- +44 141 628 6080
- info@tutorsglobe.com

18,76,764

Questions

Asked

21,311

Experts

9,67,568

Questions

Answered

Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!

Submit Assignment
## Transmission lines, Physics tutorial

Transmission lines:The transmission line is system of material boundaries that forms continuous path which can direct transmission of electromagnetic energy from power station to other stations or for transmitting energy from one point to another. Transmission line is uniform if there is no change in cross sectional geometry. Wavelengths on transmission line are compatible with size at gigahertz frequencies and capacitances and inductances are very small.

Change in potential difference per unit length i.e.

ΔV/ΔL = -LdI/dt = -j(wL + R)I

Or in limit of ΔL becoming dl

dV/dl = -(jwL + R)I

Where L is self inductance because of magnetic field around conductor V and R, is resistance of conductor. Change in current flowing in line is provided by:

dI/Dl = cdV/dt - (jwc + G)V

Where C is capacitance per unit length (formed due to finite distance between conductors of line) and G is conductance per unit length (whose existence is because of dielectric losses of dielectric medium in between conductors). If we suppose that conductors have zero resistance and that they are divided by perfect insulator (in which case transmission line becomes lossless), equations which are basic equations of transmission line becomes,

dV/dl = -LdI/dt

dI/dl = -CdV/dt

Differentiating equations with l and t respectively we have

d

^{2}V/dl^{2}= -L(d^{2}I/dldt)d

^{2}I/dtdx = -Cd^{2}V/dt^{2}(1/LC)(d

^{2}V/dl^{2}) = d^{2}V/dt^{2}(1/LC)(d

^{2}I/dl^{2}) = d^{2}I/dt^{2}Equations are familiar wave equations of voltage and current respectively mean that both voltage and current propagate as waves along transmission lines. Velocity of propagation is V = 1/√LC. It can be illustrated that characteristic impedance of the transmission line z

_{0}= √LC.Parallel wire and coaxial cable transmission line:Two major examples of transmission lines are (i) parallel wire and (ii) coaxial cable.

To get propagation velocity, V and characteristics impedance of any transmission line inductance per unit length, L and capacitance per unit length should be determined.

a) Parallel wire transmission lines: Recall that electric field E of conductor is given by Gauss law for electric as,

E = λ/(πrε

_{0}εr = q/(2πrε_{0}ε_{r}l))λ(linear charge density) = q/l, for conductor in the medium of relative permittivity (dielectric constant), ε

_{r}. For two wires, each of radius x and separated by distance y.E = E

_{1}+ E_{2}= q/2πrε_{0}ε_{r}l + q/2πrε_{0}ε_{r}l = q/πrε_{0}ε_{r}lCapacitance of pair is obtained from

C = q/V = q/

_{x}∫^{2y}Edr = [q/(q/(πε_{0}ε_{r}l_{x}∫^{2y}(dr/k)))]Giving

C/l = πε

_{0}ε_{r}/ln(2y/x) i.e. capacitance per unit length. Inductance per unit length is attained as follows: magnetic flux of two conductors (whose shape are approximately cylindrical),Φ

_{B}= 2l_{x}∫^{2y-x}μ_{0}μ_{r}Idr/2πr ≈ [μ_{0}μ_{r}Iln(2y/x)/π]But self inductance, L = Φ

_{B}/i = [μ_{0}μ_{r}Iln(2y/x)/π]Therefore velocity of propagation, V = 1/√LC = 1/√ε

_{0}ε_{r}μ_{0}μ_{r}as expected. Characteristics impedance,z

_{0}= √L/C = (μ_{0}μ_{r}/π2ε_{0}ε_{r})1/2ln(2y/x)b) Coaxial cable transmission line: the capacitance for coaxial cable is obtained following procedure utilized for the parallel wire except that ∫Edr is from a to x and b to y, a being radius of central conductor and b, distance between centres of both conductors. Capacitance per unit length is therefore

C/l = 2πε

_{0}ε_{r}/ln(y/x)Inductance of coaxial cable is attained as follows: flux of magnetic field, Φ

_{B}, through closed circuit formed by joining conductors at the end of section of cable of length, l, is l_{x}∫^{y}(μ_{0}μ_{r}I/2πr)dr = (μ_{0}μ_{r}/2π)Illn(y/x)Self inductance, L = Φ

_{B}/I = L = (μ_{0}μ_{r}/2π)ln(y/x)Velocity of propagation, V = 1/√LC

Substituting V = 1/√ε

_{0}ε_{r}μ_{0}μ_{r}, and characteristics impedance,z

_{0}= √L/C(μ_{0}μ_{r}/4π^{2}ε_{0}ε_{r})1/2 lny/xEquivalent circuit:In Figure 1 above is illustrated two infinitely long lines carrying equal and opposite current. Part between point X and Y is subsection which comprises of impedance z =R+ jωL and admittance y= G+ jωC. Resistance, R and inductance, L are due to length and diameter of conductors while capacitance exists by virtue of close separation of conductors. Conductors being separated by the imperfect insulator or dielectric necessitate leakage of current. This signifies shunt conductance that together with shunt capacitance provides admittance. 2 conductors can therefore be represented in which case resistance, R and conductance, G are negligible. Each subsection is the equivalent circuit and infinitely loss line is considered as cascade of infinite number of such circuits.

Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)Expand your confidence, grow study skills and improve your grades.

Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, one-on-one online tutoring.

Using an advanced developed tutoring system providing little or no wait time, the students are connected on-demand with a tutor at www.tutorsglobe.com. Students work one-on-one, in real-time with a tutor, communicating and studying using a virtual whiteboard technology. Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.

Free to know our price and packages for online physics tutoring. Chat with us or submit request at info@tutorsglobe.com