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0
22
3
/
2
4(
)
iR ds
dB
sR
µ
π
=
+
.
(a) Clearly, considered as a function of
s
(but thinking of “
ds
” as some finitesized
constant value), the above expression is maximum for
s
= 0.
Its value in this case is
2
max
0
/4
dB
i ds
R
=
.
(b) We want to find the
s
value such that
max
/10
dB
dB
=
. This is a nontrivial algebra
exercise, but is nonetheless straightforward. The result is
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This note was uploaded on 06/08/2010 for the course PHY 1356 taught by Professor Bonamente during the Spring '10 term at UAA.
 Spring '10
 Bonamente

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