Graphics Programs Reference
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and in this case, the true target range is
ct r 1
2
R
=
---------
(1A.9)
Finally, if
, then the target is in the first ambiguity. It follows that
t 1
=
t 2
t r 2
==
t 1
t 2
(1A.10)
and
R t r 2
=
2
(1A.11)
Since a pulse cannot be received while the following pulse is being transmit-
ted, these times correspond to blind ranges. This problem can be resolved by
using a third PRF. In this case, once an integer is selected, then in order to
guarantee that the three PRFs are relatively prime with respect to one another.
In this case, one may choose
N
,
, and
f r 1
=
NN
(
+
1
) f rd
f r 2
=
NN 2
(
+
) f rd
.
f r 3
=
(
N
+
1
) N
(
+
2
) f rd
1A.4. Resolving Doppler Ambiguity
The Doppler ambiguity problem is analogous to that of range ambiguity.
Therefore, the same methodology can be used to resolve Doppler ambiguity. In
this case, we measure the Doppler frequencies
and
instead of
and
f d 1
f d 2
t 1
.
t 2
If
, then we have
f d 1
>
f d 2
(
f d 2
–
f d 1
)
+
f r 2
M
=
-------------------------------------
(1A.12)
f r 1
–
f r 2
And if
,
f d 1
<
f d 2
f d 2
–
f d 1
M
=
-------------------
(1A.13)
f r 1
–
f r 2
and the true Doppler is
f d
=
Mf r 1
+
f d 1
(1A.14)
f d
=
Mf r 2
+
f d 2
Finally, if
, then
f d 1
=
f d 2
f d
==
f d 1
f d 2
(1A.15)
Again, blind Doppler can occur, which can be resolved using a third PRF.
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