[SI-LIST] Re: Open Termination - Sandor
- From: steve weir <weirsp@xxxxxxxxxx>
- To: si_monkey2@xxxxxxxxx
- Date: Fri, 17 Sep 2004 16:27:00 -0700
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 <weirsp@xxxxxxxxxx> 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 <weirsp@xxxxxxxxxx> 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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