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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