# Antenna Noise Temperature Online Exam Quiz

Antenna Noise Temperature GK Quiz. Question and Answers related to Antenna Noise Temperature. MCQ (Multiple Choice Questions with answers about Antenna Noise Temperature

### Effective noise temperature Te in terms of noise figure is __

**Options**

A : Te=To (F-1)

B : Te=To/(F-1)

C : Te=To/(F+1)

D : Te=To (F+1)

### Expression for noise figure F related to the effective noise temperature Te is __

**Options**

A : F=1+\frac{T_e}{T_o}

B : F=1+\frac{T_0}{T_e}

C : F=1-\frac{T_e}{T_o}

D : F=1-\frac{T_0}{T_e}

### Find the effective noise temperature if noise figure is 3 at room temperature (290K)?

**Options**

A : 290K

B : 580K

C : 289K

D : 195K

### The induction and radiation fields are equal at a distance of ___

**Options**

A : ?/4

B : ?/6

C : ?/8

D : ?/2

### The ratio of radiation intensity in a given direction from antenna to the radiation intensity over all directions is called as __

**Options**

A : Directivity

B : Radiation power density

C : Gain of antenna

D : Array Factor

### The relation between vector magnetic potential and current density is given by __

**Options**

A : ?.A=J

B : ??A=H

C : ?2A=-?J

D : ?2A=??H

### What is the overall efficiency of a lossless antenna with reflection coefficient 0.15?

**Options**

A : 0.997

B : 0.779

C : 0.669

D : 0.977

### What is the relation between noise temperature introduced by beam TB and the antenna temperature TA when the solid angle obtained by the noise source is greater than antenna solid angle?

**Options**

A : TA= TB

B : TA > TB

C : TA < TB

D : TA ? TB

### What should be the noise figure value at which the effective noise temperature equals to room temperature?

**Options**

A : 2

B : 1

C : 0

D : 1/T_(o )

### Which expression suits best when the solid angle obtained by the noise source is less than antenna solid angle?

**Options**

A : PA ?A=PB ?B and ?TA=\frac{\Omega_B}{\Omega_A} T_B

B : PA ?B=PB ?A and ?TA=\frac{\Omega_B}{\Omega_A} T_B

C : ?TA=\frac{\Omega_A}{\Omega_B} T_B and PA ?B=PB ?A

D : ?TA=\frac{\Omega_A}{\Omega_B} T_B and PA ?A=PB ?B