Waveguides and Guided Waves MCQ Quiz - Objective Question with Answer for Waveguides and Guided Waves - Download Free PDF

Concept:

The cut-off frequency of a rectangular waveguide in dominant TE₁₀ mode is given by:

\( f_c = \frac{c}{2a} \), where:

c = speed of light = \( 3 \times 10^8~m/s \), and a = broader dimension of the waveguide

Calculation:

Given:

Dimensions of waveguide = 1 cm × 0.5 cm

So, a = 1 cm = 0.01 m

Using the formula, \( f_c = \frac{3 \times 10^8}{2 \times 0.01} \)

\( f_c = \frac{3 \times 10^8}{0.02} = 1.5 \times 10^{10}~Hz = 15~GHz \)

Correct Answer: 2) 15 GHz

Concept:

In optical communication, a star coupler is used to distribute optical signals among multiple ports. For an \( N \times N \) **bi-directional star coupler**, the number of **3-dB couplers** needed is given by the formula:

\( \frac{N}{2} \times \log_2 N \)

Given:

N = 64

Calculation:

\( \log_2 64 = 6 \)

\( \text{Number of 3-dB couplers} = \frac{64}{2} \times 6 = 32 \times 6 = 192 \)

Correct Answer: 3) 192

Concept:

The number of propagating modes in a multimode step-index fiber is approximately given by, \( M = \frac{V^2}{2} \)

Where V is the normalized frequency (V-number), and is given by, \( V = \frac{2\pi a}{\lambda} \cdot NA \)

NA is the numerical aperture and is defined as, \( NA = n_1 \cdot \sqrt{2\Delta} \)

Where, n₁ = core refractive index, Δ = relative refractive index difference, λ = operating wavelength, a = core radius

Calculation:

Given:

n₁ = 1.5, Δ = 2% = 0.02, λ = 1300 nm = 1.3 × 10^-6 m, M = 1000

Calculate NA: \( NA = 1.5 \cdot \sqrt{2 \cdot 0.02} = 1.5 \cdot \sqrt{0.04} = 1.5 \cdot 0.2 = 0.3 \)

Now, \( M = \frac{V^2}{2} \Rightarrow V^2 = 2M = 2000 \Rightarrow V = \sqrt{2000} \approx 44.72 \)

From, \( V = \frac{2\pi a}{\lambda} \cdot NA \)

\( 44.72 = \frac{2\pi a}{1.3 \times 10^{-6}} \cdot 0.3 \Rightarrow a = \frac{44.72 \cdot 1.3 \times 10^{-6}}{2\pi \cdot 0.3} \approx \frac{58.136 \times 10^{-6}}{1.884} \approx 30.85 \times 10^{-6}~m \)

Core diameter = 2a ≈ \( 61.7~\mu m \)

Correct Option:

The closest answer is: 3) 62 µm

Concept:

The maximum digital transmission rate over an optical fiber is limited by pulse dispersion, which spreads the signal in time and can cause inter-symbol interference (ISI).

For Return-to-Zero (RZ) data transmission, the bit interval must be at least equal to the total pulse spreading to avoid overlap between bits.

Calculation:

Given:

Fiber length = 10 km, Pulse spreading constant = 10 ns/km

Total pulse spread = \(10~\text{km} \times 10~\text{ns/km} = 100~\text{ns}\)

In unipolar RZ transmission, the maximum bit rate is the reciprocal of the bit period.

\( \text{Bit rate} = \frac{1}{\text{Bit interval}} = \frac{1}{100~\text{ns}} = 10~\text{Mbps} \)

Explanation:

Permittivity and Relative Permittivity

Definition: Permittivity is a measure of how much electric field is 'permitted' to pass through a material. It quantifies the ability of a material to store electrical energy in an electric field. The absolute permittivity (ε) of a material is the product of the relative permittivity (εr) and the permittivity of free space (ε0), where ε0 is a constant value approximately equal to 8.854 × 10^-12 F/m (farads per meter).

Relative permittivity, also known as the dielectric constant, is a dimensionless quantity that represents the ratio of the permittivity of a material to the permittivity of free space. It indicates how much better the material is at storing electrical energy compared to a vacuum. Mathematically, it is expressed as:

εr = ε / ε0

Working Principle: When an electric field is applied to a material, the material's molecules polarize, aligning themselves with the field. This polarization reduces the overall field within the material, allowing it to store more energy. The relative permittivity reflects the extent of this polarization effect. A higher relative permittivity means the material can store more electrical energy, while a lower relative permittivity means it stores less.

Correct Option Analysis:

The correct option is:

Option 2: reduces

When the actual permittivity (ε) of a material decreases, the relative permittivity (εr) also reduces. This relationship can be understood by revisiting the formula for relative permittivity:

εr = ε / ε0

If the absolute permittivity (ε) decreases while the permittivity of free space (ε0) remains constant, the ratio εr will decrease. This means the material's ability to store electrical energy compared to a vacuum diminishes, leading to a reduction in the relative permittivity.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: increases

This option is incorrect because a reduction in actual permittivity would not lead to an increase in relative permittivity. Since relative permittivity is directly proportional to the actual permittivity, a decrease in one leads to a decrease in the other.

Option 3: becomes zero

This option is incorrect because while a reduction in actual permittivity will lower the relative permittivity, it cannot become zero unless the actual permittivity itself becomes zero. Since actual permittivity is a positive value for all materials, relative permittivity will also remain positive and will not reach zero.

Option 4: remains the same

This option is incorrect because relative permittivity is a function of actual permittivity. If actual permittivity changes, relative permittivity will also change. Therefore, it cannot remain the same if actual permittivity is reduced.

Conclusion:

Understanding the relationship between actual permittivity and relative permittivity is crucial for analyzing the dielectric properties of materials. As explained, relative permittivity reduces when actual permittivity decreases because of their direct proportionality. This fundamental concept is essential in various fields of electrical engineering and material science, where the dielectric properties of materials play a significant role in the design and application of electronic components and systems.

Concept:

• Electric field \(\left( {\vec E} \right)\) & magnetic field \(\left( {\vec H} \right)\) are both orthogonal/transverse to each other as well as for the direction of propagation, called transverse electromagnetic wave (TEM).
• E & H combination called uniform plane wave because E & H has some magnitude through any transverse plane.

Intrinsic Impedance \(\left( \eta \right) = \frac{E}{H}\)

And also \(\eta = \sqrt {\frac{{j\omega \mu }}{{\sigma + j\omega \varepsilon }}} \)

But for free space

\( \Rightarrow \eta = \frac{E}{H} = \sqrt {\frac{\mu }{\varepsilon }} \)

Pointing vector is a vector whose direction is the direction of wave propagation pointing vector \( = \vec E \times \vec H\) (Hence it is a direction of wave propagation)

Where,

E = Electric filed

B = magnetic field

H = Magnetic field

μ0 = Permeability of free space = 4π x 10-7 H / m

ϵ0 = Permittivity of free space = 8.85 x 10-12 F/m

The correct answer is total internal reflection.

Key Points

• Optical fiber:
  • The working of an optical fiber is based on total internal reflection. Hence, option 2 is correct. 
  • Optical fibers consist of many long high-quality composite glass/quartz fibers. Each fiber consists of a core and cladding.
  • The refractive index of the core (μ1) material is higher than that of the cladding (μ2).
  • When the light is incident on one end of the fiber at a small angle, the light passes inside, undergoes repeated total internal reflections along with the fiber, and finally comes out.
  • The angle of incidence is always larger than the critical angle of the core material concerning its cladding.
  • Even if the fiber is bent, the light can easily travel through along the fiber.
• 

• Microwave tubes are used for high power/frequency generation.
• Tubes generate and amplify high levels of microwave power more cheaply than solid-state devices.

Microwave tubes are classified into two types:

Linear-Beam Tube:

• Klystrons and Traveling-Wave Tube(TWT) are examples of linear beam tubes with a focused electron beam (as in CRT).

Crossed-Field Tube:

• In these tubes, both the Magnetic and electric fields are at right angles to each other.

• The Magnetron is an example of cross-field tubes.

• Magnetrons are the cross-field tubes in which the electric and magnetic fields cross, i.e. run perpendicular to each other.

​

Step-index fiber:

1. The refractive index of the core is uniform throughout and undergoes on abrupt change at the core-cladding boundary.

2. The path of light propagation is zig-zag in a manner

Additional Information

Optical fibres are transparent fibres and act as a light pipe to transmit light between its two ends. They are made up of silicon dioxide.

Refracting index is changing between the cladding and the core in optical fibre

Graded-index fiber:

1. The refractive index of the core is made to vary gradually such that it is maximum at the center of the core.

2. The path of light is helical in manner.

This type of fiber optics works when the wavelength is much smaller than the core radius.

So, the refractive index of the core remains constant for step-index fiber.

Difference between step-index and graded-index fiber.

The data rate for optical fiber is 1 Gbps.

• In an optical fiber, the information is passed through light, which must not escape outside of it.
• This phenomenon of confining the light inside the optical fiber is termed as Total internal reflection.
• For this, the construction and material used to ensure that the total internal reflection of light takes place to prevent the escape of it.

The data rate of different communication systems is:

Concept:

The dominant mode in a particular waveguide is the mode having the lowest cut-off frequency.

The cut-off frequency for a rectangular waveguide with dimension ‘a (length)’ and ‘b (width)’ is given as:

\(\Rightarrow {{f}_{c\left( mn \right)}}=\frac{c}{2}\sqrt{{{\left( \frac{m}{a} \right)}^{2}}+{{\left( \frac{n}{b} \right)}^{2}}}\)

'm' and 'n' represents the possible modes.

Application:

For a standard rectangular waveguide a > b;

The minimum frequency is obtained when m = 1 and n = 0,

\(i.e.~{{f}_{c\left( 10 \right)}}=\frac{c}{2}\sqrt{\frac{{{m}^{2}}}{{{a}^{2}}}}=\frac{c}{2a}\)

• For a rectangular waveguide with b > a, TE₀₁ will be the dominant mode with the lowest cut-off frequency.
• TEM mode cannot exist in Hollow conductor waveguide
• Circular and rectangular are hollow waveguides hence there is no TEM mode in them, they can support only TE or TM but not TEM mode
• The transmission line, parallel plate waveguide, coaxial cable can have TEM mode

Waveguides only allow frequencies above cut-off frequency and do not pass below the cut-off frequencies.

Hence it acts as a high pass filter.

The cut off frequency is given as:

\({{\rm{\lambda }}_{\rm{C}}} = \frac{2}{{\sqrt {{{\left( {\frac{{\rm{m}}}{{\rm{a}}}} \right)}^2} + {{\left( {\frac{{\rm{n}}}{{\rm{b}}}} \right)}^2}} }}\)

Where a and b are the dimensions of the waveguide (a>b)

• Single-mode step indexed fibers are widely used for wideband communications are preferred for long-distance communication.
• Single-mode step-index fiber is used to eliminate modal dispersion during optical communication.
• In this fiber, a light ray can travel on only one path so minimum refraction takes place hence, no pulse spreading permits high pulse repetition rates.

Advantages of single-mode fiber:

1) Low signal loss

2) No modal dispersion

3) Does not suffer from modal dispersion

4) Can be used for higher bandwidth applications

5) Long-distance applications

6) Cable TV ends

7) High speed local and wide area network

Single-mode means the fiber enables one type of light mode to be propagated at a time. This is explained with the help of the following diagram:

Single-mode fiber core diameter is much smaller than multimode fiber.

2) For single-mode fiber, the B.W ranges from 50 to 100 GHz/km

Multimode fibers:

Fibers that carry more than one mode are called multimode fibers. There are two types of multimode fibers:

1) Step Index

2) Graded Index

The comparison of the refractive index profile for step and graded fibers are respectively shown as:

The multimode step-index multimode fiber suffers from Modal dispersion.

• Rays of light enter the fiber with different angles to the fiber axis. The limit is the fiber’s acceptance angle.
• Rays that enter with a shallow angle travel a more direct path and arrive sooner than those that enter at steeper angles.
• This arrival of different modes of light at different times is called modal dispersion.

• Optical fibers are cylindrical solid glass material acting as waveguides made of two concentric layers of very pure glass.
• The core (the interior layer) with refractive index n₁ serves as the medium for light propagation, while the cladding (the exterior layer) has a lower refractive index n₂ where n₁ > n₂ assuring that light rays are reflected the core.
• Since the cladding does not absorb any light from the core, the light wave can travel great distances.

Explanation:

• The working principle of optical fibers is Total Internal Reflection.
• Optical fiber mostly used for communication purposes with negligible loss of energy.
• The “Total Internal Reflection” of light is the boundary between transparent media of two different refractive indices.
• At present, Optical fiber cables are used for communication like sending images, voice messages, etc.
• The designing of this cable is done with Plastic or glass so that data can be transmitted effectively and quickly than other modes of communications

Optical Fiber:

• In an optical fiber, the information is passed through light, which must not escape outside of it.
• This phenomenon of confining the light inside the optical fiber is termed Total internal reflection.
• For this, the construction and material used to ensure that the total internal reflection of light takes place to prevent the escape of it.

• In any type of optical fiber, the refractive index of the core is always greater than the refractive index of the cladding.
• It is this property of core and cladding which makes light propagate inside the fiber. 
• Thus, In a step-index optical fiber refractive index of the core is higher than the cladding

Principle:

• When light travels from a high refractive index medium to a low-refractive-index medium, it is refracted away from the normal as shown:

• At a certain angle θi, there is no refracted wave and the wave is totally internally reflected  (θr = 90°).
• This angle is called a critical angle.
• Inside an optical fiber, we have a high refractive index core (n1) and low refractive index cladding (n2).
• This results in the propagation of waves inside a fiber through total internal reflection phenomenon.