Electromagnetics Test 2

Concept:

Maxwell equation for electrostatic states that:

\(\nabla \cdot \vec D = {\rho _v}\)

\(\vec D\) = Electric flux density

ρ_v = volume density

Application:

Given:

\(\vec D = 10x{\hat a_y} - 4y{\hat a_y} + kz{\hat a_z}\frac{{\mu C}}{{{m^2}}}\)

Also ρ_v = 0

Solving for \(\nabla \cdot \vec D = 0\), we get:

\(\frac{{\partial {D_x}}}{{\partial x}} + \frac{{\partial {D_y}}}{{\partial y}} + \frac{{\partial {D_z}}}{{\partial z}} = 0\)

\(\frac{\partial }{{\partial x}}\left( {10x} \right) + \frac{\partial }{{\partial y}}\left( { - 4y} \right) + \frac{\partial }{{\partial z}}\left( {kz} \right) = 0\)

10 – 4 + k = 0

k = - 6 C/m³

Concept:

\({{Z}_{in}}={{Z}_{0}}\left( \frac{{{Z}_{L}}+j{{Z}_{0}}\tan \beta l}{{{Z}_{0}}+j{{Z}_{L}}\tan \beta l} \right)\)

Z_L = Load impedance

Z₀ = Characteristic impedance

\(\because \beta =\frac{2\pi }{\lambda } \), at a quarter \(\left( \frac{\lambda }{4} \right)\) distance away, the impedance will be:

\({{Z}_{in}}={{Z}_{0}}\left. \frac{\left( {{Z}_{L}}+j{{Z}_{0}}\tan \left( \frac{2\pi }{\lambda }.\frac{\lambda }{4} \right) \right)}{\left( {{Z}_{0}}+j{{Z}_{L}}\tan \left( \frac{2\pi }{\lambda }.\frac{\lambda }{4} \right) \right)} \right)\)

\({{Z}_{in}}=\frac{Z_{0}^{2}}{{{Z}_{L}}}\)

VSWR is defined as:

\(VSWR = \frac{{1 + {{\rm{Γ }}_L}}}{{1 + {{\rm{Γ }}_L}}}\)

Γ = Reflection coefficient at the load

Calculation:

\(1 = \frac{{1 + {{\rm{Γ }}_L}}}{{1 + {{\rm{Γ }}_L}}}\)

Γ = 0 (No reflection)

An SWR of 1 indicates that the load is to be perfectly matched, i.e. the load impedance must equal the characteristic impedance.

∴ The impedance at the input of the quarter-wave transformer must equal the characteristic impedance of the line.

With Z_L = 200 Ω and Z₀ = 300 Ω, we can write:

\(300=\frac{Z_{0}^{2}}{{200}}\)

Z₀² = 300 × 200

\(Z_0 ^2=60000\)

Z₀ = 245 Ω

Concept:

According to Maxwell’s law magnetic field in the air can be represented by the following relation:

∇ × H = J

Calculation:

\(\nabla \times H = \left[ {\begin{array}{*{20}{c}} x&y&z\\ {\frac{\partial }{{\partial {\rm{x}}}}}&{\frac{\partial }{{\partial {\rm{y}}}}}&{\frac{\partial }{{\partial {\rm{z}}}}}\\ {{H_x}}&{{H_y}}&{{H_z}} \end{array}} \right]\)

\( = \left[ {\begin{array}{*{20}{c}} x&y&z\\ {\frac{\partial }{{\partial {\rm{x}}}}}&{\frac{\partial }{{\partial {\rm{y}}}}}&{\frac{\partial }{{\partial {\rm{z}}}}}\\ {\frac{{7x}}{{{y^2}}}{H_0}}&{ - \frac{{8y}}{{{x^2}}}{H_0}}&0 \end{array}} \right]\)

= \(\left( {\frac{\partial }{{\partial {\rm{y}}}}\;0 + \frac{\partial }{{\partial {\rm{z}}}}\;\frac{{8y}}{{{x^2}}}{H_0}} \right)\hat x - \left( {\frac{\partial }{{\partial {\rm{x}}}}\;0 - \frac{\partial }{{\partial {\rm{z}}}}\;\frac{{7x}}{{{y^2}}}{H_0}} \right)\hat y + \left( {\frac{\partial }{{\partial {\rm{x}}}}\left( { - \frac{{8y}}{{{x^2}}}{H_0}} \right)\; - \frac{\partial }{{\partial {\rm{y}}}}\frac{{7x}}{{{y^2}}}{H_0}\;} \right)\hat z\)

\(=0+0+ \left[ {\frac{\partial }{{\partial {\rm{x}}}}\left( { - \frac{{8y}}{{{x^2}}}{H_0}} \right)\; - \frac{\partial }{{\partial {\rm{y}}}}\frac{{7x}}{{{y^2}}}{H_0}\;} \right]\hat z\)

\( = \frac{\partial }{{\partial {\rm{x}}}}\left( { - \frac{{8y}}{{{x^2}}}{H_0}} \right) = - \frac{{8y{H_0}\left( { - 2} \right)}}{{{x^3}}}\)

\( = \frac{{16y{H_0}}}{{{x^3}}}\)

\(\frac{\partial }{{\partial {\rm{y}}}}\frac{{7x}}{{{y^2}}}{H_0} = \frac{{7x{H_0}\left( { - 2} \right)}}{{{y^3}}}\)

\( = - \frac{{14x{H_0}}}{{{y^3}}}\)

\(\nabla \times H = \frac{{16y{H_0}}}{{{x^3}}} + \frac{{14x{H_0}}}{{{y^3}}}\)

putting x = 1, y = 1 we get, 

Given the instantaneous electric field in the free space is:

E_s = (5a_y – 6a_x) cos (ωt – 50z) V / m

So, the phasor form of electric field intensity will be:

E_s = (5a_y – 6a_x) e^-j50z V / m

The phasor form of the magnetic field is given in the terms of electric field intensity as:

\({H_s} = \frac{1}{{{\eta _0}}}\left( {{a_k}} \right) \times \left( E \right)\)

Where a_k is the unit vector in the direction of wave propagation and η₀ is the intrinsic impedance in free space.

With a_k = a_z, we can write:

\({H_s} = \frac{1}{{{\eta _0}}}\left( {{a_z}} \right) \times \left( {5{a_y} - 6{a_x}} \right){e^{ - j50z}}V/m\)

\( = \frac{1}{{{\eta _0}}}\left( { - 5{a_x} - 6{a_y}} \right){e^{ - j50z}}V/m\)

\(H_s = - \frac{1}{{{\eta _0}}}\left( {5{a_x} + 6{a_y}} \right){e^{ - j50z}}V/m\)

The reflection coefficient at the load is defined as:

\({\rm{\Gamma }} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}},\) where Z_L = load impedance and Z₀ = characteristic impedance.

Also, VSWR \(= \frac{{1 + \left| {\rm{\Gamma }} \right|}}{{1 - \left| {\rm{\Gamma }} \right|}}\)

Clearly, VSWR is a function of Z_L and Z₀.

For the given transmission line, Z₀ = Z_t and Z_L = Z_L

So, VSWR is constant and it doesn’t change with the length of the line, In the smith chart, this is represented as a constant VSWR circle.

The skin depth in a dissipative media is given by:

\(\delta = \frac{1}{\alpha } = \sqrt {\frac{2}{{\omega \mu \sigma }}} \)

While wave velocity is given by:

\(\nu = \frac{\omega }{\beta }\)

For a conductor attenuation constant α = phase constant β

α = β

wave velocity is inversely proportional to attenuation α. or directly proportional to the skin depth 

Hence wave velocity is medium B is 2ν.

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 f_c is the cut-off frequency.

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

∴ λ_g > 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 f_c 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:

The input impedance in a transmission line is given by:

\({Z_{in}} = {Z_0}.\frac{{{Z_L} + j{Z_0}tanβ l}}{{{Z_0} + j{Z_L}tanβ l}}\)

'l' = Length of the line from the Load

β = 2π/λ = Wavenumber

Z₀ = Characteristic Impedance of the Transmission Line

Z_L = Load Impedance

Analysis:

Given, the load impedance is short-circuited, i.e.

Z_L = 0

So, the input impedance for a lossless line is given as:

\({Z_{in}} = {Z_0}\left( {\frac{{{Z_L} + j{Z_0}\tan β l}}{{{Z_0} + j{Z_L}\tan β l}}} \right)\)

\(Z_{in} = j{Z_0}\tan β l\)

Now:

For l < λ / 4:

\(β l < \frac{π }{2}\)

∴ tan βl will be positive, and therefore, Z_in is inductive. (a → 2)

For \(\frac{\lambda }{4} < l < \frac{\lambda }{2}\)

\(\frac{π }{2} < β l < π \)

tan βl will be negative and therefore, Z_in will be capacitive (b → 1)

For \(l = \frac{\lambda }{4}\)

\(β l = \frac{π }{2}\)

tan βl = ∞ and therefore Z_in = ∞ (c → 4)

For \(l = \frac{\lambda }{2}\)

βl = π 

tan βl = 0 and therefore, Z_in = 0 (d → 3)

Concept:

For a reflected wave to be linearly polarised one of the components of the wave (E₁₁ or E_⊥) must completely transmit into polystyrene without reflection.

\(\tan {{\rm{θ }}_{{\rm{B}}11}} = \sqrt {\frac{{{\epsilon_{{{\rm{r}}_2}}}}}{{{\epsilon_{\rm{r}}}_1}}} \)

θ_B11 = Brewster angle for parallel polarisation

Calculation:

Putting on the respective value, we get:

\(\tan {{\rm{θ }}_{{\rm{B}}11}} = \sqrt {\frac{{2.7}}{1}} \)

Optical communication utilizes the principle of total internal reflection by which light signals can be transmitted from one place to another with a negligible loss of energy

The advantages of optical fibers are:

• Optical fibers have greater information-carrying capacity due to large bandwidth
• Optical fibers are free from electromagnetic interference and offer high signal security
• Optical fibers suffer less attenuation than the coaxial cable and twisted wire cables.
• TEM mode means the electric field and magnetic field are zero in the direction of propagation. All field components will vanish in this mode. Therefore, the rectangular waveguide does not support TEM mode.
• The waveguide propagation constant defines the phase and amplitude of each component of the wave as it propagates along the waveguide.

The factor for each component can be expressed as:

e^{[jω – γz]}

z is the direction of propagation.

If γ is real then the phase of each component is constant and amplitude decreases exponentially with z. In such a case, no significant propagation takes place hence propagation constant γ will be imaginary.

The numerical gain of the antenna is

G_t = 10^36/10 = 4000.

The power density in the main beam of the antenna at a distance of R = 20 m is

Comparing with the standard equation of electric field in waveguide

\(\frac{{mπ }}{a} = 40π\)

m = 2

\(\frac{{nπ }}{b} = 100π\)

n = 3

The mode is TE₂₃

For lossless dielectric α = 0

\(\gamma = \sqrt {{{\left( {\frac{{mπ }}{a}} \right)}^2} + {{\left( {\frac{{nπ }}{b}} \right)}^2} - {ω ^2}uϵ} \)

γ = jβ

Squaring both sides

\(- {β ^2} = {\left( {\frac{{mπ }}{a}} \right)^2} + {\left( {\frac{{nπ }}{b}} \right)^2} - {ω ^2}\mu ϵ\)

\({ω ^2}\mu ϵ = {\left( {\frac{{mπ }}{a}} \right)^2} + {\left( {\frac{{nπ }}{b}} \right)^2} + {β ^2}\)

\(\mu ϵ = \frac{{{ϵ_r}}}{{{c^2}}}\)

\({ϵ_r} = \frac{{{c^2}}}{{{ω ^2}}}\left[ {{β ^2} + {{\left( {\frac{{mπ }}{a}} \right)}^2} + {{\left( {\frac{{nπ }}{b}} \right)}^2}} \right]\)

putting,  ω = 2.4 π × 10¹⁰ rad/sec and β = 52.9 π rad/m. 

ϵ_r = 2.25

Form the equation

ω = 2.4 π × 10¹⁰ rad/sec

β = 52.9 π rad/sec

ϵ_r = 2.25

The cutoff frequency for the mode is:

\({f_{23}} = \frac{c}{{2\sqrt {{ϵ_r}} }}\sqrt {{{\left( {\frac{2}{a}} \right)}^2} + {{\left( {\frac{3}{b}} \right)}^2}} \)

Substituting the values of a and b

Concept:

The reflection coefficient is defined as:

\(\Gamma = \frac{E_r}{E_i}\)

Where Er is the reflected electric field and Ei is the incident electric field.

In terms of the intrinsic impedance of medium 2 and medium 1, the reflection coefficient is given by:

\(|\Gamma|=\frac{η_2-η_1}{η_2+η_1}\)

η2 and η1 are the intrinsic impedance of medium 2 and medium 1 respectively.

Calculation:

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

\({\left| {\rm{\Gamma }} \right|^2} = \frac{{20}}{{100}}\)

\(\left| {\rm{\Gamma }} \right| = \sqrt {0.2} = \pm 0.447\)

Where Γ is the reflection coefficient at the medium interface. Therefore, we have:

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

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

With ε_r1 = μ³_r1, and ε_r2 = μ³_r2, 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, \(\frac{{{\varepsilon _{r2}}}}{{{\varepsilon _{r1}}}} = {\left( {\frac{{{\mu _{r2}}}}{{{\mu _{r1}}}}} \right)^3}\)

\(\frac{{{\varepsilon _{r2}}}}{{{\varepsilon _{r1}}}} = 0.056\;or\;17.9\)

\(I = \smallint \bar J.\overline {ds} = \mathop \smallint \limits_0^{16} \mathop \smallint \limits_0^{2\pi } \frac{{40}}{{{p^2} + 4}}\rho \;d\rho d\phi\)

\(t = {\rho ^2} + 4\)

dt = 2ρ dρ

\(\frac{{dt}}{2} = \rho d\rho \)

\(I = \frac{{40 \times 2\pi }}{2}\mathop \smallint \limits_4^{260} \left( {\frac{{dt}}{t}} \right)\)

\(= 40 \times \pi \times \left[ {\ln p} \right]_4^{260}\)

Calculation:

\({{P}_{received}}=\frac{{{P}_{t.}}{{G}_{dt.}}{{A}_{er}}}{4\pi {{r}^{2}}}\)

A_er = Effective Aperture of the Receiver Antenna, and is given by:

\({{A}_{er}}=\frac{{{\lambda }^{2}}}{4\pi }.{{G}_{r}}\)

So, P_reccived = \(\frac{{{P}_{t}}~{{G}_{dt}}~{{G}_{r}}}{{{\left( \frac{{{4}_{n}}R}{\lambda } \right)}^{2}}}~;~where~{{\left( \frac{4\pi R}{\lambda } \right)}^{2}}=Path~loss\)

Now, the Gain of the aperture antenna (horn or parabolic) is proportional to \({{\left( \frac{D}{\lambda } \right)}^{2}},~i.e.\)

\({{G}_{dt}}\propto {{\left( \frac{D}{\lambda } \right)}^{2}}\) ; D = Dimension of Antenna.

For the Given Antennas which are the same, the distance R and P_t are both the same as well.

So, \({{G}_{dt}}\propto \frac{1}{{{\lambda }^{2}}}\)

Taking all into consideration, we can say that;

P_received ∝ G_dt .G_r.λ ²

P_received ∝ \(\frac{1}{f^2}\)

∴ We will get

\(\Rightarrow \frac{{{P}_{r1}}}{{{P}_{r2}}}=\frac{f_{2}^{2}}{f_{1}^{2}}\)

Given f₁ = 10 GHz and f₂ = 20 GHz,

\(\frac{{{P}_{r1}}}{{{P}_{r2}}}={{\left( \frac{20}{10} \right)}^{2}}\)

Taking log on both sides;

\(10\log \left( \frac{{{P}_{r1}}}{{{P}_{r2}}} \right)=10\log \left( {4} \right)~\)

(P_r1)_dB ­– (P_r2)_dB = 10log 4

⇒ (P_r2)_dB = (P_r1)_dB - 6 dB

So, the received power decreases by 6dB.

Concept:

The magnetic boundary conditions are given by:

H_1t – H_2t = J_s

B_1n = B_2n

Analysis:

Since the boundary surface of the two medium is z = 0, so the normal component B_1n and tangential component B_1t of magnetic flux density in medium 1 are

B_1n = a_x

And B_1t = 0.4 a_x + 0.8 a_y

As the normal component of magnetic flux density is uniform at the boundary of two medium So, the normal component of magnetic flux density in the medium 2 is

B_2n = B_1n = a_x         ---(1)

Now for determining the tangential component of field in medium 2, we first calculate the tangential component of magnetic field intensity in medium 1 which is given as:

\({H_{1t}} = \frac{{{B_{1t}}}}{{{\mu _1}}}\)

Where μ1 is the permeability of medium 1.

With μ₁ = 4μ₀, we can write:

\( = \frac{1}{{4{\mu _0}}}\left( {0.4\;{a_x} + 0.8\;{a_y}} \right) \)             

\(= \frac{{0.1{a_x} + 0.2\;{a_y}}}{{{\mu _0}}}\)      

Again from the boundary condition, the tangential component of magnetic field intensity in the two mediums are related as:

a_n × (H_1t – H_2t) = K   

where H_2t and H_1t are the tangential components of magnetic field intensity in medium 2 and medium 1 respectively, K is the surface current density

at the boundary interface of the two mediums and a_n is the unit vector normal to the boundary interface. So we have

\({a_z} \times \left[ {\frac{{0.1{a_x} + 0.2{a_y}}}{{{\mu _0}}} - \left( {{H_{2tx}}{a_x} + {H_{2ty}}{a_y}} \right)} \right] \)

\(= \frac{1}{{{\mu _0}}}\left( {0.2{a_x} - 0.4{a_y}} \right)\)

\(\left( {\frac{{0.1}}{{{\mu _0}}} - {H_{2tx}}} \right){a_y} - \left( {\frac{{0.2}}{{{\mu _0}}} - {H_{2ty}}} \right){a_x} \)

\(= \frac{1}{{{\mu _0}}}\left( {0.2\;{a_x} - 0.4\;{a_y}} \right)\)

Comparing the x and y-components we get:

\({H_{2tx}} = \frac{{0.1}}{{{\mu _0}}} + \frac{{0.4}}{{{\mu _0}}} = \frac{{0.5}}{{{\mu _0}}}\)

\({H_{2ty}} = \frac{{0.2}}{{{\mu _0}}} + \frac{{0.2}}{{{\mu _0}}} = \frac{{0.4}}{{{\mu _0}}}\) 

Therefore the tangential component of magnetic field intensity in medium 2 is

\({H_{2t}} = \frac{{0.5}}{{{\mu _0}}}{a_x} + \frac{{0.4}}{{{\mu _0}}}{a_y}\)

And the tangential component of magnetic flux density in medium 2 is

B_2t = μ₂ H_2t = a_x + 0.8 a_y

Thus the net magnetic flux density in medium 2 is

B₂ = B_2t + B_2n = a_x + 0.8 a_y + a_z

Concept:

The location of voltage maxima is given by

\({{Z}_{max}}=\frac{λ }{4\pi }\left( {{\theta }_{\text{ }\!\!\Gamma\!\!\text{ }}}+2n\pi \right)\)

Successive maxima differ by λ/2

Calculations:

Given θ_Γ = -150°

Length of line = 500 m

λ/2 = 75 m 

Two voltage maxima occur at a distance of 75 m. 

λ = 150 m

75 × 6 = 450 m 

Hence, clearly, at least 6 maxima occur on the line. 

Remaining length = 500 - 450 = 50 m. 

\(\frac{λ}{3} = 50 \) m 

Now, we know 

λ/2 = 360° 

λ/3 = 240° 

∴ If we start at -150°, while travelling 50m (240°) 

we will cover one more maxima.

Hence there are total 7 (6+1) maxima’s on the line. 

Concept:

The Amplitude of wave of distance ‘z’ in a medium is:

E(z) = E_oe^-αz

where α = attenuation constant.

The value of attenuation constant varies for different medium and depends on the ratio of  

Calculation:

\(\frac{\text{ }\!\!\sigma\!\!\text{ }}{\text{ }\!\!\omega\!\!\text{ }\epsilon }=\frac{5\times {{10}^{-4}}\times 36\text{ }\!\!\pi\!\!\text{ }}{2\text{ }\!\!\pi\!\!\text{ }\times 3\times {{10}^{9}}\times {{10}^{-9}}\times 9}\)
= 3.32 × 10^-4 ≪ 1

Hence the medium is low-loss dielectric

For low loss dielectric

\({\rm{\alpha }} = \frac{{\rm{\sigma }}}{2}\sqrt {\frac{{\rm{u}}}{\epsilon}} \)

\( = \frac{{\rm{\sigma }}}{2}\frac{{120{\rm{\pi }}}}{{\sqrt {{\epsilon_{\rm{r}}}} }}{\rm{\;}}\)

\(= \frac{{5 \times {{10}^{ - 4}} \times 120{\rm{\pi }}}}{{2 \times \sqrt 9 }} = 0.0314\left( {{\rm{Np}}/{\rm{m}}} \right){\rm{\;}}\)

Given at a Distance z, Amplitude

E(z) = 1 mv/m = 10^-3 V/m

E₀ = 10 v/m

10^-3 = 10e^-0.0314 (z)

ln (10^-4) = -0.0314.(z)

Concept:

\({{Z}_{i/p}}={{Z}_{0}}\left[ \frac{{{Z}_{L}}+j{{Z}_{0}}\tan \beta l}{{{Z}_{0}}+j{{Z}_{L}}\tan \beta l} \right]\)

\(wavelength~\left( λ \right)=\frac{\nu }{f}\)

Note: The impedance repeats itself after every λ/2 length from the load.

Calculations:

Phase velocity

v = 0.6c

Frequency, f = 2 MHz

The wavelength will be:

\(λ =\frac{v}{f}=\frac{0.6\times 3\times {{10}^{8}}}{2\times {{10}^{6}}}\)

= 90 m

We  observe that the length is half of the wavelength, i.e.

\(l=\frac{λ }{2}\)

For a 45 m long line, the input impedance is the same as the load impedance, i.e.

Concept:

For lossless network

|S₁₁|² + |S₂₁|² + |S₃₁|² + |S₄₁|² = 1

Calculation:

(0.1)² + (0.8)² + (0.3)² + |a|² = 1

0.74 + |a|² = 1

|a|² = 0.26

Concept:

The maximum angle over which light rays entering the fiber will be trapped in its core is known as angle of acceptance θ_a

\({\theta _a} = {\sin ^{ - 1}}\left( {\sqrt {n_1^2 - n_2^2} } \right)\)

n₁ is the refractive index of core and n₂ is the refractive index of the cladding.

The number of modes propagating in a step-index fiber can be estimated as:

\(N = \frac{{{V^2}}}{2}\)

Where V is the mode volume given as

\(V = \frac{{\pi d}}{\lambda }\sqrt {n_1^2 - n_2^2} \)

Where d is the fiber core diameter and λ is the wavelength of optical source.

The length of the fiber at which the power reduces to a certain % is given as:

\(l = \frac{{10}}{\alpha }\log \frac{{P\left( 0 \right)}}{{P\left( l \right)}}\)

Where α is the attenuation constant, P(0) is the input power and P(l) is the power at length l.

Given:

n₁ = 1.6

n₂ = 1.57

d = 100 μm

λ = 1 μm

α = 0.3 dB/km

Calculation:

Let us go by options one by one

Option 1), 2) \({\theta _a} = {\sin ^{ - 1}}(\sqrt {n_1^2 - n_2^2} )\)

= 17.96°

Option 3) The mode volume:

\(V = \frac{{\pi d}}{\lambda }\sqrt {n_1^2 - n_2^2} \)

\( = \frac{{3.14 \times 100 \times 0.308}}{1} = 96.712\)

Number of modes:

\(N = \frac{{{V^2}}}{2} = \frac{{{{96.712}^2}}}{2} = 4676\)