[SI-LIST] Re: Open Termination - Sandor
- From: Sitar Moniker <si_monkey2@xxxxxxxxx>
- To: weirsp@xxxxxxxxxx
- Date: Sat, 18 Sep 2004 11:38:01 -0700 (PDT)
Some experts think its OK to ridicule learners. In my experience often times
they use ridicule as exit strategy. Is there some one who could help me with
this conservation of momentum if that is the issue here. Thanks, Steve.
steve weir <weirsp@xxxxxxxxxx> wrote:Correctly.
Steve
At 03:59 PM 9/17/2004 -0700, Sitar Moniker wrote:
>Great, your V's are not communicating with my E's. May be the clue is in
>the momentum. When you suggested momentum is conserved in this context,
>how should I visualize what was happening?
>
>steve weir wrote:
>At 01:48 PM 9/17/2004 -0700, Sitar Moniker wrote:
>Thank you. I wish I were at the beach playing with rubber bands and
>weights, if not sailing or fishing. Since the discussion is about 2V at
>open end, let us consider your case 2). I assume you've implied (At the
>far end the initial pulse > 1X the step.) voltage doubling at the open end
>to 2V where V here is somewhat less than the step depending on the source
>impedance. Immediately after this reflection, you've explained the next
>reflection at the source end.
>Star, start with 0V everywhere on the line, and a perfectly matched source
>termination.
>Step the input from 0V to 1V. Source difference = +1V.
>Multiply difference by Zsource / ( Zsource + Zline ), Since Zsource =
>Zline, multiply by 0.5. Now, a 0.5V step propagates down the line.
>At the end of the line, the 0.5V step encounters a discontinuity.
>Momentum is conserved. So a 0.5V step superimposes on the incident 0.5V
>step at the far end and begins propagating up the line.
>
>So we have only the original step of 0.5V that has now propagated to the
>far end of the line, and due to the open, a reflection step back towards
>the source of 0.5V that superimposes on whatever it encounters on the
>line. Everywhere back to the source it encounters 0.5V, bringing the
>total to 1.0V. Finally, the reflected wave encounters the termination
>impedance of the source. Two things are notable:
>
>1) The source impedance equals that of the line. All energy in the
>reflected wave is exactly absorbed.
>2) The reflected wave amplitude equals the original driving value of
>1.0V. There is no potential difference for the termination to absorb.
>
>A reflection at the source will only occur when the source termination
>does not match the impedance of the line. I have explained what happens
>for both the over and underdamped cases.
>
>Steve
>
>Wait, you are too fast. I am still at the open end trying to understand
>what happens immediately after first reflection. The first E component
>between the trace and return path to discover there is an open end turns
>around and superposes with the next E component to double voltage at that
>point. Assume no fringing. This 2E component is now at the open end
>looking toward the source and it finds the third E component approaching.
>Does superposition continue? If it doesn't, why?
>
>I'm sorry if my question was not clear enough.
>
>steve weir wrote:
> Star, think about conservation of momentum and energy, or just go spend a
> day at the beach, hook a weight to a rubber band or hit the physics
> books. The answer to your question is very simple-
>
>
> Case 1, the source impedance is higher than the line-
> Initial launch at the termination to line I/F is < 1/2 the difference
> between the line steady-state and the driver step output. The initial
> return wave is less than 1X amplitude. A new reflection is less than 1/2
> the difference between the return and the driver step. So over time the
> voltage at the termination works its way towards the step as a negative
> exponential.
>
>
> Case 2. the source impedance is lower than the line-
> Initial launch is > 1/2 the difference. At the far end the initial
> pulse >
> 1X the step. Now the return wave towards the source is greater than the
> source value. This wave encounters the source impedance which is less
> than
> the line and a negative reflection of the difference between the return
> wave and the driver step results. The process repeats, with the second
> return wave being less than the original step, but closer to the step
> value
> than the first return. A decaying exponential ringing wave form results.
>
>
> Steve
> At 08:53 AM 9/17/2004 -0700, Sitar Moniker wrote:
> > >From your explanation in previous mail, what I understand is the E
> field
> > (though you called it energy wave) rolls down the line happily and
> > suddenly finds itself standing in a situation where it can't jump out
> > into the air (infinite impedance discontinuity) nor just sit there and
> > roadblock junior E fields rushing from behind. The clever E field turns
> > around (reflection) and hugs the oncoming E field (superposition).
> Thus,
> > we have total voltage at the open equal to 2V and a E field standing
> > there corresponding to 2V.
> >By now, the E field guy standing at the open end prefers to stand
> looking
> >toward source end(reflection mode) and use superposition to his
> advantage.
> >My question is why there are no more superpositions to make it 3V, 4V
> etc.
> >even though the driver is continuously pumping a constant V and more and
> >more E field guys approaching the open end.
> >
> >What happens if we make the source end also open (by yanking the source)
> >the instant the voltage at the load end turned 2V?
> >
> >Open end case is as ideal as it gets. So, let us keep everything else
> >related to this as ideal as they could be. The answers may be trivial to
> >the experts but I don't know how to get a hang of these concepts.
> >
> >
> >Sandor Daranyi wrote:If you imagine the wave
> >reflected at the open end, the
> >direction of travel may be reversed but it still sees
> >the same line impedance, so the voltage amplitude
> >won't change. After superposition, 1+1=3D2. In real life
> >of course, there are always losses, so it will be
> >less, but the reflection (or rather its effect on the
> >signal) is something that is really easy to observe
> >with an oscilloscope. As always, in real life it's a
> >bit more complicated because there may be some ringing
> >on the signal and perhaps several reflections, due to
> >other "imperfections" beside an unterminated line.
> >These imperfections are anything that can make the
> >transmission line nonideal. E.g. when it has
> >stubs/branches hanging off it, finite impedances
> >connected to it (a device inputs perhaps) somewhere
> >along the line, line geometry changes etc. In general:
> >impedance variations.
> >
> >Now you could say, yes, but what happens when the
> >wavefront reaches the beginning of the transmission
> >line and gets reflected again, wouldn't that result in
> >a higher voltage on the line? Well, the thing to note
> >here is that there is a driver hanging off the source
> >end. In ideal case, the driver will be a voltage
> >source or maybe a current source. If it's a current
> >source, it has infinite impedance and you do have a
> >reflection again, so you end up with higher and higher
> >voltages but going back to looking at the energy, it
> >is not surprising, since the current source will keep
> >pumping energy in there! If you have an ideal voltage
> >source driving, it'll have zero impedance. Just like
> >infinite impedance, zero impedance can not "swallow"
> >energy, so it will be reflected back, but the
> >difference is that voltage is reflected back with the
> >opposite phase in this case.
> >
> >If you are interested in a more practical case, the
> >basic things to consider are that real life drivers
> >have finite source impedances, they have
> >nonlinearities and limited V & I ranges. =20
> >
> >So, with all the signal integrity problems they can
> >cause, are reflections always bad? Not necessarily.
> >For example PCI uses reflected wave switching which
> >allows the use of relatively weak drivers.
> >
> >Cheers,
> >
> >Sandor
> >
> >-----Original Message-----
> >From: Sitar Moniker [mailto:si_monkey2@xxxxxxxxx]=20
> >Sent: Friday, 17 September 2004 5:04
> >To: sandord@xxxxxxxxxxxxx
> >Cc: si-list@xxxxxxxxxxxxx
> >Subject: Re: [SI-LIST] Re: Open Termination
> >
> >Thank you. It definitely helps to understand in terms of energy. As
> you men=
> >tioned, due to voltage superposition, the voltage at the open end
> gets do=
> >ubled. Extending this idea, will there be an occasion (ideal or
> otherwise=
> >) when the open end voltage becomes 3X or 4X. If so, under what
> condition=
> >s? If not, what limits this from happening?
> >
> >Sandor Daranyi wrote:
> >Hello Sitar,
> >
> >Others have already given good information regarding your questions
> but sin=
> >=3D
> >ce you asked for something intuitive, maybe I can add something
> worthwhil=
> >=3D
> >e. The warning is that intuitive explanations can often be
> oversimplific=3D
> >ations.
> >
> > > -----Original Message-----
> > > From: Sitar Moniker
> > > Sent: Wednesday, 15 September 2004 7:48
> > > To: si-list@xxxxxxxxxxxxx
> > > Subject: [SI-LIST] Open Termination
> > >=3D20
> > > In SI and HSDD books, they assume that open termination offers
> infinite
> > > impedance. My understanding is open end line is exposed to air
> which has
> > > about 377 ohms.
> > >=3D20
> > > 1) How do you get infinite impedance for unterminated line?
> >
> >The easiest way to see what happens is in terms of energy. (Looking
> at what=
> >=3D
> >happens to energy is a very useful techni que when trying to
> understand s=
> >=3D
> >omething like this.) The driver pumps energy into the transmission
> line. =
> >=3D
> >The resulting energy wave travels down the line at close to the speed
> of =
> >=3D
> >light and eventually reaches the end. The energy COULD continue to
> travel=
> >=3D
> >in air but you would need a coupling mechanism between the
> transmission =3D
> >line and free air. If there is no coupling, it is irrelevant what
> would h=
> >=3D
> >appen in air, or indeed that it's 377ohms. If energy can't be
> transferred=
> >=3D
> >, then by definition the impedance must be infinite. If it weren't,
> there=
> >=3D
> >would be a finite impedance and energy could be transferred.
> >
> >Nothing is perfect in the real world, so there can be some little
> coupling =
> >=3D
> >e.g. due to losses and radiation from the transmission line, so the
> menti=
> >=3D
> >oned infinite impedance is only an approximation, but for most
> practical =
> >=3D
> >cases, quite a valid one.
> >
> >Of course, if you have an antenna at the end of the transmission
> line, sudd=
> >=3D
> >enly you do have a coupling mechanism and the story changes.
> Everything g=
> >=3D
> >ets a bit trickier though, because now you would have things like
> radiati=
> >=3D
> >on efficiency and the antenna itself would behave like an impedance
> trans=
> >=3D
> >former between the transmission line and free air.
> >
> > > 2) Is there any intuitive way to show that voltage at the open end
> > > doubles- other than using math: ref. coeft. =3D3D +1?
> >
> > >From the previous answer, you may already see what happens here.
> When the f=
> >=3D
> >ace of the energy wavefront reaches the end of the transmission line
> and =
> >=3D
> >more energy still keeps arriving, it won't jump into air and it won't
> pil=
> >=3D
> >e up there, either. Instead, it will get reflected and the reflected
> wave=
> >=3D
> >will get superimposed on the arriving part of the wave. Why twice the
> vo=3D
> >ltage? Due to voltage superposition, which should sound familiar from
> bas=
> >=3D
> >ic electronic theory (a quick "voltage superposition" google search
> threw=
> >=3D
> >up: http://www.allaboutcircuits.com/vol_1/chpt_10/6.html).
> >
> >Sandor
> >
> >------
> >Sandor Daranyi
> >Snr Design Engineer
> >
> >
> >
> >
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