ESI VideoBridge 2150, Руководство по обслуживанию

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M e a s u Leme n t beg i n s w he n a s i new a ve s i g n a l a t a spe c i f i ed f Le q u e n c y i s
app l i ed t o t he u n k n ow n compone n t . The o L i g i n o f t h i s s i newa ve s i gna l
i s t he s i ne g e ne La t o L ( F i g u Le 3 - 3 ) w h i c h c o n s i s t s o f a p L e s e t coun t e L
c h a i n , a s e c o n d c o u n t e L , a 5 1 2
x
8 b i t p L e p L o g L a m m e d ROM , a
d i g i ta l- t o- a na l o g c o n v e L t e L ( DAC ) , a n d a c U L Le n t- to-vo l ta g e bu f fe r
a mp l i f i e r . The p L e se t c ou n t e r c h a i n ( U 2 2 , U 2 4 , U 2 5 ) i s a se t of
t h Le e c h i p s c on f i g u re d t o pe L f orm a d i v i de - by -N f u n c t i on . I t d i v i de s
t h e
7 .
6 8 M H z ma s t e L c l oc k i n t o one o f t h e 3 0 2 3 f Leq u e n c i e s t ha t a re
p L e s e t t a b l e by t h e m i c L oc ompu t e r . T h e new f r eque n c y , a c t u a l l y a
squa rewa v e s i g na l , g o e s i n t o a se cond co u n t e r c ha i n made o f U 1 6 and
U 1 9 . T h e s e c o n d c o u n t e r i s c o n n e c te d t o U 1 7 , a r e a d o n l y memoLY
( ROM ) , i n s u c h a w ay t ha t a s t he c ou n t e r ' s ou t p u t l i n e s t o g g l e , t hey
r i pp l e t h r ou g h a l l t he a d d re s s e s o n t he ROM .
F o r e a c h a d d r e s s t h e
ROM ou t pu t s 8 b i t s o f i n f o rma t i on w h i c h a re f e d t o t he D/A c o n ve L t e r .
Twe l ve comp l e t e s i newa ve s i g na l s a re s t o L e d i n t he 7 4 S 4 7 2 ROM i n tw o
sepa ra te l o ca t i o n s o f 2 5 6 a d d L e s s l o c a t i o ns e a c h . T h e se l o ca t i ons a Le
Le f e r red t o a s t he X 5 b l o c k ( t e n s i newave s i g na l s ) a n d t h e X l b l oc k
( tw o s i n ew a ve s i g na l s , s e e F i g u re 3- 3 ) .
For t e s t f re q ue n c i e s a bove 1 0k H z w h i c h d i v i de 6 0 , 0 0 0 e v e n l y , a n d f o r
a l l f r e qu e n c i e s a t or be l ow 1 0 k H z , t h e X l m e mo ry b l oc k L e s e L v e s 1 2 8
a d d re ss l o ca t i on s f o r t h e 0 0 s i n ewave a n d 1 2 8 l o ca t i on s f o r t h e 1 8 0 0
s i newave .
E a c h a d d L e s s l o ca t i on c on ta i n s 8 b i t s o f i n f o rma t i on .
NOTE :
7 . 6 8 M H z d i v i de d by 1 2 8 s i newave s t e p s equ a l s t he 6 0 k H z
ba s e f r e q u e n c y u se d t o d e t e rm i ne t h e s e t e s t f re q u e n c i e s .
6 0 k H z
N
=
Te s t F Le q u e n c y
W h e re : N
=
1 t o 3 0 0 0
F o r a l l t e s t f r e q u e n c i e s
a b o v e 1 0 k H z w h i c h d o n o t d i v i d e 6 0 , 0 0 0
e v e n l y , t h e X 5 m e m o r y b l ock p ro v i d e s a n a ve ra g e o f 1 2 8 / 5 o r 2 5 . 6
s t e p s
f o r
f i ve 0 0 s i n ew a v e s a n d a n a v e ra g e o f 2 5 . 6 s te p s f O L f i ve
1 8 0 0 s i n ewave s . T h i s e n a b l e s t he ROM t o s to r e f i ve t i m e s a s many
cy c l e s ( t houg h ea c h w i t h f ewe L s t eps ) a s t h e X l l o ca t i on .
3 - 6
e l s l i 2 1 5 0 / 2 1 6 0 5 /8 5