Roger Sparks wrote:

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.

vf1(t, x) means that vf1 is a function of (varies with) both t and x.
Change t and/or x, and vf1 changes. This is not phasor notation, which I
specifically avoided. It is a simple description of the voltage at any
time and position. Plug t and x (and w, which stays fixed for the whole
analysis) into the argument of the sine function, poke it into your
pocket calculator, hit the sin key, and you have the voltage at that
time and place.

(Link http://campus.murraystate.edu/tsm/ts...h6_5/ch6_5.htm)

There are only two brief illustrations of a phasor (figs. 6-25 and 6-26)
in that document, which is otherwise just fine. The remainder of it is a
discussion of time functions like I've used.

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).

That's correct except for the description of wt. wt does not describe
the location of the wave in any way - it only tells how the value of the
wave changes with time. w converts the time, in seconds, to an angle in
radians which the sine function can be applied to.

How is "x" located on the transmission line?

x is the distance from the input end of the line. It should be in the
same units as wt. I rather carelessly specified that it be in electrical
degrees, but that would be only if wt is also converted to degrees.

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.

Yes, that's correct.

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.

The change from -x to +x is due to the change in direction of
propagation. The input end of the line is at x = 0. In the examples, the
line is an integral number of wavelengths long, so the far end of the
line is x = n*2*pi radians (or n*360 degrees) where n is an integer. So
at those points, sin(wt - x) and sin(wt + x) are the same value.

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.

Uh, oh.

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)

This looks ok so far, but with x = 0 at the input end of the line, as I
described. Let me show you -- start TLVis1, demo 1, and pause it just as
the initial wave reaches the end. (The line is two wavelengths, by the
way, not 3 as I incorrectly said in my introductory posting.) You've
stopped it at wt = 4*pi radians, or 720 degrees, using a frequency of 1
Hz for simplicity. (I'll use degrees since most of us think better in
degrees than radians.) What is the value of the voltage 1/4 wave from
the input end of the line (x = 90 degrees)? sin(wt - x) = sin(720 - 90
deg.) = sin(-90 deg) = -1, which you can see is correct. What's the
value 1/4 wave from the far end of the line (x = 630 deg.)? sin(720 -
630) = sin(90) = +1, which you can see is also correct.

Now start the analysis again and run it until the reflected wave reaches
the source, and pause it again. You're now at wt = 1440 degrees. What's
the value of the reflected wave at x = 90 degrees, that is, 90 degrees
from the input end of the line? sin(wt + x) = sin(1440 + 90) = sin(90) =
+1. 90 degrees from the far end (x = 630 deg), the value is sin(1440 +
630) = -1. These are what what the graph shows. You'll find that at
every point, at every time, the value of the forward wave is k * sin(wt
- x) and the reverse wave k * sin(wt + x) with x = 0 at the line input.
(k changes at each reflection). Please let me know if you see any time
or position for which this isn't true.

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.

That's correct.

Mathematically, this is accomplished by multiplying either the forward or reflective part of the equation a factor of -1.

It's done by applying a reflection coefficient of -1 to the impinging
wave to result in a reflected wave which is equal and opposite in phase.

The effect is to shift the phase of both forward and reflected waves by 90 degrees for a total of 180 degrees.

No. The analysis using superposition treats the forward wave and reverse
waves separately. Reflection has no effect on the forward wave, and it
creates a reflected wave. The reflected wave is the inverse of the
forward wave if the reflection coefficient is -1.

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.

I'm not quite sure what you're doing here. If the impinging wave is
sin(wt - x), the reflected wave is -sin(wt + x) upon hitting a short
circuit. If the impinging wave is sin(wt + x), its reflection is -sin(wt
- x).

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.

I'm not sure, but it looks like you're analyzing the sum of one forward
and one reflected wave, at the far end of the line. At wt = 0, there is
no voltage at the far end of the line. Perhaps you're also referencing
the time to the reflection point, that is, t = 0 is when the wave reflects?

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

Those are correct values. You get the same result when you measure x
from the input end of the line and time from when the source is turned on.

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

By which I believe you mean the far end of the line.

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.

Where this would run into trouble would be on a line which is not an
integral number of wavelengths long.

Certainly, you could use any point on or off the line as the x
reference, and any time t as the t reference, as long as you modify the
equations appropriately, as I believe you've shown. Your equations have
reversed the sign of x and the roles of forward and reflected waves, to
arrive at the same result when the two are summed.

The equations I presented use the input end of the line as the x
reference and the time of source turn-on as the t reference. The
equations were simpler because of the chosen line lengths, but apply to
all line lengths when the appropriate phase terms are added.


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.

This is an interesting alternative analysis. Is there any point on the
line at any time for which it predicts a different result than my
analysis? Does it correctly show the voltage of each forward and
reflected wave individually as well as its sum? If so, then it's an
equally valid analysis for this condition of integral wavelength lines.

If I get the time I'll try to extend either the analysis or program, or
both, to non-integral-wavelength line lengths.

Roy Lewallen, W7EL