Electromagnetics Test 1

As per Maxwell’s equation,

\(\Rightarrow \bar{\nabla }\times \bar{H}=\bar{J}+\frac{\partial \bar{D}}{\partial t}\)

\(\Rightarrow \mathop{\iint }_{s}\left( \bar{\nabla }\times \bar{H} \right).d\bar{s}=\mathop{\iint }_{s}\left( \bar{J}+\frac{\partial \bar{D}}{\partial t} \right).d\bar{s}\)

Using Stokes Theorem, we can write:

\(\mathop{\oint }_{C}\bar{H}.d\bar{l}=\mathop{\iint }_{s}\left( \bar{J}+\frac{\partial \bar{D}}{\partial t} \right).d\bar{s}\)

Concept:

The distance between two adjacent maxima or minima points for standing waves is given by:

\(d=\frac{λ}{2}\)

λ = wavelength of the operating wave

Also, frequency and wavelength are related by the relation:

c = ν λ 

c = Speed of light = 3 × 10¹⁰ cm/s

Calculation:

d = 37.5 - 12.5 = 25 cm

\(25=\frac{λ}{2}\)

λ = 50 cm

\(ν =\frac{3\times10^{10}}{50}~Hz\)

∴ The operating frequency = 600 MHz

Concept:

The boundary condition for electrostatic fields are defined as:

E1t = E2t (Tangential components are equal across the boundary surface)

Also, the normal component satisfies the following relation:

D1n – D2n = ρs

For a charge-free boundary, ρs = 0, the above expression becomes:

D1n – D2n = 0

D1n = D2n

Calculation:

As the dielectric slab occupies the region 0 < z < d and the field intensity in the free space is in +a_z direction, the field will be normal to the boundary of the plane dielectric slab.

So from the boundary condition, the field normal to the surface are related as:

ε E_in = ε₀ E

With ε = 4ε₀, we can write:

\({E_{in}} = \frac{{{{\rm{\varepsilon }}_0}}}{{4{{\rm{\varepsilon }}_0}}}{E_0}{a_z} = \frac{{{E_0}}}{4}{a_z}\)

∴ The electric field intensity (Ein) inside the dielectric slab will be \(\frac{E}{4}{a_z}\) 

\({D_{in}} = {\rm{\varepsilon \;}}{{\rm{E}}_{{\rm{in}}}} = 4{{\rm{\varepsilon }}_0}\frac{{{E_0}}}{4}{a_z}\)

\({D_{in}} = {{\rm{\varepsilon }}_0}{E_0}{a_z}\)

And the electric flux density (Din) inside the dielectric slab will remain unchanged.

Concept:

The general equation of magnetic field intensity is given by:

H = H0 cos (ωt - βz)ax A / m

The phase velocity is defined as the rate at which the phase of the wave propagates in space.

Mathematically this is calculated as:

\({v_p} = \frac{ω }{β }\)

Calculation:

Given magnetic field intensity in the non-magnetic medium is:

H = 1.5 cos (10⁹ t – 5z) a_x A / m

Comparing both the expressions, we get:

ω = 10⁹ rad / sec

And β = 5

So, the phase velocity of the wave in the medium will be:

\({v_p} = \frac{ω }{β } = \frac{{{{10}^9}}}{5} = 2 \times {10^8}m/s\)

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.

The power received P_r = P_avg A_e

P_avg = Time average power

\({P_r} = \frac{{E_0^2}}{{2\;\eta }}{A_e}\)

Concept:

The general equation of electric field intensity of an EM wave propagating in a_x direction in a medium is given as:

E = E₀ cos (ωt - βx) a_y A/m

ω = Angular velocity defined as:

\(\omega =\frac{2\pi }{T}\)

T = Time period

Calculation:

Given:

E = 9 cos (4 × 10⁸ t - βx) az V / m

Comparing it with the given expression of electric field intensity, we get:

ω = 4 × 10⁸ rad / s

So, the time period of the wave in the air will be:

\(T = \frac{{2\pi }}{ω } = \frac{{2\pi }}{{4 \times {{10}^8}}}\)

= 15.71 ns

Since in one time period, the wave covers one wavelength (λ), ∴ The time taken by the wave to travel λ / 4 distance will be:

Concept:

The velocity of propagation is given by:

\(v = \frac{1}{{\sqrt {LC} }}\)

Characteristic Impedance:

\({Z_o} = \sqrt {\frac{L}{C}} \)

Calculations:

\(v = \frac{1}{{\sqrt {LC} }}\)

\(L = \frac{1}{{C{v^2}}} \)

\(L= \frac{1}{{\left( {30 \times {{10}^{ - 12}}} \right){{\left( {2.5 \times {{10}^8}} \right)}^2}}} \)

\(L= 533\;nH/m\)

Characteristic Impedance will be:

\({Z_o} = \sqrt {\frac{L}{C}} = \sqrt {\frac{{533 \times {{10}^{ - 9}}}}{{30 \times {{10}^{ - 12}}}}} \)

\(Z_0= 133.29\;{\rm{\Omega }}\)

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:

The electric field at a distance is given by:

E = E_oe^-αd

Where ‘α’ is the attenuation constant

For conductors, the attenuation constant is the inverse of the skin depth, i.e. the attenuation constant is given by:

\({\rm{\alpha }} = \sqrt {{\rm{\pi f}}{{\rm{\mu }}_0}{\rm{\sigma }}}\)

Calculations:

If the Amplitude of election filed intensity is E₀ on one side of the sheet, then the magnitude of the Electric field intensity on the other side is 10^-6 E_o

\({\rm{\alpha }} = \sqrt {{\rm{\pi f}}{{\rm{\mu }}_0}{\rm{\sigma }}}\)

\(= \sqrt {{\rm{\pi }} \times {{10}^6} \times 4{\rm{\pi }} \times {{10}^{ - 7}} \times 3.7 \times {{10}^7}} \)

= 1.21 × 10⁴ Np/m

The electric field is related as:

10^-6 E_o = E_o e^-αd

Taking log both sides

-6 ln 10 = -αd

\({\rm{d}} = \frac{{6{\rm{ln}}\left( {10} \right)}}{{\rm{\alpha }}} = 0.00142{\rm{\;m}}\)

Concept:

The reflection coefficient for a wave propagating from medium 1 to medium 2 is defined as:

\({\rm{Γ }} = \frac{{{\eta _2} - {\eta _1}}}{{{\eta _2} + {\eta _1}}}\)   ---(1)

Where η1 and η2 are the intrinsic impedance of the two mediums respectively.

For a given Electric field Ei incident at the interface, the reflected wave is given by:

\({E_r} = {\rm{Γ }}{E_i}\)

Analysis:

Since 20% of the energy in the incident wave is reflected at the boundary, we can write:

\((\frac{E_r}{E_i})^2 = Γ^2 =0.2\)

\(\left| {\rm{Γ }} \right| = \sqrt {0.2} = \pm 0.447\), i.e. for both Γ = +0.447 and Γ = - 0.447, the reflected power will be 20% of the incident power.

Using equation (1), we can write:

\(\frac{{{\eta _2} - {\eta _1}}}{{{\eta _2} + {\eta _1}}} = \pm 0.447\)

\(\frac{{{\eta _0}\sqrt {\frac{{{\mu _{r2}}}}{{{\varepsilon _{r2}}}}} - {\eta _0}\sqrt {\frac{{{\mu _{r1}}}}{{{\varepsilon _{r1}}}}} }}{{{\eta _0}\sqrt {\frac{{{\mu _{r2}}}}{{{\varepsilon _{r2}}}}} + {\eta _0}\sqrt {\frac{{{\mu _{r1}}}}{{{\varepsilon _{r1}}}}} }} = \pm 0.447\)

With εr1 = μ3r1, and εr2 = μ3r2, the above expression becomes:

\(\frac{{\sqrt {\frac{{{\mu _{r2}}}}{{\mu _{r2}^3}}} - \sqrt {\frac{{{\mu _{r1}}}}{{\mu _{r1}^3}}} }}{{\sqrt {\frac{{{\mu _{r2}}}}{{\mu _{r2}^3}}} + \sqrt {\frac{{{\mu _{r1}}}}{{\mu _{r1}^3}}} }} = \pm 0.447\)     

\(\frac{{{\mu _{r1}} - {\mu _{r2}}}}{{{\mu _{r1}} + {\mu _{r2}}}} = \pm 0.447\)

\(\frac{{{\mu _{r1}}}}{{{\mu _{r2}}}} = \frac{{1 \pm 0.447}}{{1 \mp 0.447}}\)

\(\frac{{{\mu _{r1}}}}{{{\mu _{r2}}}} = 2.62\;or\;0.38\)

So, \(\frac{{{\varepsilon _{r2}}}}{{{\varepsilon _{r1}}}} = {\left( {\frac{{{\mu _{r2}}}}{{{\mu _{r1}}}}} \right)^3}\)

\({P_r} = \frac{{{P_t}\left( {{G_r}} \right)\left( {{G_t}} \right)}}{{{d ^2}{4 \pi}}}\)

Pr= Power received 

Pt = Power of transmitting antenna 

Gr= Gain of receiving antenna

Gt= Gain of transmitting antenna

d= distance between the two antenna

\(Directivity = \frac{{4\pi {A_e}}}{{{\lambda ^2}}}\)

Directivity = 1.63 for half wave dipole

\({P_r} = \frac{{{P_t}\left( {\frac{{4\pi {A_{eT}}}}{{{\lambda ^2}}}} \right)\left( {\frac{{4\pi {A_{eR}}}}{{{\lambda ^2}}}} \right)}}{{4\pi {d^2}}}\)

Where Ae is the effective area of the antenna'

\(=\frac{{80\left( {A_e^2} \right)4\pi }}{{{d^2}{\lambda ^4}}}\)

\(\lambda = \frac{C}{f} = \frac{{3 \times {{10}^8}}}{{100 \times {{10}^6}}} = 3m.\)

substituting the values

For the optical power

\(\alpha = \frac{{10}}{L}\log \left( {\frac{{{P_{in}}}}{{{P_{out}}}}} \right)\) 

\(L = \left( {\frac{{10}}{{1.41}}} \right)\log \left( {\frac{1}{{0.5}}} \right)\) 

\(L = \left( {\frac{{10}}{{1.41}}} \right)\log \left( 2 \right)\) 

L = 2.134 km

Concept:

The voltage at load is superposition of 2 waves

1) Forward travelling wave

2) Reflected wave

If V⁺ is the amplitude of forward travelling wave

Then voltage at load

V_L = V⁺ + V_R

V_L = V⁺ + Γ V⁺

V_L = V⁺ (1 + Γ)

Where V_R = Reflected wave voltage

Also:

V_max = |V⁺| (1 + (Γ(z)))

\(=\left| {{V}^{+}} \right|\left( 1+\frac{SWR-1}{SWR+1} \right)\)

\(=\left| {{V}^{+}} \right|\left( \frac{2~S\omega R}{S\omega R+1} \right)\)

Similarly,

\({{V}_{min}}=\left| {{V}^{+}} \right|\left( 1-\text{ }\!\!\Gamma\!\!\text{ }\left( z \right) \right)\)

\(=\left| {{V}^{+}} \right|\left( \frac{2}{SWR+1} \right)\)

Calculations:

Reflection coefficient (Γ_L)

\({{\text{ }\!\!\Gamma\!\!\text{ }}_{L}}=\frac{{{Z}_{L}}-{{Z}_{0}}}{{{Z}_{L}}+{{Z}_{0}}}\)

\(=\frac{50-100}{50+100}\)

\(=\frac{-1}{3}\)

\(=\frac{1}{3}{{e}^{j\pi }}\)

\(\left| {{\text{ }\!\!\Gamma\!\!\text{ }}_{L}} \right|=\frac{1}{3}~,~{{\theta }_{L}}=-\pi \)

\(SWR=\frac{1+\left| {{\text{ }\!\!\Gamma\!\!\text{ }}_{L}} \right|}{1-\left| {{\text{ }\!\!\Gamma\!\!\text{ }}_{L}} \right|}=\frac{3+1}{3-1}=2\)

To calculate minimum and maximum voltage first calculate forward voltage amplitude.

\({{V}_{L}}={{V}^{+}}\left( 1+{{\text{ }\!\!\Gamma\!\!\text{ }}_{L}} \right)\)

\({{V}^{+}}=\frac{{{V}_{L}}}{\left( 1+{{\text{ }\!\!\Gamma\!\!\text{ }}_{L}} \right)}=\frac{50}{\left( 1-\frac{1}{3} \right)}=\frac{50}{\frac{2}{3}}\)

= 75 V

\({{V}_{max}}=\left| {{V}^{+}} \right|\left( \frac{2~SWR}{SWR+1} \right)\)

\(=75\left( \frac{2\times 2}{2+1} \right)\)

In a waveguide the EM energy (signals) travel at the speed equal to group velocity:

\({V_g} = {\frac{c}{\sqrt\epsilon_r}}\sqrt {{{1-\left( {\frac{{{f_c}}}{f}} \right)}^2}}\)

\({f_c} = \frac{c}{{\sqrt 4 \times 2 \times {{10}^{ - 2}}}}\sqrt {{{\left( {\frac{2}{{2.5}}} \right)}^2} + {{\left( {\frac{1}{1}} \right)}^2}}\)

= 9.6 GHz

For 13 GHz pulse

\({v_g} = {\frac{3}{2}} \times {10^8}\sqrt {1 - {{\left( {\frac{{9.6}}{{13}}} \right)}^2}}\)

= 1.01 × 10⁸m/s

For 18 GHz pulse

\({v_g} = {\frac{3}{2}} \times {10^8}\sqrt {1 - {{\left( {\frac{{9.6}}{{18}}} \right)}^2}}\)

= 1.27 × 10⁸m/s

The time delay between two pulses at the end of 200 m tunnel is

\({\rm{\Delta }}t = \frac{{200}}{{1.01 \times {{10}^8}}} - \frac{{200}}{{1.27\times {{10}^8}}}\)