Parallel Plane Waveguide MCQ || Parallel Plane Waveguide Questions & Answers

Answer.3. $\sqrt {\frac{\mu }{ \in }}$

Explanation

Electric field (E) & magnetic field (H) combination called uniform plane wave because E & H has some magnitude through any transverse plane.

Intrinsic Impedance (η) = 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 =E→×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

Answer.3. 1

Explanation

For parallel plate waveguide, the propagation constant is defined as:

β = mπ/a

a = distance between the plates

Where the minimum value of m = 1

Cut-off frequency for the parallel plate waveguide is given by:

fc = m/2a

The wavelength will be:

λc = 2a/m

Answer.4. 4.5 × 10⁸ m/s

Explanation

Group velocity × Phase velocity = c²

2 × 10⁸ × phase velocity = (3 × 10⁸)²

V_P = (3 × 10¹⁶)/(2 × 10⁸ )

V_P = 4.5 × 10⁸ m/s

Answer.4. Greater than the velocity of light in free space

Explanation

The phase velocity of waves propagating in a hollow metal waveguide is Greater than the velocity of light in free space.

Phase velocity is defined as:

V_p = ω/β

β is the phase constant defined as:

$\begin{array}{l} \beta = \sqrt {{\omega ^2}\mu \in – {{\left( {\frac{{m\pi }}{a}} \right)}^2}} \\ \\ {V_p} = \frac{\omega }{{\sqrt {{\omega ^2}\mu \in – {{\left( {\frac{{m\pi }}{a}} \right)}^2}} }}\\ \\ {V_p} = \frac{1}{{\sqrt {\mu \in – {{\left( {\frac{{m\pi }}{{a\omega }}} \right)}^2}} }}\\ \\ {V_p} = \frac{{\frac{1}{{\sqrt {\mu \in } }}}}{{\sqrt {1 – {{\left( {\frac{{m\pi }}{{a\omega \sqrt {\mu \in } }}} \right)}^2}} }} \end{array}$

Using $c = \frac{1}{{\sqrt {\mu C} }}$

where c =  speed of light, the above expression becomes:

${V_p} = \frac{c}{{\sqrt {1 – {{\left( {\frac{{m\pi C}}{{a\omega }}} \right)}^2}} }};$

Also

ω_c = mπc/a

${V_p} = \frac{C}{{\sqrt {1 – {{\left( {\frac{{{\omega _c}}}{\omega }} \right)}^2}} }}$

Using Sinθ = ω_c/ω

$\begin{array}{l} {V_p} = \frac{c}{{\sqrt {1 – {{\sin }^2}\theta } }}\\ \\ {V_p} = \frac{c}{{\cos \theta }}; \end{array}$

Since -1 ≤ cos θ ≤ 1

∴ V_p > c

Answer.2. 1.73

Explanation

The cutoff frequency of TE_mn mode is given by:

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

for TE₁₀ mode

${f_{{c_{10}}}} = \frac{c}{2}\left( {\frac{1}{a}} \right)$ ——- (1)

for TE11 mode

${f_{{c_{11}}}} = \frac{c}{2}\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}}$ —— (2)

Dividing equation 1 by equation 2 we get

$\begin{array}{l} \frac{1}{2} = \frac{{\frac{c}{2}\left( {\frac{1}{a}} \right)}}{{\frac{c}{2}\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} }}\\ \\ = \frac{1}{2} = \frac{{\frac{c}{{2a}}}}{{\frac{c}{2}\sqrt {\frac{{{a^2} + {b^2}}}{{ab}}} }}\\ \\ = \frac{b}{{\sqrt {{b^2} + {a^2}} }}\\ \\ \frac{1}{2} = \frac{1}{{\sqrt {1 + {{\left( {\frac{a}{b}} \right)}^2}} }}\\ \\ {\left( {\frac{a}{b}} \right)^2} = {\left( {\frac{w}{h}} \right)^2} \end{array}$

= 4 − 1 = 3

ω/h = √3 = 1.732

Answer.2. During wave propagation in Z-direction, the components of H and E are transverse to the direction of propagation of the waves.

Explanation

Transverse Electromagnetic waves:

• In an electromagnetic wave, electric and magnetic field vectors are perpendicular to each other and at the same time are perpendicular to the direction of propagation of the wave.
• This nature of electromagnetic wave is known as Transverse nature.
• Maxwell proved that both the electric(E) and magnetic fields(H) are perpendicular to each other in the direction of wave propagation. He considered an electromagnetic wave propagating along the positive z-axis
• The electric and magnetic fields propagate sinusoidal with the z-axis.

Answer.2. 0.75

Explanation

Dominant mode of rectangular waveguide is TE₁₀

The cutoff frequency is given as:

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

Given:

(fc)_TE01 = 2(fc)_{TE10 ——-1}

Operating frequency:

f = (f_c)_TE10 + 25% of (f_c)_TE10

f = 1.25(f_c)_TE10

For air filled wavelength C, we can write:

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

For TE₀₁ mode

m = 0, n = 1

(fc)_TE10 = c/2a

For TE₁₀ mode

m = 1; n = 0

(fc)_TE10 = c/2a ——2

From equation (1)

(f_c)_TE01 = 2(f_c)_TE10

c/2b = 2c/2a

a = 2b

operating frequency:

f = (1.25) (f_c)_TE01 = 2(f_c)_TE10

$f = \left( {1.25} \right)\left[ {\frac{c}{{2a}}} \right]$

=  1.25c/4b

Operating wavelength

λ = c/f = 4b/1.25 = 3.2b

Guide wavelength is given by:

$\begin{array}{l} {\lambda _g} = \frac{\lambda }{{\sqrt {1 – {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} }}\\ \\ 4 = \frac{{3.2b}}{{\sqrt {1 – {{\left( {\frac{1}{{1.25}}} \right)}^2}} }} \end{array}$

4 = 3.2b/0.6

b = 4 × 0.6/3.2

b = 0.75 cm

Answer.3. The phase velocity of waves tends to infinite.

Explanation

Phase velocity is given as

${v_p} = \frac{c}{{\sqrt {1 – {{\left( {\frac{{{f_c}}}{f}} \right)}^2}{\rm{\;}}} }}$

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

We can also write phase velocity as:

Vp = c/cosθ

Where θ is the angle with which the wave enters the waveguide.

Given, f = fc

$\begin{array}{l} {v_p} = \frac{c}{{\sqrt {1 – {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} }}\\ \\ {v_p} = \frac{c}{{\sqrt {1 – 1} \:}} = \frac{c}{0} = \infty \end{array}$

v_p = ∞ means that,

cos θ = 0, i.e. θ = 90°.

Answer.3. 1.5 × 10⁸ m/s

Explanation

The velocity (V_p) of a plane electromagnetic wave is given as-

${v_p} = \frac{c}{{\sqrt {{\mu _r}{\varepsilon _r}} }}$

$c = \frac{1}{{\sqrt {{\mu _o}{\varepsilon _o}} }}$ = 3 × 10⁸m/sec

μ0 = permeability in free space 4π × 10^-7 H/m

εo = permittivity in free space 8.854 × 10^-12 C²/Nm²

Given:

μ_r = 4

Vp = 3 × 10⁸/√4

Vp = 1.5 x 10⁸ m/sec

Answer.2. 8.19 GHz ≤ f ≤ 12.45 GHz

Explanation

Cut-off frequency of (rectangular) waveguide

$ \Rightarrow {f_c} = \frac{C}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \:$ ——1

Dominant mode in rectangular waveguide (a > b) is TE₁₀ and the cutoff frequency is defined as;

${f_{{c_{\left( {10} \right)}}}} = \frac{C}{2}\sqrt {\frac{1}{{{a^2}}}} = \frac{C}{{2a}}$

Calculation:

${f_{{c_{\left( {10} \right)}}}} = \frac{C}{{2a}} = \frac{{3 \times {{10}^{10}}}}{{2 \times 2.29}} = 6.554\:Hz$

Desired operating frequency = 25% more than fc₍₁₀₎

(1 + 0.25)fc₍₁₀₎ = 1.25fc₍₁₀₎ = 1.25 × 6.55

= 8.199 Hz

Using equation (1), we calculate the different cut-off frequencies for higher modes to find the next dominant mode,

f_c for TE₀₁ mode = 14.74 Hz

f_c for TE₁₁ mode = 16.4 Hz

and

f_c for TE₂₀ mode = 13.144 Hz

So, TE₂₀ is the next higher cut-off frequency,

95% of 13.144 Hz

⇒ 12.484 GHz

Answer.2. No change in v and halving of Z

Explanation

$\begin{array}{l} Z = \sqrt {\frac{L}{C}} \\ \\ Z = \sqrt {\frac{{\frac{{\mu h}}{w}}}{{\frac{{Ew}}{h}}}} \\ \\ Z = \sqrt {\frac{\mu }{E}} \:\frac{h}{w} \end{array}$

We observe that If ‘h’ is halved, z is halved.

V = E_y h

I = H_x ⋅ w

Answer.2. No change in v and halving of Z

Explanation

Given

a = 8 cm

b = 4 cm

f = 3.4 GHz

The cut-off frequency will be:

fc = c/2a = 3 × 10¹⁰/ 2 × 8

Sinθ = fc/f

sinθ = 1.875/3.4 = 0.5514

∴ cos θ = 0.83419

Phase velocity is given by:

Vp = c/cosθ

vp/c = 1/cosθ

vp/c = 1.198

Answer.4. 15.8 GHz

Explanation

Given, a = 3 cm

b = 1 cm

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

\[\begin{array}{l} {f_c} = \frac{{3 \times {{10}^8}}}{2}\sqrt {\frac{1}{{{{\left( {3 \times {{10}^{ – 2}}} \right)}^2}}} + \frac{1}{{{{\left( {1 \times {{10}^{ – 2}}} \right)}^2}}}} \\ \\ {f_c} = 150 \times {10^6}\sqrt {11111.11} \:\: \end{array}\]

fc = 150 × 10⁶ × 105.41

fc = 15.81 GHz