cómo funciona la transformada de fourier
Post on 16-Jun-2015
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La Transformada de Fourier
Dr. José Enrique Alvarez Estradahttp://www.software.org.mx/~jalvarez/
Las ideas deJean Baptiste Fourier
cualquier señal puedeformarse sumando
funciones senode diferentes frecuencias
a diferentes amplitudes
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*1
*2
*0
*1
+
-1
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*1
*2
*0
*1
+
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1
-1
-0.5
0
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1
-2
-1
0
1
2+
-1
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0
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1
-1
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0
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1
-2
-1
0
1
2+
+
-1
-0.5
0
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1
-1
-0.5
0
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1
-2
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0
1
2+
+
-3.5
-2.5
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-0.5
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1.5
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3
-1
-0.5
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-1
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-2
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1
2+
+
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
-3.5
-2.5
-1.5
-0.5
0.5
1.5
2.5
3.5
3
¿Cuánto sen(3t)?
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-0.5
0
0.5
1
-1
-0.5
0
0.5
1
-3.5
-2.5
-1.5
-0.5
0.5
1.5
2.5
3.5
3
¿Cuánto sen(3t)?
¿Cuánto sen(4t)?
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1
-3.5
-2.5
-1.5
-0.5
0.5
1.5
2.5
3.5
3
¿Cuánto sen(3t)?
¿Cuánto sen(4t)?
¿Cuánto sen(6t)?
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1
-1
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0
0.5
1
-3.5
-2.5
-1.5
-0.5
0.5
1.5
2.5
3.5
3
¿Cuánto sen(3t)?
¿Cuánto sen(4t)?
¿Cuánto sen(6t)?
¿Cuánto sen(12t)?
Así que la palabra transformarsignifica cambiar el dominio de la señal,
pasando del dominio del tiempo aldominio de la frecuencia.
Fourier
-3.5
-2.5
-1.5
-0.5
0.5
1.5
2.5
3.5
3 Fourier
tiempo
-3.5
-2.5
-1.5
-0.5
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1.5
2.5
3.5
3
0 1 2 3 4 5 6 7 8 9 10 11 120
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Fourier
tiempo frecuencia
Por simplicidad, trabajaremos conuna versión discreta de la señal
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
2.00
3.00
4.00
5.00
fenómenoanalógicoa estudiar
ConversorA/D
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
ConversorA/D
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
20 Hz
Frecuenciade muestreo (Fs)
ConversorA/D
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
2.00
3.00
4.00
5.00
20 Hz
versióndiscreta
(muestras)
10 muestras
ConversorA/D
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
2.00
3.00
4.00
5.00
20 Hz
tamaño deventana
“N”
-2.63 -2.62 -2.63 -4.25 0.00 4.25
10 muestras
2.63 2.62 4.25 0.00
ConversorA/D
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
2.00
3.00
4.00
5.00
20 Hz ventana “m”
-2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00
-2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.000 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
-2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00
ConversorD/A
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
-2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00
ConversorD/A
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
2.00
3.00
4.00
5.00
Hasta aquí, tenemos grabadoun archivo WAV...
Pero un WAV ocupa mucho espacio,porque guarda todas las muestras.
¿Y si sólo almacenamos algunas de sus características?
-2.63 -2.62 -2.63 -4.25 0.00 4.25
10 muestras
2.63 2.62 4.25 0.00
ConversorA/D
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
2.00
3.00
4.00
5.00
20 Hz 1/Fs
periodo demuestreo
“T”
-2.63
n=510Hz
-2.62 -2.63 -4.25 0.00 4.25
10 muestras
2.63 2.62 4.25 0.00
ConversorA/D
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
2.00
3.00
4.00
5.00
20 Hz 1/Fs
“n” de la frecuenciacuyo aporte sequiere conocer
Fourier-2.63
n=510Hz
-2.62 -2.63 -4.25 0.00 4.25
10 muestras
2.63 2.62 4.25 0.00
ConversorA/D
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
2.00
3.00
4.00
5.00
20 Hz 1/Fs
Transformadade Fourier
Fourier-2.63
10Hz
-2.62 -2.63 -4.25 0.00 4.25
10 muestras
2.63 2.62 4.25 0.00
ConversorA/D
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
2.00
3.00
4.00
5.00
20 Hz 1/Fs
AFn
Aporte de la frecuencia que
se quiere conocer
Hagamos un ejemplo
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-5
-4
-3
-2
-1
0
1
2
3
4
5
# de muestra
Inte
nsi
da
d
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
-5
-4
-3
-2
-1
0
1
2
3
4
5
# de muestra
Inte
nsi
da
d
Esta señal se muestrearáa una frecuencia Fs = 20Hz
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-5
-4
-3
-2
-1
0
1
2
3
4
5
# de muestra
Inte
nsi
da
d
0 0.001 4.252 2.633 2.624 4.255 0.006 -4.257 -2.638 -2.629 -4.25
10 0.0011 4.2512 2.6313 2.6214 4.2515 0.00
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
-5
-4
-3
-2
-1
0
1
2
3
4
5
# de muestra
Inte
nsi
da
d
¿Qué frecuencias básicasdebemos analizar para
reconstruir la señalen la ventana?
La frecuencia de muestreo
debe ser al menos el doblede la frecuencia máximaque se requiere analizar.
F s≥2 Fmáx
La frecuencia de muestreo
debe ser al menos el doblede la frecuencia máximaque se requiere analizar.
la frecuencia máxima analizable
no puede ser mayor que la mitadde la frecuencia de muestreo
Fmáx≤F s
2
la frecuencia máxima analizable
no puede ser mayor que la mitadde la frecuencia de muestreo
Así que, paranuestro ejemplo...
Fmáx⩽20 Hz
2⩽10 Hz
Por tanto, los valores de “n”y las frecuenciasa analizar serán:
n=0⇒F0=010
20=0 Hz
n=0⇒F0=0
1020=0 Hz
n=1⇒F1=1
1020=2 Hz
n=0⇒F0=0
1020=0 Hz
n=1⇒F1=1
1020=2 Hz
n=2⇒F2=2
1020=4 Hz
n=0⇒F0=0
1020=0 Hz
n=1⇒F1=1
1020=2 Hz
n=2⇒F2=2
1020=4 Hz
n=3⇒F3=3
1020=6 Hz
n=0⇒F0=0
1020=0 Hz
n=1⇒F1=110
20=2 Hz
n=2⇒F2=2
1020=4 Hz
n=3⇒F3=310
20=6 Hz
n=4⇒F4=4
1020=8 Hz
n=0⇒F0=010
20=0 Hz
n=1⇒F1=1
1020=2Hz
n=2⇒F2=210
20=4 Hz
n=3⇒F3=3
1020=6 Hz
n=4⇒F4=410
20=8 Hz
n=5⇒F5=510
20=10 Hz
¿Cuánto aporta n=0, sen(0t),para formar la señal?
sen(0t)0 0.00001 0.00002 0.00003 0.00004 0.00005 0.00006 0.00007 0.00008 0.00009 0.0000
10 0.000011 0.000012 0.000013 0.000014 0.000015 0.000016 0.000017 0.000018 0.000019 0.000020 0.0000
t
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
El valor de sen(0t)en los 21 instantes
de muestreo
t sen(0t) m[t]0 0.0000 0.001 0.0000 4.252 0.0000 2.633 0.0000 2.624 0.0000 4.255 0.0000 0.006 0.0000 -4.257 0.0000 -2.638 0.0000 -2.629 0.0000 -4.25
10 0.0000 0.0011 0.0000 4.2512 0.0000 2.6313 0.0000 2.6214 0.0000 4.2515 0.0000 0.0016 0.0000 -4.2517 0.0000 -2.6318 0.0000 -2.6219 0.0000 -4.2520 0.0000 0.00
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
El valor de las21 muestras
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
¿Cómo podemoscomparar una señal
con la otra?
t sen(0t) m[t]0 0.0000 0.001 0.0000 4.252 0.0000 2.633 0.0000 2.624 0.0000 4.255 0.0000 0.006 0.0000 -4.257 0.0000 -2.638 0.0000 -2.629 0.0000 -4.25
10 0.0000 0.0011 0.0000 4.2512 0.0000 2.6313 0.0000 2.6214 0.0000 4.2515 0.0000 0.0016 0.0000 -4.2517 0.0000 -2.6318 0.0000 -2.6219 0.0000 -4.2520 0.0000 0.00
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
¿Y si multiplicamos ambas?El resultado nos dará
el área de un rectángulopor cada muestra
sen(0t) Producto0 0.0000 0.00 0.0001 0.0000 4.25 0.0002 0.0000 2.63 0.0003 0.0000 2.62 0.0004 0.0000 4.25 0.0005 0.0000 0.00 0.0006 0.0000 -4.25 0.0007 0.0000 -2.63 0.0008 0.0000 -2.62 0.0009 0.0000 -4.25 0.000
10 0.0000 0.00 0.00011 0.0000 4.25 0.00012 0.0000 2.63 0.00013 0.0000 2.62 0.00014 0.0000 4.25 0.00015 0.0000 0.00 0.00016 0.0000 -4.25 0.00017 0.0000 -2.63 0.00018 0.0000 -2.62 0.00019 0.0000 -4.25 0.00020 0.0000 0.00 0.000
t m[t]
sen(0t) Producto0 0.0000 0.00 0.0001 0.0000 4.25 0.0002 0.0000 2.63 0.0003 0.0000 2.62 0.0004 0.0000 4.25 0.0005 0.0000 0.00 0.0006 0.0000 -4.25 0.0007 0.0000 -2.63 0.0008 0.0000 -2.62 0.0009 0.0000 -4.25 0.000
10 0.0000 0.00 0.00011 0.0000 4.25 0.00012 0.0000 2.63 0.00013 0.0000 2.62 0.00014 0.0000 4.25 0.00015 0.0000 0.00 0.00016 0.0000 -4.25 0.00017 0.0000 -2.63 0.00018 0.0000 -2.62 0.00019 0.0000 -4.25 0.00020 0.0000 0.00 0.000
t m[t]
Si ambas se parecen,las áreas de los
rectángulos serán grandes.
Si no se parecen,serán pequeñas.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
sen(0t) Producto0 0.0000 0.00 0.0001 0.0000 4.25 0.0002 0.0000 2.63 0.0003 0.0000 2.62 0.0004 0.0000 4.25 0.0005 0.0000 0.00 0.0006 0.0000 -4.25 0.0007 0.0000 -2.63 0.0008 0.0000 -2.62 0.0009 0.0000 -4.25 0.000
10 0.0000 0.00 0.00011 0.0000 4.25 0.00012 0.0000 2.63 0.00013 0.0000 2.62 0.00014 0.0000 4.25 0.00015 0.0000 0.00 0.00016 0.0000 -4.25 0.00017 0.0000 -2.63 0.00018 0.0000 -2.62 0.00019 0.0000 -4.25 0.00020 0.0000 0.00 0.000
t m[t]
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Como las señalesno se parecen, las
áreas de los rectángulosson nulas.
t sen(0t) m[t] Producto0 0.0000 0.00 0.0001 0.0000 4.25 0.0002 0.0000 2.63 0.0003 0.0000 2.62 0.0004 0.0000 4.25 0.0005 0.0000 0.00 0.0006 0.0000 -4.25 0.0007 0.0000 -2.63 0.0008 0.0000 -2.62 0.0009 0.0000 -4.25 0.000
10 0.0000 0.00 0.00011 0.0000 4.25 0.00012 0.0000 2.63 0.00013 0.0000 2.62 0.00014 0.0000 4.25 0.00015 0.0000 0.00 0.00016 0.0000 -4.25 0.00017 0.0000 -2.63 0.00018 0.0000 -2.62 0.00019 0.0000 -4.25 0.00020 0.0000 0.00 0.000
Sumatoria: 0.000Normalizada: 0.000
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
La sumatoria de lasáreas de los
rectángulos equivalea 0 unidades.
t sen(0t) m[t] Producto0 0.0000 0.00 0.0001 0.0000 4.25 0.0002 0.0000 2.63 0.0003 0.0000 2.62 0.0004 0.0000 4.25 0.0005 0.0000 0.00 0.0006 0.0000 -4.25 0.0007 0.0000 -2.63 0.0008 0.0000 -2.62 0.0009 0.0000 -4.25 0.000
10 0.0000 0.00 0.00011 0.0000 4.25 0.00012 0.0000 2.63 0.00013 0.0000 2.62 0.00014 0.0000 4.25 0.00015 0.0000 0.00 0.00016 0.0000 -4.25 0.00017 0.0000 -2.63 0.00018 0.0000 -2.62 0.00019 0.0000 -4.25 0.00020 0.0000 0.00 0.000
Sumatoria: 0.000Normalizada: 0.000
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Luego sen(0t) no aporta nadaa la señal original.
¿Cuánto aporta n=1, sen(2t),para formar la señal?
sen(2t)0 0.00001 0.58782 0.95113 0.95114 0.58785 0.00006 -0.58787 -0.95118 -0.95119 -0.5878
10 0.000011 0.587812 0.951113 0.951114 0.587815 0.000016 -0.587817 -0.951118 -0.951119 -0.587820 0.0000
t
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
t sen(2t) m[t]0 0.0000 0.001 0.5878 4.252 0.9511 2.633 0.9511 2.624 0.5878 4.255 0.0000 0.006 -0.5878 -4.257 -0.9511 -2.638 -0.9511 -2.629 -0.5878 -4.25
10 0.0000 0.0011 0.5878 4.2512 0.9511 2.6313 0.9511 2.6214 0.5878 4.2515 0.0000 0.0016 -0.5878 -4.2517 -0.9511 -2.6318 -0.9511 -2.6219 -0.5878 -4.2520 0.0000 0.00
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
sen(2t) Producto0 0.0000 0.00 0.0001 0.5878 4.25 2.4982 0.9511 2.63 2.5013 0.9511 2.62 2.4924 0.5878 4.25 2.4985 0.0000 0.00 0.0006 -0.5878 -4.25 2.4987 -0.9511 -2.63 2.5018 -0.9511 -2.62 2.4929 -0.5878 -4.25 2.498
10 0.0000 0.00 0.00011 0.5878 4.25 2.49812 0.9511 2.63 2.50113 0.9511 2.62 2.49214 0.5878 4.25 2.49815 0.0000 0.00 0.00016 -0.5878 -4.25 2.49817 -0.9511 -2.63 2.50118 -0.9511 -2.62 2.49219 -0.5878 -4.25 2.49820 0.0000 0.00 0.000
t m[t]
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
t sen(2t) m[t] Producto0 0.0000 0.00 0.0001 0.5878 4.25 2.4982 0.9511 2.63 2.5013 0.9511 2.62 2.4924 0.5878 4.25 2.4985 0.0000 0.00 0.0006 -0.5878 -4.25 2.4987 -0.9511 -2.63 2.5018 -0.9511 -2.62 2.4929 -0.5878 -4.25 2.498
10 0.0000 0.00 0.00011 0.5878 4.25 2.49812 0.9511 2.63 2.50113 0.9511 2.62 2.49214 0.5878 4.25 2.49815 0.0000 0.00 0.00016 -0.5878 -4.25 2.49817 -0.9511 -2.63 2.50118 -0.9511 -2.62 2.49219 -0.5878 -4.25 2.49820 0.0000 0.00 0.000
Sumatoria: 39.957Normalizada: 1.998
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
La sumatoria de las áreas delos rectángulos equivale a
casi 40 unidades. Normalizada representa casi 2.
t sen(2t) m[t] Producto0 0.0000 0.00 0.0001 0.5878 4.25 2.4982 0.9511 2.63 2.5013 0.9511 2.62 2.4924 0.5878 4.25 2.4985 0.0000 0.00 0.0006 -0.5878 -4.25 2.4987 -0.9511 -2.63 2.5018 -0.9511 -2.62 2.4929 -0.5878 -4.25 2.498
10 0.0000 0.00 0.00011 0.5878 4.25 2.49812 0.9511 2.63 2.50113 0.9511 2.62 2.49214 0.5878 4.25 2.49815 0.0000 0.00 0.00016 -0.5878 -4.25 2.49817 -0.9511 -2.63 2.50118 -0.9511 -2.62 2.49219 -0.5878 -4.25 2.49820 0.0000 0.00 0.000
Sumatoria: 39.957Normalizada: 1.998
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
Luego sen(2t) aporta 2a la señal original.
¿Cuánto aporta n=2, sen(4t),para formar la señal?
sen(4t)0 0.00001 0.95112 0.58783 -0.58784 -0.95115 0.00006 0.95117 0.58788 -0.58789 -0.9511
10 0.000011 0.951112 0.587813 -0.587814 -0.951115 0.000016 0.951117 0.587818 -0.587819 -0.951120 0.0000
t
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
t sen(4t) m[t]0 0.0000 0.001 0.9511 4.252 0.5878 2.633 -0.5878 2.624 -0.9511 4.255 0.0000 0.006 0.9511 -4.257 0.5878 -2.638 -0.5878 -2.629 -0.9511 -4.25
10 0.0000 0.0011 0.9511 4.2512 0.5878 2.6313 -0.5878 2.6214 -0.9511 4.2515 0.0000 0.0016 0.9511 -4.2517 0.5878 -2.6318 -0.5878 -2.6219 -0.9511 -4.2520 0.0000 0.00
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
t sen(4t) m[t] Producto0 0.0000 0.00 0.0001 0.9511 4.25 4.0422 0.5878 2.63 1.5463 -0.5878 2.62 -1.5404 -0.9511 4.25 -4.0425 0.0000 0.00 0.0006 0.9511 -4.25 -4.0427 0.5878 -2.63 -1.5468 -0.5878 -2.62 1.5409 -0.9511 -4.25 4.042
10 0.0000 0.00 0.00011 0.9511 4.25 4.04212 0.5878 2.63 1.54613 -0.5878 2.62 -1.54014 -0.9511 4.25 -4.04215 0.0000 0.00 0.00016 0.9511 -4.25 -4.04217 0.5878 -2.63 -1.54618 -0.5878 -2.62 1.54019 -0.9511 -4.25 4.04220 0.0000 0.00 0.000
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
t sen(4t) m[t] Producto0 0.0000 0.00 0.0001 0.9511 4.25 4.0422 0.5878 2.63 1.5463 -0.5878 2.62 -1.5404 -0.9511 4.25 -4.0425 0.0000 0.00 0.0006 0.9511 -4.25 -4.0427 0.5878 -2.63 -1.5468 -0.5878 -2.62 1.5409 -0.9511 -4.25 4.042
10 0.0000 0.00 0.00011 0.9511 4.25 4.04212 0.5878 2.63 1.54613 -0.5878 2.62 -1.54014 -0.9511 4.25 -4.04215 0.0000 0.00 0.00016 0.9511 -4.25 -4.04217 0.5878 -2.63 -1.54618 -0.5878 -2.62 1.54019 -0.9511 -4.25 4.04220 0.0000 0.00 0.000
Sumatoria: 0.000Normalizada: 0.000
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
La sumatoria de las áreas delos rectángulos equivale a
0 unidades.
t sen(4t) m[t] Producto0 0.0000 0.00 0.0001 0.9511 4.25 4.0422 0.5878 2.63 1.5463 -0.5878 2.62 -1.5404 -0.9511 4.25 -4.0425 0.0000 0.00 0.0006 0.9511 -4.25 -4.0427 0.5878 -2.63 -1.5468 -0.5878 -2.62 1.5409 -0.9511 -4.25 4.042
10 0.0000 0.00 0.00011 0.9511 4.25 4.04212 0.5878 2.63 1.54613 -0.5878 2.62 -1.54014 -0.9511 4.25 -4.04215 0.0000 0.00 0.00016 0.9511 -4.25 -4.04217 0.5878 -2.63 -1.54618 -0.5878 -2.62 1.54019 -0.9511 -4.25 4.04220 0.0000 0.00 0.000
Sumatoria: 0.000Normalizada: 0.000
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
Luego sen(4t) no aporta nadaa la señal original.
¿Cuánto aporta n=3, sen(6t),para formar la señal?
sen(6t)0 0.00001 0.95112 -0.58783 -0.58784 0.95115 0.00006 -0.95117 0.58788 0.58789 -0.9511
10 0.000011 0.951112 -0.587813 -0.587814 0.951115 0.000016 -0.951117 0.587818 0.587819 -0.951120 0.0000
t
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
t sen(6t) m[t]0 0.0000 0.001 0.9511 4.252 -0.5878 2.633 -0.5878 2.624 0.9511 4.255 0.0000 0.006 -0.9511 -4.257 0.5878 -2.638 0.5878 -2.629 -0.9511 -4.25
10 0.0000 0.0011 0.9511 4.2512 -0.5878 2.6313 -0.5878 2.6214 0.9511 4.2515 0.0000 0.0016 -0.9511 -4.2517 0.5878 -2.6318 0.5878 -2.6219 -0.9511 -4.2520 0.0000 0.00
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
sen(6t) Producto0 0.0000 0.00 0.0001 0.9511 4.25 4.0422 -0.5878 2.63 -1.5463 -0.5878 2.62 -1.5404 0.9511 4.25 4.0425 0.0000 0.00 0.0006 -0.9511 -4.25 4.0427 0.5878 -2.63 -1.5468 0.5878 -2.62 -1.5409 -0.9511 -4.25 4.042
10 0.0000 0.00 0.00011 0.9511 4.25 4.04212 -0.5878 2.63 -1.54613 -0.5878 2.62 -1.54014 0.9511 4.25 4.04215 0.0000 0.00 0.00016 -0.9511 -4.25 4.04217 0.5878 -2.63 -1.54618 0.5878 -2.62 -1.54019 -0.9511 -4.25 4.04220 0.0000 0.00 0.000
t m[t]
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
t sen(6t) m[t] Producto0 0.0000 0.00 0.0001 0.9511 4.25 4.0422 -0.5878 2.63 -1.5463 -0.5878 2.62 -1.5404 0.9511 4.25 4.0425 0.0000 0.00 0.0006 -0.9511 -4.25 4.0427 0.5878 -2.63 -1.5468 0.5878 -2.62 -1.5409 -0.9511 -4.25 4.042
10 0.0000 0.00 0.00011 0.9511 4.25 4.04212 -0.5878 2.63 -1.54613 -0.5878 2.62 -1.54014 0.9511 4.25 4.04215 0.0000 0.00 0.00016 -0.9511 -4.25 4.04217 0.5878 -2.63 -1.54618 0.5878 -2.62 -1.54019 -0.9511 -4.25 4.04220 0.0000 0.00 0.000
Sumatoria: 19.992Normalizada: 1.000
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
La sumatoria de las áreas delos rectángulos equivale a
casi 20 unidades.Normalizada representa casi 1.
t sen(6t) m[t] Producto0 0.0000 0.00 0.0001 0.9511 4.25 4.0422 -0.5878 2.63 -1.5463 -0.5878 2.62 -1.5404 0.9511 4.25 4.0425 0.0000 0.00 0.0006 -0.9511 -4.25 4.0427 0.5878 -2.63 -1.5468 0.5878 -2.62 -1.5409 -0.9511 -4.25 4.042
10 0.0000 0.00 0.00011 0.9511 4.25 4.04212 -0.5878 2.63 -1.54613 -0.5878 2.62 -1.54014 0.9511 4.25 4.04215 0.0000 0.00 0.00016 -0.9511 -4.25 4.04217 0.5878 -2.63 -1.54618 0.5878 -2.62 -1.54019 -0.9511 -4.25 4.04220 0.0000 0.00 0.000
Sumatoria: 19.992Normalizada: 1.000
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
Luego sen(6t) aporta 1a la señal original.
¿Cuánto aporta n=4, sen(8t),para formar la señal?
sen(8t)0 0.00001 0.58782 -0.95113 0.95114 -0.58785 0.00006 0.58787 -0.95118 0.95119 -0.5878
10 0.000011 0.587812 -0.951113 0.951114 -0.587815 0.000016 0.587817 -0.951118 0.951119 -0.587820 0.0000
t
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
t sen(8t) m[t]0 0.0000 0.001 0.5878 4.252 -0.9511 2.633 0.9511 2.624 -0.5878 4.255 0.0000 0.006 0.5878 -4.257 -0.9511 -2.638 0.9511 -2.629 -0.5878 -4.25
10 0.0000 0.0011 0.5878 4.2512 -0.9511 2.6313 0.9511 2.6214 -0.5878 4.2515 0.0000 0.0016 0.5878 -4.2517 -0.9511 -2.6318 0.9511 -2.6219 -0.5878 -4.2520 0.0000 0.00
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
sen(8t) Producto0 0.0000 0.00 0.0001 0.5878 4.25 2.4982 -0.9511 2.63 -2.5013 0.9511 2.62 2.4924 -0.5878 4.25 -2.4985 0.0000 0.00 0.0006 0.5878 -4.25 -2.4987 -0.9511 -2.63 2.5018 0.9511 -2.62 -2.4929 -0.5878 -4.25 2.498
10 0.0000 0.00 0.00011 0.5878 4.25 2.49812 -0.9511 2.63 -2.50113 0.9511 2.62 2.49214 -0.5878 4.25 -2.49815 0.0000 0.00 0.00016 0.5878 -4.25 -2.49817 -0.9511 -2.63 2.50118 0.9511 -2.62 -2.49219 -0.5878 -4.25 2.49820 0.0000 0.00 0.000
t m[t]
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
t sen(8t) m[t] Producto0 0.0000 0.00 0.0001 0.5878 4.25 2.4982 -0.9511 2.63 -2.5013 0.9511 2.62 2.4924 -0.5878 4.25 -2.4985 0.0000 0.00 0.0006 0.5878 -4.25 -2.4987 -0.9511 -2.63 2.5018 0.9511 -2.62 -2.4929 -0.5878 -4.25 2.498
10 0.0000 0.00 0.00011 0.5878 4.25 2.49812 -0.9511 2.63 -2.50113 0.9511 2.62 2.49214 -0.5878 4.25 -2.49815 0.0000 0.00 0.00016 0.5878 -4.25 -2.49817 -0.9511 -2.63 2.50118 0.9511 -2.62 -2.49219 -0.5878 -4.25 2.49820 0.0000 0.00 0.000
Sumatoria: 0.000Normalizada: 0.000
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
La sumatoria de las áreas delos rectángulos equivale a
0 unidades.
t sen(8t) m[t] Producto0 0.0000 0.00 0.0001 0.5878 4.25 2.4982 -0.9511 2.63 -2.5013 0.9511 2.62 2.4924 -0.5878 4.25 -2.4985 0.0000 0.00 0.0006 0.5878 -4.25 -2.4987 -0.9511 -2.63 2.5018 0.9511 -2.62 -2.4929 -0.5878 -4.25 2.498
10 0.0000 0.00 0.00011 0.5878 4.25 2.49812 -0.9511 2.63 -2.50113 0.9511 2.62 2.49214 -0.5878 4.25 -2.49815 0.0000 0.00 0.00016 0.5878 -4.25 -2.49817 -0.9511 -2.63 2.50118 0.9511 -2.62 -2.49219 -0.5878 -4.25 2.49820 0.0000 0.00 0.000
Sumatoria: 0.000Normalizada: 0.000
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
Luego sen(8t)no aporta nadaa la señal original.
¿Cuánto aporta n=5, sen(10t),para formar la señal?
sen(10t)0 0.00001 0.00002 0.00003 0.00004 0.00005 0.00006 0.00007 0.00008 0.00009 0.0000
10 0.000011 0.000012 0.000013 0.000014 0.000015 0.000016 0.000017 0.000018 0.000019 0.000020 0.0000
t
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
t sen(10t) m[t]0 0.0000 0.001 0.0000 4.252 0.0000 2.633 0.0000 2.624 0.0000 4.255 0.0000 0.006 0.0000 -4.257 0.0000 -2.638 0.0000 -2.629 0.0000 -4.25
10 0.0000 0.0011 0.0000 4.2512 0.0000 2.6313 0.0000 2.6214 0.0000 4.2515 0.0000 0.0016 0.0000 -4.2517 0.0000 -2.6318 0.0000 -2.6219 0.0000 -4.2520 0.0000 0.00
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
sen(10t) Producto0 0.0000 0.00 0.0001 0.0000 4.25 0.0002 0.0000 2.63 0.0003 0.0000 2.62 0.0004 0.0000 4.25 0.0005 0.0000 0.00 0.0006 0.0000 -4.25 0.0007 0.0000 -2.63 0.0008 0.0000 -2.62 0.0009 0.0000 -4.25 0.000
10 0.0000 0.00 0.00011 0.0000 4.25 0.00012 0.0000 2.63 0.00013 0.0000 2.62 0.00014 0.0000 4.25 0.00015 0.0000 0.00 0.00016 0.0000 -4.25 0.00017 0.0000 -2.63 0.00018 0.0000 -2.62 0.00019 0.0000 -4.25 0.00020 0.0000 0.00 0.000
t m[t]
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5t sen(10t) m[t] Producto0 0.0000 0.00 0.0001 0.0000 4.25 0.0002 0.0000 2.63 0.0003 0.0000 2.62 0.0004 0.0000 4.25 0.0005 0.0000 0.00 0.0006 0.0000 -4.25 0.0007 0.0000 -2.63 0.0008 0.0000 -2.62 0.0009 0.0000 -4.25 0.000
10 0.0000 0.00 0.00011 0.0000 4.25 0.00012 0.0000 2.63 0.00013 0.0000 2.62 0.00014 0.0000 4.25 0.00015 0.0000 0.00 0.00016 0.0000 -4.25 0.00017 0.0000 -2.63 0.00018 0.0000 -2.62 0.00019 0.0000 -4.25 0.00020 0.0000 0.00 0.000
Sumatoria: 0.000Normalizada: 0.000
La sumatoria de las áreas delos rectángulos equivale a
0 unidades.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-5
-4
-3
-2
-1
0
1
2
3
4
5t sen(10t) m[t] Producto0 0.0000 0.00 0.0001 0.0000 4.25 0.0002 0.0000 2.63 0.0003 0.0000 2.62 0.0004 0.0000 4.25 0.0005 0.0000 0.00 0.0006 0.0000 -4.25 0.0007 0.0000 -2.63 0.0008 0.0000 -2.62 0.0009 0.0000 -4.25 0.000
10 0.0000 0.00 0.00011 0.0000 4.25 0.00012 0.0000 2.63 0.00013 0.0000 2.62 0.00014 0.0000 4.25 0.00015 0.0000 0.00 0.00016 0.0000 -4.25 0.00017 0.0000 -2.63 0.00018 0.0000 -2.62 0.00019 0.0000 -4.25 0.00020 0.0000 0.00 0.000
Sumatoria: 0.000Normalizada: 0.000
Luego sen(10t) no aporta nadaa la señal original.
¡Son demasiados cálculos!
Se necesita crear un método abreviado
que requiera menos esfuerzo.
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
La Transformadade Fourier...
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
-2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00
...de una ventana “m”de muestras...
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
n=12 Hz ...para la “n” de la
frecuencia cuyo aporte se quiere conocer...
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
10 muestras ...con un tamañode ventana de
“N” muestras...-2.63 -2.62 -2.63 -4.25 0.00 4.25 2.63 2.62 4.25 0.00
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
1/Fs...y cuyo periodo de
muestreo fue “T”...
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
...se calcula como...
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
¡Difícil de entender!
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
?
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
?
Para facilitar el cálculo...
aplicando micélebre fórmula...
exj=cos(x)+ sen(x) j
exj=cos(x)+ sen(x) j
exj=cos(x)+ sen(x) j
mi Transformadaqueda como...
exj=cos(x)+ sen(x) j
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
-5
-4
-3
-2
-1
0
1
2
3
4
5
# de muestra
Inte
nsi
da
d
Muestra Valor0 0.001 4.252 2.633 2.624 4.255 0.006 -4.257 -2.638 -2.629 -4.25
10 0.0011 4.2512 2.6313 2.6214 4.2515 0.0016 -4.2517 -2.63
En vez de operar sobretodo el conjunto de datos,elegiremos una ventana “m” de tamaño “N”
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
-5
-4
-3
-2
-1
0
1
2
3
4
5
# de muestra
Inte
nsi
da
d
Ventanade análisis
N = 10 muestrasm[0..9]
Muestra Valor0 0.001 4.252 2.633 2.624 4.255 0.006 -4.257 -2.638 -2.629 -4.25
10 0.0011 4.2512 2.6313 2.6214 4.2515 0.0016 -4.2517 -2.63
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
-5
-4
-3
-2
-1
0
1
2
3
4
5
# de muestra
Inte
nsi
da
d
la ventanase recorre a
m[1..10]
Muestra Valor0 0.001 4.252 2.633 2.624 4.255 0.006 -4.257 -2.638 -2.629 -4.25
10 0.0011 4.2512 2.6313 2.6214 4.2515 0.0016 -4.2517 -2.63
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
-5
-4
-3
-2
-1
0
1
2
3
4
5
# de muestra
Inte
nsi
da
d
m[2..11]
Muestra Valor0 0.001 4.252 2.633 2.624 4.255 0.006 -4.257 -2.638 -2.629 -4.25
10 0.0011 4.2512 2.6313 2.6214 4.2515 0.0016 -4.2517 -2.63
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
-5
-4
-3
-2
-1
0
1
2
3
4
5
# de muestra
Inte
nsi
da
d
m[3..12]
Muestra Valor0 0.001 4.252 2.633 2.624 4.255 0.006 -4.257 -2.638 -2.629 -4.25
10 0.0011 4.2512 2.6313 2.6214 4.2515 0.0016 -4.2517 -2.63
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
-5
-4
-3
-2
-1
0
1
2
3
4
5
# de muestra
Inte
nsi
da
d
m[4..13]
Muestra Valor0 0.001 4.252 2.633 2.624 4.255 0.006 -4.257 -2.638 -2.629 -4.25
10 0.0011 4.2512 2.6313 2.6214 4.2515 0.0016 -4.2517 -2.63
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
-5
-4
-3
-2
-1
0
1
2
3
4
5
# de muestra
Inte
nsi
da
d
m[5..14]
Muestra Valor0 0.001 4.252 2.633 2.624 4.255 0.006 -4.257 -2.638 -2.629 -4.25
10 0.0011 4.2512 2.6313 2.6214 4.2515 0.0016 -4.2517 -2.63
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
-5
-4
-3
-2
-1
0
1
2
3
4
5
# de muestra
Inte
nsi
da
d
m[6..15]
Muestra Valor0 0.001 4.252 2.633 2.624 4.255 0.006 -4.257 -2.638 -2.629 -4.25
10 0.0011 4.2512 2.6313 2.6214 4.2515 0.0016 -4.2517 -2.63
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
-5
-4
-3
-2
-1
0
1
2
3
4
5
# de muestra
Inte
nsi
da
d
usaremos éstacomo ejemplo:
m[7..16]
Muestra Valor0 0.001 4.252 2.633 2.624 4.255 0.006 -4.257 -2.638 -2.629 -4.25
10 0.0011 4.2512 2.6313 2.6214 4.2515 0.0016 -4.2517 -2.63
¿Cuánto aporta n=0, sen(0t),para formar la señal en la ventana?
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 0N = 10T = 0.05
n = 0N = 10T = 0.05
0123456789
k
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 0N = 10T = 0.05
0 0.00001 0.00002 0.00003 0.00004 0.00005 0.00006 0.00007 0.00008 0.00009 0.0000
k sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 0N = 10T = 0.05
0 1.0000 0.00001 1.0000 0.00002 1.0000 0.00003 1.0000 0.00004 1.0000 0.00005 1.0000 0.00006 1.0000 0.00007 1.0000 0.00008 1.0000 0.00009 1.0000 0.0000
k cos(-2πnk/N) sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 0N = 10T = 0.05
0 -2.63 1.0000 0.00001 -2.62 1.0000 0.00002 -4.25 1.0000 0.00003 0.00 1.0000 0.00004 4.25 1.0000 0.00005 2.63 1.0000 0.00006 2.62 1.0000 0.00007 4.25 1.0000 0.00008 0.00 1.0000 0.00009 -4.25 1.0000 0.0000
k m[kT] cos(-2πnk/N) sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 0N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 1.0000 -2.6200 0.0000 0.00002 -4.25 1.0000 -4.2500 0.0000 0.00003 0.00 1.0000 0.0000 0.0000 0.00004 4.25 1.0000 4.2500 0.0000 0.00005 2.63 1.0000 2.6300 0.0000 0.00006 2.62 1.0000 2.6200 0.0000 0.00007 4.25 1.0000 4.2500 0.0000 0.00008 0.00 1.0000 0.0000 0.0000 0.00009 -4.25 1.0000 -4.2500 0.0000 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 0N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 1.0000 -2.6200 0.0000 0.00002 -4.25 1.0000 -4.2500 0.0000 0.00003 0.00 1.0000 0.0000 0.0000 0.00004 4.25 1.0000 4.2500 0.0000 0.00005 2.63 1.0000 2.6300 0.0000 0.00006 2.62 1.0000 2.6200 0.0000 0.00007 4.25 1.0000 4.2500 0.0000 0.00008 0.00 1.0000 0.0000 0.0000 0.00009 -4.25 1.0000 -4.2500 0.0000 0.0000
Suma: 0.0000 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 0N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 1.0000 -2.6200 0.0000 0.00002 -4.25 1.0000 -4.2500 0.0000 0.00003 0.00 1.0000 0.0000 0.0000 0.00004 4.25 1.0000 4.2500 0.0000 0.00005 2.63 1.0000 2.6300 0.0000 0.00006 2.62 1.0000 2.6200 0.0000 0.00007 4.25 1.0000 4.2500 0.0000 0.00008 0.00 1.0000 0.0000 0.0000 0.00009 -4.25 1.0000 -4.2500 0.0000 0.0000
Suma: 0.0000 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
√02+ 02=0
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 0N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 1.0000 -2.6200 0.0000 0.00002 -4.25 1.0000 -4.2500 0.0000 0.00003 0.00 1.0000 0.0000 0.0000 0.00004 4.25 1.0000 4.2500 0.0000 0.00005 2.63 1.0000 2.6300 0.0000 0.00006 2.62 1.0000 2.6200 0.0000 0.00007 4.25 1.0000 4.2500 0.0000 0.00008 0.00 1.0000 0.0000 0.0000 0.00009 -4.25 1.0000 -4.2500 0.0000 0.0000
Suma: 0.0000 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
110
.0=0
√02+ 02=0
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 0N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 1.0000 -2.6200 0.0000 0.00002 -4.25 1.0000 -4.2500 0.0000 0.00003 0.00 1.0000 0.0000 0.0000 0.00004 4.25 1.0000 4.2500 0.0000 0.00005 2.63 1.0000 2.6300 0.0000 0.00006 2.62 1.0000 2.6200 0.0000 0.00007 4.25 1.0000 4.2500 0.0000 0.00008 0.00 1.0000 0.0000 0.0000 0.00009 -4.25 1.0000 -4.2500 0.0000 0.0000
Suma: 0.0000 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
110
.0=0la señal sen(0t)
aportó 0 de amplituda la señal analizada
√02+ 02=0
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
¿Cuánto aporta n=1, sen(2t),para formar la señal en la ventana?
n = 1N = 10T = 0.05
0123456789
k
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 1N = 10T = 0.05
0 0.00001 -0.58782 -0.95113 -0.95114 -0.58785 0.00006 0.58787 0.95118 0.95119 0.5878
k sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 1N = 10T = 0.05
0 1.0000 0.00001 0.8090 -0.58782 0.3090 -0.95113 -0.3090 -0.95114 -0.8090 -0.58785 -1.0000 0.00006 -0.8090 0.58787 -0.3090 0.95118 0.3090 0.95119 0.8090 0.5878
k cos(-2πnk/N) sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 1N = 10T = 0.05
0 -2.63 1.0000 0.00001 -2.62 0.8090 -0.58782 -4.25 0.3090 -0.95113 0.00 -0.3090 -0.95114 4.25 -0.8090 -0.58785 2.63 -1.0000 0.00006 2.62 -0.8090 0.58787 4.25 -0.3090 0.95118 0.00 0.3090 0.95119 -4.25 0.8090 0.5878
k m[kT] cos(-2πnk/N) sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 1N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 0.8090 -2.1196 -0.5878 1.54002 -4.25 0.3090 -1.3133 -0.9511 4.04203 0.00 -0.3090 0.0000 -0.9511 0.00004 4.25 -0.8090 -3.4383 -0.5878 -2.49815 2.63 -1.0000 -2.6300 0.0000 0.00006 2.62 -0.8090 -2.1196 0.5878 1.54007 4.25 -0.3090 -1.3133 0.9511 4.04208 0.00 0.3090 0.0000 0.9511 0.00009 -4.25 0.8090 -3.4383 0.5878 -2.4981
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 1N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 0.8090 -2.1196 -0.5878 1.54002 -4.25 0.3090 -1.3133 -0.9511 4.04203 0.00 -0.3090 0.0000 -0.9511 0.00004 4.25 -0.8090 -3.4383 -0.5878 -2.49815 2.63 -1.0000 -2.6300 0.0000 0.00006 2.62 -0.8090 -2.1196 0.5878 1.54007 4.25 -0.3090 -1.3133 0.9511 4.04208 0.00 0.3090 0.0000 0.9511 0.00009 -4.25 0.8090 -3.4383 0.5878 -2.4981
Suma: -19.0025 6.1678
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 1N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 0.8090 -2.1196 -0.5878 1.54002 -4.25 0.3090 -1.3133 -0.9511 4.04203 0.00 -0.3090 0.0000 -0.9511 0.00004 4.25 -0.8090 -3.4383 -0.5878 -2.49815 2.63 -1.0000 -2.6300 0.0000 0.00006 2.62 -0.8090 -2.1196 0.5878 1.54007 4.25 -0.3090 -1.3133 0.9511 4.04208 0.00 0.3090 0.0000 0.9511 0.00009 -4.25 0.8090 -3.4383 0.5878 -2.4981
Suma: -19.0025 6.1678
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
√−19.00252+ 6.16782≈20
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 1N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 0.8090 -2.1196 -0.5878 1.54002 -4.25 0.3090 -1.3133 -0.9511 4.04203 0.00 -0.3090 0.0000 -0.9511 0.00004 4.25 -0.8090 -3.4383 -0.5878 -2.49815 2.63 -1.0000 -2.6300 0.0000 0.00006 2.62 -0.8090 -2.1196 0.5878 1.54007 4.25 -0.3090 -1.3133 0.9511 4.04208 0.00 0.3090 0.0000 0.9511 0.00009 -4.25 0.8090 -3.4383 0.5878 -2.4981
Suma: -19.0025 6.1678
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
√−19.00252+ 6.16782≈20
110
.20=2
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 1N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 0.8090 -2.1196 -0.5878 1.54002 -4.25 0.3090 -1.3133 -0.9511 4.04203 0.00 -0.3090 0.0000 -0.9511 0.00004 4.25 -0.8090 -3.4383 -0.5878 -2.49815 2.63 -1.0000 -2.6300 0.0000 0.00006 2.62 -0.8090 -2.1196 0.5878 1.54007 4.25 -0.3090 -1.3133 0.9511 4.04208 0.00 0.3090 0.0000 0.9511 0.00009 -4.25 0.8090 -3.4383 0.5878 -2.4981
Suma: -19.0025 6.1678
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
√−19.00252+ 6.16782≈20
110
.20=2la señal sen(2t)
aportó 2 de amplituda la señal analizada
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
¿Cuánto aporta n=2, sen(4t),para formar la señal en la ventana?
n = 2N = 10T = 0.05
0123456789
k
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 2N = 10T = 0.05
0 0.00001 -0.95112 -0.58783 0.58784 0.95115 0.00006 -0.95117 -0.58788 0.58789 0.9511
k sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 2N = 10T = 0.05
0 1.0000 0.00001 0.3090 -0.95112 -0.8090 -0.58783 -0.8090 0.58784 0.3090 0.95115 1.0000 0.00006 0.3090 -0.95117 -0.8090 -0.58788 -0.8090 0.58789 0.3090 0.9511
k cos(-2πnk/N) sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 2N = 10T = 0.05
0 -2.63 1.0000 0.00001 -2.62 0.3090 -0.95112 -4.25 -0.8090 -0.58783 0.00 -0.8090 0.58784 4.25 0.3090 0.95115 2.63 1.0000 0.00006 2.62 0.3090 -0.95117 4.25 -0.8090 -0.58788 0.00 -0.8090 0.58789 -4.25 0.3090 0.9511
k m[kT] cos(-2πnk/N) sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 2N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 0.3090 -0.8096 -0.9511 2.49182 -4.25 -0.8090 3.4383 -0.5878 2.49813 0.00 -0.8090 0.0000 0.5878 0.00004 4.25 0.3090 1.3133 0.9511 4.04205 2.63 1.0000 2.6300 0.0000 0.00006 2.62 0.3090 0.8096 -0.9511 -2.49187 4.25 -0.8090 -3.4383 -0.5878 -2.49818 0.00 -0.8090 0.0000 0.5878 0.00009 -4.25 0.3090 -1.3133 0.9511 -4.0420
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 2N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 0.3090 -0.8096 -0.9511 2.49182 -4.25 -0.8090 3.4383 -0.5878 2.49813 0.00 -0.8090 0.0000 0.5878 0.00004 4.25 0.3090 1.3133 0.9511 4.04205 2.63 1.0000 2.6300 0.0000 0.00006 2.62 0.3090 0.8096 -0.9511 -2.49187 4.25 -0.8090 -3.4383 -0.5878 -2.49818 0.00 -0.8090 0.0000 0.5878 0.00009 -4.25 0.3090 -1.3133 0.9511 -4.0420
Suma: 0.0000 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 2N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 0.3090 -0.8096 -0.9511 2.49182 -4.25 -0.8090 3.4383 -0.5878 2.49813 0.00 -0.8090 0.0000 0.5878 0.00004 4.25 0.3090 1.3133 0.9511 4.04205 2.63 1.0000 2.6300 0.0000 0.00006 2.62 0.3090 0.8096 -0.9511 -2.49187 4.25 -0.8090 -3.4383 -0.5878 -2.49818 0.00 -0.8090 0.0000 0.5878 0.00009 -4.25 0.3090 -1.3133 0.9511 -4.0420
Suma: 0.0000 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
√02+ 02=0
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 2N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 0.3090 -0.8096 -0.9511 2.49182 -4.25 -0.8090 3.4383 -0.5878 2.49813 0.00 -0.8090 0.0000 0.5878 0.00004 4.25 0.3090 1.3133 0.9511 4.04205 2.63 1.0000 2.6300 0.0000 0.00006 2.62 0.3090 0.8096 -0.9511 -2.49187 4.25 -0.8090 -3.4383 -0.5878 -2.49818 0.00 -0.8090 0.0000 0.5878 0.00009 -4.25 0.3090 -1.3133 0.9511 -4.0420
Suma: 0.0000 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
√02+ 02=0
110
.0=0
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 2N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 0.3090 -0.8096 -0.9511 2.49182 -4.25 -0.8090 3.4383 -0.5878 2.49813 0.00 -0.8090 0.0000 0.5878 0.00004 4.25 0.3090 1.3133 0.9511 4.04205 2.63 1.0000 2.6300 0.0000 0.00006 2.62 0.3090 0.8096 -0.9511 -2.49187 4.25 -0.8090 -3.4383 -0.5878 -2.49818 0.00 -0.8090 0.0000 0.5878 0.00009 -4.25 0.3090 -1.3133 0.9511 -4.0420
Suma: 0.0000 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
√02+ 02=0
110
.0=0la señal sen(4t)
aportó 0 de amplituda la señal analizada
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
¿Cuánto aporta n=3, sen(6t),para formar la señal en la ventana?
n = 3N = 10T = 0.05
0123456789
k
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 3N = 10T = 0.05
0 0.00001 -0.95112 0.58783 0.58784 -0.95115 0.00006 0.95117 -0.58788 -0.58789 0.9511
k sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 3N = 10T = 0.05
0 1.0000 0.00001 -0.3090 -0.95112 -0.8090 0.58783 0.8090 0.58784 0.3090 -0.95115 -1.0000 0.00006 0.3090 0.95117 0.8090 -0.58788 -0.8090 -0.58789 -0.3090 0.9511
k cos(-2πnk/N) sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 3N = 10T = 0.05
0 -2.63 1.0000 0.00001 -2.62 -0.3090 -0.95112 -4.25 -0.8090 0.58783 0.00 0.8090 0.58784 4.25 0.3090 -0.95115 2.63 -1.0000 0.00006 2.62 0.3090 0.95117 4.25 0.8090 -0.58788 0.00 -0.8090 -0.58789 -4.25 -0.3090 0.9511
k m[kT] cos(-2πnk/N) sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 3N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 -0.3090 0.8096 -0.9511 2.49182 -4.25 -0.8090 3.4383 0.5878 -2.49813 0.00 0.8090 0.0000 0.5878 0.00004 4.25 0.3090 1.3133 -0.9511 -4.04205 2.63 -1.0000 -2.6300 0.0000 0.00006 2.62 0.3090 0.8096 0.9511 2.49187 4.25 0.8090 3.4383 -0.5878 -2.49818 0.00 -0.8090 0.0000 -0.5878 0.00009 -4.25 -0.3090 1.3133 0.9511 -4.0420
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 3N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 -0.3090 0.8096 -0.9511 2.49182 -4.25 -0.8090 3.4383 0.5878 -2.49813 0.00 0.8090 0.0000 0.5878 0.00004 4.25 0.3090 1.3133 -0.9511 -4.04205 2.63 -1.0000 -2.6300 0.0000 0.00006 2.62 0.3090 0.8096 0.9511 2.49187 4.25 0.8090 3.4383 -0.5878 -2.49818 0.00 -0.8090 0.0000 -0.5878 0.00009 -4.25 -0.3090 1.3133 0.9511 -4.0420
Suma: 5.8625 -8.0966
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 3N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 -0.3090 0.8096 -0.9511 2.49182 -4.25 -0.8090 3.4383 0.5878 -2.49813 0.00 0.8090 0.0000 0.5878 0.00004 4.25 0.3090 1.3133 -0.9511 -4.04205 2.63 -1.0000 -2.6300 0.0000 0.00006 2.62 0.3090 0.8096 0.9511 2.49187 4.25 0.8090 3.4383 -0.5878 -2.49818 0.00 -0.8090 0.0000 -0.5878 0.00009 -4.25 -0.3090 1.3133 0.9511 -4.0420
Suma: 5.8625 -8.0966
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
√5.86252+ (−8.0966)
2≈10
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 3N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 -0.3090 0.8096 -0.9511 2.49182 -4.25 -0.8090 3.4383 0.5878 -2.49813 0.00 0.8090 0.0000 0.5878 0.00004 4.25 0.3090 1.3133 -0.9511 -4.04205 2.63 -1.0000 -2.6300 0.0000 0.00006 2.62 0.3090 0.8096 0.9511 2.49187 4.25 0.8090 3.4383 -0.5878 -2.49818 0.00 -0.8090 0.0000 -0.5878 0.00009 -4.25 -0.3090 1.3133 0.9511 -4.0420
Suma: 5.8625 -8.0966
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
√5.86252+ (−8.0966)
2≈10
110
.10=1
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 3N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 -0.3090 0.8096 -0.9511 2.49182 -4.25 -0.8090 3.4383 0.5878 -2.49813 0.00 0.8090 0.0000 0.5878 0.00004 4.25 0.3090 1.3133 -0.9511 -4.04205 2.63 -1.0000 -2.6300 0.0000 0.00006 2.62 0.3090 0.8096 0.9511 2.49187 4.25 0.8090 3.4383 -0.5878 -2.49818 0.00 -0.8090 0.0000 -0.5878 0.00009 -4.25 -0.3090 1.3133 0.9511 -4.0420
Suma: 5.8625 -8.0966
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
√5.86252+ (−8.0966)
2≈10
110
.10=1la señal sen(6t)
aportó 1 de amplituda la señal analizada
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
¿Cuánto aporta n=4, sen(8t),para formar la señal en la ventana?
n = 4N = 10T = 0.05
0123456789
k
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 4N = 10T = 0.05
0 0.00001 -0.58782 0.95113 -0.95114 0.58785 0.00006 -0.58787 0.95118 -0.95119 0.5878
k sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 4N = 10T = 0.05
0 1.0000 0.00001 -0.8090 -0.58782 0.3090 0.95113 0.3090 -0.95114 -0.8090 0.58785 1.0000 0.00006 -0.8090 -0.58787 0.3090 0.95118 0.3090 -0.95119 -0.8090 0.5878
k cos(-2πnk/N) sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 4N = 10T = 0.05
0 -2.63 1.0000 0.00001 -2.62 -0.8090 -0.58782 -4.25 0.3090 0.95113 0.00 0.3090 -0.95114 4.25 -0.8090 0.58785 2.63 1.0000 0.00006 2.62 -0.8090 -0.58787 4.25 0.3090 0.95118 0.00 0.3090 -0.95119 -4.25 -0.8090 0.5878
k m[kT] cos(-2πnk/N) sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 4N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 -0.8090 2.1196 -0.5878 1.54002 -4.25 0.3090 -1.3133 0.9511 -4.04203 0.00 0.3090 0.0000 -0.9511 0.00004 4.25 -0.8090 -3.4383 0.5878 2.49815 2.63 1.0000 2.6300 0.0000 0.00006 2.62 -0.8090 -2.1196 -0.5878 -1.54007 4.25 0.3090 1.3133 0.9511 4.04208 0.00 0.3090 0.0000 -0.9511 0.00009 -4.25 -0.8090 3.4383 0.5878 -2.4981
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 4N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 -0.8090 2.1196 -0.5878 1.54002 -4.25 0.3090 -1.3133 0.9511 -4.04203 0.00 0.3090 0.0000 -0.9511 0.00004 4.25 -0.8090 -3.4383 0.5878 2.49815 2.63 1.0000 2.6300 0.0000 0.00006 2.62 -0.8090 -2.1196 -0.5878 -1.54007 4.25 0.3090 1.3133 0.9511 4.04208 0.00 0.3090 0.0000 -0.9511 0.00009 -4.25 -0.8090 3.4383 0.5878 -2.4981
Suma: 0.0000 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 4N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 -0.8090 2.1196 -0.5878 1.54002 -4.25 0.3090 -1.3133 0.9511 -4.04203 0.00 0.3090 0.0000 -0.9511 0.00004 4.25 -0.8090 -3.4383 0.5878 2.49815 2.63 1.0000 2.6300 0.0000 0.00006 2.62 -0.8090 -2.1196 -0.5878 -1.54007 4.25 0.3090 1.3133 0.9511 4.04208 0.00 0.3090 0.0000 -0.9511 0.00009 -4.25 -0.8090 3.4383 0.5878 -2.4981
Suma: 0.0000 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
√02+ 02=0
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 4N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 -0.8090 2.1196 -0.5878 1.54002 -4.25 0.3090 -1.3133 0.9511 -4.04203 0.00 0.3090 0.0000 -0.9511 0.00004 4.25 -0.8090 -3.4383 0.5878 2.49815 2.63 1.0000 2.6300 0.0000 0.00006 2.62 -0.8090 -2.1196 -0.5878 -1.54007 4.25 0.3090 1.3133 0.9511 4.04208 0.00 0.3090 0.0000 -0.9511 0.00009 -4.25 -0.8090 3.4383 0.5878 -2.4981
Suma: 0.0000 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
110
.0=0
√02+ 02=0
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 4N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 -0.8090 2.1196 -0.5878 1.54002 -4.25 0.3090 -1.3133 0.9511 -4.04203 0.00 0.3090 0.0000 -0.9511 0.00004 4.25 -0.8090 -3.4383 0.5878 2.49815 2.63 1.0000 2.6300 0.0000 0.00006 2.62 -0.8090 -2.1196 -0.5878 -1.54007 4.25 0.3090 1.3133 0.9511 4.04208 0.00 0.3090 0.0000 -0.9511 0.00009 -4.25 -0.8090 3.4383 0.5878 -2.4981
Suma: 0.0000 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
110
.0=0la señal sen(8t)
aportó 0 de amplituda la señal analizada
√02+ 02=0
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
¿Cuánto aporta n=5, sen(10t),para formar la señal en la ventana?
n = 5N = 10T = 0.05
0123456789
k
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 5N = 10T = 0.05
0 0.00001 0.00002 0.00003 0.00004 0.00005 0.00006 0.00007 0.00008 0.00009 0.0000
k sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 5N = 10T = 0.05
0 1.0000 0.00001 -1.0000 0.00002 1.0000 0.00003 -1.0000 0.00004 1.0000 0.00005 -1.0000 0.00006 1.0000 0.00007 -1.0000 0.00008 1.0000 0.00009 -1.0000 0.0000
k cos(-2πnk/N) sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 5N = 10T = 0.05
0 -2.63 1.0000 0.00001 -2.62 -1.0000 0.00002 -4.25 1.0000 0.00003 0.00 -1.0000 0.00004 4.25 1.0000 0.00005 2.63 -1.0000 0.00006 2.62 1.0000 0.00007 4.25 -1.0000 0.00008 0.00 1.0000 0.00009 -4.25 -1.0000 0.0000
k m[kT] cos(-2πnk/N) sen(-2πnk/N)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 5N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 -1.0000 2.6200 0.0000 0.00002 -4.25 1.0000 -4.2500 0.0000 0.00003 0.00 -1.0000 0.0000 0.0000 0.00004 4.25 1.0000 4.2500 0.0000 0.00005 2.63 -1.0000 -2.6300 0.0000 0.00006 2.62 1.0000 2.6200 0.0000 0.00007 4.25 -1.0000 -4.2500 0.0000 0.00008 0.00 1.0000 0.0000 0.0000 0.00009 -4.25 -1.0000 4.2500 0.0000 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 5N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 -1.0000 2.6200 0.0000 0.00002 -4.25 1.0000 -4.2500 0.0000 0.00003 0.00 -1.0000 0.0000 0.0000 0.00004 4.25 1.0000 4.2500 0.0000 0.00005 2.63 -1.0000 -2.6300 0.0000 0.00006 2.62 1.0000 2.6200 0.0000 0.00007 4.25 -1.0000 -4.2500 0.0000 0.00008 0.00 1.0000 0.0000 0.0000 0.00009 -4.25 -1.0000 4.2500 0.0000 0.0000
Suma: -0.0200 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 5N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 -1.0000 2.6200 0.0000 0.00002 -4.25 1.0000 -4.2500 0.0000 0.00003 0.00 -1.0000 0.0000 0.0000 0.00004 4.25 1.0000 4.2500 0.0000 0.00005 2.63 -1.0000 -2.6300 0.0000 0.00006 2.62 1.0000 2.6200 0.0000 0.00007 4.25 -1.0000 -4.2500 0.0000 0.00008 0.00 1.0000 0.0000 0.0000 0.00009 -4.25 -1.0000 4.2500 0.0000 0.0000
Suma: -0.0200 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
√−0.02002+ 0.00002≈0
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 5N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 -1.0000 2.6200 0.0000 0.00002 -4.25 1.0000 -4.2500 0.0000 0.00003 0.00 -1.0000 0.0000 0.0000 0.00004 4.25 1.0000 4.2500 0.0000 0.00005 2.63 -1.0000 -2.6300 0.0000 0.00006 2.62 1.0000 2.6200 0.0000 0.00007 4.25 -1.0000 -4.2500 0.0000 0.00008 0.00 1.0000 0.0000 0.0000 0.00009 -4.25 -1.0000 4.2500 0.0000 0.0000
Suma: -0.0200 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
110
.0=0
√−0.02002+ 0.00002≈0
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
n = 5N = 10T = 0.05
0 -2.63 1.0000 -2.6300 0.0000 0.00001 -2.62 -1.0000 2.6200 0.0000 0.00002 -4.25 1.0000 -4.2500 0.0000 0.00003 0.00 -1.0000 0.0000 0.0000 0.00004 4.25 1.0000 4.2500 0.0000 0.00005 2.63 -1.0000 -2.6300 0.0000 0.00006 2.62 1.0000 2.6200 0.0000 0.00007 4.25 -1.0000 -4.2500 0.0000 0.00008 0.00 1.0000 0.0000 0.0000 0.00009 -4.25 -1.0000 4.2500 0.0000 0.0000
Suma: -0.0200 0.0000
k m[kT] cos(-2πnk/N) m[kT]*cos(...) sen(-2πnk/N) m[kT]*sen(...)
110
.0=0la señal sen(10t)
aportó 0 de amplituda la señal analizada
√−0.02002+ 0.00002≈0
1N∑k=0
N−1
m [kT ](cos(−2π nkN
)+ sen(−2π nkN
) j)
Reuniendo todas las componentespodemos expresar analíticamente que...
Señal (t )=2.sen(2t)+ 1.sen (6t)
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
Si representamos los resultadosen un Ecualizador Gráfico...
y = 2*sen(2t)
y = 1*sen(6t)
Pero, ¿cómo y por quéfunciona la Transformada de Fourier?
Fourier (m,n ,N ,T )=1N ∑
k=0
N−1
m [kT ]e−2πn k
Nj
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