Graphics Programs Reference
In-Depth Information
Example:
A certain radar uses two PRFs to resolve range ambiguities. The desired
unambiguous range is
. Choose
. Compute
,
,
R
u
=
100
Km
N
=
59
f
r
1
f
r
2
, and
.
R
u
1
R
u
2
Solution:
First let us compute the desired PRF,
f
rd
3 0
8
c
2
R
u
×
200
f
rd
=
---------
=
-----------------------
=
1 . 5
KHz
10
3
×
It follows that
f
r
1
=
Nf
rd
=
5()1500
(
)
=
88.5
KHz
f
r
2
=
(
N
+
1
)
f
rd
=
(
59
+
1
) 1500
(
)
=
90
KHz
3 0
8
c
2
f
r
1
×
R
u
1
=
---------
=
----------------------------------
=
1 . 6 9 5
Km
10
3
2
×
88.5
×
3 0
8
c
2
f
r
2
×
2 0 0
3
.
R
u
2
=
---------
=
------------------------------
=
.
Km
×
×
Example:
Consider a radar with three PRFs;
,
, and
f
r
1
=
15
KHz
f
r
2
=
18
KHz
. Assume
. Calculate the frequency position of each
f
r
3
=
21
KHz
f
0
=
9
GHz
PRF for a target whose velocity is
. Calculate
(Doppler frequency)
550
ms
⁄
f
d
for another target appearing at
,
, and
for each PRF.
8
KHz
2
KHz
17
KHz
Solution:
The Doppler frequency is
2 09 0
9
2
vf
0
c
×
×
×
f
d
=
-------
=
------------------------------------------
=
KHz
3 0
8
×
Then by using Eq. (1A.14)
where
, we can write
n
i
f
ri
+
f
di
=
f
d
i
=
123
,,
n
1
f
r
1
+
f
d
1
=
15
n
1
+
f
d
1
=
33
n
2
f
r
2
+
f
d
2
=
18
n
2
+
f
d
2
=
33
n
3
f
r
3
+
f
d
3
=
21
n
3
+
f
d
3
=
33
We will show here how to compute
, and leave the computations of
and
n
1
n
2
to the reader. First, if we choose
, that means
, which
n
3
n
1
=
0
f
d
1
=
33
KHz
cannot be true since
cannot be greater than
. Choosing
is also
f
d
1
f
r
1
n
1
=
1
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