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r.r.a.a WARNING!!!
On 25 Dec 2007, 04:26, "Dave" wrote:
or FAQ depending on how you look at it... I should probably repeat this regularly on here. This newsgroup should NOT be used as a reference source for concepts or equations regarding fields, waves, transmission lines, or other physical phenomena. *Please consult published text books and peer reviewed journals for analysis of technical questions. *The regular contributors in this group have a wide variety of misconceptions and erroneous views which they frequently throw in as if they were well known facts. On the lighter side, it can be fun now and then to throw them a simple problem and watch them swarm around like a kicked hornet nest. Did you ever think that your post would last this long? Obviously the regular contributors in this group cannot handle the truth and thus will not consult anything.Now the experts are argueing over the term SWR a very, very, deep discussion revealing things unknown to the amateur community at this time. No need for books if you quest is an arguement. Thus you are a model member of this newsgroup.What goes around comes around. Your buddy Art |
r.r.a.a WARNING!!!
art wrote:
On 25 Dec 2007, 04:26, "Dave" wrote: or FAQ depending on how you look at it... I should probably repeat this regularly on here. This newsgroup should NOT be used as a reference source for concepts or equations regarding fields, waves, transmission lines, or other physical phenomena. Please consult published text books and peer reviewed journals for analysis of technical questions. The regular contributors in this group have a wide variety of misconceptions and erroneous views which they frequently throw in as if they were well known facts. On the lighter side, it can be fun now and then to throw them a simple problem and watch them swarm around like a kicked hornet nest. Did you ever think that your post would last this long? Obviously the regular contributors in this group cannot handle the truth and thus will not consult anything.Now the experts are argueing over the term SWR a very, very, deep discussion revealing things unknown to the amateur community at this time. No need for books if you quest is an arguement. Thus you are a model member of this newsgroup.What goes around comes around. Your buddy Art Hi Art, The arrogance and false superiority evidenced in Dave's post can only be learned in a university. Dave is an old regular on this newsgroup who has posted under many different names over the years. Pay him no attention. 73, Tom Donaly, KA6RUH |
r.r.a.a WARNING!!!
On 6 Jan, 17:04, "Tom Donaly" wrote:
art wrote: On 25 Dec 2007, 04:26, "Dave" wrote: or FAQ depending on how you look at it... I should probably repeat this regularly on here. This newsgroup should NOT be used as a reference source for concepts or equations regarding fields, waves, transmission lines, or other physical phenomena. *Please consult published text books and peer reviewed journals for analysis of technical questions. *The regular contributors in this group have a wide variety of misconceptions and erroneous views which they frequently throw in as if they were well known facts. On the lighter side, it can be fun now and then to throw them a simple problem and watch them swarm around like a kicked hornet nest. Did you ever think that your post would last this long? Obviously the regular contributors in this group cannot handle the truth and thus will not consult anything.Now the experts are argueing over the term SWR a very, very, deep discussion revealing things unknown to the amateur community at this time. No need for books if you quest is an arguement. Thus you are a model member of this newsgroup.What goes around comes around. Your buddy Art Hi Art, * * * * *The arrogance and false superiority evidenced in Dave's post can only be learned in a university. Dave is an old regular on this newsgroup who has posted under many different names over the years. Pay him no attention. 73, Tom Donaly, KA6RUH- Hide quoted text - - Show quoted text - But Tom, this newsgroup is for auguments. Nobody is responding to the science of antennas. Nobody has the knoweledge to refute statements supplied How about just you and I discuss radiation and how it is created? That sort of thing may interest people other than the instant experts. For instance, I would love a counter discussion to what I have theorised even if it shows me to be in error, but this requires more than handwaving . I know you are knoweledgable about the things I have discussed where as others are still back in the old days and thus cannot contribute on a modern basis.. If we make a "field" all will follow and leave the puffed up pretend experts in the dust. At least the thread would NOT consist mainly of one person which would be a change.On top of that they would read a civil discussion that has not happened in a long while. As for Dave he has not had what is known as a education,Witness his denials of static versus dynamic fields, where I know of no written text book that alignes with him. You just can't fake an engineering structured education. Best regards Art |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
On Sun, 6 Jan 2008 21:07:11 -0500, "AI4QJ" wrote:
But you said (in CAPS), "8:1 as evidenced by CURRENT on the wire. :-), You cannot make a SWR measurement on a receive antenna any other way." Sorry if the CAPS made my response look terse but I was only repeating your word verbatim. Hi Dan, All very true, but the meaning to you is what you are questioning I am quite sure, and as I am quite comfortable with what I meant (I wrote it after all), just what is your question? Given that, just as with the voltage VSWR, the current standing wave is merely a depiction of the envelope of maximum amd minimum current values at the various points along distance kx, how do you measure the standing wave current on the wire? It is reported with every other aspect of the antenna by EZNEC. You should acquaint yourself with the common reports provided by it or other modelers in the NEC field. Do you use a current loop and measure the maxima and minima of a great number of points on a line and then plot the ISWR outer envelope on graph paper? My point is that standing wave current does not travel through the wire, it merely oscillates at different max/min amplitudes on each of the infinite number of points on the line. It cannot be measured directly with a current loop. Traveling? You've got yourself twisted around the axle. I've measured these phenomenon professionally, to NBS standards across the full range of RF out to 12GHz. Although the technique can be heavily invested with up-front work, and certain methods must be chosen with care, conceptually it is quite simple. The VSWR meter on the ham rig is merely looking at the balance of forward and reflected "power" and it is calibrated to read it out as VSWR (or SWR). It may as well say "ISWR"; it is all the same thing. But it is not measured by sensing either voltage ot current going into the antenna...it measures the delta power. 99% of correspondents here have never had any experience with determining SWR beyond the meter you just described. SWR was being measured long before its invention, and you would be hard pressed to find one in a laboratory (except as a customer's item to be tested). Now, if you would simply take my advice to heart: strip away the static and ask the question that is plaguing you. 73's Richard Clark, KB7QHC |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
Richard Clark wrote:
It is reported with every other aspect of the antenna by EZNEC. TravWave.EZ is not really an antenna. It is a loaded 1/4WL single- wire transmission line designed to display the nature of traveling waves. Take a look at the current phase vs the current phase of StndWave.EZ. Both files are available from my web page. -- 73, Cecil http://www.w5dxp.com |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
On Sun, 06 Jan 2008 23:18:04 -0600, Cecil Moore
wrote: Richard Clark wrote: It is reported with every other aspect of the antenna by EZNEC. TravWave.EZ is not really an antenna. It is more an artificial ground toaster. It bears no more relationship to an antenna than its applicability to Standing/Traveling Waves does to the Prince of Orange. |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
Richard Clark wrote:
On Sun, 06 Jan 2008 23:18:04 -0600, Cecil Moore wrote: Richard Clark wrote: It is reported with every other aspect of the antenna by EZNEC. TravWave.EZ is not really an antenna. It is more an artificial ground toaster. It bears no more relationship to an antenna than its applicability to Standing/Traveling Waves does to the Prince of Orange. Your motive with all this handing waving is unclear. It's only purpose is to illustrate traveling waves on a single piece of wire. You can make the wire 1/4WL or any other length. And the same thing can be illustrated by a model of a terminated rhombic antenna in free space. http://www.w5dxp.com/rhombicT.EZ This is a terminated rhombic in free space. The termination resistor is 880 ohms and the 880 ohm feedpoint SWR is 1.032:1. Take a look at the current with and without the phase. Then compare that traveling-wave current to the standing-wave current on a 1/2WL dipole. Or simply learn enough math to tell the difference between Io*cos(kx)*cos(wt) and Io*cos(kx+wt). -- 73, Cecil http://www.w5dxp.com |
Standing morphing to travelling waves, and other stupid notions
Jim Kelley wrote:
Cecil Moore wrote: Sorry Jim, but [Io*cos(kx)*cos(wt)] and [Io*cos(kx+wt)] *ARE* different. Evidently, you can't recognize a trig identity when you see one. One really should take a look at the math before waving one's hands and opening one's mouth in ignorance. Please enlighten us as to exactly what trig "identity" will make the following terms equal. E1*e^j(wt-kx) ?=? E2*e^j(wt-kx) + E2*e^j(wt+kx) Seems to me the only condition for which they are equal is when E2=0, i.e. when reflections (and therefore standing waves) don't exist. -- 73, Cecil http://www.w5dxp.com |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
On Mon, 07 Jan 2008 08:09:44 -0600, Cecil Moore
wrote: Your motive with all this handing waving is unclear. It's only purpose is to illustrate traveling waves on a single piece of wire. Three pieces of wire, one of them yours (a piece of trash as you describe it; and two of mine, following the conventions of their designers). The demonstration is remarkable enough to accomplish what you set out to do, clear up the confusion about standing waves on a Rhombic, and then later a Beverage |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
On Mon, 07 Jan 2008 08:09:44 -0600, Cecil Moore
wrote: http://www.w5dxp.com/rhombicT.EZ This is a terminated rhombic in free space. The termination resistor is 880 ohms and the 880 ohm feedpoint SWR is 1.032:1. Which, of course, has nothing to do with Standing Waves ON THE ANTENNA. On wire 1, the variation that is swinging along the line like any other Standing Wave antenna of length greater than a quarter wave. The value is higher than your "source SWR," of course. Take a look at the current with and without the phase. Swinging 180 degrees at 6 or 7 degrees per segment. Exactly like my earlier reports in contrast to your proclamations. Then compare that traveling-wave current to the standing-wave current on a 1/2WL dipole. Or simply learn enough math to tell the difference between Io*cos(kx)*cos(wt) and Io*cos(kx+wt). I was there weeks ago ahead of you where the formula applies (on the wire, not an EZNEC report of source SWR). It was more interesting to compare your Rhombic to itself in a closer to ground situation (elevated 12 feet above a real, high accuracy ground). SWR there, out of the gate (not at the source) and on the antenna wire itself (where traveling waves would be presumed to inhabit the design): SWR: 1.15 Further down the wires, radiation loss does ameliorate this SWR: 1.08 Still in excess of your Source SWR, which, obviously has nothing to do with STANDING/TRAVELING WAVES ON THE WIRE. Nice of you to confirm every point I made in refuting your claims. Others are free to observe every classic indication of Standing Waves upon a "Traveling Wave" antenna, complete with phase shift as I had reported some time ago. |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
Richard Clark wrote:
The demonstration is remarkable enough to accomplish what you set out to do, clear up the confusion about standing waves on a Rhombic, and then later a Beverage There's no confusion. Terminated rhombics and Beverages are traveling-wave antennas. Their currents are primarily: I(x,t) = Io*cos(kx-wt) A 1/2WL dipole is a standing-wave antenna. It's current is primarily: I(x,t) = Io*cos(kx)*cos(wt) There's no confusion except in your mind. -- 73, Cecil http://www.w5dxp.com |
Standing morphing to travelling waves. was r.r.a.a LaughRiot!!!
On Mon, 07 Jan 2008 17:51:15 GMT
Cecil Moore wrote: Richard Clark wrote: The demonstration is remarkable enough to accomplish what you set out to do, clear up the confusion about standing waves on a Rhombic, and then later a Beverage There's no confusion. Terminated rhombics and Beverages are traveling-wave antennas. Their currents are primarily: I(x,t) = Io*cos(kx-wt) A 1/2WL dipole is a standing-wave antenna. It's current is primarily: I(x,t) = Io*cos(kx)*cos(wt) There's no confusion except in your mind. -- 73, Cecil http://www.w5dxp.com 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? -- Roger Sparks |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
Richard Clark wrote:
Cecil Moore wrote: http://www.w5dxp.com/rhombicT.EZ This is a terminated rhombic in free space. The termination resistor is 880 ohms and the 880 ohm feedpoint SWR is 1.032:1. Which, of course, has nothing to do with Standing Waves ON THE ANTENNA. The SWR on the antenna has nothing to do with the standing waves on the antenna?????? Good grief, Richard, it makes sense now. You don't even know what a standing-wave antenna is - so allow me to explain. An antenna has a characteristic impedance range. If the terminating resistor is in the center of that range, *reflections are reduced to a negligible value*. That's why terminated antennas are (usually) classified as *traveling-wave antennas* - because they minimize reflections. The ends of a dipole are (surprise) an open-circuit. All of the forward wave is reflected. That's why a dipole is a *standing-wave antenna*. There is nothing except radiation that can reduce the reflections. Sorry, nothing else you could have to say would make any difference with a gross misconception like that. -- 73, Cecil http://www.w5dxp.com |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
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 |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
On 7 Jan, 10:12, Cecil Moore wrote:
Richard Clark wrote: Cecil Moore wrote: http://www.w5dxp.com/rhombicT.EZ This is a terminated rhombic in free space. The termination resistor is 880 ohms and the 880 ohm feedpoint SWR is 1.032:1. Which, of course, has nothing to do with Standing Waves ON THE ANTENNA. The SWR on the antenna has nothing to do with the standing waves on the antenna?????? Good grief, Richard, it makes sense now. You don't even know what a standing-wave antenna is - so allow me to explain. An antenna has a characteristic impedance range. If the terminating resistor is in the center of that range, *reflections are reduced to a negligible value*. That's why terminated antennas are (usually) classified as *traveling-wave antennas* - because they minimize reflections. The ends of a dipole are (surprise) an open-circuit. All of the forward wave is reflected. That's why a dipole is a *standing-wave antenna*. There is nothing except radiation that can reduce the reflections. Sorry, nothing else you could have to say would make any difference with a gross misconception like that. -- 73, Cecil *http://www.w5dxp.com Cecil, Can you help me understand that the end of a dipole is an open circuit? Nothing falls out because the current changes direction ala AC because the end of a dipole equates with the time that requires the current changes direction. So where is the open circuit? I also cannot see why radiation reduces reflection if the circuit can be completed with iron! Best regards Art |
Standing morphing to travelling waves. was r.r.a.a LaughRiot!!!
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 |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
Cecil Moore wrote:
Richard Clark wrote: It is reported with every other aspect of the antenna by EZNEC. TravWave.EZ is not really an antenna. It is a loaded 1/4WL single- wire transmission line designed to display the nature of traveling waves. Take a look at the current phase vs the current phase of StndWave.EZ. Both files are available from my web page. Hi Cecil Are the formulas used for the calculations in Eznec compatible with what you are trying to prove? - 73 de Mike N3LI - |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
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 |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
Michael Coslo wrote:
Are the formulas used for the calculations in Eznec compatible with what you are trying to prove? EZNEC seems to be reporting the envelope, not the instantaneous values but that is sufficient for what I am trying to prove. "Trying to prove"? Shirley, you jest. What I am saying was proven more than a century ago. What the heck has happened to the US educational system since I went to Texas A&M during the 50s? The differences between standing-wave antennas and traveling-wave antennas have been understood for more than a century. That there is so much ignorance exhibited on this newsgroup is not my fault. -- 73, Cecil http://www.w5dxp.com |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
Roy Lewallen wrote:
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. In the past, you have absolutely rejected any and all equally valid analysis. Are you now changing your mind? -- 73, Cecil http://www.w5dxp.com |
Standing morphing to travelling waves. was r.r.a.a LaughRiot!!!
On Mon, 07 Jan 2008 12:43:24 -0800
Roy Lewallen wrote: 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. No, it would work for any line length. The line length is the largest x value picked. For instance, to see the voltages at 597 degrees for a line 597 degrees long, make the x line at least 597 degrees long. Because the phase completely rotates every 360 degrees, a line 597-360 = 237 degrees long will have the same voltage relationship. Assume we want to compute the total voltage at the input when the input is at peak applied voltage (as if from in ideal source) at +90 degrees. Using the 597 degree long line, wt at the end of the line will be 90 + 597 = 687 degrees. We are examining the line at 597 degrees so x would be 597 degrees. For 687,597, vtot(wt,x) = sine(687 - 597) + sine(687 + 597) = sine(327 - 237) + sine(327 + 237) = sine(90) + sine(564) = sine(90) + sine(204) = 1 + (-.406) = 0.594v I hope you come up with the same value. 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? Not that I know of. I think you are not really using x, but x plus line length in your equations. This continues to have me confused. 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. So far as I have tested, yes. There is no reason to use the reflection point as the reference unless it simplifies the equations or improves understanding. 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 Using the reflection point as the zero reference seems to correspond with an observation you made about the end of the line controlling the SWR. 73, Roger, W7WKB |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
On Mon, 07 Jan 2008 18:12:59 GMT, Cecil Moore
wrote: feedpoint SWR is 1.032:1. Which, of course, has nothing to do with Standing Waves ON THE ANTENNA. |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
On Mon, 7 Jan 2008 18:26:16 -0500, "AI4QJ" wrote:
"Richard Clark" wrote in message Now, if you would simply take my advice to heart: strip away the static and ask the question that is plaguing you. OK. Thank you for your patience. You said "The SWR on the wire is load- based and is equal to 8:1 as evidenced by CURRENT on the wire. :-) Hi Dan, Well, in fact I don't recall having said anything about "Load-Based," but it is immaterial to your question(s). You cannot make a SWR measurement on a receive antenna any other way." You are still trapped in a loose thread of associations. We will get back to that in a moment (I hate doing this, trying to intuit what you are after). The specific comment you recite above is wholly out of context, and is being joined to a new one. This is why I ask you to strip out the noise and ask the question that is bothering you. My very simple question: How do I measure that ISWR current? Specifically, as you have already suggested, with a loop pickup. However, in my career I have never encountered its use as a precision device when measuring voltage is so much easier. EZNEC does not report voltage (although it could be computed, I will leave that to others as there is no significant information obtained in that exercise). Please describe possible meters and probes and where on the receive antenna do I measure this current? I've commented on that already. But this tied back into receive and you will not find a satisfactory, bench answer for that which is practical (in the sense of $). At the input? For waves on a line you go to the line. EZNEC provides this quite clearly and even provides a graphic solution if you are not into the actual numbers. At the nodes? At the zero crosses? How do you calculate ISWR? Is it Imax/Imin? That is the conventional way to observe Standing Waves. I am only trying to conceptualize what you mean. Do you have to block the reverse current somehow and measure forward current and vice versa? What does this do to the standing wave? What does any direct current measurement do to the ISWR itself? Dan, you've got yourself wrapped around the axle and here I have to force a solution to your problem that you only vaguely offer through a chain of disconnected references above. First, you spend an inordinate amount of effort quoting to the point of receiving, and yet none of your questions appear to be aware of the significance of receiving with these breed of (Traveling Wave) antennas. Cecil's wire, one foot off the ground, is strongly influenced by its proximity. I demonstrated that. However, Cecil divorces this effect while embracing his goal of explaining away the confusion about Traveling Wave antennas. This one foot high monstrosity is most closely allied with the Beverage antenna which was developed with ground in mind and even here, Cecil corrupts this concept to turn his model into a Traveling Wave transmission line. In effect, his is a bait and switch argument as he abandoned the "explanation" 350 posts ago and never returned until I forced the argument. Cecil, under protest to my proofs of Standing Wave on these Traveling Wave, then supplied a citation that heads this very thread: "Because the Beverage is a traveling wave, terminated antenna, it has no standing waves resulting from radio signals." It contains FROM radio signals. FROM is an externality. In other words the antenna must be externally excited to test the validity of this specific statement. Further, this statement has its own internal logic in that the antenna (the Beverage) is a RECEIVE antenna. Further, this statement has its own internal logic in that a Beverage antenna employs ground as an active element. Hence, all internal logical characteristics are consistent with the statement about externality. Any reference to source SWR is wholly inappropriate. EZNEC is perfectly capable of measuring a receive antenna, if you can supply an antenna to excite it. I did so explicitly. The reports offered by EZNEC will give you a source SWR reading like Cecil quoted, but that is entirely unrelated to the receive antenna performance. I did not report that reading for that reason, Cecil did - for whatever reason, but a reason that bore no relation to Standing/Traveling waves on the test antenna. EZNEC will report all currents on all wires everywhere. When the Beverage is excited by a source 100 kilometers away, those currents in the wires defining the Beverage exhibit classic Standing Waves. This is quite simple. Now as to your enquiry in how to measure those currents. That is not practical at the nanoAmpere levels, but software does it quite easily. If I were tasked to do it, I would immediately transform this into the voltage model, modulate the source with an unique pattern, and perform a synchronous detection of the levels involved along the length of the line. A trivial concept that is difficult in practice. Now, I have pounded out a lot of words in an effort to eke out your question. Was it answered? 73's Richard Clark, KB7QHC |
r.r.a.a WARNING!!!
On 6 Jan, 18:39, "AI4QJ" wrote:
"art" wrote in message ... "Did you ever think that your post would last this long? Obviously the regular contributors in this group cannot handle the truth and thus will not consult anything.Now the experts are argueing over the term SWR a very, very, deep discussion revealing things unknown to the amateur community at this time. [...]" Hello Art, the concept of SWR is extremely misunderstood even by people with degrees in electronics engineering. It is assumed to be simple, yet many people get it wrong. Indeed, a good understanding is essential for antenna development. I do not fault anyone for not understanding the concept because standing waves, simple as they may seem, are actually expressed as the product of a cos wave over distance and a sine wave over time. Many things are happening over the length of the antenna as the function is operating. If you think of it, it is the essence of space-time and it may be productive for you to consider it even more broadly in your own hypotheses, more broadly that is by possibly incorporating the mechancial SWR analogues to voltage/current SWR's and who knows what new ideas may come to mind with your model. I don't understand a lot of this talk about waves bouncing which is fortunate. Icould not possibly stay on a thread where everybody is talking past each other and then changing the subject as they didn't understand the subject in the first place. In ham radio nothing is believed if it is contrary to the norm.This bouncing wave thing will never come to closure as all participants are deaf. As far as me getting involved all the answers involved in my description of radio are known facts in the scientific world and fully coroberated. Heck they are even corroberated by existing antenna computor programs and actual tests. I can't see how these waves fit in with classical science so it must be another invented science that it referes to. Now if the trend changed to debate the voracity of existing accepted data is proved to be incorrect then they would have my attention but the group is not competant enough for that trail. Regards Art |
Standing morphing to travelling waves, and other stupid notions
Cecil Moore wrote: Jim Kelley wrote: Cecil Moore wrote: Sorry Jim, but [Io*cos(kx)*cos(wt)] and [Io*cos(kx+wt)] *ARE* different. Evidently, you can't recognize a trig identity when you see one. One really should take a look at the math before waving one's hands and opening one's mouth in ignorance. Please enlighten us as to exactly what trig "identity" will make the following terms equal. E1*e^j(wt-kx) ?=? E2*e^j(wt-kx) + E2*e^j(wt+kx) Seems to me the only condition for which they are equal is when E2=0, i.e. when reflections (and therefore standing waves) don't exist. The purpose of pointing out the trigonometric relationship between the sum of sines and the product of sine and cosine was to illustrate that, contrary to your assertion, there isn't a difference in the waves. The traveling waves can either be written mathematically as two separate traveling waves, or as one standing wave. It makes no difference; the waves are the same in either case irrespective of how you choose to describe them mathematically. Do you grasp the meaning here, or not? Thanks, ac6xg |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
Richard Clark wrote:
. . . Specifically, as you have already suggested, with a loop pickup. However, in my career I have never encountered its use as a precision device when measuring voltage is so much easier. EZNEC does not report voltage (although it could be computed, I will leave that to others as there is no significant information obtained in that exercise). . . . As a general rule, the voltage can't be computed because of the spacing between the wire and whatever reference point you're measuring voltage to. The voltage depends on the path taken between the two points (conceptually, how you arrange the voltmeter leads), so there's no single answer. That's why EZNEC computes only current. This is also one of the several problems with treating an antenna like a transmission line. The similarity between the two is just enough to tempt people to take the analogy farther than it holds. Roy Lewallen, W7EL |
Standing morphing to travelling waves. was r.r.a.a LaughRiot!!!
On Mon, 7 Jan 2008 20:13:32 -0500
"AI4QJ" wrote: "Roger Sparks" wrote in message ... On Mon, 07 Jan 2008 12:34:03 -0600 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. Try this: Pick a distance x = pi/3. Now take the sheet of paper perpendicular to the dipole wire. Move it at a constant velocity in the perpendicular direction with a pen recorder tracing the magnitude of the voltage on the antenna to the paper as it moves. There will be a cosine function drawn on the paper. Given: I = Io*cos(kx)*cos(wt) Thanks for the examples. You use the term "I" which is usually the current, but the math makes sense for voltage. To find the distance x = pi/3, I assume you mean from one end of the dipole, back toward the dipole center? At pi/3 radians, cos(pi/3) = 0.5 The cosine function drawn on the paper moving at right angles to the antenna will be: I = 0.5*Io*cos(kt). Now move the paper to x = pi/2. cos(pi/2) = 0 The function drawn on the paper will be I = 0. Now move the paper to x = pi cos(pi) = -1 The function drawn on the paper, at RIGHT ANGLES tio the antenna will be: I = -Io*cos(wt) As cos(wt) rotates between 0 and 2pi, cos(wt) moves between +1 and -1. The voltage will always be the negative of the initial voltage times the cosine of the instantaneous angle wt. That is how a standing wave operates. OK, I think I understand. Thanks for sending the examples. 73, Roger W7WKB |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
AI4QJ wrote:
"Richard Clark" wrote in message Now, if you would simply take my advice to heart: strip away the static and ask the question that is plaguing you. OK. Thank you for your patience. You said "The SWR on the wire is load- based and is equal to 8:1 as evidenced by CURRENT on the wire. :-) That was an ignorant mistake. Like a rank novice, Richard assumed a 50 ohm SWR but the antenna is *NOT* a 50 ohm antenna. The acutal SWR is barely above 1.0:1. -- 73, Cecil http://www.w5dxp.com |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
Richard Clark wrote:
Cecil, under protest to my proofs of Standing Wave on these Traveling Wave, then supplied a citation that heads this very thread: "Because the Beverage is a traveling wave, terminated antenna, it has no standing waves resulting from radio signals." It contains FROM radio signals. That was a quote from one of my references. But I suspect that a Beverage antenna must be excited in its directional dimension for that to be true. It probably would have been more accurate to say it has no reflections rather than to say it has no standing waves. -- 73, Cecil http://www.w5dxp.com |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
AI4QJ wrote:
"Cecil Moore" wrote in message There's no confusion. Terminated rhombics and Beverages are traveling-wave antennas. Their currents are primarily: I(x,t) = Io*cos(kx-wt) A 1/2WL dipole is a standing-wave antenna. It's current is primarily: I(x,t) = Io*cos(kx)*cos(wt) People need to understand the huge difference in the 2 waveforms that you present. I agree with you but what do you do with people who choose to remain ignorant? Allow their ignorance to dominate the newsgroup simply because they think they already know everything? -- 73, Cecil http://www.w5dxp.com |
Standing morphing to travelling waves, and other stupid notions
Jim Kelley wrote:
The purpose of pointing out the trigonometric relationship between the sum of sines and the product of sine and cosine was to illustrate that, contrary to your assertion, there isn't a difference in the waves. Jim, please go ask the head of your math department if there's any difference in those equations. That you don't see any difference is just extreme ignorance on your part. Set cos(kx-wt) = cos(kx)*cos(wt) and wrestle with the trig identities until you alleviate your ignorance. You are extremely wrong and ignorant of mathematics. You will realize that fact when you are unable to prove your assertions even to yourself. Hint: cos(kx+wt) = cos(kx)*cos(wt) - sin(kx)*sin(wt) You are obviously missing half of the terms when you say there "isn't a difference in the waves". (FYI, anyone who knows anything about mathematics is laughing at you.) -- 73, Cecil http://www.w5dxp.com |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
Roy Lewallen wrote:
As a general rule, the voltage can't be computed because of the spacing between the wire and whatever reference point you're measuring voltage to. The voltage depends on the path taken between the two points (conceptually, how you arrange the voltmeter leads), so there's no single answer. That's why EZNEC computes only current. This is also one of the several problems with treating an antenna like a transmission line. The similarity between the two is just enough to tempt people to take the analogy farther than it holds. Just because the voltage is hard to measure doesn't mean it isn't proportional to the E-field. -- 73, Cecil http://www.w5dxp.com |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
AI4QJ wrote:
"Cecil, Can you help me understand that the end of a dipole is an open circuit? Nothing falls out because the current changes direction ala AC because the end of a dipole equates with the time that requires the current changes direction. So where is the open circuit? I also cannot see why radiation reduces reflection if the circuit can be completed with iron!" OK Cecil, I'm waiting for your answer on this one! ;-) The end of a dipole is like an open circuit transmission line. The traveling wave hits the open-circuit and is reflected. The forward current and reflected current are equal in magnitude at the end point and 180 degrees out of phase. Therefore, the current is zero. That's logical. The forward voltage and reflected voltage are in phase. Therefore, the voltage is at a maximum at the ends of the dipole. That's known. That voltage can be calculated to a certain accuracy. At the feedpoint of a dipole, the forward wave energy is about 20% higher than the reflected wave energy. That 20% difference between forward wave and reflected wave is the amount of power lost to losses and radiation. A #14 horizontal wire 30 feet in the air has a characteristic impedance of 600 ohms. That's all you need to know to perform the voltage and current calculations for a particular power input to the antenna. -- 73, Cecil http://www.w5dxp.com |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
On Mon, 7 Jan 2008 21:03:35 -0500, "AI4QJ" wrote:
Well, you said the only way to measure SWR on a receive antenna is by measuring current (sort of as an aside). I wanted to know the significance of that. Now I know that the reason you said that was to support your SWR calculation for a receive antenna using EZNEC, which is based on current (method of moments). You were not saying it in terms of making physical measurements but only to support the accuracy of your statement (i.e. there's no better way to determine SWR on a receiving antenna than to measure current, and, EZNEC is based on current). I know this was not your main point, it was just an aside, but I don't agree with it; however, that in itself does not diminish your main point. Hi Dan, It is merely a response to the "framing" of a specific expectation within the context of Cecil's citation. The citation demanded an externality, I supplied a stimulus external to the antenna. However, this does not mean that there is no other indicator of Standing Waves on a Traveling Wave antenna, and it does not make this exotic testing the only proof of Standing Waves on a Traveling Wave antenna. There is, after all, the concept of reciprocity. If you look at the reciprocal actions offered by exciting a Traveling Wave antenna with a source directly attached to it (and Cecil's last example proves this), you find Standing Waves. This may confound the SWR meter, but then that meter doesn't indicate what is on the wires, it indicates what is impressed upon the finals. Now, if you want to discard EZNEC (which for some odd reason you seem to approach method of moments with a sneer), conventional methods would still bear out the same results. Lord knows I've sat at the bench doing it the conventional way for thousands of measurements. I've probably made more physical measurements in a day, than anyone here has in a lifetime. Others, don't bore us with indignities about all your SWR meter readings in reply to that last statement. :-) So now to the shoe you dropped: I know this was not your main point, it was just an aside, but I don't agree with it What was my main point, and how is yours conflict with it? Is yours a philosophical triviality so common to these threads, or does it come with physical measurements experience? 73's Richard Clark, KB7QHC |
Standing morphing to travelling waves
My daughter's surgery went well.
I will be away from my computer for a week. -- 73, Cecil http://www.w5dxp.com |
Standing morphing to travelling waves
Cecil Moore wrote:
My daughter's surgery went well. I will be away from my computer for a week. I cannot express the relief I feel that all has gone well for your daughterj--and yourself ... We will be here when you get back, take your time and take care OM ... Warmest regards, JS |
Standing morphing to travelling waves, and other stupid notions
Cecil Moore wrote:
That you don't see any difference is just extreme ignorance on your part. Set cos(kx-wt) = cos(kx)*cos(wt) and wrestle with the trig identities until you alleviate your ignorance. Cecil, There are an infinite number of different functions one could write to describe an infinite number of different possible wave shapes. And none of them would necessarily be mathematically equivalent to another. But when you write the equation for the superposition of traveling waves and claim that resultant standing wave is a different kind of electromagnetic wave, that is a misguided point of view. The equation for a standing wave is simply a different way of writing the sum of two traveling waves. Being that there is only one kind to chose from, there cannot be a different kind of electromagnetic wave. Is it impossible for you to acknowledge this simple point in a gentlemanly fashion? You are extremely wrong and ignorant of mathematics. Any 'extreme wrongness' notwithstanding, what I don't know about mathematics could fill a book. On the other hand, you almost inevitably end up lying and turning technical discussions into personal attacks. Ask anyone. ac6xg |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
On 7 Jan, 22:19, Cecil Moore wrote:
AI4QJ wrote: "Cecil, Can you help me understand that the end of a dipole is an open circuit? Nothing falls out because the current changes direction ala AC because the end of a dipole equates with the time that requires *the current changes direction. So where is the open circuit? I also cannot see why radiation reduces reflection if the circuit can be completed with iron!" OK Cecil, I'm waiting for your answer on this one! ;-) The end of a dipole is like an open circuit transmission line. The traveling wave hits the open-circuit and is reflected. snip. -- STOP STOP STOOOOOOOOOP RIGHT THERE Thank you Lord for this opportinity you have given meto bring Ham radio back to reality after Terman went to heaven Gentleman this is a relativistic statement where Einstein tried to get a unification theory \before he died. He failed.You all know your houses are wired for AC, usually a ring circuit. Yes 50/60 is frequency just like 10 metrrs. Now go homw and cut the wire in the kitchen to make it an open circuit Now ham r5adio says the cuyrrent will turn around where you cutt it and it will go back. Well science has not seen a trace of this new frequency of 100, 110 0r 220 line double its frequency to twice what it was before. On top of that the lights went out on the circuit before and after the wire was cut which means the current stopped when it was cut. Same for a antenna which usually has a higher frequency. No the current does not turn back or fall of the end. Right from the get go the current knows that it must travel forward regards until the time of half a period comes about. It cannot turn around and collide with current that is obeying the rules. When it first started its travels up the radiator the current travelled on or above the skin that encloses the conductor knowing it can only referse at a specific time. When the current6 reaches the top it will turn around if the time alloted has also exspired. If this hasn't happenned it sees that the inside of the antenna has nothing but decaying particles on its inside which is really just a resistor. The current flows down the center of this resistance. True it is not radiating at this point as it is on the wrongside of this decaying skin. Fortunately forward time expires just at the time the current reached the bottom end of the radiator at the same time half a period expired. So the current retraces its path up the center and down the outside of the antenna where it can again radiate. It reaches its starting point at exactly the same time as two haf periods have elavsed which is one cycle. Yes the cuurent traveled a full circuit which it must do as we know current flow today. No open circuits and no current flow each and every way because it was forced somehow to change direction and ofcourse all the lights went out. All of this started because physicsts could not reconcile the laws of electricity or Maxwell with those of Newton and ofcourse Gauss. This thinking continues to this day even tho Einstein died without proving things. Was this thread really worth it just to carry on the legacy of Terman and Einstein regardless of the facts? Now look at the URL i have supplied on another thread and note how a diamagnetic field is somehow different from what you have been taught. Note especially that the lines of force do not follow the same direction that you expected. If a paricle gets carried by these special lines of force the particle becomes ejected AWAY from the field. Now.Is all this "blabbering" or is it worth pursueing in the absence of Einstein? Can't we get back to classical physics that the masters of which Newton, Maxwell and Gauss stood on the shoulders of. Their science and our science is the same. Let us not kick the legs from under them by inventing new sciences so that our difficulties in understanding can be resolved. Read the URL PLEASE . I have found for you a description which also has picture as well as words and mull over the implications of what I place before you. Art Unwin KB9MZ...xg (uk) 73, Cecil *http://www.w5dxp.com |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
"AI4QJ" wrote in message ... The VSWR meter on the ham rig is merely looking at the balance of forward and reflected "power" and it is calibrated to read it out as VSWR (or SWR). It may as well say "ISWR"; it is all the same thing. But it is not measured by sensing either voltage ot current going into the antenna...it measures the delta power. how do you measure 'power'?? you don't. and no swr meter in the world measures 'power'. they all take samples of voltage and/or current and drive a simple meter circuit that just happens to be calibrated in units of watts because thats what most users of cheap meters want to see. they could just as easily be calibrated in volts or amps referenced to 50 ohms |
Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
On Tue, 08 Jan 2008 23:20:44 GMT, "Dave" wrote:
they could just as easily be calibrated in volts or amps referenced to 50 ohms Which is power. |
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