ETTC'2003 - SEE

GPS time binary data is about 1ms, a time interval counter is needed in the ground station. Considering
the equipment complexity and working toward minimum parts cost, the clock frequency of the time
interval counter is 200MHz. The ground station clock quantification error ε q is 5ns.
4) Telemetry receiver ε d
The error ε d resulted from the telemetry receiver is affected by the selection of the telemetry system.
In the PCM system, ε d is about one fiftieth of the chip width. If the bit rate is 2Mbps, ε d is about
10ns. In the PPK (Pulse Position Keying) system, ε d is affected by the signal pulse jitter ε t , the
pulse delay resulted by the signal level change ∆ tds
, the pulse delay resulted by the difference of the
'
decision level ∆tds and the RF pulse jitter σ df .
ε t = tr S N
(4)
where tr is the rise time of the pulse signal, S N is the square root of the receiver’s
Signal-to-Noise. For the reliability of the transferred data, the S/N is about 16dB to 17dB. Then
1
ε t = tr
(5)
7
In general, the relationship between the system bandwidth and the rise time is = 0.
5 B .
rt ( k −1)
r 1
∆ tds
=
(6)
k1
where r is the relative decision level, k is the ratio of the actual signal level to the ideal one.
2 , ∆t = 0.
5rt
. = 0.
6 ,
1 = k ds
r
1
∆ t = 0.
3t
. It means that if the actual signal level is twice larger
r ds r
than the ideal one, ∆tds is three times larger than ε t . For this reason, the system gain should be well
controlled to keep the received signal level invariable.
'
∆tds = rtr
k −1)
(7)
( 2
where k is the ratio of the changed decision level to the original one. If the decision level changed
2
10%, then 2 1.
1 , . The error
= k '
∆tds ≈ 0.
42ε
t ε d resulted from the telemetry receiver in one
ground station is
ε = ε + ∆t
+ ∆t
+ σ
(8)
d
2
t
2
ds
In the 3∆t system, the ground stations receive the same telemetry signal and calculate the target’s
position by using the time difference of arrival. The RF pulse jitter contributes nothing. The ε t , ∆ tds
'
and ∆t
of each ground station are independent, then in determining the time difference these errors
ds
will be enlarged. The error
ε d is
'2
ds
2
df
2 2 '2
ε d = 2ε t + 2∆t
ds + 2∆t
ds
(9)
If the bandwidth of PPK system is 9MHz, r = . 6,
k = 2,
k = 1.
1 , S N is about 16dB to
17dB, ≈ 56ns
, ≈ 8ns
t r
5) Doppler shift ε
f
ε , ≈ 16.
8ns
t
t ds
0 1 2
∆ , ∆ 3.
36ns
,
t r
'
tds ≈ d ≈ 26.
7ns
ε 。