161. 161. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 0.8090 -2.1196 -0.5878 1.5400 -4.25 0.3090 -1.3133 -0.9511 4.0420 0.00 -0.3090 0.0000 -0.9511 0.0000 4.25 -0.8090 -3.4383 -0.5878 -2.4981 2.63 -1.0000 -2.6300 0.0000 0.0000 2.62 -0.8090 -2.1196 0.5878 1.5400 4.25 -0.3090 -1.3133 0.9511 4.0420 0.00 0.3090 0.0000 0.9511 0.0000 -4.25 0.8090 -3.4383 0.5878 -2.4981 Suma: -19.0025 6.1678 √−19.0025 + 6.1678 ≈20 2 n=1 N = 10 T = 0.05 1 . 20=2 10 2 )
  162. 162. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 0.8090 -2.1196 -0.5878 1.5400 -4.25 0.3090 -1.3133 -0.9511 4.0420 0.00 -0.3090 0.0000 -0.9511 0.0000 4.25 -0.8090 -3.4383 -0.5878 -2.4981 2.63 -1.0000 -2.6300 0.0000 0.0000 2.62 -0.8090 -2.1196 0.5878 1.5400 4.25 -0.3090 -1.3133 0.9511 4.0420 0.00 0.3090 0.0000 0.9511 0.0000 -4.25 0.8090 -3.4383 0.5878 -2.4981 Suma: -19.0025 6.1678 √−19.0025 + 6.1678 ≈20 2 n=1 N = 10 T = 0.05 ) 1 . 20=2 10 2 la señal sen(2t) aportó 2 de amplitud a la señal analizada
  163. 163. ¿Cuánto aporta n=2, sen(4t), para formar la señal en la ventana?
  164. 164. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=2 N = 10 T = 0.05 )
  165. 165. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=2 N = 10 T = 0.05 sen(-2πnk/N) 0.0000 -0.9511 -0.5878 0.5878 0.9511 0.0000 -0.9511 -0.5878 0.5878 0.9511 )
  166. 166. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=2 N = 10 T = 0.05 cos(-2πnk/N) 1.0000 0.3090 -0.8090 -0.8090 0.3090 1.0000 0.3090 -0.8090 -0.8090 0.3090 sen(-2πnk/N) 0.0000 -0.9511 -0.5878 0.5878 0.9511 0.0000 -0.9511 -0.5878 0.5878 0.9511 )
  167. 167. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=2 N = 10 T = 0.05 m[kT] cos(-2πnk/N) -2.63 1.0000 -2.62 0.3090 -4.25 -0.8090 0.00 -0.8090 4.25 0.3090 2.63 1.0000 2.62 0.3090 4.25 -0.8090 0.00 -0.8090 -4.25 0.3090 sen(-2πnk/N) 0.0000 -0.9511 -0.5878 0.5878 0.9511 0.0000 -0.9511 -0.5878 0.5878 0.9511 )
  168. 168. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=2 N = 10 T = 0.05 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 0.3090 -0.8096 -0.9511 2.4918 -4.25 -0.8090 3.4383 -0.5878 2.4981 0.00 -0.8090 0.0000 0.5878 0.0000 4.25 0.3090 1.3133 0.9511 4.0420 2.63 1.0000 2.6300 0.0000 0.0000 2.62 0.3090 0.8096 -0.9511 -2.4918 4.25 -0.8090 -3.4383 -0.5878 -2.4981 0.00 -0.8090 0.0000 0.5878 0.0000 -4.25 0.3090 -1.3133 0.9511 -4.0420 )
  169. 169. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=2 N = 10 T = 0.05 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 0.3090 -0.8096 -0.9511 2.4918 -4.25 -0.8090 3.4383 -0.5878 2.4981 0.00 -0.8090 0.0000 0.5878 0.0000 4.25 0.3090 1.3133 0.9511 4.0420 2.63 1.0000 2.6300 0.0000 0.0000 2.62 0.3090 0.8096 -0.9511 -2.4918 4.25 -0.8090 -3.4383 -0.5878 -2.4981 0.00 -0.8090 0.0000 0.5878 0.0000 -4.25 0.3090 -1.3133 0.9511 -4.0420 Suma: 0.0000 0.0000 )
  170. 170. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 0.3090 -0.8096 -0.9511 2.4918 -4.25 -0.8090 3.4383 -0.5878 2.4981 0.00 -0.8090 0.0000 0.5878 0.0000 4.25 0.3090 1.3133 0.9511 4.0420 2.63 1.0000 2.6300 0.0000 0.0000 2.62 0.3090 0.8096 -0.9511 -2.4918 4.25 -0.8090 -3.4383 -0.5878 -2.4981 0.00 -0.8090 0.0000 0.5878 0.0000 -4.25 0.3090 -1.3133 0.9511 -4.0420 Suma: 0.0000 0.0000 √ 0 + 0 =0 2 n=2 N = 10 T = 0.05 2 )
  171. 171. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 0.3090 -0.8096 -0.9511 2.4918 -4.25 -0.8090 3.4383 -0.5878 2.4981 0.00 -0.8090 0.0000 0.5878 0.0000 4.25 0.3090 1.3133 0.9511 4.0420 2.63 1.0000 2.6300 0.0000 0.0000 2.62 0.3090 0.8096 -0.9511 -2.4918 4.25 -0.8090 -3.4383 -0.5878 -2.4981 0.00 -0.8090 0.0000 0.5878 0.0000 -4.25 0.3090 -1.3133 0.9511 -4.0420 Suma: 0.0000 0.0000 √ 0 + 0 =0 2 n=2 N = 10 T = 0.05 2 1 . 0=0 10 )
  172. 172. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 0.3090 -0.8096 -0.9511 2.4918 -4.25 -0.8090 3.4383 -0.5878 2.4981 0.00 -0.8090 0.0000 0.5878 0.0000 4.25 0.3090 1.3133 0.9511 4.0420 2.63 1.0000 2.6300 0.0000 0.0000 2.62 0.3090 0.8096 -0.9511 -2.4918 4.25 -0.8090 -3.4383 -0.5878 -2.4981 0.00 -0.8090 0.0000 0.5878 0.0000 -4.25 0.3090 -1.3133 0.9511 -4.0420 Suma: 0.0000 0.0000 √ 0 + 0 =0 2 n=2 N = 10 T = 0.05 ) 2 1 . 0=0 10 la señal sen(4t) aportó 0 de amplitud a la señal analizada
  173. 173. ¿Cuánto aporta n=3, sen(6t), para formar la señal en la ventana?
  174. 174. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=3 N = 10 T = 0.05 )
  175. 175. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=3 N = 10 T = 0.05 sen(-2πnk/N) 0.0000 -0.9511 0.5878 0.5878 -0.9511 0.0000 0.9511 -0.5878 -0.5878 0.9511 )
  176. 176. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=3 N = 10 T = 0.05 cos(-2πnk/N) 1.0000 -0.3090 -0.8090 0.8090 0.3090 -1.0000 0.3090 0.8090 -0.8090 -0.3090 sen(-2πnk/N) 0.0000 -0.9511 0.5878 0.5878 -0.9511 0.0000 0.9511 -0.5878 -0.5878 0.9511 )
  177. 177. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=3 N = 10 T = 0.05 m[kT] cos(-2πnk/N) -2.63 1.0000 -2.62 -0.3090 -4.25 -0.8090 0.00 0.8090 4.25 0.3090 2.63 -1.0000 2.62 0.3090 4.25 0.8090 0.00 -0.8090 -4.25 -0.3090 sen(-2πnk/N) 0.0000 -0.9511 0.5878 0.5878 -0.9511 0.0000 0.9511 -0.5878 -0.5878 0.9511 )
  178. 178. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=3 N = 10 T = 0.05 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 -0.3090 0.8096 -0.9511 2.4918 -4.25 -0.8090 3.4383 0.5878 -2.4981 0.00 0.8090 0.0000 0.5878 0.0000 4.25 0.3090 1.3133 -0.9511 -4.0420 2.63 -1.0000 -2.6300 0.0000 0.0000 2.62 0.3090 0.8096 0.9511 2.4918 4.25 0.8090 3.4383 -0.5878 -2.4981 0.00 -0.8090 0.0000 -0.5878 0.0000 -4.25 -0.3090 1.3133 0.9511 -4.0420 )
  179. 179. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=3 N = 10 T = 0.05 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 -0.3090 0.8096 -0.9511 2.4918 -4.25 -0.8090 3.4383 0.5878 -2.4981 0.00 0.8090 0.0000 0.5878 0.0000 4.25 0.3090 1.3133 -0.9511 -4.0420 2.63 -1.0000 -2.6300 0.0000 0.0000 2.62 0.3090 0.8096 0.9511 2.4918 4.25 0.8090 3.4383 -0.5878 -2.4981 0.00 -0.8090 0.0000 -0.5878 0.0000 -4.25 -0.3090 1.3133 0.9511 -4.0420 Suma: 5.8625 -8.0966 )
  180. 180. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 -0.3090 0.8096 -0.9511 2.4918 -4.25 -0.8090 3.4383 0.5878 -2.4981 0.00 0.8090 0.0000 0.5878 0.0000 4.25 0.3090 1.3133 -0.9511 -4.0420 2.63 -1.0000 -2.6300 0.0000 0.0000 2.62 0.3090 0.8096 0.9511 2.4918 4.25 0.8090 3.4383 -0.5878 -2.4981 0.00 -0.8090 0.0000 -0.5878 0.0000 -4.25 -0.3090 1.3133 0.9511 -4.0420 Suma: 5.8625 -8.0966 √ 5.8625 + (−8.0966) ≈10 2 n=3 N = 10 T = 0.05 2 )
  181. 181. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 -0.3090 0.8096 -0.9511 2.4918 -4.25 -0.8090 3.4383 0.5878 -2.4981 0.00 0.8090 0.0000 0.5878 0.0000 4.25 0.3090 1.3133 -0.9511 -4.0420 2.63 -1.0000 -2.6300 0.0000 0.0000 2.62 0.3090 0.8096 0.9511 2.4918 4.25 0.8090 3.4383 -0.5878 -2.4981 0.00 -0.8090 0.0000 -0.5878 0.0000 -4.25 -0.3090 1.3133 0.9511 -4.0420 Suma: 5.8625 -8.0966 √ 5.8625 + (−8.0966) ≈10 2 n=3 N = 10 T = 0.05 1 .10=1 10 2 )
  182. 182. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 -0.3090 0.8096 -0.9511 2.4918 -4.25 -0.8090 3.4383 0.5878 -2.4981 0.00 0.8090 0.0000 0.5878 0.0000 4.25 0.3090 1.3133 -0.9511 -4.0420 2.63 -1.0000 -2.6300 0.0000 0.0000 2.62 0.3090 0.8096 0.9511 2.4918 4.25 0.8090 3.4383 -0.5878 -2.4981 0.00 -0.8090 0.0000 -0.5878 0.0000 -4.25 -0.3090 1.3133 0.9511 -4.0420 Suma: 5.8625 -8.0966 √ 5.8625 + (−8.0966) ≈10 2 n=3 N = 10 T = 0.05 ) 1 .10=1 10 2 la señal sen(6t) aportó 1 de amplitud a la señal analizada
  183. 183. ¿Cuánto aporta n=4, sen(8t), para formar la señal en la ventana?
  184. 184. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=4 N = 10 T = 0.05 )
  185. 185. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=4 N = 10 T = 0.05 sen(-2πnk/N) 0.0000 -0.5878 0.9511 -0.9511 0.5878 0.0000 -0.5878 0.9511 -0.9511 0.5878 )
  186. 186. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=4 N = 10 T = 0.05 cos(-2πnk/N) 1.0000 -0.8090 0.3090 0.3090 -0.8090 1.0000 -0.8090 0.3090 0.3090 -0.8090 sen(-2πnk/N) 0.0000 -0.5878 0.9511 -0.9511 0.5878 0.0000 -0.5878 0.9511 -0.9511 0.5878 )
  187. 187. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=4 N = 10 T = 0.05 m[kT] cos(-2πnk/N) -2.63 1.0000 -2.62 -0.8090 -4.25 0.3090 0.00 0.3090 4.25 -0.8090 2.63 1.0000 2.62 -0.8090 4.25 0.3090 0.00 0.3090 -4.25 -0.8090 sen(-2πnk/N) 0.0000 -0.5878 0.9511 -0.9511 0.5878 0.0000 -0.5878 0.9511 -0.9511 0.5878 )
  188. 188. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=4 N = 10 T = 0.05 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 -0.8090 2.1196 -0.5878 1.5400 -4.25 0.3090 -1.3133 0.9511 -4.0420 0.00 0.3090 0.0000 -0.9511 0.0000 4.25 -0.8090 -3.4383 0.5878 2.4981 2.63 1.0000 2.6300 0.0000 0.0000 2.62 -0.8090 -2.1196 -0.5878 -1.5400 4.25 0.3090 1.3133 0.9511 4.0420 0.00 0.3090 0.0000 -0.9511 0.0000 -4.25 -0.8090 3.4383 0.5878 -2.4981 )
  189. 189. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=4 N = 10 T = 0.05 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 -0.8090 2.1196 -0.5878 1.5400 -4.25 0.3090 -1.3133 0.9511 -4.0420 0.00 0.3090 0.0000 -0.9511 0.0000 4.25 -0.8090 -3.4383 0.5878 2.4981 2.63 1.0000 2.6300 0.0000 0.0000 2.62 -0.8090 -2.1196 -0.5878 -1.5400 4.25 0.3090 1.3133 0.9511 4.0420 0.00 0.3090 0.0000 -0.9511 0.0000 -4.25 -0.8090 3.4383 0.5878 -2.4981 Suma: 0.0000 0.0000 )
  190. 190. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 -0.8090 2.1196 -0.5878 1.5400 -4.25 0.3090 -1.3133 0.9511 -4.0420 0.00 0.3090 0.0000 -0.9511 0.0000 4.25 -0.8090 -3.4383 0.5878 2.4981 2.63 1.0000 2.6300 0.0000 0.0000 2.62 -0.8090 -2.1196 -0.5878 -1.5400 4.25 0.3090 1.3133 0.9511 4.0420 0.00 0.3090 0.0000 -0.9511 0.0000 -4.25 -0.8090 3.4383 0.5878 -2.4981 Suma: 0.0000 0.0000 √ 0 + 0 =0 2 n=4 N = 10 T = 0.05 2 )
  191. 191. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 -0.8090 2.1196 -0.5878 1.5400 -4.25 0.3090 -1.3133 0.9511 -4.0420 0.00 0.3090 0.0000 -0.9511 0.0000 4.25 -0.8090 -3.4383 0.5878 2.4981 2.63 1.0000 2.6300 0.0000 0.0000 2.62 -0.8090 -2.1196 -0.5878 -1.5400 4.25 0.3090 1.3133 0.9511 4.0420 0.00 0.3090 0.0000 -0.9511 0.0000 -4.25 -0.8090 3.4383 0.5878 -2.4981 Suma: 0.0000 0.0000 √ 0 + 0 =0 2 n=4 N = 10 T = 0.05 2 1 . 0=0 10 )
  192. 192. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 -0.8090 2.1196 -0.5878 1.5400 -4.25 0.3090 -1.3133 0.9511 -4.0420 0.00 0.3090 0.0000 -0.9511 0.0000 4.25 -0.8090 -3.4383 0.5878 2.4981 2.63 1.0000 2.6300 0.0000 0.0000 2.62 -0.8090 -2.1196 -0.5878 -1.5400 4.25 0.3090 1.3133 0.9511 4.0420 0.00 0.3090 0.0000 -0.9511 0.0000 -4.25 -0.8090 3.4383 0.5878 -2.4981 Suma: 0.0000 0.0000 √ 0 + 0 =0 2 n=4 N = 10 T = 0.05 ) 2 1 . 0=0 10 la señal sen(8t) aportó 0 de amplitud a la señal analizada
  193. 193. ¿Cuánto aporta n=5, sen(10t), para formar la señal en la ventana?
  194. 194. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=5 N = 10 T = 0.05 )
  195. 195. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=5 N = 10 T = 0.05 sen(-2πnk/N) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 )
  196. 196. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=5 N = 10 T = 0.05 cos(-2πnk/N) 1.0000 -1.0000 1.0000 -1.0000 1.0000 -1.0000 1.0000 -1.0000 1.0000 -1.0000 sen(-2πnk/N) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 )
  197. 197. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=5 N = 10 T = 0.05 m[kT] cos(-2πnk/N) -2.63 1.0000 -2.62 -1.0000 -4.25 1.0000 0.00 -1.0000 4.25 1.0000 2.63 -1.0000 2.62 1.0000 4.25 -1.0000 0.00 1.0000 -4.25 -1.0000 sen(-2πnk/N) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 )
  198. 198. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=5 N = 10 T = 0.05 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 -1.0000 2.6200 0.0000 0.0000 -4.25 1.0000 -4.2500 0.0000 0.0000 0.00 -1.0000 0.0000 0.0000 0.0000 4.25 1.0000 4.2500 0.0000 0.0000 2.63 -1.0000 -2.6300 0.0000 0.0000 2.62 1.0000 2.6200 0.0000 0.0000 4.25 -1.0000 -4.2500 0.0000 0.0000 0.00 1.0000 0.0000 0.0000 0.0000 -4.25 -1.0000 4.2500 0.0000 0.0000 )
  199. 199. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 n=5 N = 10 T = 0.05 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 -1.0000 2.6200 0.0000 0.0000 -4.25 1.0000 -4.2500 0.0000 0.0000 0.00 -1.0000 0.0000 0.0000 0.0000 4.25 1.0000 4.2500 0.0000 0.0000 2.63 -1.0000 -2.6300 0.0000 0.0000 2.62 1.0000 2.6200 0.0000 0.0000 4.25 -1.0000 -4.2500 0.0000 0.0000 0.00 1.0000 0.0000 0.0000 0.0000 -4.25 -1.0000 4.2500 0.0000 0.0000 Suma: -0.0200 0.0000 )
  200. 200. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 -1.0000 2.6200 0.0000 0.0000 -4.25 1.0000 -4.2500 0.0000 0.0000 0.00 -1.0000 0.0000 0.0000 0.0000 4.25 1.0000 4.2500 0.0000 0.0000 2.63 -1.0000 -2.6300 0.0000 0.0000 2.62 1.0000 2.6200 0.0000 0.0000 4.25 -1.0000 -4.2500 0.0000 0.0000 0.00 1.0000 0.0000 0.0000 0.0000 -4.25 -1.0000 4.2500 0.0000 0.0000 Suma: -0.0200 0.0000 √−0.0200 + 0.0000 ≈0 2 n=5 N = 10 T = 0.05 2 )
  201. 201. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 -1.0000 2.6200 0.0000 0.0000 -4.25 1.0000 -4.2500 0.0000 0.0000 0.00 -1.0000 0.0000 0.0000 0.0000 4.25 1.0000 4.2500 0.0000 0.0000 2.63 -1.0000 -2.6300 0.0000 0.0000 2.62 1.0000 2.6200 0.0000 0.0000 4.25 -1.0000 -4.2500 0.0000 0.0000 0.00 1.0000 0.0000 0.0000 0.0000 -4.25 -1.0000 4.2500 0.0000 0.0000 Suma: -0.0200 0.0000 √−0.0200 + 0.0000 ≈0 2 n=5 N = 10 T = 0.05 1 . 0=0 10 2 )
  202. 202. 1 N N −1 k k ∑ m[kT ] cos(−2 π n N )+ sen(−2 π n N ) j k =0 ( k 0 1 2 3 4 5 6 7 8 9 m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...) -2.63 1.0000 -2.6300 0.0000 0.0000 -2.62 -1.0000 2.6200 0.0000 0.0000 -4.25 1.0000 -4.2500 0.0000 0.0000 0.00 -1.0000 0.0000 0.0000 0.0000 4.25 1.0000 4.2500 0.0000 0.0000 2.63 -1.0000 -2.6300 0.0000 0.0000 2.62 1.0000 2.6200 0.0000 0.0000 4.25 -1.0000 -4.2500 0.0000 0.0000 0.00 1.0000 0.0000 0.0000 0.0000 -4.25 -1.0000 4.2500 0.0000 0.0000 Suma: -0.0200 0.0000 √−0.0200 + 0.0000 ≈0 2 n=5 N = 10 T = 0.05 ) 1 . 0=0 10 2 la señal sen(10t) aportó 0 de amplitud a la señal analizada
  203. 203. Reuniendo todas las componentes podemos expresar analíticamente que...
  204. 204. 5.00 4.00 3.00 2.00 1.00 0.00 -1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 -2.00 -3.00 -4.00 -5.00 Señal (t )=2.sen(2t)+ 1.sen (6t )
  205. 205. Si representamos los resultados en un Ecualizador Gráfico...
  206. 206. y = 2*sen(2t)
  207. 207. y = 1*sen(6t)
  208. 208. Pero, ¿cómo y por qué funciona la Transformada de Fourier?
  209. 209. 1 Fourier (m , n , N , T )= N N −1 ∑ m [kT ]e k=0 −2 π n k j N
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