127
As it can be seen from the figure, a periodic
autocorrelation function has ideal correlation
properties. But the product of two signals has a
drawback–notenoughdecreasingthevalueofpeak
factor
5.9H which leads to a noticeable
sensitivity to Doppler frequency shift, which is
undesirable. Therefore we consider the option, in
whichwecanfurtherreducethepeakfactor.To do
this, we can choose the smaller parts for
product
123
347 84NNNN . This will
decrease the peak factor (according to (2)) of the
signal to a value
7.4H . At the same time, it is
expectedtoreducethesidelobesofthecrosssections
ofAFwithaDopplerfrequencyshift.
а)
b)
c)
Figure3. Sections of AF of signal
1234
4 5 7 9 1260NNNNN :а) l=0; b)
l=1;c)l=2.
As it’s expected and seen in Fig. 2 above,
sensitivity to the Doppler frequency shift was
reduced.
The presented approach is effective and can be
used for obtaining signals with long periods.
Consider, for example, a signal in the form of a
product of four signals:
1234
4 5 7 9 1260 NNNNN
(AF is on
Figure3)andpeakfactorisequalto
5
H .
Itwas confirmedthat wehave receivedperiodic
signaloflongdurationwithasmallpeakfactorand
lowsensitivitytotheDopplerfrequencyshifts.
Thus, we can widely modify the peak factor of
signals (1), obtaining anideal correlation properties
in the AF zero section, and with an increase
in the
coherent part of the signal we can reduce the side
lobesofAF.
Furtherincreasingofsegmentsofresultingsignal
will lead to an increase in the peak factor. For the
individual modes of radar let’s consider the case
whenthepeakfactorisone:
1
.Inthiscase,the
formofthesignalisdeterminedaccordingto:
abb b ,where
1, 1ab
. (4)
Thus to provide ideal properties of periodic
autocorrelation function it can be used mismatched
processing that is development of an approach
described in [4]. The filter coefficients can be
calculatedaccordingtotheexpressionfound:
0
12 2
; 1
1
n
aa N a N
ww
a
(n=1
1N
). (5)
Arisingsomelossesinsignal‐to‐noiseratiocanbe
foundfromthefollowingexpression:
2
2
2
2
12 2 11
12 2 11 1
aaa N a N N a
aa N a N N a a N
(6)
If
2
2
N
a
, then there has no losses in
signal‐to‐noiseratio
1
.
If
1; a
thenthelossesinsignal‐to‐noiseratio
canbecalculatedfromtherelation:
2
2
42
31
N
NNN
(7)
Asanexamplethereisasignal(4)whenN=9,and
on the Figure 4 there is cross‐ ambiguity function
(CAF)ofit.