Hardware Reference
In-Depth Information
Substituting this value in
U
we obtain
K
K
(1
−
1
/n
)
−
1)
K
1
/n
+
U
lub
−
U
s
=
n
−
n
=
nK
1
/n
K
1
/n
+
K
1
/n
=
−
−
n
=
n
(
K
1
/n
=
−
1);
that is,
=
U
s
+
n
(
K
1
/n
U
lub
−
1)
.
(5.10)
Now, noting that
C
s
T
s
T
1
−
T
s
2
T
s
R
s
−
1
U
s
=
=
=
2
we have
R
s
=(2
U
s
+1)
.
Thus,
K
can be rewritten as
K
=
3
2
R
s
=
+
1
2
U
s
+2
2
U
s
+1
,
and finally
U
lub
=
U
s
+
n
U
s
+2
2
U
s
+1
1
.
1
/n
−
(5.11)
Taking the limit as
n
→∞
, we find the worst-case bound as a function of
U
s
to be
given by
U
lub
=
U
s
+ln
U
s
+2
2
U
s
+1
.
lim
n→∞
(5.12)
A plot of Equation (5.12) as a function of
U
s
is shown in Figure 5.11. For comparison,
the RM bound is also reported in the plot. Notice that for
U
s
<
0
.
4 the presence of
DS worsens the RM bound, whereas for
U
s
>
0
.
4 the RM bound is improved.
Deriving Equation (5.12) with respect to
U
s
, we can find the absolute minimum value
of
U
lub
:
2
U
s
+5
U
s
−
∂U
lub
∂U
s
=1+
(2
U
s
+1)
(
U
s
+2)
(2
U
s
+1)
2(
U
s
+2)
(2
U
s
+1)
2
−
1
(
U
s
+ 2)(2
U
s
+1)
.
=
The value of
U
s
that minimizes the above expression is
=
√
33
−
5
U
s
0
.
186
,
4
is
U
lub
so the minimum value of
U
lub
0
.
652.
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