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Right CHOICE is (c) 11.19dB

Explanation: Given Pt=5W, Pr=150μW, f=500MHz, R=100m and Gtin dB=10dB

 Gtin dB=10log10 Gt=10dB

 Gt=10

⇨ \(\lambda = \FRAC{c}{f} = \frac{3×10^8}{500Mhz} = 0.6m\)

From Friss transmission equation, \(\frac{P_r}{P_t} = \frac{G_t G_r \lambda^2}{(4\pi R)^2} \)

⇨ \(G_r=\frac{P_r (4\pi R)^2}{P_t G_t \lambda^2} = \frac{150\mu(4\pi ×100)^2}{5×10×0.6^2}=13.16\)

⇨ GR in dB=10log10 Gr=10log10 13.16=11.19dB

The correct choice is (b) 0

To elaborate: When the receiving and transmitting antennas POLARIZATION is not MATCHED, the Friss transmission equation includes a polarization loss factor given by cos^2θ. Since one is vertically polarized and other is horizontally polarized, the ANGLE difference is 900. PLF=cos^2θ=0

∴ \(\FRAC{P_r}{P_t} = PLF\frac{G_t G_r \lambda^2}{(4\pi R)^2} =0\)

So, no power is RECEIVED.

Correct answer is (b) False

Easiest explanation: Since Gpmax=ηrGdmax and the value of RADIATION efficiency LIES in range 0 to 1, The maximum POWER gain will be always less than or equal to the maximum directive gain of antenna.

Correct answer is (b) Independent, independent

To EXPLAIN I would say: Due to symmetric DISTRIBUTION in isotropic source, there will be no θ, ф components. Only radial component will be PRESENT. \( P_{TOTAL \,rad}=∯ Usinθdθd∅=4πU (For \,θ: 0 \,to \,π, \phi: 0 \,to\, 2π).\)

∴ Radiation INTENSITY \(U=\frac{P_{total \,rad}}{4π}\)

Correct option is (B) NEAR field

To EXPLAIN I would say: Induction field is also KNOWN as ‘Near field’, VARIES inversely with r^2. Electrostatic field varies inversely with r^3. Far field is also known as Radiation field, varies inversely with r.

Correct answer is (d) \(2π×\frac{Total \,ENERGY\, stored\, by \,antenna}{energy\, radiated \,per \,CYCLE}\)

The explanation: Quality FACTOR Q=\(2π×\frac{Total \,energy\, stored\, by \,antenna}{energy\, radiated \,per \,cycle}\). Higher Q antennas will have LOW BANDWIDTH.

Correct choice is (C) 0≤η≤1

Explanation: RADIATION efficiency ηr=\(\FRAC{R_r}{R_r+R_l}\)

Rr+Rl>Rr So \(\frac{R_r}{R_r+R_l}\) < 1 and If Rl=0 then ηr=1

Therefore, the value of efficiency lies in the range 0 to 1.

Right option is (c) 1

Best explanation: Directive GAIN \(G_d = \FRAC{P_{d(\theta,\emptyset)}}{P_{AVG}} = \frac{P_{d(\theta,\emptyset)}}{P_r/4\pi r^2} = \frac{P_{d(\theta,\emptyset)^{r^2}}}{P_r/4\pi} = \frac{U_{d(\theta,\emptyset)}}{U_{avg}} \)

 ∴ \(G_d = \frac{U_{(\theta,\emptyset)}}{U_{avg}}\)=1.

Right option is (c) \(\eta_r=\frac{R_r}{R_r+R_l}\)

Explanation: Radiation efficiency is DEFINED as the ratio of power radiated to the total input power to the antenna. Total input power is the sum of the radiated power PRAND the OHMIC losses Pl.

\(\eta_r = \frac{P_r}{P_{in}} = \frac{P_r}{P_r+P_l} = \frac{R_r I_{rms}^2}{I_{rms}^2 R_r+I_{rms}^2 R_l} = \frac{R_r}{R_r+R_l} \)

Right OPTION is (a) FNBW

Easiest EXPLANATION: The angular distance between two successive nulls of main lobe is called First NULL Beam WIDTH. Half-power beam width is the angular distance when 50% of power is radiated. FBR is the Front-to-Back ratio DEFINED as ratio of power radiated at 0° to power radiated at 180°.

Correct answer is (d) HIGH, low

The best I can explain: The current distribution follows a triangular pattern. At the center it is maximum and NEARLY zero at ends.Amount of power radiates will depend on the wavelength λ. For vertical antennas, with wavelength N\(\frac{\LAMBDA}{2}\) the current distribution at center is high.

Right OPTION is (a) 0.73

To ELABORATE: Radiation resistance \(R_{rad}=80π^2(\frac{l}{\lambda})^2=80π^2 (\frac{\lambda/12}{\lambda})^2\)=5.48Ω

Antenna EFFICIENCY \(\eta=\frac{R_{rad}}{R_{rad}+R_{LOSS}}=\frac{5.48}{5.48+2}=0.73\)

Correct option is (a) True

The explanation is: Directivity is DEFINED as ratio of MAXIMUM power density in desired direction to the average power radiated in all directions. This is simply, maximum DIRECTIVE GAIN. Maximum Directive gain is obtained if maximum radiation intensity is present in desired direction.

Right choice is (a) \(A_e = \frac{dL^2\eta}{4R_{rad}}\)

The BEST I can explain: Maximum EFFECTIVE aperture is ratio of maximum POWER received to the AVERAGE power density. \(A_e=\frac{P_{Rmax}}{P_{avg}}\)

The received power is maximum when load equal to complex conjugate of network resistance.

⇨ \(I_{total}=\frac{V_{oc}}{2R_{rad}}\)

⇨ \(P_{Rmax}=I_{rms}^2 R_{rad}= (\frac{I_{total}}{\sqrt 2})^2 R_{rad}=\frac{V_{oc}^2}{8R_{rad}}=\frac{\mid E_\theta\mid^2 dL^2}{8R_{rad}} and P_{avg}=\frac{\mid E_\theta\mid^2}{2\eta} \)

\(A_e=\frac{P_{Rmax}}{P_{avg}}=\frac{dL^2\eta}{4R_{rad}}\)

Correct OPTION is (a) Reactive near-field

Easiest explanation: The portion of the near field immediate to the surrounding the ANTENNA is called Reactive near-field. The boundary of this REGION exists at a distance of R < 0.62√(D^3/λ) where D is the LARGEST dimension of the antenna. Radiating near field lies between reactive near field and far field. Far field is also known as Fraunhofer region.

The correct answer is (a) Power Pattern

To explain: Power pattern varies with square of magnitude of FIELD. The average power =\(\frac{\mid \overline{E}\mid^2}{2\eta}\). It is defined as the trace of received power at a CONSTANT RADIUS.

Right option is (d) End-fire ARRAY antenna has its main BEAM NORMAL to the axis containing antenna

The best explanation: End-fire array has its main beam parallel to the axis of antenna (θ=0° or 180°). For broadside antenna it is normal to the axis of antenna. Omni-directional antenna radiates power in only one DIRECTION and is non-radiating in other DIRECTIONS. So it is a special case of directional antenna.

The correct answer is (b) Load impedance must be equal complex conjugate to the antenna impedance

Easy explanation: Foe EFFECTIVE APERTURE to be maximum, the receiver power should be maximum. Maximum power transfer states that the maximum received power is obtained when the impedance of NETWORK matches with the complex conjugate of the load impedance. HENCE, Load impedance must be equal to the complex conjugate of the antenna impedance.

Right choice is (a) will decrease

The best EXPLANATION: From the Friss transmission EQUATION, the received power depends on the wavelength which is inversely PROPORTIONAL to the FREQUENCY. So the power decreases as the frequency increases.

\(\frac{P_r}{P_t} = \frac{G_t G_r λ^2}{(4πR)^2} = \frac{G_t G_r c^2}{(4πRf)^2} \)

Correct OPTION is (a) Isotropic

To EXPLAIN I would say: Isotropic radiators radiate POWER in all directions uniformly. Omni-DIRECTIONAL antenna radiates only in one direction. Directional antenna radiates maximum power in particular direction. Transducer converts one form of energy to other.

Correct option is (a) DIRECTIVITY

To explain: Directivity of antenna is defined as the ratio of radiation INTENSITY in a given direction from antenna to the radiation intensity over all directions. \(D = \FRAC{U_{max}}{U_0}.\)

 Radiation Intensity is power RADIATED from an antenna for unit solid ANGLE. \(U_0= w_r.r^2\frac{watts}{steradians}.\)

Gain of antenna is ratio of radiation intensity in given direction to the radiation intensity of isotropic radiation. Array factor is a function of geometry of array and the excitation phase.

The correct OPTION is (a) DECREASES

Explanation: As beam width of antenna INCREASES its AREA coverage broadens, THEREBY directivity decreases. Beam area and directivity are inversely proportional. \(D=\frac{4\pi}{Beam \,Area}.\)

Correct option is (a) 1.64

Easy explanation: DIRECTIVITY \(D=\FRAC{U_{max}}{U_{av}} = \frac{4\PI U_{max}}{P_{rad}} \)

Prad=36.54I0^2 Since the radiation resistance for Half-wave dipole is 36.54Ω.

Maximum radiation intensity is given by \(U_{max}=P_{max}.r^2=\frac{\mid E_\theta\mid_{max}^2}{2\eta}.r^2\)

Where \(E_\theta=\frac{j\eta I_0 e^{-jkr}cos⁡(\frac{\pi cos\theta}{2})}{2\pi rsin\theta}\)

⇨ \(D=\frac{4\pi}{1}\frac{\eta I_0^2}{8\pi^2} \frac{1}{36.54I_0^2}=1.64.\)

Right answer is (a) True

The BEST EXPLANATION: Directive gain is the RATIO of power density to the average power RADIATED.

\(G_d = \frac{P_{d(\theta,\emptyset)}}{P_{AVG}}\)

Correct choice is (b) Power GAIN

To explain I WOULD say: The ratio of power RADIATED in a particular direction to the actual power input to antenna is called Power gain. \(G_p=\frac{P_{d(\theta,\emptyset)}}{P_{in}}\). Directive gain is the ratio of power radiated in desired direction to the average power radiated from the antenna. PARTIAL directivity is the part of radiation INTENSITY in a particular polarization to radiation intensity in all directions.

The correct OPTION is (d) Far FIELD

The explanation is: The region of the field that angular distribution is independent of the distance from the antenna is CALLED far field. This region is present at d distance greater than 2D^2/λ. IDEALLY the outer boundary is taken at infinity.

Right option is (a) Low, high

The best EXPLANATION: Side lobe ratio is ratio of POWER density in side lobes to MAIN lobe. A receiving antenna is said to be efficient if side lobes are minimized and receives most of the transmitted signal. So it should have low SLR and high SNR.

Right option is (c) INCREASES by FACTOR 4

The explanation: From Friss transmission equation, \(P_r=P_t\frac{G_t G_r \lambda^2}{(4πR)^2}\)

\(\frac{P_{r1}}{P_{r2}} = \frac{R_2^2}{R_1^2} = \frac{(R/2)^2}{R^2} = \frac{1}{4}\)

Pr2=4Pr1.

Correct answer is (b) False

Best explanation: Friss transmission is used to FIND the receiver power in ANTENNA when power is transmitted from ANOTHER antenna. These are separated by a FAR ZONE distance.

The correct answer is (b) capacitive

The explanation: Antenna is a series R,L,C EQUIVALENT circuit.For FREQUENCIES less than resonant frequency, the impedance is capacitive.

The correct ANSWER is (a) 3.65kΩ

To elaborate: For a half-wave dipole the radiation RESISTANCE is 73Ω.

 \(I_{rms}=\FRAC{I_m}{\sqrt{2}}=\frac{10}{\sqrt{2}}=5\sqrt{2}\)

POWER radiated Prad=Irms^2 Rrad=(5√2)^2×73=3650Ω=3.65kΩ

Right ANSWER is (a) DECREASES

Explanation: As the DIRECTIVITY increases, the beam area decreases. So, the coverage area of beam decreases and takes more TIME to scan a target for target detection.

Right option is (a) 30mV/m

Explanation: For ISOTROPIC radiator, power per unit area = \(\frac{P_t}{4πr^2}\) — (1)

From pointing THEOREM, \(P=\frac{\mid E_{max}^2\mid}{2\eta}\), where η=120π. —– (2)

Equating (1) & (2) and Erms=Emax/√2, we get

\(E_{rms}=\frac{\SQRT{30P_t}}{r}=\frac{\sqrt{30×3000}}{10000}=30mV/m.\)

The correct answer is (a) Gpmax = ηrGdmax

The explanation is: MAXIMUM POWER GAIN is obtained when there are no ohmic LOSSES. Gpmax=\(\FRAC{U_{max}}{P_{in}/4π}\)

Maximum directive gainGdmax=\(\frac{U_{max}}{P_r/4π}\, and\, \eta_r=\frac{P_r}{P_{in}}\)

∴Gpmax=ηr Gdmax

Right option is (a) R » 2D^2/λ

Best EXPLANATION: The transmitting and receiving antennas are in a FAR ZONE to each other. So the separation DISTANCE between them is R » 2D^2/λ.

The CORRECT answer is (d) 0.977

Explanation: For a LOSSLESS antenna, the RADIATION efficiency ecd=1.

Overall efficiency of antenna is given by EO =ecd (1-\(\mid\gamma^2\mid)\)=1×(1-(0.15^2))=0.977.

Correct choice is (a) directivity

To explain: The RATIO of maximum power density in the desired direction to the average power radiated from the antenna is called Directivity. The ratio of power radiated in a PARTICULAR direction to the actual power INPUT to antenna is called Power gain. Directive gain is the ratio of power radiated in desired direction to the average power radiated from the antenna.Maximum Directive gain is called as Directivity. Partial directivity is the PART of radiation intensity in a particular polarization to radiation intensity in all directions.

The correct OPTION is (c) 4/π

The best I can explain: BEAM AREA ΩA≈ θ1r θ2rwhere θ1r, θ2r are half-power beam widths in radians.

Directivity \(D=\frac{4\pi}{\Omega_A}=\frac{4\pi}{\theta_{1R}\theta_{2r}}=\frac{4\pi}{\pi.\pi} = 4/\pi\)

The correct answer is (a) low

Best explanation: Fractional Bandwidth is the ratio of the frequency range of ANTENNA to the CENTER frequency. If Fractional bandwidth is high, then Bandwidth of the antenna is also more. So the quality FACTOR decreases as it is INVERSELY proportional to the bandwidth.

Right choice is (a) \(\frac{P_r}{P_t} = \frac{G_t G_r\lambda^2}{(4πR)^2}\)

EXPLANATION: Friss TRANSMISSION equation is used to calculate the power received by the receiving ANTENNA when transmitted from other antenna separated by a DISTANCE R. the equation is given by \(\frac{P_r}{P_t} = \frac{G_t G_r \lambda^2}{(4πR)^2}.\)

The CORRECT answer is (C) 10MHZ

The EXPLANATION is: Bandwidth \(Q=\frac{f_0}{Q}=\frac{200MHz}{20}=10MHz\)