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Why LNA is placed closed to antenna of outdoor unit?

Mumbai University > EXTC > Sem 8 > Satellite Communication and Networks

Marks: 5M

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Case i: Consider the effect of inserting a lossy waveguide (cable) between the antenna & the LNA

enter image description here

The overall system noise temperature referred to the input,

$T_{s}=T_{ant}+T_{e1}+\frac{(L-1)T_0}{G1}+\frac{L(F-1)T_0}{G1}$

L represents the power loss of the cable.

The power loss is simply the ratio of input power to output power and is always greater than unity.

(L-1)T0 represents the equivalent noise temperature of the cable referred to its input.

For the system shown in Fig. a, the receiver noise figure is 12 dB, the cable loss is 5 dB, the LNA gain is 50 dB, and its noise temperature 150 K. The antenna noise temperature is 35 K. The noise temperature referred to the input is

$T_{s}=T_{ant}+T_{e1}+\frac{(L-1)T_0}{G1}+\frac{L(F-1)T_0}{G1}$

$T_{s}=35+150+\frac{(3.16-1)290}{10^5}+\frac{3.16(15.85-1)290}{10^5}=185K$

Case ii: Consider the effect of inserting a lossy waveguide (cable) between the LNA and the receiver.

enter image description here

In this case the cable precedes the LNA, and therefore, the equivalent noise temperature referred to the cable input is

$T_s=T_{ant}+(L-1)T_0+L\times T_{ef}+\frac{L(F-1)T_0}{G1}$

$T_s=35+(3.16-1)\times 290+3.16\times 150+\frac{3.16(15.85-1)290}{10^{5}}=1136K$

It is seen that in case ii the noise temperature is higher (1136K) than that in case i (185K). For communication system the equivalent noise temperature should be less. Therefore the LNA must be placed ahead of the cable, which is why one sees amplifiers mounted right at the dish in satellite receive systems.

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