Graphics Programs Reference
In-Depth Information
v
()
y(t)
x(t)
Σ
Σ
delay,
T
+
+
-
+
w
()
1
K
Figure 7.8. MTI recursive filter.
Solving for the transfer function
H
()
Y
()
X
()
=
⁄
yields
1
1
z
H
()
=
--------------------
(7.21)
1
1
Kz
The modulus square of
H
()
is then equal to
1
zz
1
(
1
z
) 1
(
z
)
2
(
+
)
H
()
2
=
----------------------------------------------
=
---------------------------------------------------
(7.22)
1
K
2
z
1
(
1
Kz
) 1
(
Kz
)
(
1
+
)
Kz
(
+
)
j
ω
T
Using the transformation
z
=
yields
zz
1
+
=
2
cos
ω
T
(7.23)
Thus, Eq. (7.22) can now be rewritten as
21
(
cos
ω
T
)
)
2
He
j
ω
T
(
=
-------------------------------------------------------
(7.24)
K
2
(
1
+
)
2
K
cos
(
ω
T
)
Note that when , Eq. (7.24) collapses to Eq. (7.10) (single line can-
celer).
Fig. 7.9
shows a plot of Eq. (7.24) for
K
=
0
K
=
0.25 0.7 0.9
,
,
. Clearly, by
changing the gain factor
K
one can control the filter response.
In order to avoid oscillation due to the positive feedback, the value of
K
1
should be less than unity. The value is normally equal to the number
of pulses received from the target. For example,
(
1
K
)
K
=
0.9
corresponds to ten
pulses, while
K
=
0.98
corresponds to about fifty pulses.
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