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Re-Normalizing the Smith Chart (Changing the SWR into thesame...
Dr. Slick wrote:
"Therefore, the reflected voltage can never be greater than the input voltage for a passive network, and the reflection coeficient can never be greater than 1 for such a case." A reflection coefficient implies that transmission lines long in terms of wavelength may qualify as passive networks. At the open-circuit end of a long transmission line, current is interrupted by the open circuit. Having nowhere else to go, energy in the current wave must be accepted by the voltage wave which doubles on the spot. 2X the incident voltage is the sum of the voltage in the incident wave and the voltage in the new reflected wave as they are in-phase. The reflected voltage is no greater than the incident voltage. It is equal to the incident voltage. It`s the sum that doubles. Best regards, Richard Harrison, KB5WZI |
Dr. Slick wrote:
"Therefore, the reflected voltage can never be greater than the input voltage for a passive network, and the reflection coefficient can never be greater than 1 for such a case." Terman, despite Reg`s disdain for experts, seems to agree with Dr. Slick. Here is Terman`s gist: 1. The reflected wave is identical to the incident wave except it travels toward the generator. 2. Ereflected / I reflected = -Zo. Just as Eforward / I forward = Zo 3. Line loss causes the reflected wave to decline as it travels toward the generator. 4. Phase of the reflected wave drops back as distance back from the load increases. 5. Volts at the load are the sum of the incident and reflected wave volts. Likewise for currents. 6. E/I at the load equals Zload. 7. The vector ratio Ereflected / E incident erquals rho, the reflection coefficient.. 8. In a lossless line, rho is the same everywhere on the line. 9. The effect of a reactive load is merely to displace the SWR pattern on a transmission line. There is no opportunity in the stated conditions on a transmission line for a reflected voltage to exceed the incident voltage or for the reflection coefficient to exceed one (1). Best regards, Richard Harrison, KB5WZI |
Keith wrote:
"Not an "of course" at all." Terman and Bird Electronic Corporation say: Usually, power delivered by the transmitter and to the load equals Forward Power less Reflected Power. Best regards, Richard Harrison, KB5WZI |
Deacon Dave wrote:
"Is it possible for a resonant condition to exist " (with a short cable)? If the cable is too short to enforce Zo, the source for Zload is not necessarily a resistance and the load for the source can be resonant. A short-circuited transmission line, shorter than 1/4-wavelength, is an inductive reactance. An open-circuited transmission line, shorter than 1/4-wavelength, is a capacitive reactance. A transmission line is not required for mismatch between source and load. If a source can only supply a certain voltage to current ratio, and it`s different from the immutable Zload, you have a mismatch and a surplus of volts or amps which are rejected by the load. The simple analysis of a series-resonant situation involving line inductance and a load capacitance is that volts across either is (I)(Z). The Z`s are about equal at resonanvce. Trivia: 1/8-wavelength of line has a reactance which is equal to Zo. Best regards, Richard Harrison, KB5WZI |
On Tue, 19 Aug 2003 15:43:54 GMT, Dave Shrader
wrote: Richard I am very familiar with all you write. But I have one question: if the 'transmission line' is very short so that the 'significant portion of a wavelength' criteria is not met, then the capacitive reactance of the line and any inductive reactance in the load and the effective Q of the circuit may all come into play. Is it possible for a resonant condition to exist? If so, does the reflection coefficient have any significance? Deacon Dave, W1MCE Hi Dave, Depends on two things. The degree of resolution you demand and the possibility of end effects (would it be easier to go around instead of through?). When I was at the Metrologist Bench, I would be asked "how accurate can you measure this?" My response was "How much can you afford?" 73's Richard Clark, KB7QHC |
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Reading Keith`s posting, somebody wrote:
"The power is Vavg"Iavg"cos(theta)" This is mistaken. The average value of a sine wave is 0.637 times the peak value. We use 0.707 toimes the peak value of a sine wave, the rms value, which is the effective value, that is, it is as effective as d-c in making power calculations. The average power must be the same as d-c. This is peak volts times peak amps divided by two, or 1/2 the peak power. 0,707 x 0.707 = 0.5, so we use rms, not avg. volts and amps to calculate average power, which is the same in capability when producing light and heat from resistive devices. Best regards, Richard Harrison, KB5WZI |
Roy Lewallen wrote:
"But the avreage value of a sine wave, assuming no DC offset, is zero, not 0.637 times the peak value." Yes. The value of any undistorted sine wave which is symmetrical about the zero axis is zero over one complete cycle. 0.637 and 0.707 are half-cycle values which are quoted as the average and rms fractions of the peak value of a sine wave in references containing such trivia. I just remember tthat the average in this case is about 0.9x the rms value. Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
Keith wrote: "It does go back to "double think" unless you can explain how energy can flow when power is zero." Think! The only way the voltage can go to zero on a good transmission line which is energized is by interference between an incident wave and a reflected wave. Zero volts on the line is merely a manifestation of VSWR. No disagreement from me as long as you replace 'incident wave' and 'reflected wave' with the more precise 'incident voltage wave' and 'reflected voltage wave'. This aligns with the precise use of V in VSWR. Adopting Roy's more precise notation for power P(t) = V(t) * I(t) (1) The question is: Do you accept this expression as describing power? a) If not, then there is no "double think", but a lot of electrical engineering will have to be tossed as well. b) If you do accept (1), then in a case where V(t) is zero for all time, power must be zero as well. V(t) is zero for all time at a current maximum in a shorted transmission line, so the power (energy flowing) must be zero as well. c) If you accept (1), but also claim that there is energy flowing when V(t) is zero, then "double think" is an appropriate description. I think these are the only 3 options. If you choose b), then I think we are in agreement. If you choose c), then ... I'm not sure what the 'then' is. If you choose a), we can explore all the difficulties that will arise when (1) is not true and with some effort you may arrive at b). Forward (incident) power and reverse (reflected) power are both on the line. A zero voltage on the line requires a complete load reflection so that the reflection volts are as strong as the forward volts. The sentence with 'volts' is correct. The sentence with 'power' leads to a great deal of difficulty as described above. A directional coupler, at the very spot where a slotted line probe would sense zero volts, would show you have full power (with its volts and amps) coming and going. Many people use directional couplers as a reason to stay out of camp b), but this necessarily means they are in a) or c). The first step to enlightenment is to briefly set aside directional couplers and 'Bird watt' meters, and realize that in a choice between a), b) or c), b) is the only place it makes sense to be. Then go back and figure out how directional couplers are not inconsistent with b). In reality there is not zero volts in the incident wave or in the reflected wave. There`s full voltage coming and going. The volts just happen to be out-of-phase at this point. Yes, indeed. But there is no power. ....Keith |
Richard Harrison wrote:
Keith wrote: "Yes, indeed. But there is no power." Actually I wrote much more than that and the stuff which preceded was much more important. Keith`s perception is flawed. He is addressing r-f with a d-c mindset. R-F power flow does not stop at an SWR zero-voltage point. Like "Old Man River" it just keeps rolling along. P(t) = V(t) * I(t) is much more general than DC or sinusoidal RF. It works for any signal shape you choose, even non-repetitive ones. It certainly works for my electic company who integrate P(t) and regularly send me a bill. The zero-voltage point exists because the phase relationship between two oppositely traveling waves is fixed, and the zero-volts point is where the vectors cancel. You must simultaneously sense both waves to find a zero. Sensing either wave alone finds no dip in voltage. Carramba! From this response it is clear that you are not yet in group b). But is it: group a) P(t) is not always equal to V(t) * I(t); or group c) energy can flow when P(t) is a constant 0 ? Or maybe my list is incomplete and there is a 4th option which I missed. May I respectfully suggest that if you are having difficulty selecting which option applies to your thinking (and yet can't produce a fourth option), that you try to discover the source of your discomfort. From there, enlightenment may arise. For myself, I recognized the "double think" and so rejected c). P(t) = V(t) * I(t) seemed so fundamental that it had to be accepted thus I was forced to reject a). This left only b). Although it rejects some of the often stated and accepted explanations about transmission lines, it was the lesser of the evils. Some thinking revealed the weaknesses in the 'often stated and accepted' explanations and now all is consistent. This process took some time, prompted and assisted by the never ending arguments which go on in this group. So I do thank those mis-guided souls who argue endlessly and the patient answerers who respond mostly for the benefit of the lurker. But be that as it may, do try and figure out whether option a), b) or c) best describes your thinking. Only an increased understanding can arise. ....Keith |
Richard Harrison wrote:
Keith wrote: "Yes, indeed. But there is no power." Keith`s perception is flawed. He is addressing r-f with a d-c mindset. R-F power flow does not stop at an SWR zero-voltage point. Like "Old Man River" it just keeps rolling along. The zero-voltage point exists because the phase relationship between two oppositely traveling waves is fixed, and the zero-volts point is where the vectors cancel. Using Keith's "logic", he would also be forced to assert that in a bright ring-dark ring light interference pattern, the energy in the bright rings is trapped between the dark rings and just sits there and circulates. That's one reason why the field of optics is so far ahead of RF transmission lines. Everyone in optics understands interference patterns which is exactly what RF standing waves are. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
wrote:
In reality there is not zero volts in the incident wave or in the reflected wave. There`s full voltage coming and going. The volts just happen to be out-of-phase at this point. Yes, indeed. But there is no power. Power is the same as irradiance in optics. When total V=0, it is simply the result of destructive interference. Perhaps this quote from _Optics_, by Hecht, will enlighten you. "The principle of conservation of energy makes it clear that if there is constructive interference at one point, the "extra" energy at that location must have come from elsewhere. There must therefore be destructive interference somewhere else." The voltage goes to zero because two voltage waves are engaged in destructive interference. The current goes to maximum because two current waves are engaged in constructive interference. The momentum in the voltage waves simply transfers to the current waves and they just keep on rolling along. There is no mechanism of physics existing at that point to change the momentum of the waves. Believing that no energy crosses a superposed V=0 boundary is just a wet dream. The current is at an absolute maximum point so plenty of charge carriers are crossing that boundary. -- 73, Cecil http://www.qsl.net/w5dxp "One thing I have learned in a long life: that all our science, measured against reality, is primitive and childlike ..." Albert Einstein -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Wed, 20 Aug 2003 11:36:57 -0500, W5DXP
wrote: That's one reason why the field of optics is so far ahead of RF transmission lines. Everyone in optics understands interference patterns which is exactly what RF standing waves are. Hi Cecil, And you have yet to confirm it (or refute it) at the bench. Is your interest more rhetorical than actual? We engaged in a rather lengthy interchange to this topic, up to the point of how you could evidence this for yourself. My methods and data support your statement's sense. I offered only one proviso of your rig being able to withstand the demands of such mismatch and you offered your SGC could do that easily. As is stands left at that, the remaining task should occupy no more than 10 Minutes of 16 readings and note taking. 73's Richard Clark, KB7QHC |
W5DXP wrote:
wrote: In reality there is not zero volts in the incident wave or in the reflected wave. There`s full voltage coming and going. The volts just happen to be out-of-phase at this point. Yes, indeed. But there is no power. Power is the same as irradiance in optics. When total V=0, it is simply the result of destructive interference. Perhaps this quote from _Optics_, by Hecht, will enlighten you. "The principle of conservation of energy makes it clear that if there is constructive interference at one point, the "extra" energy at that location must have come from elsewhere. There must therefore be destructive interference somewhere else." My knowledge of optics is insufficient to comment on any analogies you choose to draw. Fortunately, a knowledge of optics is unnecessary to understand circuits and transmission lines. The voltage goes to zero because two voltage waves are engaged in destructive interference. The current goes to maximum because two current waves are engaged in constructive interference. The momentum in the voltage waves simply transfers to the current waves and they just keep on rolling along. There is no mechanism of physics existing at that point to change the momentum of the waves. Believing that no energy crosses a superposed V=0 boundary is just a wet dream. This puts you in group a) P(t) is not always equal to V(t) * I(t); or group c) "double think". Care to think about which and comment? The current is at an absolute maximum point so plenty of charge carriers are crossing that boundary. Yes indeed, but current by itself is not energy. Remember P(t) = V(t) * I(t) [unless you choose option a)] Both volts and amps are simultaneously necessary for power. ....Keith |
W5DXP wrote:
It happens all the time in optics and no optics engineer would be silly enough to assert that the bright ring energy is trapped and circulating between the dark rings. Nor would he be silly enough to assert that energy first goes to the dark ring and then turns around and goes to bright ring. ac6xg |
Richard Clark wrote:
W5DXP wrote: That's one reason why the field of optics is so far ahead of RF transmission lines. Everyone in optics understands interference patterns which is exactly what RF standing waves are. And you have yet to confirm it (or refute it) at the bench. Is your interest more rhetorical than actual? Actually, I confirmed it in my back yard some 15 years ago. It wasn't rocket science. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
wrote:
My knowledge of optics is insufficient to comment on any analogies you choose to draw. Fortunately, a knowledge of optics is unnecessary to understand ... transmission lines. Equally unfortunately, that's just a delusion of yours. Care to think about which and comment? I have no idea what you are talking about. Yes indeed, but current by itself is not energy. Hmmmmm, I^2*Z0 is not power? (Somebody get the net). -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Wed, 20 Aug 2003 18:25:17 -0500, W5DXP
wrote: Richard Clark wrote: W5DXP wrote: That's one reason why the field of optics is so far ahead of RF transmission lines. Everyone in optics understands interference patterns which is exactly what RF standing waves are. And you have yet to confirm it (or refute it) at the bench. Is your interest more rhetorical than actual? Actually, I confirmed it in my back yard some 15 years ago. It wasn't rocket science. Hi Cecil, I see you haven't got a clue what I wrote. Never mind. 73's Richard Clark, KB7QHC |
Richard Clark wrote:
W5DXP wrote: Richard Clark wrote: And you have yet to confirm it (or refute it) at the bench. Is your interest more rhetorical than actual? Actually, I confirmed it in my back yard some 15 years ago. It wasn't rocket science. I see you haven't got a clue what I wrote. Never mind. On the contrary, my bench was located in my back yard in Queen Creek, AZ at the time - during the dry season, of course. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Jim Kelley wrote:
W5DXP wrote: Energy cannot reverse its momentum without reversing its momentum. If no other bright rings exist, then it must reverse its momentum. He wouldn't be silly enough to say that, either. :-) Because he probably wouldn't even bother speaking to someone who necessitated such an obvious statement. He would probably just call the guys in the white coats. :-) -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Richard Clark wrote:
W5DXP wrote: On the contrary, my bench was located in my back yard in Queen Creek, AZ at the time - during the dry season, of course. Even more remote from the discussion. Will we be regaled about the splinters in the bench next? May I remind you, Richard, that it was you who brought up the subject of bench testing (as if testing cannot be done without a bench). I merely responded to your requirement for a bench. My bench was a card table in my back yard. That's exactly where I did my variable length transmission line testing - in my back yard in Queen Creek, AZ on a pleasant spring or fall day (I forget which). It wasn't boiling and it wasn't freezing so it must have been spring or fall. :-) -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Keith wrote:
"Are you sure you want to discard all thoughts of the instantaneous? Certainly not, but it has little application to power in transmission line problems. Power is the rate of transferring energy or the rate of doing work. Electrical power is measured in joules per seconds or more succinctly in watts. What is the value in watts or joules per second when seconds equal zero? I venture an answer: It is the V x I x cos. theta at that instant, but since work is power x time, it won`t do anything for you in zero seconds. Best regards, Richard Harrison, KB5WZI |
Keith wrote: "Are you sure you want to discard all thoughts of the instantaneous? Certainly not, but it has little application to power in transmission line problems. Power is the rate of transferring energy or the rate of doing work. Electrical power is measured in joules per seconds or more succinctly in watts. What is the value in watts or joules per second when seconds equal zero? I venture an answer: It is the V x I x cos. theta at that instant, but since work is power x time, it won`t do anything for you in zero seconds. Best regards, Richard Harrison, KB5WZI If Zeno were around today, he could prove that - based on Cecil's idea that, as dt goes to zero, energy transfer also goes to zero - there can be no transfer of energy in a transmission line, since any number times zero is still zero. Of course, we can all see the fallacy in that argument, can't we? |
Tdonaly wrote:
If Zeno were around today, he could prove that - based on Cecil's idea that, as dt goes to zero, energy transfer also goes to zero - there can be no transfer of energy in a transmission line, since any number times zero is still zero. Of course, we can all see the fallacy in that argument, can't we? What's the fallacy? If dt=0, then time stands still, and of course, nothing happens and nothing moves. -- 73, Cecil, W5DXP |
Tdonaly wrote: If Zeno were around today, he could prove that - based on Cecil's idea that, as dt goes to zero, energy transfer also goes to zero - there can be no transfer of energy in a transmission line, since any number times zero is still zero. Of course, we can all see the fallacy in that argument, can't we? What's the fallacy? If dt=0, then time stands still, and of course, nothing happens and nothing moves. -- 73, Cecil, W5DXP So you think it's impossible to send energy from one place to another via a transmission line? Hmmm. I guess Achilles never did catch that turtle. 73, Tom Donaly, KA6RUH |
Tdonaly wrote:
So you think it's impossible to send energy from one place to another via a transmission line? In zero time, yes. Isn't that what the speed of light limit is all about? -- 73, Cecil, W5DXP |
On Fri, 22 Aug 2003 11:16:03 -0700, W5DXP
wrote: Tdonaly wrote: So you think it's impossible to send energy from one place to another via a transmission line? In zero time, yes. Isn't that what the speed of light limit is all about? Hi Cecil, At the speed of light (in any media) power is transferred in zero time by definition. Of course if you are Achilles, outside of the power's frame of reference, you have already lost the chase. :-) 73's Richard Clark, KB7QHC |
In circuits involving purely sinusoidal V and I of the same frequency,
the power waveform is actually a true sinusoidal function, except with a D.C. offset. It doesn't at all resemble the output from a full wave rectifier. The D.C. offset is the average value, and the frequency of the sine portion is twice the frequency of V or I. Roy Lewallen, W7EL Jim Kelley wrote: Richard Harrison wrote: What is the value in watts or joules per second when seconds equal zero? I venture an answer: It is the V x I x cos. theta at that instant, but since work is power x time, it won`t do anything for you in zero seconds. I think you have a slight misconception about the meaning of instantaneous power. AC power is a pseudo-sinusoidal function with respect to time, like that of full-wave rectifier. The function has a value, an instantaneous amplitude, at any time t which represents the rate at which energy in Joules is moving past a point x at time t. It may not be a terribly useful thing to know, but it isn't a ficticious quantity. 73, ac6xg |
Tdonaly wrote: So you think it's impossible to send energy from one place to another via a transmission line? In zero time, yes. Isn't that what the speed of light limit is all about? -- 73, Cecil, W5DXP Cecil, as delta t goes to zero, the quantity dx/dt doesn't necessarily also go to zero. If it did, no one would ever again have to get a permanent headache studying calculus. 73, Tom Donaly, KA6RUH |
Roy Lewallen wrote: In circuits involving purely sinusoidal V and I of the same frequency, the power waveform is actually a true sinusoidal function, except with a D.C. offset. It doesn't at all resemble the output from a full wave rectifier. The D.C. offset is the average value, and the frequency of the sine portion is twice the frequency of V or I. Yes, thanks Roy. I've had absolute value circuits on the brain all this week. Nevertheless, instantaneous power is simply the instantaneous amplitude at time t of the (sin^2(wt))/2 function. 73, ac6xg Roy Lewallen, W7EL Jim Kelley wrote: Richard Harrison wrote: What is the value in watts or joules per second when seconds equal zero? I venture an answer: It is the V x I x cos. theta at that instant, but since work is power x time, it won`t do anything for you in zero seconds. I think you have a slight misconception about the meaning of instantaneous power. AC power is a pseudo-sinusoidal function with respect to time, like that of full-wave rectifier. The function has a value, an instantaneous amplitude, at any time t which represents the rate at which energy in Joules is moving past a point x at time t. It may not be a terribly useful thing to know, but it isn't a ficticious quantity. 73, ac6xg |
Jim Kelley wrote:
Roy Lewallen wrote: In circuits involving purely sinusoidal V and I of the same frequency, the power waveform is actually a true sinusoidal function, except with a D.C. offset. It doesn't at all resemble the output from a full wave rectifier. The D.C. offset is the average value, and the frequency of the sine portion is twice the frequency of V or I. Yes, thanks Roy. I've had absolute value circuits on the brain all this week. Nevertheless, instantaneous power is simply the instantaneous amplitude at time t of the (sin^2(wt))/2 function. 73, ac6xg Only if the voltage and current are in phase. Here's the more general solution (cosines could be used instead with equal validity): Given that v = V * sin(wt + phiv) i = I * sin(wt + phii) Then p = v * i = VI * sin(wt + phiv) * sin(wt + phii) The product of the sines can be transformed via a simple trig identity to give p = VI * 1/2[cos(phiv - phii) - cos(2wt + phiv + phii)] The first term in the brackets is D.C. -- it's time-independent. The second term is a pure sine wave. So the result is a pure sine wave with a D.C. offset. I've described the meaning and significance of the power waveform in at least one earlier posting on this newsgroup. If anyone is interested who can't find it on Google, I'll look it up and post the subject and date. Roy Lewallen, W7EL |
I apologize if my response seemed argumentative. It wasn't intended that
way. Certainly, sin^2(wt) has the same shape as the power waveform I derived -- the only difference is its fixed D.C. term. And I certainly agree that letting delta t approaching zero doesn't make any function of t become zero at that point. And just as the analysis I've presented is in your first year college electronics book, so is the point about delta t in everyone's high school or first semester college calculus book. But it's evident that some number of participants in this thread have either forgotten, never seen, or never understood those basic principles. And quite a few people either don't have any textbooks, don't understand them, or are unwilling to open and read them. Hence the postings containing information that you or I could find in moments. Roy Lewallen, W7EL Jim Kelley wrote: You seem to be looking for an argument any way you can, Roy. ;-) Sin^2(wt)/2 is the general form of any equation with the shape you described in your previous post. Furthermore, instantaneous power can be evaluated at any time t, irrespective of relative phase. The point is simply that instantaneous power isn't necessarily zero as a result of delta t's approaching zero. Given that v = V * sin(wt + phiv) i = I * sin(wt + phii) Then p = v * i = VI * sin(wt + phiv) * sin(wt + phii) The product of the sines can be transformed via a simple trig identity to give p = VI * 1/2[cos(phiv - phii) - cos(2wt + phiv + phii)] The first term in the brackets is D.C. -- it's time-independent. The second term is a pure sine wave. So the result is a pure sine wave with a D.C. offset. I've described the meaning and significance of the power waveform in at least one earlier posting on this newsgroup. If anyone is interested who can't find it on Google, I'll look it up and post the subject and date. Yes. It's also in my first year college electronics book. Thanks and 73, AC6XG |
I apologize if my response seemed argumentative. It wasn't intended that way. Certainly, sin^2(wt) has the same shape as the power waveform I derived -- the only difference is its fixed D.C. term. And I certainly agree that letting delta t approaching zero doesn't make any function of t become zero at that point. And just as the analysis I've presented is in your first year college electronics book, so is the point about delta t in everyone's high school or first semester college calculus book. But it's evident that some number of participants in this thread have either forgotten, never seen, or never understood those basic principles. And quite a few people either don't have any textbooks, don't understand them, or are unwilling to open and read them. Hence the postings containing information that you or I could find in moments. Roy Lewallen, W7EL Cecil seemed to indicate that he thought delta t going to zero meant that t was perpetually zero. I know he knows better than that. 73, Tom Donaly, KA6RUH |
GEE!!
If dV/dt = 0 then I must have maximum voltage. This has meaning. Hmmm ... if dP/dt = 0 then I must have maximum 'PEAK' power. This might have meaning IF I splatter all over the band. But for SWR purposes it means nothing ... I think ... am I confused? Deacon Dave, W1MCE + + + Richard Harrison wrote: Cecil, W5DXP wrote: "If dt=0, then time stands still,---." Yes, it is a spot frozen in time, a snapshot of slope at one instant. What really interests us is average power over a half cycle or more, not instantaneous power, energy`s rate of change, or power`s rate of change. Best regards, Richard Harrison, KB5WZI |
Deacon Dave wrote:
"---am I confused?" If Dave is confused, I suffer the same confusion. The slope at the maximum of a sine wave is zero. The slope is maximum at zero crossings of an undistorted sine wave. For the power sine wave, though the fact that a minus times a minus is a plus results in 2x the voltage frequency, dP/dt=0 at maxima. A question raised in this thread is, how can energy, which is joules per second times seconds, be zero when the number of seconds is zero? The answer seems obvious. Zero times anything is zero. Best regards, Richard Harrison, KB5WZI |
Tdonaly wrote:
as delta t goes to zero, the quantity dx/dt doesn't necessarily also go to zero. If it did, no one would ever again have to get a permanent headache studying calculus. I am familiar with limits. Some make sense and some don't. The impedance, frequency, and SWR of a transmission line with an SWR doesn't make sense as V goes to zero and I goes to zero. Any old piece of transmission laying in the yard has zero volts and zero amps. What is the SWR? What is the frequency? What is the Z0 of the line? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Jim Kelley wrote:
Nevertheless, instantaneous power is simply the instantaneous amplitude at time t of the (sin^2(wt))/2 function. And of what benefit is that value to the average ham operator? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Richard Harrison wrote:
Cecil, W5DXP wrote: "If dt=0, then time stands still,---." Yes, it is a spot frozen in time, a snapshot of slope at one instant. What really interests us is average power over a half cycle or more, not instantaneous power, energy`s rate of change, or power`s rate of change. Yeeaaahhhhh! We are not enamored with tits on a boar hog, are we Richard? :-) -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Tdonaly wrote:
Cecil seemed to indicate that he thought delta t going to zero meant that t was perpetually zero. If delta-t ever gets to zero, time stands still. All you can allow delta-t to do is to approach zero. Once it reaches zero the ballgame is over. Limit delta-t to a minimum of a yoctosecond and everything will be perfectly OK. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
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