Cutoff Frequency MCQ Quiz in தமிழ் - Objective Question with Answer for Cutoff Frequency - இலவச PDF ஐப் பதிவிறக்கவும்

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:

Waveguides only allow frequencies above the cut-off frequency to pass through. It blocks or attenuates the frequencies below the cut-off frequencies.

The cut off frequency is mathematically calculated as:

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

\(v=\frac{1}{\sqrt{\mu \epsilon}}=\frac{1}{\sqrt{\mu_r\mu_0\epsilon_r\epsilon_0}}=\frac{c}{\sqrt{\mu_r\epsilon_r}}\)

c = 3 × 10¹⁰ cm/sec

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

Where a and b are the dimensions of the waveguide.

Calculation:

Since in the given rectangular waveguide a > b so, the dominant mode is TE₁₀ and the cutoff frequency for the dominant mode will be:

\({f_c} =\frac{c}{2\sqrt{\mu_r\epsilon_r}}\sqrt {{{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{0}{b}} \right)}^2}} \)

\({f_c} =\frac{3× 10^{10}}{2\sqrt{4}}× \frac{1}{4}\)

f_c = 0.1875 × 10¹⁰ Hz

fc = 1.875 GHz

Concept:

• Waveguides only allow frequencies above the cut-off frequency to pass through.
• A wave mode propagates in a waveguide only if its frequency is greater than the cutoff frequency.
• If there is no propagating mode inside the waveguide then the energy in the propagating mode is zero.
• So, the average power flow down the waveguide below cutoff frequency is zero.

∴ Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I)

The cut off frequency is mathematically calculated as:

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

Where a and b are the dimensions of the waveguide.

Concept:

Waveguides only allow frequencies above the cut-off frequency to pass through. It blocks or attenuates the frequencies below the cut-off frequencies.

The cut off frequency is mathematically calculated as:

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

Where a and b are the dimensions of the waveguide

Calculation:

f_C = 9 GHz for TE₂₀ mode

f_C = 12 GHz for TE₁₁ mode

For TE₂₀ mode:

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

\(9 \times {10^9} = \frac{{3\; \times \;{{10}^8}}}{a}\)

a = 3.33 cm

For T_E11 mode:

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

\(\frac{{12\; \times \;{{10}^9}\; \times \;2}}{{3\; \times \;{{10}^8}}} = \sqrt {{{\left( {\frac{1}{a}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2}} \)

\(\frac{{12\; \times\; {{10}^9}\; \times \;2}}{{3\; \times\; {{10}^8}}} = \sqrt {{{\left( {\frac{1}{a}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2}} \)

\(6400 = {\left( {\frac{1}{a}} \right)^2} + {\left( {\frac{1}{b}} \right)^2}\)

\(b = \frac{1}{{74.161}}\;m\)

b = 1.348 cm

Concept:

The frequency of the dominant mode in a rectangular waveguide is given by:

\({f_0} = \frac{1}{{2\sqrt {\mu \epsilon} }}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \)

'm' and 'n' is the mode of operation for the rectangular waveguide.

Calculation:

For an air-filled rectangular waveguide:

\(c = \frac{1}{\sqrt {\mu \varepsilon}}\)

With a > b, TE₁₀ will be the dominant mode of the waveguide with the cut-off frequency as:

\({\left( {{f_0}} \right)_{10}} = \frac{c}{2} \times \frac{1}{a} \)

\((f_0)_{10}= \frac{c}{{2a}}\)

Now, the next higher-order mode of the the waveguide will be TE₀₁. ∴ The cut-ff frequency will be:

\({\left( {{f_0}} \right)_{01}} = \frac{c}{{2b}}\)

For the condition that the design frequency is 10% larger than the cutoff frequency of dominant mode while being 10% lower than the cutoff frequency, we can write:

\(f = 1.1{\left( {{f_0}} \right)_{10}} =0.9{\left( {{f_0}} \right)_{01}}\)   ---(1)

Since the operating frequency of the waveguide is: f = 5 GHz = 5 × 10⁹ Hz, we get:

\({\left( {{f_0}} \right)_{10}} = \frac{{5 \times {{10}^9}}}{{1.1}}\)

\(\frac{c}{{2a}} = \frac{{5 \times {{10}^9}}}{{1.1}}\)

\(a = \frac{{\left( {3 \times {{10}^8}} \right) \times 1.1}}{{2 \times \left( {5 \times {{10}^9}} \right)}} \)

a = 3.3 cm

Similarly \({\left( {{f_0}} \right)_{01}} = \frac{{5 \times {{10}^9}}}{{0.9}}\)

\(\frac{c}{{2b}} = \frac{{5 \times {{10}^9}}}{{0.9}}\)

\(b = \frac{{\left( {3 \times {{10}^8}} \right) \times 0.9}}{{2 \times \left( {5 \times {{10}^9}} \right)}}\)

b = 2.7 cm

Concept:

if a > b, TE₁₀ is the dominant mode

Cut-off frequency:

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

Wave impedance:

\({\eta _{TE}} = \frac{\eta }{{\sqrt {1 - {{\left( {\frac{{{f_c}}}{f}} \right)}^2}\;} }}\)

The reflection coefficient (Γ) is given by:

\(Γ = \frac{{{\eta _2} - {\eta _1}}}{{{\eta _2} + {\eta _1}}}\;\)

Also, the VSWR is related to the reflection coefficient as:

\(VSWR = \frac{{1 + \left| \tau \right|}}{{1 - \left| \tau \right|}}\)

Calculation:

Since a > b, the dominant mode is TE₁₀

In free space, we can write:

\({f_{c1}} = \frac{{\left( {3 \times {{10}^8}} \right)}}{{2 \times 0.05}} = 3\;GHz\)

\({\eta _1} = \frac{{377}}{{\sqrt {1 - {{\left( {\frac{3}{8}} \right)}^2}\;} }} = 406.7\;\Omega \)

In the dielectric medium, we can write:

\({{f}_{c2}}=\frac{\frac{\left( 3\times {{10}^{8}} \right)}{\sqrt{{{\epsilon }_{r}}{{\mu }_{r}}~}}}{2\times 0.05}=\frac{\left( 1.5\times {{10}^{8}} \right)}{2\times 0.05}\)

= 1.5 GHz

\({{\eta }_{2}}=\frac{\frac{377}{\sqrt{{{\epsilon }_{r}}{{\mu }_{r}}~}}}{\sqrt{1-{{\left( \frac{3}{8} \right)}^{2}}~}}=\frac{188.5}{\sqrt{1-{{\left( \frac{3}{8} \right)}^{2}}~}}\)

= 203.3

The reflection coefficient (Γ) will be:

\(\Gamma=\frac{203.3-406.7}{406.7+203.3}\)

= -0.33

Now, the VSWR at the interface will be:

\(VSWR=\frac{1+\left| -0.33 \right|}{1-\left| 0.33 \right|}\)

= 1.98

Concept:

For rectangular waveguide TE₁₀ is the dominant mode.

The cutoff frequency for TE_mn mode is given by

\({f_{{c_{mn}}}} = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \) 

Calculation:

For TE₀₁ mode

Cutoff frequency for TE₀₁ is:

\(f{c_{01}}\;\; = \frac{c}{2}\sqrt {{{\left( {\frac{0}{a}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2}} \) 

\(= \frac{c}{2} \times \frac{1}{b}\left[ {b = \frac{a}{2}} \right]\) 

\(= \frac{c}{2} \times \frac{1}{a} \times 2\) 

= c/a

For TE₂₀ mode:

\(f{c_{20}}\; = \frac{c}{2}\sqrt {{{\left( {\frac{2}{a}} \right)}^2} + {{\left( {\frac{0}{b}} \right)}^2}} \) 

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

For TE₁₁ mode:

\(f{c_{11}}\; = \frac{c}{2}\sqrt {{{\left( {\frac{1}{a}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2}} \) 

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

\(\Rightarrow \frac{c}{a}\left( {\frac{{\sqrt 5 }}{2}} \right)\) 

fc₀₁ = fc₂₀ < fc₁₁

Concept:

The cutoff frequency of hollow waveguide is given as

\({f_{{c_{mn}}}} = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \) 

Calculation:

TE₁₀ is the dominant mode

f₁₀ = 0.75 f₀ = 0.75 × 6 = 4.5 GHz

\(f{c_{10}} = \frac{c}{{2a}}\) 

\(a = \frac{c}{{2{f_{10}}}} = \frac{{3 \times {{10}^8}}}{{2 \times 4.5 \times {{10}^9}}} = 3.33\;cm\) 

For next higher mode:

f_c = 1.25 f₀ = 1.25 (f) = 7.5 GHz

Let us consider next higher mode

TE₀₁

\(f{c_{01}} = \frac{c}{{2b}}\) 

\(b = \frac{c}{{2{f_{01}}}} = \frac{{3 \times {{10}^8}}}{{2 \times 7.5 \times {{10}^9}}} \Rightarrow 2\;cm\) 

\(b > \frac{a}{2}\;satisfied\) 

TE₁₁

\(fc = \frac{c}{ \propto }\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} \) 

\(7.5 \times {10^9} = \frac{{3 \times {{10}^8}}}{{2 \times {{10}^{ - 2}}}}\sqrt {{{\left( {\frac{3}{{10}}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2}} \) 

\({\left( {\frac{{75 \times 2}}{{300}}} \right)^2} = {\left( {\frac{3}{{10}}} \right)^2} + {\left( {\frac{1}{b}} \right)^2}\) 

\({\left( {\frac{1}{ \propto }} \right)^2} - {\left( {\frac{3}{{10}}} \right)^2} = {\left( {\frac{1}{b}} \right)^2}\) 

\({\left( {\frac{4}{{10}}} \right)^2} = {\left( {\frac{1}{b}} \right)^2}\) 

b = 2.5 cm

a > b > a/2

For TE₀₂

\({f_c} = \frac{c}{2} \times \left( {\frac{2}{b}} \right)\) 

\({f_c} = \frac{c}{b}\) 

\(7.5 \times {10^9} = \frac{{3 \times {{10}^{10}}}}{a}\) 

\(b = \frac{{3 \times {{10}^{10}}}}{{7.5 \times {{10}^9}}} = \frac{{30}}{{7.5}} = 4\) 

(b > a) Not-possible

So we have two options not to select b as either 2 cm or 2.5 cm

When b = 2 cm

\({f_{{c_{01}}}} = \frac{c}{{2b}} = 7.5\;GHz\) 

\({f_{{c_{11}}}} = \frac{c}{2}\sqrt {{{\left( {\frac{3}{{10}}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}} \) 

\(= \frac{{3 \times {{10}^{10}}}}{2}\sqrt {\frac{{9 + 25}}{{100}}}\) 

\(\Rightarrow \frac{3}{2} \times {10^{10}} \times \sqrt {34} \times \frac{1}{{10}}\) 

= 1.5 × √34 GHz

= 8.74 GHz

TE₀₁ remains dominant

When b = 2.5 cm

f_c11 = 7.5 GHz

\(f{c_{01}} = \frac{c}{{2b}} = \frac{{3 \times {{10}^8}}}{{2.5 \times 2 \times {{10}^{ - 2}}}} = 6\;GHz\) 

The cutoff frequency of next higher mode is 6 GHz which is not desirable.