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Correct OPTION is (b) ld, LDE, lr

Explanation: The lengths of reflector, driven element, director is in the orderlr > lde > ld.

So, the increasing ORDER is ld, lde, lr .

Right option is (a) Only thickness of CONDUCTORS

Explanation: The shortening effect on the multi conductors DEPENDS on the thickness of the conductor whereas in WIRES it depends on the length and the FREQUENCY of OPERATION.

Correct choice is (c) n^2Z

For EXPLANATION: For a multi conductor FOLDED dipole with n CONDUCTORS, the IMPEDANCE is n^2Z.

Where value of radiation RESISTANCE Z depends on the length of the dipole used.

The correct CHOICE is (c) Both SUPER directive and super gain

The EXPLANATION: A Yagi-Uda antenna provides high directivity by increasing reflectors and gain due to the directors. Directors will increase the FORWARD gain of the antenna. So it is both super directive and super gain antenna.

The correct option is (a) HF-UFH

The best I can explain: Yagi-Uda antenna operates mostly in the HF to UFH BAND frequency. IT ranges from 3MHZ to 3GHz.

VLF-MF: 3 kHz to 3MHz

LF-HF: 30 kHz to 30MHz

UHF –EHF: 300MHZ to 300GHz.

Right option is (a) Director

For explanation I WOULD say: Directors add the field of the DRIVEN element and will EXCITE the next parasitic element. Directors will increase the gain of the antenna in the forward direction. REFLECTORS will add fields of the driven element in the direction from reflector to driven element.

Right answer is (C) Driven ELEMENT

For explanation I would say: The dipole to which the POWER is applied directly from the feeder in the Yagi-Uda antenna is called driven element. Directors add the field of the driven element and will excite the next parasitic element. REFLECTORS will increase the DIRECTIVITY of antenna.

Correct choice is (d) Radii, radiation resistance and space between conductors.

For explanation I WOULD say: If the radii of two conductors are r1 (FEED arm), r2 (non feed arm) and separated by a distance, then the IMPEDANCE transformation for folded dipole is given by

\(Z = Z’\left(1+\frac{log⁡(\frac{a}{r1})}{log⁡(\frac{a}{r2})}\right)^2\) and Z’ is the radiation resistance of the dipole USED (which DEPENDS on the length of the dipole).

So the overall impedance depends on radii, radiation resistance of the dipole and space between conductors.

The correct ANSWER is (a) True

The explanation: In unequal conductor folded dipole, the diameters of the two conductors connecting TOGETHER are unequal. If the radii of two conductors are r1 (feed arm), r2 (NON feed arm) and SEPARATED by a distance, then the impedance transformation for folded dipole is given by

\(Z = Z’\LEFT(1+\frac{log⁡(\frac{a}{r1})}{log⁡(\frac{a}{r2})}\right)^2.\)

The correct choice is (a) 0.95

Easiest explanation: For a 3 ELEMENT Yagi-Uda array,

DRIVEN element length (feet) = 475/f (MHZ) = 475/500 = 0.95 feet.

Correct answer is (a) 0.525λ

Explanation: The reflector length is approximately 5% greater than the DRIVEN ELEMENT length.

So, l = 0.525λ is correct as it is the only OPTION which is greater than 0.5λ.

The correct answer is (a) Multi conductor folded DIPOLE

Easiest EXPLANATION: In Multi conductor folded dipole, the two or more conductors are connected together. This INCREASES the impedance and ALSO the BANDWIDTH.

Right CHOICE is (b) End-fire

Explanation: RADIATION PATTERN of a Yagi-Uda antenna is end-fire. It has its main beam PARALLEL to the axis of antenna (boom). The ADDITION of directors will increase the gain of antenna while reflectors will increase the directivity of antenna.

Right answer is (d) 292Ω

To explain I would SAY: Half WAVE FOLDED dipole impedance is 4 TIMES the half wave dipole.

⇨ Z = 4 × 73 = 292Ω.

Right option is (a) Greater

Best EXPLANATION: Reflector is used to reflect all the energy towards the RADIATION direction. It increases the front to back lobe ratio. So ALWAYS the reflector length is greater than the driven element length.

Right choice is (a) True

Best explanation: Folded dipole PRODUCES flatter IMPEDANCE v/s frequency COMPARED to single dipole. So, folded dipole is USED in Yagi-Uda to OBTAIN wider frequency range.

Right answer is (C) Parasitic element

Explanation: We use REFLECTORS and DIRECTORS which are passive elements ALSO KNOWN as the parasitic elements to increase the directivity and gain if the antenna.

Correct ANSWER is (b) \(73\left(1+\frac{log⁡(\frac{a}{r1})}{log⁡(\frac{a}{r2})}\right)^2\)

To elaborate: If the radii of two CONDUCTORS are r1 (FEED arm), r2 (non feed arm) and separated by a DISTANCE, then the impedance transformation for folded dipole is given by

\(Z = Z’\left(1+\frac{log⁡(\frac{a}{r1})}{log⁡(\frac{a}{r2})}\right)^2\)

For a half dipole Z’=73Ω

∴ \(Z = 73\left(1+\frac{log⁡(\frac{a}{r1})}{log⁡(\frac{a}{r2})}\right)^2\)

Right choice is (a) 0.4m

Best EXPLANATION: λ = \(\frac{c}{F} = \frac{3×10^8}{150MHz}=2M \)

The DISTANCE between directors of Yagi-Uda antenna is 0.2λ

⇨ D=0.2λ=0.2(2)=0.4m

The CORRECT CHOICE is (a) 0.2λ

To elaborate: The distance between directors of Yagi-Uda antenna is 0.2λ. The directors will INCREASE the GAIN.

Correct answer is (b) Gain

The best EXPLANATION: REFLECTORS will INCREASE the DIRECTIVITY of antenna by reflecting all energy towards RADIATION direction of antenna. Addition of directors leads to focus the beam in the forward direction. So, directors will increase the gain of antenna.

Right OPTION is (b) 657

Explanation: For a half WAVE FOLDED dipole of 3 wires the impedance = n^2×73=9×73=657

Right answer is (a) 0.91

The BEST explanation: For a 3 element Yagi-Uda ARRAY,

Director element LENGTH (feet) = 455/f (MHZ) = 455/500 = 0.91 feet.

Right answer is (d) Wire

To elaborate: FOLDED DIPOLE ANTENNA belongs to wire antenna. It is a dipole antenna with two ends folded BACK and connected to each other forming a loop.

Correct option is (B) Smaller

The BEST I can EXPLAIN: The length of the director is less than the DRIVEN element. It is shorter than half wavelength of the dipole.Director is USED to increase the gain of antenna.