Motorola BT100 - Manual (Page 28)

Amplifier Setup 3-3
BT100 Installation and Operation Manual
STARLINE Forward Equalizers
Select the appropriate model SFE-100-* to compensate for cable attenuation versus frequency
and to obtain the proper output tilt. The BT100 is equipped with the BDR-1G interstage
equalizer and flatness board, which compensates for cable attenuation. You must compensate
for any cable or passive slope beyond that of the BDR-1G by selecting and installing the
appropriate SFE-100-* cable equalizer.
Equalizers are available in 1 dB steps from 0 dB to 22 dB. The following examples describe how
to choose the correct equalizer.
Example 1
The amplifier location includes 20 dB of cable (at 1 GHz) between its input and the preceding
amplifier. Consider cable loss only. Exclude any flat loss due to splitters or other passive
devices. The internal equalizer, model BDR-1G, compensates for approximately 14 dB of cable.
Subtract this cable length from the 20 dB of this example (20
− 14 = 6). The SFE-100-6 is the
proper equalizer in this case. With this equalizer installed, the amplifier reproduces the output
tilt of the last upstream amplifier.
When selecting an equalizer, choose the next lower value if the exact value is not available or in
cases where the calculated value makes two choices possible.
Example 2
The BT100 amplifier is used in a link following a fiber node. The fiber receiver output is flat and
connects to the input of the amplifier through 18 dB of cable plus passive loss. Determine which
is the proper equalizer to achieve 11 dB output tilt from the BT100.
Calculate the equalizer value using the following equation:
SLOPEeq = TILTout + SIGlo – SIGhi – SLOPEieq
where: SLOPEeq = required SFE-100-* slope
TILTout = required amplifier output tilt
SIGlo
= signal input level at 54 MHz (channel 2)
SIGhi = signal input level at 1 GHz
SLOPEieq = interstage equalizer slope (14 dB)
From various references, such as manufacturer’s catalogs, you can determine that at an
operating frequency of 1003 MHz, 18 dB of cable produces approximately 3.6 dB of loss at
54 MHz. This suggests that the channel 2 signal input to the BT100 is 14.4 dB greater
(18
− 3.6 = 14.4) than it is at 1003 MHz. Our example assumes that the high-end frequency level
into the BT100 is +12 dBmV.
Substituting this information into the above equation provides the following result:
11 dB + 26.4 dB − 12 dBmV − 14 dB = 11.4 dB
The slope of the required equalizer is 11.4 dB. Table 3-1 and the graph in Figure 3-1 show that
11.4 dB of slope is caused by approximately 14 dB of cable at 1003 MHz. Therefore the correct
equalizer is model SFE-100-14.
When selecting an equalizer, choose the next lower value if the exact value is not available or in
cases where the calculated value makes two choices possible.