On Mon, 07 Jan 2008 12:34:03 -0600
Cecil Moore wrote:

Roger Sparks wrote:
What is the common reference point of x,t? In other words, when both x and t are zero, where on the antenna is the zero location?

By convention, x=0 and t=0 are at the feedpoint with the same
phase as the feedpoint signal. Thus, at the feedpoint, the
current is equal, but at no other point until one wavelength
is reached.
--
73, Cecil http://www.w5dxp.com

Thanks Cecil.

This provides a great entry to post my conclusions after studying over the weekend. I use some of Roy's work in this but only as an example of how equations are written, not as an example of accuracy. Thanks Roy, for providing the material.


On Tue, 01 Jan 2008 23:13:18 -0800
Roy Lewallen wrote:

Second analysis: +0.5 input reflection coefficient


clip.......

so that the initial voltage at the line input is simply sin(wt). At any
point that the forward wave has reached, then,

vf1(t, x) = sin(wt - x)

The open far end of the line has a reflection coefficient of +1, so as
before the returning wave is:

vr1(t, x) = sin(wt + x)

(The change from -x to +x is due to the reversal of direction of
propagation. You'll see this with every reflection.) At any time between
when vf1 reflects from the output end (t = 2*pi/w) and when it reaches
the input end of the line (t = 4*pi/w), the total voltage at any point
the returning wave has reached is

vtot = vf1(t, x) + vr1(t, x) = sin(wt - x) + sin(wt + x) [Eq. 1]


Roy's work is only provided as an example of the use of mathematical notation.

The notation "vf1(t, x)" is not fully explained in the text, so any
reader who is unfamiliar with the notation must do some additional
study. This describes my situation, so after some study, this is what
I think these terms mean, focusing on the two coordinates, wt and x.

Roy is using phasor math here, a subject comprehensively covered
elsewhere. (Link http://campus.murraystate.edu/tsm/ts...h6_5/ch6_5.htm)
The sum of "wt" and "x" denote the phase angle of the wave, measured in
degrees, both referenced to zero. Two angles are needed, one to represent the physical location of each wave as a function of time, the second to locate the point of examination of the wave. "wt" locates the wave in time and physical space . "x" locates the examination point in physical space (on a transmission line in this case).

How is "x" located on the transmission line?

My automatic assumption was that "x" was based at the input to the
line. Thus, if the input was on the left side of the page, the line
would extend to the right, and x would become larger when moving
right. This led to a contradiction when I considered how the wave
reflects from the right side upon encountering a discontinuity.
Suddenly, a positive reflected wave represented by sine(-x) was made
sine(x) at an open circuit. At position sin(90), the wave reversed and
became sine(-90). It was if a factor of -1 was applied to the
analysis, which was not the case here.

After considerable thought and time, I realized that the origin of "x"
was not at the input point, but was at the reflection point. Then the
notation made perfect sense.

Forward wave-
input-_____________________________________|Reflectio n point
++x - increasing x - Reflected wave 0

Scale both the sine wave and physical line in degrees. The wave will repeat every 360 degrees (2pi radians). The x scale can be as long (in degrees or radians) as desired

Think of the sine wave as an curve traced on a MOVING sheet of paper. The paper/curve moves in the direction of the forward wave. As the sine wave trace passes the zero point, the wave has a(unit) magnitude, described as sin(wt). Pick any wt and match wt to 0 on the physical scale. Then go to the x location of interest to find the magnitude of the sine wave at that location for one instant of time (wt).

If a reflected wave is present, think of the same wave picture but moving in the opposite direction. When the reflection point is an open circuit, the reflected wave is generated by the forward wave so the two waves will always have a 1 to 1 corespondence at the zero point (point of reflection). The two waves are related by

For the forward wave, vf(t, x) = sin(wt - x)
For the reflected wave, vr(t, x) = sin(wt + x)

For the total voltage at any point (open circuit case).
vtot = vf(t, x) + vr(t, x)
= sin(wt - x) + sin(wt + x)

When the reflection point is a short circuit, the voltage becomes zero at the short. The sum of forward and reflected wave voltages (at the short) will always be zero. Mathematically, this is accomplished by multiplying either the forward or reflective part of the equation a factor of -1. The effect is to shift the phase of both forward and reflected waves by 90 degrees for a total of 180 degrees. We will apply the factor of -1 to the forward wave to maintain polarity consistancy with the open circuit case.

vfS(t, x) = -sine(wt - x) For short circuit reflection.

Notice that if x = 0,

vtotS(t, x) = -vfS(t, x) + vrS(t, x)
= -sine(wt) + sine(wt)
= 0

For practice, let's analyze a transmission line fed at one end, open at the other. The relfections will have a voltage equal to the forward voltage at all times so we will use the equation
vtot = sin(wt - x) + sin(wt + x)

Assuming one volt peak for the sine wave amplitude (a unit amplitude), find the voltage at points vtot(wt, x),(0,0),(90,0),(180,0),(270,0),(360,0). Notice that every voltage will be computed at the reflection point, x = 0.

For 0,0, vtot(wt,x) = sine(0 - 0) + sine(0 + 0) = 0
For 90,0 vtot(wt,x) = sine(90 - 0) + sine(90 + 0) = 2v
For 180,0, vtot(wt,x) = sine(180 - 0) + sine(180 + 0) = 0
For 270,0, vtot(wt,x) = sine(270 - 0) + sine(270 + 0) = -2v
For 360,0, vtot(wt,x) = sine(360 - 0) + sine(360 + 0) = 0

In this example, one complete sine wave cycle has passed the x origin.

For the next example, let's assume we want to compute the voltage at the point located 45 degrees from the reflection point at the same time that the voltage at the end is at a maximum. The maximum points for voltage at the end occurs when the 90 and 270 degree phases reach the reflection point. Lets look at the 90 degree case.

For 90,45, vtot(wt,x) = sine(90 - 45) + sine(90 + 45)
= sine(45) + sine(135)
= 0.707 + 0.707
= 1.414v
For 90,135, vtot(wt,x) = sine(90 - 135) + sine(90 + 135) = 0
= sine(-45) + sine(225)
= sine(225) + sine(225)
= -0.707 + (-0.707)
= -1.414v

We can continue this process to find the voltage envelope for any location/phase relationship desired. It works equally well for the short circuit case.

Let's look at the case for short circuited 0 degree case, where the phase is 0 degrees, and examination point is 90 degrees.

For 0,90 (short circuit),
vtotS(wt,x) = - sine(0 - 90) + sine(0 + 90)
= - sine(-90) + sine(90)
= - sine(270) + sine(90)
= -(-1) + 1
= 2

Now advance the phase at 45 degrees, and examine x = 90 degrees

For 45,90 (short circuit)
vtotS(wt,x) = - sine(45 - 90) + sine(45 + 90)
= - sine(-45) + sine(135)
= - sine(315) + sine(135)
= -(-0.707) + (0.707)
= 1.414

Now advance the phase to 90 degrees, examine at x = 90 degrees.
For 90,90, (short circuit),
vtotS(wr,x)= - sine(90 - 90) + sine(90 + 90)
= - sine(0) + sine(180)
= - 0 + 0
= 0

This discussion focused on voltage. Similar notation is used in equations for current and power.

73, Roger, W7WKB