Transmission lines

Transverse Electromagnetic Waves: Propagation of energy in a transmission line takes place such that electric and magnetic fields transverse to one another and also to direction of propagation. The resultant wave is termed as TEM wave.

Consider a small section n of a parallel wire.

Length of this line is dx.

Voltage at input is V, at the other end is V+dV; Similarly current is I, I+dI.

Assume Primary Line Constants of the line are R,L,G and C

Note: w= 2.pi.f

For small dx, dI is zero.

Potential drop across the line is V – (V+dV) = R.dx.I + jwL.dx.I

à -V’ = I(R+jwL) ….. eq.1

Note: V’ = (d/dx)V

In a similar method, assuming dV is zero,

à -I’ = V(G+iwC) ….. eq.2

Differentiate eq.1,2

V”= V(R+jwL)(G+jwC) ….. eq.3

I” = I(R+jwL)(G+jwC) …..eq.4

Let (gamma)^2 = (R+jwL)(G+jwC)

Gamma = (alpha) + j(beta)

alpha is attenuation constant

beta is phase constant

gamma is propagation constant

Now eq.3,4 becomes

V” = V. (gamma)^2

I” = I. (gamma)^2

Solving the above equations,

V=A.exp(-gamma.x) + B.exp(gamma.x) …..eq.5

I= C.exp(-gamma.x) + D.exp(gamma.x) …..eq.6

Note: exp(x) = e^x

The first terms in eq.5,6 are called incident component(magnitude of V or I decreases from source towards load, whereas the second terms are called reflected component(magnitude of V or I decreases form load towards source)

Hypothetical infinite line: Voltage at distant end approaches zero(i.e no reflected component)

At x=0, V=Vs

Substitute in eq.5, Vs=A+B

At x=infinity, V=0

à B=0, V= Vs.exp(-gamma.x)

V’=-gamma.Vs.exp(-gamma.x)=-(R+jwL)I

Simplifying,

I=(Vs.exp(-gamma.x))/Z0

Where Z0=[(R+jwL)/(G+jwC)]^(1/2)

Z0 the input impedance of such infinite line is commonly referred to as Characteristic Impedance of the line. Z0 and propagation constant are termed as secondary constants (or coefficients) of the line

Line Terminated in a Load Impedance Zr:

At a distance x from the source, voltage and current are Vx, Ix.

-V’=(R+jwL).Ix

Substitute the above after differentiating eq.5,6 and then simplifying,

Voltage, current at source are Vs=A+B, Is=(A-B)/Z0

Vx=Vs.cosh(gamma.x)-Is.Z0.sinh(gamma.x) …..eq.7

Ix=Is.cosh(gamma.x)-(Vs/Z0)sinh(gamma.x) …..eq.8

For load impedance Zr, length of transmission line is l.

Ix=Ir, Vx=Vr such that Vr=Ir.Zr

Substitute the above in eq.7,8, and solve for Vs/Is

Input Impedance Zin = Vs/Is = Z0.N/D

Where N = Zr.cosh(gamma.l) + Z0.sinh(gamma.l)

D= Z0.cosh(gamma.l) + Zr.sinh(gamma.l)

Line Terminated in Load Impedance Z0:

Zr=Z0 in N,D

à Zin=Z0

A line terminated in its characteristic impedance has input impedance equal to Z0. In such a line, there is no reflected component and at any point x distant from the signal source, the voltage and current are same as that for infinite length transmission line.

Low frequency transmission line:

R is very bigger than wL

G is very lesser than wC

Z0 = (R/jwC)^(1/2)

High frequency line:

R is very lesser than wL

G is very lesser than wC

Z0 = (L/C)^(1/2)

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