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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