Electromagnetics Test 5

Concept:

Guided wavelength:

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

Where λ is the free space wavelength, f is the operating frequency and fc is the cut-off frequency.

∵ \(\sqrt{1-{{\left( \frac{{{f}_{c}}}{f} \right)}^{2}}~}\) is always less than 1

∴ λg > λ always

∴ Statement 1 is incorrect.

Phase velocity:

\({{v}_{p}}=\frac{c}{\sqrt{1-{{\left( \frac{{{f}_{c}}}{f} \right)}^{2}}~}}\)

Where c is the free space velocity, f is the operating frequency and fc is the cut-off frequency.

We can readily see that the phase velocity is a non-linear function of the operating frequency.

∴ Statement 2 is incorrect

The intrinsic/waveguide impedance is defined as the ratio of the electric field and the magnetic field phasor (complex amplitude), i.e.

\(η =\frac{E}{H}=\frac{j\omega \mu }{\gamma }\)

For free-space, the intrinsic impedance is a real quantity, i.e.

η  = η0 = 120π

∴ Statement 3 is incorrect

For TE mode:

Wave impedance

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

For TM mode:

Wave impedance

\({{\eta }_{TM}}=\eta \sqrt{1-{{\left( \frac{{{f}_{c}}}{f} \right)}^{2}}~}\)

Where η is the free space impedance

∴ Wave impedance can be greater or less than the intrinsic impedance. It depends on the mode of propagation.

∴ Statement 4 is Correct.

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

m and n are mode numbers TEmn

Calculation:

Given:

f = √2 × f_c

For TE ₃₂ mode,

\({{f}_{c\left( min \right)}}=\frac{c}{2}\sqrt{{{\left( \frac{3}{6} \right)}^{2}}+{{\left( \frac{2}{4} \right)}^{2}}}\)

\({{f}_{c\left( min \right)}}=\frac{c}{2√2}\)

f = √2 × f_c

\(f=\frac{√2c}{2√2}\)

c = 3 × 10¹⁰ cm/sec

f = 15 GHz

Concept:

The cutoff frequency for dominant mode is given as:

\({f_0} = \frac{1}{{2√ {{\mu _0}{\varepsilon _0}} }}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \)

a and b are the dimensions of the waveguide

Note:

1) If the frequency is less than the cut-on frequency, the wave is evanescent and will not propagate.

2) Degenerate modes are the different modes that have the same cut-off frequency.

Analysis:

Given, the dimension of waveguide a = b = 3 cm and so the dominant mode is either TE₀₁ or TE₁₀ mode (as both will give the same frequency for a square waveguide)

So, for TE₀₁ or TE₁₀ mode, the cut-off frequency will be:

\( = \frac{{3 \times {{10}^8}}}{2} \times \frac{1}{{3 \times {{10}^{ - 2}}}}\)

f₀ = 5 GHz

The frequency of the next higher mode (TE₁₁) for a = b will be:

\({f_0} = \frac{1}{{2√ {{\mu _0}{\varepsilon _0}} }}\sqrt {{{\left( {\frac{1}{a}} \right)}^2} + {{\left( {\frac{1}{a}} \right)}^2}} \)

f₀ = 5 √2 GHz = 7.07 GHz

∴ At 6 GHz, the dominant (TE₁₀ or TE₀₁) mode will propagate.

Statement 1 is correct.

At 4 GHz, no modes will propagate. So the modes are evanescent at 4 GHz.

Statement 2 is correct.

At 11 GHz, along with the dominant mode, TE₁₁ mode with a frequency of 7.07 GHz will also propagate.

Statement 3 is incorrect.

Since the Degenerate modes are the different modes that have the same cut off frequency, at 7 GHz, the frequency TE₀₁ and TE₁₀ propagate that has the same cut off frequency i.e. Degenerate modes propagate at 7 GHz.

Statement 4 is correct.

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:

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:

If a > b then the dominant mode of the rectangular waveguide is TE10

TM0n, TMm0, TM00 modes are not possible in a rectangular waveguide.

The cut-off frequency in the rectangular waveguide is given as:

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

Where a and b are the dimensions of the rectangular waveguide and m and n are half-cycle variations in the x and y directions respectively.

\(v=\frac{1}{\sqrt{\mu \epsilon }}=\frac{c}{\sqrt{{{\mu }_{r}}{{\epsilon }_{r}}}}\)   c is the velocity of light

Calculation:

1) When filled with air:

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

\(=\frac{3\times {{10}^{8}}}{2a}~\sqrt{{{m}^{2}}+\left( \frac{{{a}^{2}}}{{{b}^{2}}} \right){{n}^{2}}}\)

\(=6\sqrt{{{m}^{2}}+6.25~{{n}^{2}}}\) GHz

f < 9 GHz

For TE modes

TE01, fc = 15 GHz

∴ TE0n modes will not propagate.

TE10, fc = 6 GHz

TE20, fc = 12 GHz

∴ only TE10 mode will propagate

TE11, fc = 16.15 GHz

∴ no TEmn where m≠0 and n≠0 will propagate

For TM modes:

TM11, fc = 16.15 GHz

∴ no TM mode will propagate

Total 1 TE mode and 0 TM mode.

2) When filled with the material:

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

\(=\frac{3\times {{10}^{8}}}{4a}~\sqrt{{{m}^{2}}+\left( \frac{{{a}^{2}}}{{{b}^{2}}} \right){{n}^{2}}}\)

\(=3\sqrt{{{m}^{2}}+6.25~{{n}^{2}}}\) GHz

f < 9 GHz

For TE modes:

TE01, fc = 7.5 GHz

TE02, fc = 15 GHz

∴ TE01 mode will only propagate.

TE10, fc = 3 GHz

TE20, fc = 6 GHz

TE30, fc = 9 GHz

∴ only TE10 and TE20 modes will propagate

TE11, fc = 8.078 GHz

TE12, fc = 15.3 GHz

TE21, fc = 9.6 GHz

∴ only TE11 mode will propagate

Similarly, for TM modes:

TM11, fc = 8.078 GHz

∴ only TM11 mode will propagate

Total 4 TE mode and 1 TM mode.

Given,

f = 1.2 f_c

The group velocity V_g

v = Velocity of wave in air

\({v_g} = 3 \times {10^8}\sqrt {1 - {{\left( {\frac{{{f_c}}}{{1.2\;{f_c}}}} \right)}^2}}\)

≃ 1.66 × 10⁸ m/s

∴ Time taken for the excitation to reach the other end is

\(\begin{array}{l} t = \frac{L}{{{v_g}}}\\ t = \frac{1}{{{1.66\times 10^8}}} \end{array}\)

≃ 6 ns