Question 1 |

An antenna with a directive gain of 6\text{ dB}
is radiating a total power of \text{16 kW}. The amplitude of the electric field in free space at a distance of \text{8 km}
from the antenna in the direction of \text{6 dB}
gain (rounded off to three decimal places) is ______ \text{V/m}
.

0.244 | |

0.417 | |

1.254 | |

2.365 |

Question 1 Explanation:

\begin{aligned} G_{d}(\mathrm{~dB}) &=10 \log _{10}\left(G_{d}\right) \\ \Rightarrow\qquad 6 &=10 \log _{10} G_{d} \\ \Rightarrow\qquad G_{d} &=10^{0.6} \\ G_{d} &=3.981 \\ E_{m} &=\frac{\sqrt{60 G_{d} P_{\text {rad }}}}{r} \\ E_{m} &=\frac{\sqrt{60\left(16 \times 10^{3}\right)(3.981)}}{8 \times 10^{3}}=0.244(\mathrm{~V} / \mathrm{m}) \end{aligned}

Question 2 |

For an infinitesimally small dipole in free space, the electric field E_\theta in the far field
is proportional to (e^{-jkr}/r) \sin \theta, where k = 2 \pi /\lambda. A vertical infinitesimally small
electric dipole (\delta l \lt \lt \lambda) is placed at a distance h(h \gt 0) above an infinite ideal
conducting plane, as shown in the figure. The minimum value of h, for which one
of the maxima in the far field radiation pattern occurs at \theta =60^{\circ}, is

\lambda | |

0.5\lambda | |

0.25\lambda | |

0.75\lambda |

Question 2 Explanation:

\begin{aligned}\left | \text{Total E }\right |&=\left | (E_{\text{single element}}) \right | \left | (A.F.) \right | \\ \left | (A.F.) \right |&=\frac{\sin\left ( N\frac{\psi }{2} \right )}{\sin (\frac{\psi }{2})}=\frac{\sin \left ( 2\frac{\psi }{2} \right )}{\sin \left ( \frac{\psi }{2} \right )} \\ & =\frac{2\sin \left ( \frac{\psi }{2} \right )\cos (\frac{\psi }{2})}{\sin \frac{\psi }{2}} \\ & =2\cos (\frac{\psi }{2})\\ \left | A.F_{N} \right | &=\frac{A.F}{A.F_{max}}=\frac{2\cos (\frac{\psi }{2})}{2} \\ &=\left | \cos \left ( \frac{\psi }{2} \right ) \right | \\ \text{where, }\psi &=\beta d\cos \theta =\frac{2\pi }{\lambda }(2h)\cos \theta \\ \left.\begin{matrix} \left | A.F_{N} \right |_{\theta =60^{\circ}} \end{matrix}\right|& =\left | \cos (\frac{2\pi }{\lambda }h\cos 60^{\circ}) \right |\\ &=\left | \cos \left ( \frac{\pi h}{\lambda } \right ) \right |\\ \cos \theta \text{ is maximum ,where } \theta &=n\pi \;\; n=0,1,2..\\ \frac{\pi h}{\lambda }&=n\pi \Rightarrow h=n\lambda \\ \Rightarrow \; \text{For } n=1, \;\; h_{min}&=\lambda \end{aligned}

Question 3 |

Radiation resistance of a small dipole current element of length l at a frequency of 3 GHz is 3 ohms. If the length is changed by 1%, then the percentage change in the radiation resistance, rounded off to two decimal places, is _____ %.

0.86 | |

2.01 | |

2.54 | |

2.68 |

Question 3 Explanation:

Radiation resistance of a small dipole current element of length 'I' is

\begin{array}{l} R_{\mathrm{rad}}=80 \pi^{2}\left(\frac{l}{\lambda}\right)^{2} \Rightarrow R \propto l^{2} \\ \frac{R_{2}}{R_{1}}=\left(\frac{l_{2}}{l_{1}}\right)^{2} \end{array}

If length is changed by 1% then percentage change in the radiation resistance.

\frac{R_{2}}{R_{1}}=\left(\frac{1.01 l}{l}\right)^{2}=1.0201

\text { Percentage change in radiation resistance }

\begin{array}{l} =\frac{R_{2}-R_{1}}{R_{1}} \times 100 \\ =0.0201 \times 100=2.01 \% \end{array}

\begin{array}{l} R_{\mathrm{rad}}=80 \pi^{2}\left(\frac{l}{\lambda}\right)^{2} \Rightarrow R \propto l^{2} \\ \frac{R_{2}}{R_{1}}=\left(\frac{l_{2}}{l_{1}}\right)^{2} \end{array}

If length is changed by 1% then percentage change in the radiation resistance.

\frac{R_{2}}{R_{1}}=\left(\frac{1.01 l}{l}\right)^{2}=1.0201

\text { Percentage change in radiation resistance }

\begin{array}{l} =\frac{R_{2}-R_{1}}{R_{1}} \times 100 \\ =0.0201 \times 100=2.01 \% \end{array}

Question 4 |

A half wavelength dipole is kept in the x-y plane and oriented along 45^{\circ} from the x-axis. Determine the direction of null in the radiation pattern for 0\leq \phi \leq \pi. Here the angle \theta (0\leq \theta \leq \pi)
is measured from the z-axis, and the angle \phi(0\leq \phi \leq 2\pi) is measured from the x-axis in the x-y plane.

\theta =90^{\circ},\; \; \phi =45^{\circ} | |

\theta =45^{\circ},\; \; \phi =90^{\circ} | |

\theta =90^{\circ},\; \; \phi =135^{\circ} | |

\theta =45^{\circ},\; \; \phi =135^{\circ} |

Question 4 Explanation:

x-y plane means Z=0 with Z=r \cos \theta (spherical coordinates) \theta=90^{\circ}

With null points along the axis of the dipole it is \phi=45^{\circ} and 225^{\circ}

With null points along the axis of the dipole it is \phi=45^{\circ} and 225^{\circ}

Question 5 |

Consider a wireless communication link between a transmitter and a receiver located in free space, with finite and strictly positive capacity. If the effective areas of the transmitter and the receiver antennas, and the distance between them are all doubled, and everything else remains
unchanged, the maximum capacity of the wireless link

increases by a factor of 2 | |

decreases by a factor of 2 | |

remains unchanged | |

decreases by a factor of \sqrt{2} |

Question 5 Explanation:

As per Friis free space propagation equation,

W_{r}=\frac{W_{t} A_{e r} A_{e t}}{(\lambda d)^{2}}

when A_{e r} and A_{e t} are doubled and d also doubled, W_{r} is same. Hence capacity is also same.

W_{r}=\frac{W_{t} A_{e r} A_{e t}}{(\lambda d)^{2}}

when A_{e r} and A_{e t} are doubled and d also doubled, W_{r} is same. Hence capacity is also same.

There are 5 questions to complete.