Exam (elaborations)
Ejercicios sonido e imagen
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Recopilación de ejercicios de clase y exámenes de sonido e imagen
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,Tema 3:Ejemplo 3.1 :.
Hallamos DBW :
Wo-_1W-bWz1Wo-_2-DWz11W-_2W-DWz-_2W@10.
(¥ ) ) = 10 -
log
log ( 4¥ = 3,010 DBW =3 DBW
NOTA :
si no
ponemos las unidades solo 3dB un nivel
ya
no es
,
.
,
Ejemplo 3.2 : .
×
"
Da 90dB
referencia
-
Wo -10 W
porque se usa como
Laelolog (¥ ) -10 log (0%4%7)=90 dBr
.
-
Para pasarlo a DBW , tomamos como
referencia Wo -1W
Lw -10
log ( Q%w -30413¥ =
Para pasarlo a dBm ,
tomamos como
referencia Wo = 1mW
Lw -10
log ( 0,91¥) -0dB# -
, Ejemplo 3.3 .
:
calculamos Wr Wi
y
"" "
www.logf# 4¥ log 1¥ ) Equo
-
-
.
Wp = Wo 10L " ""
W, 1mW 1010dB ml? 10mW
¿
.
=
-
-
www.eolwr/1O=bWz=1mW.pGdBmtEpoomW--
Ejemplo 3.4 .
:
÷. . .. . . ..
.. , .µ .. .. . .. . . . . ...
"
si usamos esa sustitución
""
""" ? yp"
pz-pro.ro pq-fr.eo-st.ro
E- Portó
""
pj-fz.ro spieóooleoypa
-
} PFI 4+4=8 Pratt =
zvzpamm
Problema 3.1 :
El valor de referencia para DBV es tu
y para
DBV 0,774 V ,
por
lo tanto DBU siempre será mayor que dB al tener una
referencia
menor .
Ejemplo :
• × = 1
DBV
eologl eolog (F) DBV dBm
eolog ( Í÷ ) -10 log (¥744 -423dBno
-
-
- = .
•
F- 2
§ en ambos casos
DBV
eologl to
log É) -402dBV dBm
eolog ( YÍ ) -10 log (Éa ) -925dB u es
mayor
-
-
- = .
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