Electromagnetics Test 4

Concept:

The reflection coefficient gives the percentage of incident voltage reflected from the load.

Mathematically, this is defined as:

\(Γ = \frac{{\left( {{Z_L} - {Z_o}} \right)}}{{({Z_L} + {Z_o})}}\)

Γ = Reflection coefficient

Z_L = Load Impedance

Z₀ = Characteristic Impedance.

Calculation:

Load impedance, Z_L = 300 Ω

Characteristic impedance, Z₀ = 150 Ω

So, the reflection coefficient at the load terminal will be:

\({{\rm{\Gamma }}_L} = \frac{{300 - 150}}{{300 + 150}} = \frac{1}{3}\)

And the reflection coefficient at the generator end will be:

\({{\rm{\Gamma }}_g} = \frac{{{Z_g} - {Z_0}}}{{{Z_g} + {Z_0}}}\)

Where Z_g is the internal impedance of the generator.

Since, it is given that the internal resistance of the generator is zero (i.e. Z_g = 0) so, we get:

Concept:

The characteristic impedance of a transmission line is defined as:

\({Z_0} = \sqrt {\frac{{R' + j\omega L'}}{{G' + j\omega C'}}} \)    ---(1)

And the propagation constant of the transmission line is defined as:

\(\gamma = \alpha + j\beta = \sqrt {\left( {R' + j\omega C'} \right)\left( {G' + j\omega C'} \right)} \)    ---(2)

Where,

α is attenuation constant

β is phase constant

R’ = Resistance per unit length of the line

G’ is conductance per unit length of the line

L’ is the inductance per unit length of the line

C’ is the capacitance per unit length of the line

Analysis:

For a lossless line:

R’ = G’ = 0

So, the characteristic impedance of a lossless transmission line using Equation (1) will be:

\({Z_0} = \sqrt {\frac{{L'}}{{C'}}} \)

And the propagation constant of a lossless transmission line using Equation (2) will be:

\(\gamma = \alpha + j\beta = j\omega \sqrt {L'C'} \)

α = 0

Therefore, the attenuation constant of the lossless line is always zero (real).

Statement (1) is correct.

A distortion less line satisfies the following condition:

\(\frac{{R'}}{{L'}} = \frac{{G'}}{{C'}}\)

So, the characteristic impedance of the distortionless line will be:

\({Z_0} = \sqrt {\frac{{L'}}{{C'}}} = \sqrt {\frac{{R'}}{{G'}}} \)

∴ The characteristic impedance of both lossless and a distortionless line is real.

And the propagation constant of the distortion less line will be:

\(\gamma = \alpha + j\beta = \sqrt {R'G'} + j\omega \sqrt {L'C'} \)

\(\alpha = \sqrt {R'G'} \ne 0\)

Therefore, the attenuation constant of distortion less line is not zero but it is real.

Statement 3 is incorrect.

The scattering matrix for the lossless network follows the symmetry S_ij = S_ji

Example of lossless scattering matrix is

\(S = \left[ {\begin{array}{*{20}{c}} {\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{{\sqrt 2 }}}\\ {\frac{1}{2}}&{\frac{1}{2}}&{\frac{{ - 1}}{{\sqrt 2 }}}\\ {\frac{1}{{\sqrt 2 }}}&{\frac{{ - 1}}{{\sqrt 2 }}}&0 \end{array}} \right]\)

Concept:

Distance between minimum and adjacent maximum in a transmission line is λ/4         …(1)

\(\lambda = \frac{c}{f}\)        …(2)

Calculations:

\(\frac{\lambda }{4} = 9 - 3 = 6\:cm\)

λ = 24 cm = 0.24 m

Concept:

Standing wave ratio is minimum when reflection coefficient is minimum.

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

\(=\frac{{{R}_{L}}+j{{X}_{L}}-{{Z}_{O}}}{{{R}_{L}}+j{{X}_{L}}+{{Z}_{O}}}\) 

\({{\left| \text{ }\!\!\Gamma\!\!\text{ } \right|}^{2}}={{\left| \frac{{{R}_{L}}+j{{X}_{L}}-{{Z}_{O}}}{{{R}_{L}}+j{{X}_{L}}+{{Z}_{O}}} \right|}^{2}}\) 

To minimize reflection coefficient

\(\frac{d}{d{{Z}_{O}}}{{\left| \text{ }\!\!\Gamma\!\!\text{ } \right|}^{2}}=0\) 

Which gives

Z²₀ = R²_L + X²_L

Student should remember this result

This is similar to matching the resistive load to a complex network in circuit

Calculation:

Z₀² = (30)² + (40)²

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

ZL = Load Impedance

Calculation:

1) Length of line, l = λ/8

Load impedance Z_L = 0 (Shorted)

The phase length will be:

\(\beta l = \left( {\frac{{2\pi }}{\lambda }} \right)\left( {\frac{\lambda }{8}} \right) = \frac{\pi }{4}\)

Therefore, the input impedance of the lossless transmission line will be:

\({Z_{in}} = {Z_0}\left( {\frac{{0 + j{Z_0}\tan \beta l}}{{{Z_0} + j0\tan \beta l}}} \right) = {Z_0}\left( {\frac{{j{Z_0}\tan \frac{\pi }{4}}}{{{Z_0}}}} \right)\)

= jZ₀ (Statement 1 is incorrect)

2) Length of line, l = λ/4

Load impedance, Z_L = 0 (Shorted)

\(\therefore \beta l = \left( {\frac{{2\pi }}{\lambda }} \right)\left( {\frac{\lambda }{4}} \right) = \frac{\pi }{2}\)   

Therefore, the input impedance of the lossless transmission line will be:

\({Z_{in}} = {Z_0}\left( {\frac{{0 + j{Z_0}\tan \beta l}}{{{Z_0} + j0\tan \beta l}}} \right)\)

\( = {Z_0}\left( {\frac{{j{Z_0}\tan \frac{\pi }{2}}}{{{Z_0}}}} \right) = j\infty \)

Statement (2) is correct.

3) Length of line, l = λ/2

Load impedance, Z_L = ∞ (Open-circuited)

\(\therefore \beta l = \left( {\frac{{2\pi }}{\lambda }} \right)\left( {\frac{\lambda }{2}} \right) = \pi \)

Therefore, the input impedance of the lossless transmission line will be:

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

\(= {Z_0}\left( {\frac{1}{{j\tan \pi }}} \right) = - j\infty \)

Statement (3) is Correct

4) Matched line have the load impedance equal to its characteristic impedance, i.e.

Z_L = Z₀

So, for the matched line the input impedance will be:

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

Statement (4) is correct.

Concept:

The location of the first voltage minima is at:

\({{Z}_{min}}=\frac{\lambda }{4\pi }\left( {{\theta }_{\text{ }\!\!\Gamma\!\!\text{ }}}+\pi \right)\)

Where θ_Γ is the ∠ of the reflection coefficient.

Calculation:

The reflection coefficient at load is

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

\(=\frac{100-\left( 50+j50 \right)}{100+\left( 50+j50 \right)}\)

\(=\frac{50-j50}{150+j50}\)

\(=\frac{1-j1}{3+j1}=\frac{(1-j)(3-j)}{10}= \frac{1-2j}{5}\)

\(\Rightarrow \frac{1}{\sqrt{5}}{{e}^{-j0.3524\pi }}\)

θ­_L = -0.3524 π

first voltage minimum

\({{Z}_{min}}=\frac{\lambda }{4\pi }\left( {{\theta }_{\text{ }\!\!\Gamma\!\!\text{ }}}+\pi \right)\)

\(=\frac{\lambda }{4\pi }\left( -0.3524~\pi +\pi \right)\)