Cutoff Frequency MCQ Quiz - Objective Question with Answer for Cutoff Frequency - Download Free PDF

Cut-off Frequency of TEM Wave

Definition: The cut-off frequency of a wave refers to the minimum frequency below which the propagation of a particular wave mode ceases in a specific medium or waveguide. For a Transverse Electromagnetic (TEM) wave, the cut-off frequency is a critical parameter that determines the conditions under which the wave can propagate through a waveguide or transmission line.

Correct Option Analysis:

The correct answer is Option 2: DC. This implies that the cut-off frequency for a TEM wave is zero, meaning that TEM waves can propagate even at very low frequencies (including DC). This unique property arises due to the nature of TEM waves, where both the electric and magnetic fields are perpendicular to each other and to the direction of wave propagation.

Why is the Cut-off Frequency for TEM Waves Zero?

• Structure of TEM Waves: TEM waves consist of electric and magnetic fields that are transverse to the direction of propagation. Unlike other wave modes (such as TE or TM modes), TEM waves do not have any longitudinal field components.
• Propagation Conditions: TEM waves do not require a minimum frequency to propagate. As a result, they can exist even at a frequency of zero, which corresponds to DC (Direct Current).
• Waveguide Design: In structures such as coaxial cables or two-wire transmission lines, the geometry allows TEM waves to propagate without requiring a specific minimum frequency.

Applications of TEM Waves:

• Used in coaxial cables for transmitting signals over a wide range of frequencies, including DC.
• Ideal for low-frequency applications due to their zero cut-off frequency.
• Widely used in RF and microwave engineering for signal transmission.

Conclusion:

The zero cut-off frequency of TEM waves makes them highly versatile and suitable for applications across a broad spectrum of frequencies, including DC. This property distinguishes TEM waves from other modes, such as TE and TM modes, which have non-zero cut-off frequencies.

Explanation:

Waveguide and Guide Wavelength

A waveguide is a physical structure that guides electromagnetic waves from one point to another. It is commonly used for high-frequency signal transmission, such as microwave and RF (radio frequency) signals.

The guide wavelength refers to the wavelength of the signal within the waveguide, which differs from the wavelength in free space due to the boundary conditions imposed by the waveguide structure.

Guide Wavelength Formula:

The relationship between the guide wavelength (λ_g), the free space wavelength (λ₀), and the cutoff wavelength (λ_c) is given by the formula:

λ_g = λ₀ / √(1 - (λ₀ / λ_c)²)

Where:

• λ_g: Guide wavelength
• λ₀: Free space wavelength (c/f, where c is the speed of light and f is the frequency)
• λ_c: Cutoff wavelength of the waveguide, determined by its dimensions and the operating mode

From this equation, it is evident that λ_g is always longer than λ₀ when the operating frequency is above the cutoff frequency (ensuring wave propagation).

The correct option is: Option 1: Guide wavelength is longer than free space wavelength.

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:

Minimum operating frequency or the cut off frequency for a rectangular waveguide is given by:

\({f_c} = \frac{c}{2}\sqrt {\frac{{{m^2}}}{{{a^2}}} + \frac{{{n^2}}}{{{b^2}}}} \)

a = length of the waveguide

b = height of the waveguide

m,n = modes of operation

Calculation:

Given, a = 4 cm

b = 3 cm

The minimum frequency in TE₁₁ is nothing but the cut-off frequency calculated as:

\({f_c} = \frac{{3 \times {{10}^8}}}{2}\sqrt {\frac{1}{{{{\left( {4 \times {{10}^{ - 2}}} \right)}^2}}} + \frac{1}{{{{\left( {3 \times {{10}^{ - 2}}} \right)}^2}}}} \)

\(f_c= \frac{{3 \times {{10}^8}}}{{2 \times {{10}^{ - 2}}}} \times \frac{5}{{4 \times 3}} = 6.25 \times {10^9}Hz\)

f_C = 6.25 GHz

Concept:

The cutoff frequency of waveguide is given by:

\({f_c} = \frac{c}{{2\sqrt {{\mu _b}{\epsilon_b}} }}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \)

where c = 3 × 108 m/s

The lowest propagating mode is TE10 cutoff frequency

\({f_c} = \frac{c}{{2a}}\)

\(a = \frac{c}{{2{f_c}}}\)

Calculations:

Given:

a = 22.86 mm and b = 10.6 mm

As a > b so,

\({f_c} = \frac{c}{{2a}}\)

\({f_c} = \frac{3\times10^8}{2\times22.86 \ mm}\)

f_c = 6.5 GHz

As input applied frequency is less than the cut-off frequency, the mode is non-propagating. The mode can be made propagating if dielectric with a proper dielectric constant is inserted because it decreases phase velocity.

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)

Concept:

• All electromagnetic waves consist of electric and magnetic fields propagating in the same direction of travel, but perpendicular to each other.
• Along the length of a normal transmission line, both electric and magnetic fields are perpendicular (transverse) to the direction of wave travel. This is known as the principal mode, or TEM (Transverse Electric and Magnetic) mode.
• This mode of wave propagation can exist only where there are two conductors, and it is the dominant mode of wave propagation where the cross-sectional dimensions of the transmission line are small compared to the wavelength of the signal.
• The cutoff frequency of TEM wave is zero. 

Concept:

The guided wavelength of a rectangular Waveguide is given by:

\({{\lambda }_{g}}=\frac{\lambda }{\sqrt{1-{{\left( \frac{{{f}_{c}}}{f} \right)}^{2}}}}\)

Where, f_c = Cut-off frequency.

f = Operating frequency

λ = Operating wavelength

Calculation:

For \(T{{E}_{10}}~mode\Rightarrow {{f}_{c1}}=\frac{c}{2}\sqrt{{{\left( \frac{m}{a} \right)}^{2}}+{{\left( \frac{n}{b} \right)}^{2}}}\)

With m = 1 and n = 0,

\(\Rightarrow {{f}_{c1}}=\frac{c}{2}\sqrt{{{\left( \frac{1}{a} \right)}^{2}}}=\frac{c}{2a}\)

\({{f}_{c1}}=\frac{3\times {{10}^{8}}}{2\times 75\times {{10}^{-3}}}Hz\)

\({{f}_{c1}}=\frac{3\times {{10}^{11}}}{150}=2~GHz\)

Similarly,

f_c2 for TE₂₀ will be,

\({{f}_{c2}}=\frac{c}{2}\sqrt{{{\left( \frac{2}{a} \right)}^{2}}}\)

\(=\frac{c}{2}\times \frac{2}{a}=\frac{c}{a}=\frac{3\times {{10}^{8}}}{75\times {{10}^{-3}}}=4~GHz\)

Given, Guide wavelength is same for both f₁ and f₂, i.e. λ_g1 = λ_g2

i.e. \(\frac{\frac{c}{{{f}_{1}}}}{\sqrt{1-{{\left( \frac{{{f}_{c1}}}{{{f}_{1}}} \right)}^{2}}}}=\frac{\frac{c}{{{f}_{2}}}}{\sqrt{1-{{\left( \frac{{{f}_{c2}}}{{{f}_{2}}} \right)}^{2}}}}\)

\(\Rightarrow \sqrt{f_{1}^{2}-f_{c1}^{2}}=\sqrt{f_{2}^{2}-f_{c2}^{2}}~\)

\(\Rightarrow \sqrt{{{\left( \sqrt{13} \right)}^{2}}-{{\left( 2 \right)}^{2}}}=\sqrt{f_{2}^{2}-{{\left( 4 \right)}^{2}}}\)

\(\sqrt{13-4}=\sqrt{f_{2}^{2}-16}\)

\( f_{2}^{2}-16=9\)

\(\Rightarrow f_{2}^{2}=25\)

f₂ = 5 GHz

Calculation:

Guided wavelength is given by:

\({\lambda _g} = \;\frac{\lambda }{{\sqrt {1 - {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} \;}}\) ; where

f_c = Cut-off frequency

And f = Operating frequency.

For waveguide R­₁ : f₁ = 2GHz and f_c1 = 1 GHz

For waveguide R₂ : f₂ = 4 GHz_. We are to find fc­₂.

Given λ_g1 = λ_g2

Equating the wavelength for Both waveguides ;

\( \Rightarrow \frac{{{\lambda _1}}}{{\sqrt {1 - {{\left( {\frac{{{f_{c1}}}}{{{f_1}}}} \right)}^2}} }} = \;\frac{{{\lambda _2}}}{{\sqrt {2 - {{\left( {\frac{{f{c_2}}}{{{f_2}}}} \right)}^2}} }}\)

\(\frac{{c/{f_1}}}{{\sqrt {1 - {{\left( {\frac{{f{c_1}}}{{{f_1}}}} \right)}^2}} }} = \frac{{c/{f_2}}}{{\sqrt {1 - {{\left( {\frac{{{f_{c2}}}}{{{f_2}}}} \right)}^2}\;} }}\)

\(\sqrt {f_1^2 - f_{c1}^2} = \;\sqrt {f_2^2 - f_{c2}^2} \;\)

\( \Rightarrow \sqrt {4 - 1} = \sqrt {16 - f_{c2}^2} \)

⇒ 3 = 16 – fc₂²

fc₂² = 13

\({f_{c2}} = \sqrt {13} \)

Concept:

For a TE_MN mode of Electromagnetic wave;

\({{\text{H}}_{\text{z}}}={{\text{H}}_{\text{z}0}}\cos \left( \frac{m\pi }{a}.x \right)\cos \left( \frac{n\pi }{b}.y \right){{e}^{-\gamma z}}.~{{e}^{j\omega z}}~.~{{\hat{a}}_{z}}\)

Calculation:

Comparing the given equation with the standard equation;

m = 1, and n = 1.

\({{H}_{z}}={{H}_{0}}\cos \left( \frac{\pi x}{\sqrt{8}} \right)\cos \left( \frac{\pi y}{\sqrt{8}} \right)~A/m~\)

Given a = b = √8 cm.

\({{f}_{c}}=\frac{c}{2}\sqrt{{{\left( \frac{1}{a} \right)}^{2}}+{{\left( \frac{1}{b} \right)}^{2}}}=\frac{c}{2}\times \frac{1}{a}\times \sqrt{2}\)

\({{f}_{C}}=\frac{3\times {{10}^{10}}}{2}\times \frac{1}{\sqrt{8}}\times \sqrt{2}\)

\({{f}_{C}}=\frac{3\times {{10}^{10}}\times \sqrt{2}}{2\times 2\sqrt{2}}=0.75\times {{10}^{10}}~cm\)

Concept:

Minimum operating frequency or the cut off frequency for a rectangular waveguide is given by:

\({f_c} = \frac{c}{2}\sqrt {\frac{{{m^2}}}{{{a^2}}} + \frac{{{n^2}}}{{{b^2}}}} \)

a = length of the waveguide

b = height of the waveguide

m,n = modes of operation

Calculation:

Given, a = 4 cm

b = 3 cm

The minimum frequency in TE₁₁ is nothing but the cut-off frequency calculated as:

\({f_c} = \frac{{3 \times {{10}^8}}}{2}\sqrt {\frac{1}{{{{\left( {4 \times {{10}^{ - 2}}} \right)}^2}}} + \frac{1}{{{{\left( {3 \times {{10}^{ - 2}}} \right)}^2}}}} \)

\(f_c= \frac{{3 \times {{10}^8}}}{{2 \times {{10}^{ - 2}}}} \times \frac{5}{{4 \times 3}} = 6.25 \times {10^9}Hz\)

f_C = 6.25 GHz

Concept:

The cut-off frequency for a rectangular waveguide is given by:

\({f_{m,\;n}} = \frac{C}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \)

C = velocity of light in free space.

m, n = mode of operation of the waveguide

a, b = dimensions of the waveguide

Calculation:

For TE_{2, 1} mode, m = 2 and n = 1

Also, a = 5 cm

b = 3 cm

The cut-off frequency will be:

\({f_{c\left( {2,\;1} \right)}} = \frac{{3\; \times \;{{10}^{10}}}}{2}\sqrt {{{\left( {\frac{2}{5}} \right)}^2} + {{\left( {\frac{1}{3}} \right)}^2}} \)

\( = 1.5 \times {10^{10}}\sqrt {0.16 + 0.11} Hz\)

= 1.5 × 10¹⁰ × 0.52

f_{c(2, 1)} = 7810.24 MHz

Concept:

The cutoff frequency of TE_mn mode in the rectangular waveguide is given as:

\({f_c} = \frac{C}{{2\pi }}\sqrt {{{\left( {\frac{{m\pi }}{a}} \right)}^2} + {{\left( {\frac{{n\pi }}{b}} \right)}^2}} \) 

f_c for TE₁₀ will be:

\({f_{c\left( {10} \right)}} = \frac{C}{{2\pi }}\sqrt {{{\left( {\frac{{1\pi }}{a}} \right)}^2}} = \frac{C}{{2a}}\)      ---(1)

f_c for TE₂₀ will be:

\({f_{c\left( {20} \right)}} = \frac{C}{{2\pi }}\sqrt {{{\left( {\frac{{2\pi }}{a}} \right)}^2}} = \frac{C}{a}\)      ---(2)

f_c for TE11 will be:

\({f_{c\left( {11} \right)}} = \frac{c}{{2\pi }}\sqrt {{{\left( {\frac{\pi }{a}} \right)}^2} + {{\left( {\frac{\pi }{n}} \right)}^2}} \)

\( = \frac{c}{2}\sqrt {{{\left( {\frac{1}{a}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2}} \;\)       ---(3)

Calculation:

Given:

\({f_c}\left( {T{E_{11}}} \right) = \frac{{{f_c}\left( {T{E_{10}}} \right) + {f_c}\left( {T{E_{20}}} \right)}}{2}\) 

From equation 1, 2, 3, we can write:

\(\frac{c}{2}\sqrt {{{\left( {\frac{1}{a}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2}} = \frac{{\frac{c}{{2a}} + \frac{c}{a}}}{2}\) 

\(\sqrt {{{\left( {\frac{1}{a}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2}} = \frac{1}{{2a}} + \frac{1}{a} = \frac{3}{{2a}}\) 

\({\left( {\frac{1}{a}} \right)^2} + {\left( {\frac{1}{b}} \right)^2} = \frac{9}{{4{a^2}}}\) 

a = √5

\({\left( {\frac{1}{b}} \right)^2} = \frac{5}{{4{a^2}}}\) 

\(b = \frac{{2a}}{{\sqrt 5 }}\) 

b = 2 cm

Concept:

The cutoff frequency of waveguide is given by:

\({f_c} = \frac{c}{{2\sqrt {{\mu _b}{\epsilon_b}} }}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \)

where c = 3 × 108 m/s

The lowest propagating mode is TE10 cutoff frequency

\({f_c} = \frac{c}{{2a}}\)

\(a = \frac{c}{{2{f_c}}}\)

Calculations:

Given:

a = 22.86 mm and b = 10.6 mm

As a > b so,

\({f_c} = \frac{c}{{2a}}\)

\({f_c} = \frac{3\times10^8}{2\times22.86 \ mm}\)

f_c = 6.5 GHz

As input applied frequency is less than the cut-off frequency, the mode is non-propagating. The mode can be made propagating if dielectric with a proper dielectric constant is inserted because it decreases phase velocity.

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:

The minimum cutoff frequency in a rectangular waveguide is for TE₁₀ 

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

The cut off frequency is given by:

\({f_c} = \frac{c}{{2a}} = \frac{{3 \times {{10}^8}}}{{2\times1 \times {{10}^{ - 2}}}} = 15\;GHz\)

It will be the same for both since the cut-off frequency is only dependent on broadside length (a).