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.

1. La Transformada de Fourier
Dr. José Enrique Alvarez Estrada
http://www.software.org.mx/~jalvarez/

2. Las ideas de
Jean Baptiste Fourier

3. cualquier señal puede
formarse sumando
funciones seno
de diferentes frecuencias
a diferentes amplitudes

4. 1
0.5
0
-0.5
-1

5. 1
0.5
0
-0.5
-1
1
0.5
0
-0.5
-1

6. 1
0.5
0
-0.5
-1
1
0.5
0
-0.5
-1
1
0.5
0
-0.5
-1

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. 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. 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. 1
0.5
0
-0.5
-1

11. 1
0.5
0
+
-0.5
-1
2
1
0
-1
-2

12. 1
0.5
0
+
-0.5
-1
2
1
0
-1
+
-2
1
0.5
0
-0.5
-1

13. 1
0.5
0
+
-0.5
-1
2
1
0
-1
+
-2
1
0.5
0
-0.5
-1

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. Y a la inversa...

16. Dada una señal compuesta,
¿cuánto tengo que agregar
de cada señal fundamental
para recrearla?

17. 3.5
2.5
1.5
0.5
-0.5
-1.5
-2.5
-3.5
3

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. 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. 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. 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. Así que la palabra transformar
significa cambiar el dominio de la señal,
pasando del dominio del tiempo al
dominio de la frecuencia.

23. Fourier

24. 3.5
2.5
1.5
0.5
3
-0.5
-1.5
-2.5
-3.5
tiempo
Fourier

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. Por simplicidad, trabajaremos con
una versión discreta de la señal

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

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. 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. 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. Hasta aquí, tenemos grabado
un archivo WAV...

39. Pero un WAV ocupa mucho espacio,
porque guarda todas las muestras.
¿Y si sólo almacenamos
algunas de sus características?

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. 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. 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. 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. Hagamos un ejemplo

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. 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. 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. 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. ¿Qué frecuencias básicas
debemos analizar para
reconstruir la señal
en la ventana?

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

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

53. la frecuencia máxima analizable
no puede ser mayor que la mitad
de la frecuencia de muestreo
Fs
F máx ≤
2

54. Así que, para
nuestro ejemplo...

55. 20 Hz
F máx ⩽
⩽10 Hz
2

56. Por tanto,
los valores de “n”
y las frecuencias
a analizar serán:

57. n=0⇒ F 0=
0
20=0 Hz
10

58. 0
20=0 Hz
10
1
n=1⇒ F 1= 20=2 Hz
10
n=0 ⇒ F 0 =

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. 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. 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. 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. ¿Cuánto aporta n=0, sen(0t),
para formar la señal?

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. 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. 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. 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. 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. 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. 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. 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. ¿Cuánto aporta n=1, sen(2t),
para formar la señal?

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. 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. 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. 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. 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. ¿Cuánto aporta n=2, sen(4t),
para formar la señal?

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. 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. 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. 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. 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. ¿Cuánto aporta n=3, sen(6t),
para formar la señal?

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. 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. 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. 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. 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. ¿Cuánto aporta n=4, sen(8t),
para formar la señal?

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. 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. 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. 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. 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. ¿Cuánto aporta n=5, sen(10t),
para formar la señal?

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. 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. 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. 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. 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. ¡Son demasiados cálculos!

103. Se necesita crear un
método abreviado
que requiera
menos esfuerzo.

104. 1
Fourier (m , n , N , T )=
N
N −1
∑ m [kT ]e
k=0
−2 π n k
j
N

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. 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. 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. 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. 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. 1
Fourier (m , n , N , T )=
N
...se calcula como...
N −1
∑ m [kT ]e
k=0
−2 π n k
j
N

111. 1
Fourier (m , n , N , T )=
N
N −1
∑ m [kT ]e
k=0
−2 π n k
j
N

112. 1
Fourier (m , n , N , T )=
N
N −1
∑ m [kT ]e
k=0
−2 π n k
j
N

113. 1
Fourier (m , n , N , T )=
N
N −1
∑ m [kT ]e
k=0
−2 π n k
j
N

114. 1
Fourier (m , n , N , T )=
N
N −1
∑ m [kT ]e
k=0
−2 π n k
j
N

115. 1
Fourier (m , n , N , T )=
N
N −1
∑ m [kT ]e
k=0
−2 π n k
j
N

116. 1
Fourier (m , n , N , T )=
N
N −1
∑ m [kT ]e
k=0
−2 π n k
j
N

117. ¡Difícil de entender!

118. ?
1
Fourier (m , n , N , T )=
N
N −1
∑ m [kT ]e
k=0
−2 π n k
j
N

119. ?
1
Fourier (m , n , N , T )=
N
N −1
∑ m [kT ]e
k=0
−2 π n k
j
N

120. Para facilitar el cálculo...

121. aplicando mi
célebre fórmula...

122. e xj =cos( x)+ sen( x) j

123. e xj =cos( x)+ sen( x) j

124. e xj =cos( x)+ sen( x) j

125. e xj =cos( x)+ sen( x) j
mi Transformada
queda como...

126. 1
N
N −1
∑
k =0
k
k
m[kT ] cos(−2 π n )+ sen(−2 π n ) j
N
N
(
)

127. 1
N
N −1
∑
k =0
k
k
m[kT ] cos(−2 π n )+ sen(−2 π n ) j
N
N
(
)

128. 1
N
N −1
∑
k =0
k
k
m[kT ] cos(−2 π n )+ sen(−2 π n ) j
N
N
(
)

129. 1
N
N −1
∑
k =0
k
k
m[kT ] cos(−2 π n )+ sen(−2 π n ) j
N
N
(
)

130. 1
N
N −1
∑
k =0
k
k
m[kT ] cos(−2 π n )+ sen(−2 π n ) j
N
N
(
)

131. 1
N
N −1
∑
k =0
k
k
m[kT ] cos(−2 π n )+ sen(−2 π n ) j
N
N
(
)

132. 1
N
N −1
∑
k =0
k
k
m[kT ] cos(−2 π n )+ sen(−2 π n ) j
N
N
(
)

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. 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. 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. 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. 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. 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. 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. 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. 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. ¿Cuánto aporta n=0, sen(0t),
para formar la señal en la ventana?

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. 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. 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. 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. 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. 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. 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. 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. 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. 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. ¿Cuánto aporta n=1, sen(2t),
para formar la señal en la ventana?

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. 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. 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. 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. 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. 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. 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. 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. 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. ¿Cuánto aporta n=2, sen(4t),
para formar la señal en la ventana?

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. 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. 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. 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. 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. 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. 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. 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. 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. ¿Cuánto aporta n=3, sen(6t),
para formar la señal en la ventana?

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. 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. 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. 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. 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. 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. 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. 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. 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. ¿Cuánto aporta n=4, sen(8t),
para formar la señal en la ventana?

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. 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. 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. 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. 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. 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. 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. 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. 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. ¿Cuánto aporta n=5, sen(10t),
para formar la señal en la ventana?

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. 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. 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. 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. 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. 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. 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. 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. 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. Reuniendo todas las componentes
podemos expresar analíticamente que...

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. Si representamos los resultados
en un Ecualizador Gráfico...

207. y = 2*sen(2t)

208. y = 1*sen(6t)

209. Pero, ¿cómo y por qué
funciona la Transformada de Fourier?

210. 1
Fourier (m , n , N , T )=
N
N −1
∑ m [kT ]e
k=0
−2 π n k
j
N