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
 -V’ = I(R+jwL) ….. eq.1
Note: V’ = (d/dx)V

In a similar method, assuming dV is zero,
 -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
 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
 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)

Differential/Balanced transmission

A pair of signal lines(true and inverted) is essential for each channel (there is additional ground return path). Noise is coupled to both wires of the pair, hence rejected by common mode rejection capability of differential amplifier.
Ground noise is also rejected by common mode rejection capability.

Single-ended transmission

Single-ended transmission is performed on one signal line, and the logical state is interpreted with respect to ground. Twisted pair cable recommended for distance more than 1 metre. E.g. EIA232
The poor noise immunity limits the distance and speed of reliable operation.

Characteristic Impedance of Micro-strip line

Z0=[L/C]^(1/2)= 377(h/w)[Er^{-1/2)]
Er=Relative permittivity of the dielectric material
h = thickness of dielectric
w = trace width

Reflections on a transmission line

On a long straight line, waves can travel in both directions. Consider a
mechanical transmission line suspended vertically and hung from the
ceiling. Reflections is seen by setting off a pulse from the bottom of
the line. After a time equal to twice the transit time, the pulse
reflects from the ceiling and returns to your hand. This takes a few
seconds. The pulse has been reflected at the top because the line is
anchored there at zero displacement for all time. In an electronic
transmission line, this is equivalent to holding the voltage at zero for
all time by using a short circuit across the line. If you look at the
direction of displacement in the pulse, it reverses on reflection. Thus,
a wave of displacement to the right is returned as a wave of
displacement to the left. In an electronic transmission line, a square
pulse of 1 volt amplitude is returned as a square pulse of -1 volt
amplitude.