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What do you mean by system noise temperature? How does it affect C/N and G/T ratio

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What do you mean by system noise temperature? How does it affect C/N ratio and G/I ratio calculate C/N ratio in db Hz given that B=36MHZ calculate C/N in dB

Marks: 10 M

Year: Dec 2012, May 2014

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The main source of noise in the satellite equipment is the noise arising from the random thermal motion of electrons in the various devices in the receiver. Thermal noise is also generated in the lossy components of the antenna and a thermal – like noise is picked – up by the antenna as radiation.

Power from a thermal noise source is given by: $P_N=kT_N B_N$

where: $TN$ noise temperature, BN Noise Bandwidth,

$K$ Boltzman constant having the value $1.38 x 10-23 J/k$

  • Antenna noise: antenna losses + sky noise (background microwave radiation)
  • Antenna noise temperature: Antenna noise temperature is the temperature of a theoretical resistor at the input of an ideal noise-free receiver that would generate the same output noise power per unit bandwidth as that at the antenna output at a specified frequency. Antenna noise temperature has contributions from several sources:

  • Vast radiation

  • Earth heating
  • The sun
  • Electrical devices
  • The antenna itself

Single Amplifier:

enter image description here

The available power gain of the amplifier is denoted as G, and the noise power output as $P_no$ Considering noise power per unit bandwidth which is noise energy in joules is given by: $N_{0}{ant}=kT_ant$ The output noise energy in $N_{o}{out}$ will be $GN_{o}{out}$ plus the contribution made by the amplifier. The summation of all the amplifier noise is referred to the input in terms of an equivalent input noise temperature for the amplifier $T_e$. Thus output could be written as:

${N_{o}{out}}=Gk(T_{ant}+T_e)$

The total noise referred to the input is

$\frac{N_(o_out )}{G} OR N_oin=k(T_ant+T_e)$

Amplifiers in Cascade:

A cascade amplifier is any amplifier constructed from a series of amplifiers, where each amplifier sends its output to the input of the next amplifier in a daisy chain.

enter image description here

For the arrangement of amplifiers shown in fig; the overall gain can be considered as:

$G=G_1 G_2$

The noise energy of amplifier 2 referred to its own inputs is imply $kT_e2$

The noise input to amplifier 2 from the preceding stages is $G_1 k(T_ant+T_e1)$, and thus the total noise energy referred to amplifier 2 inputs is: N_o2=G_1 k(T_ant+T_e1 )+kT_e2

This noise energy may be referred to amplifier 1 input by dividing by the available over gain of amplifier 1: $N_02=N_02/G_1 =k(T_ant+T_e1 )+T_e2/G_1$

A system noise temperature may now be defined as $T_S by N_01=kT_S$ And hence it will be seen that $T_S$ is given by $T_S=T_ant+T_e1+T_e2/G_1$

This result can be generalized to any number of stages in cascade giving

$T_s=T_ant+T_e1+T_e2/G_1 +T_e3/(G_1 G_2 )+⋯….$

It shows that the noise temperature of the second stage is divided by the power gain of the first stage when referred to the input. Therefore, in order to keep the overall system noise as low as possible, the first stage (usually as LNA) should have high power gain as well as noise temperature.

Gain to Noise temperature ratio $\frac{G}{T}$:

The antenna gain to noise temperature ratio G/T is a figure of merit to indicate the performance of the earth station antenna and the low noise amplifier in relation to sensitivity in receiving down link carrier from the satellite. The parameter G is the receive antenna gain referred to the input of the low noise amplifier. The parameter T is defined as the earth station system noise temperature referred also to the input of the low noise amplifier. $\frac{G}{T}=[G_rx ]-10log⁡(T_S)$

where: $G_rx= receive gain in dB$

$T_S = system noise temperature in °K$

In the link equation, by unfolding the kTB product under the logarithm, the link equation becomes:

$\frac{C}{N}=[EIRP]-[Losses]+[G]-10 log⁡(k)-10 log⁡(T_S )-10log⁡(B)$

The difference$[G]-10 log⁡(T_S )$, is the figure of merit:

$\frac{C}{N}=[EIRP]-[Losses]+\frac{G}{T_S }-10 log⁡(k)-10log⁡(B)$

where:

$k = Boltzmann constant$

$B = carrier occupied bandwidth$

We know $P_N=kTB=N_0 B$

$∴\frac{C}{N}=\frac{C}{N_0 B}$

$i.e.\frac{C}{N}=\frac{C}{N_0}-B$

Here the unit of $\frac{C}{N_0 }$ is in $dBHz, \frac{C}{N} in dB$

Hence $\frac{C}{N}=\frac{C}{N_0 }-10log⁡(36×10^6)$

Note: The Question has some discrepancy as no other specification required is given.

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