Question

A receiving antenna with a gain of 40 dB looks at a sky with a noise temperature of 15 K. The loss between the antenna and LNA input due to the feed horn is 0.4 dB, and the LNA has a noise temperature of 40 K. The \(\frac{G}{T}\) value is
1. 11.2 dB/K
2. 13.4 dB/K
3. 20.6 dB/K
4. 39.0 dB/K

Answer 1

Correct Answer - Option 3 : 20.6 dB/K

Concept:

• System Noise Temperature is the sum of all noise temperature at the receiving Antenna.

           i.e. T_s = sum of all noise Temperature in the system.

• LNA means low noise Amplifier and it is used at the receiver side to boost the received signal strength.
• value of the Receiver shows the figure of merit of a receiver, and it the ratio of receiver gain to the temperature of the system in dB per Kelvin.

Calculation:

Given that:

Receiving antenna gain = 40 dB

Antenna noise temperature = 15 K

Loss between Antenna & LNA input = 0.4 dB

LNA noise temperature = 40 K.

Total noise temperature in the system will be:

= (Antenna Noise Temperature) + (LNA Noise Temperature)

= 15 + 40

i.e. Total noise Temperature in the system = 55 K

and total receiver gain in the system = Antenna gain + loss

= 40 – 0.4

= 39.6 dB

\(\therefore \frac{G}{T}\) value = G(dB) – 10 log₁₀ (TK)

= 39.6 – 10 log₁₀ (55)

= 22.2 dB/k

\(\frac{G}{T} \approx 20.6\;\frac{{dB}}{k}\)