Two wire open line. Transmission Lines. a. Strip line. d. w. a. Coaxial Cable. h. b. Two Wire Open Line. E x = E 0 exp (i[kz - t]), H y = (E 0 /Z 0 ) exp (i[kz- t]).
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Ex = E0 exp (i[kz -t]), Hy = (E0/Z0) exp (i[kz- t]).
Ex = E0, V = E0d, Hy = E0=Z0, I=wJs=w(E0/Z0).
The electric field has only an x-component, if edge effects are
neglected, and in the first approximation this component is independent of position across the width of the strip, i.e. Ex is independent of y. Similarly, the magnetic field has only a y-component and this is independent of x and y.
Hy = 0c G(z + ct) = -Ex/Z0
The electric field has a radial component, Er, and the magnetic field has only the component H
independent of the angle .
The field distribution due to a pair of wires
When the switch is closed how long does the lamp take before it lights?
If we make it easy and let the length of the wires between battery and lamp be 3.108 m, then the time between the switch closing and lamp lighting will be approximately ______s.
Does it mean that the mechanism relies on electromagnetic energy being transported from start to finish of the structure?
If so where is the energy stored during transit?
If it is electromagnetic energy that is stored then it has to be stored in inductance for magnetic energy and capacitance for electric energy.
This means that the transmission line has to have an inductance (per unit length) and a capacitance (per unit length).Two wire open line
Low Loss line at High Frequency
Inductance and capacitance are uniformly and continuously distributed as L (Henrys/m) and C (Farads/m) respectively.
When the switch is closed and a voltage V is applied to the line through a source impedance Zs, simple reasoning shows that the C's take a finite time to charge up through the L's; thus, the voltage propagates at a finite rate towards the load i.e. volts do not reach the load instantaneously.
Let L and C be distributed inductance
and capacitance per unit length.
In time t, electric flux q = (C ·x)v ( Q = CV) is
produced in time x/u sec , and
i(amps) = q/ t = Cx• v/ t
(capacitive charging or "displacement" current)
i = Cx• v /(x/ u ) = Cvu (I)
In time t, magnetic flux linkages =(L•x)i
(from = LI) ; are produced in time x/u sec
v(volts) = /t = L•x•i/t=Liu (ind Volts)…(II)
Multiplying I and II:- vi =viCLu2
Or u = 1/ LC ms-1velocity of propagation
Dividing I by II:- i/v =vC/iL or v2/i2 =L/C
v/I = L/C……Characteristic Impedance Z0
; i2 L = v2C
½ Li2 = ½ Cv2
Note:- the equality of Electric and Magnetic field energy in unit length of transmission line.
Rate of energy input to line due to advance of wavefront, is vi watts.
resistive load, connected directly to voltage source. Rate of energy input, due to thermal dissipation in R, is vi watts (also i2R).
Rate of energy input = vi watts. Idea of a
'matched line': looks like an infinitely
Note: if Rg = Z0, maximum power if transferred from the generator to the line.
As can be seen from the intuitive picture of a transmission line, wave propagation characteristics are dependent on the inductance and capacitance of the line. Thus we need to find expressions for the L and C of typical lines.
Assume r << S
Due to conductor A,
for 1 amp Linkages per metre, axially,
Due to conductor B, an identical expression is
External field due to 1 amp in line A, at radius
Flux linkages per metre axially through circuit
(between conductor 1 & 2) is:-
Capacitance of a Twin Line (r << S)
Assign a line charge of 1 C/m to both conductors.
D at x from conductor A is;
Similarly for VB(=VA)
per unit length
For air, therefore Z0 = 120 loge(b/a), (approx )