Vaisala RVP900

Chapter 5 _______________________________________________________ Plot-Assisted Setups
VAISALA______________________________________________________________________ 159
where f is the frequency in Hertz. This is the familiar "synch" function,
whose main frequency lobe extends from 1/T
o
to 1/T
o
Hertz, and whose
total power integrated over all frequencies is 1.0.
We can now examine what the filter loss (dB
loss
) would be if this pulse
were applied to a band pass filter. The filter loss is simply the ratio of the
power that is passed by the filter, divided by the total input power (1.0 in
this case). Assume for the moment that the filter is an ideal band pass filter
centered at zero Hertz (corresponding to how S(f) was defined) and having
a bandwidth B
w
, then:
This integral can be computed for a few "interesting" filter bandwidths,
yielding filter losses of 0.44 dB, 1.11 dB, and 3.31 dB when B
w
is 2/T
o
,
1/T
o
, and 1/2T
o
respectively. These three example bandwidths correspond
to filters that pass the entire main frequency lobe, half of that lobe, and one
quarter of it.
You can experimentally verify these results using the RVP900 as follows:
- Using the Mt0 command, setup a T
o
= 0.5 µsec trigger pulse from the
RVP900 in the vicinity of range zero, and use that trigger to gate a
signal generator whose output is applied to the IFRD Burst Input. Also
setup 125 m range resolution, and a rather long 6.0 µsec impulse
response length. The long length will make the transition edges of the
matched filter as steep as possible, so that it becomes a reasonably
good approximation to the ideal band pass filter used in the above
analysis.
-Use the Pb command to verify that the burst pulse is present, and
position the triggers left and right until the pulse is centered exactly at
zero.
-Use the Ps command to examine the frequency spectrum of the pulse.
You should see a main energy lobe that is 4 MHz wide and centered
at the radars IF. There should also be weaker lobes spaced 2 MHz
apart on both sides of the main lobe. If the lobe spacing does not look
Sf
1
T
0
----------2fttcos td
T–
0
2
T
0
2






2
sin
2
fT
0


2
f
2
T
0
--------------------------==
dB
loss
10log
10
Sf fd
B–
w
2
B
w
2







–=