Seagate CHEETAH NS 10K.2 ST3450802SS, Руководство по эксплуатации продукта

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66 Cheetah NS 10K.2 SAS Product Manual, Rev. B
mitter signal at the test points on the required test loads. The transmitter uses the same settings (e.g., pre-
emphasis, voltage swing) with both the zero-length test load and the TCTF test load. The signal specifications
at IR are met under each of these loading conditions.
The TCTF is the mathematical statement of the transfer function through which the transmitter shall be capable
of producing acceptable signals as defined by a receive mask. The transmission magnitude response of the
TCTF in dB is given by the following equation for 1.5 Gbps:
for 50 MHz < f < 1.5 GHz, and:
for 1.5 GHz < f < 5.0 GHz,
where:
a) f is the signal frequency in hertz.
The transmission magnitude response of the TCTF in dB is given by the following equation for 3.0 Gbps:
for 50 MHz < f < 3.0 GHz, and:
for 3.0 GHz < f < 5.0 GHz,
where f is the signal frequency in hertz.
The TCTF is used to specify the requirements on transmitters that may or may not incorporate pre-emphasis or
other forms of compensation. A compliance interconnect is any physical interconnect with loss equal to or
greater than that of the TCTF at the above frequencies that also meets the ISI loss requirements shown in fig-
ure 19 and figure 20.
Compliance with the TCTF test load requirement is verified by measuring the signal produced by the transmit-
ter through a physical compliance interconnect attached to the transmitter.
Compliance with the zero-length test load requirement verified by measurement made across a load equiva-
lent to the zero-length load shown in figure 18.
For both test load cases, the transmitter delivers the output voltages and timing listed in table 26 at the desig-
nated compliance points. The default mask is IR for intra-cabinet TxRx connections. The eye masks are shown
in 9.5.2.1.
Figure 17 shows the compliance interconnect test load.
Figure 17. Compliance interconnect test load
S
21
20 – log
10
e
( )
6 5
,
10
6 –
f
0 5
,
× × ( )
2 0
,
10
10 –
f
× × ( )
3 3
,
10
20 –
f
2
× × ( )
+ +
( ) ×
dB
×
=
S
21
5 437
,
– dB =
S
21
20 – log
10
e
( )
6 5
,
10
6 –
f
0 5
,
× × ( )
2 0
,
10
10 –
f
× × ( )
3 3
,
10
20 –
f
2
× × ( )
+ +
( ) ×
dB
×
=
S
21
10 884
,
– dB =