Electromagnetic theory miscellaneous Easy Questions and Answers | Page - 7

Electromagnetic theory miscellaneous

The ground wave coverage of the medium wave transmitter is 100 km and in the height the first reflected ray is at 800 km. The skip distance is—

1. 900 km
1. 700 km
1. 100 km
1. 800 km

View Hint View Answer Discuss in Forum

Skip distance is the minimum distance where the reflected wave returns back to earth so a wave transmit ground wave coverage of 100 km and then reflected back to the earth at 800 km from transmitter so skip distance is 700 km.

Correct Option: B

Skip distance is the minimum distance where the reflected wave returns back to earth so a wave transmit ground wave coverage of 100 km and then reflected back to the earth at 800 km from transmitter so skip distance is 700 km.

In a broadside array of 20 isotropic radiators equally spaced at a distance of λ/2. The beamwidth between first nulls is—

1. 51.3 degrees
1. 11.46 degrees
1. 22.9 degrees
1. 102.6 degrees

View Hint View Answer Discuss in Forum

For broadside array of 20 point sources. We have general formula for the array of n point sources have been as follow
2φ = 2 sin^-1 m λ
nd

where
m = no. of null 1^st or 2^nd
d = separation between the point sources
n = no. of point sources
for first nulls we take m = 1, n = 20 is given, so we get
2φ = 2 sin^-1 λ = 2 ×5.739
20 × λ/2

= 11.478º

Correct Option: B

For broadside array of 20 point sources. We have general formula for the array of n point sources have been as follow
2φ = 2 sin^-1 m λ
nd

where
m = no. of null 1^st or 2^nd
d = separation between the point sources
n = no. of point sources
for first nulls we take m = 1, n = 20 is given, so we get
2φ = 2 sin^-1 λ = 2 ×5.739
20 × λ/2

= 11.478º

A radiowave is incident on a layer of ionosphere at an angle of 30 degree with the vertical common. If the critical frequency is 1.2 MHz the maximum usable frequency is—

1. 1.2 MHz
1. 2.4 MHz
1. 0.6 MHz
1. 1.386 MHz

View Hint View Answer Discuss in Forum

Relation between maximum usable frequency and critical frequency with incident angle is given by as follows
f_muf = f_c sec i
∴ f_muf = 1.2 × 2 × 10⁶
√3

= 1.3856 × 10⁶
= 1.385 MHz

Correct Option: D

Relation between maximum usable frequency and critical frequency with incident angle is given by as follows
f_muf = f_c sec i
∴ f_muf = 1.2 × 2 × 10⁶
√3

= 1.3856 × 10⁶
= 1.385 MHz

When a wave is propagated in a good dielectric where σ/ωε << 1 the attenuation factor and the phase shift factor are given by—

1. α = σ √ μ , β = ω √με
  2 ε
1. α = σ , β = ω √με
  2
1. α = σ √ μ √ μ , β = √ ωμε
  2ε ε 2
1. α = 1, β = 45°

View Hint View Answer Discuss in Forum

Refer synopsis

Correct Option: A

Refer synopsis

An antenna having resonant frequency f and having a Q factor of 20 can handle a bandwidth of 10 MHz then f is—

1. 10 MHz
1. 20 MHz
1. 200 MHz
1. None of these

View Hint View Answer Discuss in Forum

We have important relation of quality factor, bandwidth and cut-off frequency
Q = f ⇒ f = 20 ⇒ f = 200 MHz
BW 10 × 10⁶

where,
Q = quality factor
f = resonance frequency
BW = bandwidth

Correct Option: C

We have important relation of quality factor, bandwidth and cut-off frequency
Q = f ⇒ f = 20 ⇒ f = 200 MHz
BW 10 × 10⁶

where,
Q = quality factor
f = resonance frequency
BW = bandwidth