# [SI-LIST] Re: capacitor impedance in time domain

• From: Peter Fekete <thefekete@xxxxxxxxx>
• To: group_delay@xxxxxxxxx
• Date: Fri, 28 Jan 2005 17:08:38 -0800 (PST)

```Dear group_delay, Arpad, ?

When I answered to this topic three days ago I thought that I explained enough
in my message but apparently not.

Let?s start specifying that any voltage (and current) distribution on a
transmission line can be decomposed in two waves. One is the progressive (from
source to load) and the other is regressive.

Anywhere on the line the actual voltages and currents are superpositions of
these, more exactly: V=V(+)  + V(-)  and  I = I(+) - I(-); the voltages add and
the currents are substracted

For both waves the relationship between voltage and current is the well known
V(+)=Zc* I(+) and V(-)=Zc *I(-)

The way the problem is formulated (by the way, it looks like a homework
problem) implies that we have the progressive wave V(+) defined as trapezoidal
and we want the reflected wave ie V(-)

An important detail is that the source is connected to a 50 ohm resistor and
then to the line. Assuming the line has a 50 ohm characteristic impedance and
the source is perfect results that the V(-) wave returning to the source sees a
perfectly terminated line and will not give any more reflections.

This simplifies things and also the 50 ohm resistor and 50 ohm line will result
in half of the source voltage actually going on the line (as group_delay
noticed).

To solve the problem in time domain we don?t need to use the concept of
impedance, although if you like you can talk about it.

At the capacitor we have I=C dV/dt

Note that these V and I are not the trapezoidal ones, but they are as mentioned
V=V(+)  + V(-)  and  I = I(+) - I(-) and only V(+) and I(+) are trapezoidal.

This equation gives

V(+)/Zc  - V(-)/Zc = C (  dV(+)/dt + dV(-)/dt )

V(+) is known so I?ll group the terms:

dV(-)/dt + V(-)/CZc= dV(+)/dt ?V(+)/CZc

the term C*Zc is the ?RC constant? mentioned in some messages  = tau

so the equation for the unknown V(-) is:

dV(-)/dt + V(-)/tau = dV(+)/dt ? V(+)/tau

I solved this equation quickly for rectangular pulse V(+) and it took a while
for the trapezoidal pulse but it is doable.

For a rectangular pulse (rise time=0) V(-)=[1-2exp(-t/tau)]*0.5      for t=(0
to T) ( time origin shifted to when the pulse arrives at the capacitor) and
V(-)= [1-exp(-T/tau)]*exp[-(t-T)/tau] for t= T to infinity

Note that this gives you V(-)= - 0.5 for t=0 ie exactly the dip that takes the
V(+)=0.5 V down to zero as group_delay noticed for zero  rise time

For a trapezoidal pulse it is much uglier

V(-)= 0.5/tr * (t-2tau)+tau/tr * exp(-t/tau) for the up ramp t= 0 to tr

And so on ?

The voltage at the end of the rise time is

0.5- tau/tr[1-exp(-tr/tau)]

if you take the limit for tr ->0 you?ll find the previous result ?0.5

however for tr different from zero it is > -0.5 and does not cancel completely
the 0.5 of the V(+) , again as noticed by group_delay

With these V(-) and V(+) propagated  one can reconstruct the actual voltage at
the capacitor and the source

The problem can also be solved, as I mentioned earlier, in Laplace or Fourier
domains, in which cases one would actually use the capacitor impedance.

Hope this helped in solving the homework problem.

Peter

group_delay <group_delay@xxxxxxxxx> wrote:
i didn't quite understand some of the characters in your equation.
when you write (finite dV)/(dI = 3D 0) = 3D> infinite impedance,
what do you mean?

yes, in my case the input waveform to the tline is a trapezoidal
signal 0V to 1V. the setup is shown below.

signal source -- 50 ohm resistor -- transmission line-- capacitor to
ground.

when i monitor the voltage at the junction of the resistor and the
tline, i see it jumps to 0.5V initially which makes sense. However,
when the reflection from the end comes back, it dips down very fast
and then rises in an exponential fashion. The dip down looks
parabolic (2nd order). The dip does go to zero for a zero rise time
input, but it does not go to zero for a finite rise time signal. i
am trying to figure out the best way to calculate the amount of this
dip...

thanks for you help.

--- In si-list@xxxxxxxxxxxxxxx, "Muranyi, Arpad"
wrote:
> You will get full reflection, because the impedance
> you need to use for the equation you quoted is the
> small signal (AC) impedance. Referring to my previous
> posting a short time ago, the constant current source
> equivalent during the ramping portion of your trapezoid
> waveform has a (finite dV) / (dI =3D 0) =3D> infinite impedance.
>
> The question I have for you is this: where do you mean
> that your waveform is trapezoid? At the beginning of
> the T-line, or at the end, where the capacitor is?
> If the first, be prepared for a non trapezoid waveform
> at the capacitor, because the T-line and the cap forms
> an RC circuit, who's response is an exponential waveform.
> If the ramp is faster the RC constant you will see
> an exponential waveform, if it is slower, you will see
> a more or less trapezoid waveform.
>
> I hope this helps,
>
>
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
=3D=3D=
>
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
=3D=3D=
> =3D=3D=3D=3D=3D
>
>
> -----Original Message-----
> From: si-list-bounce@xxxx [mailto:si-list-bounce@xxxx] =
> On Behalf Of group_delay
> Sent: Friday, January 28, 2005 1:48 PM
> To: si-list@xxxx
> Subject: [SI-LIST] Re: capacitor impedance in time domain
>
>
> hi all,
> what i really want to do is find out how much waveform gets
reflected
> from the end of a lossless transmission line terminated with a
> lossless capacitor, assuming the input waveform is a trapezoidal
> signal. I know this can be computed using: gamma =3D (Zl-Zo)/
(Zl+Zo),
> but this requires you to calculate Zl for the time domain signal.
If I
> wanted to avoid it and use time domain analysis, how would I setup
the
> equation?
>
> thanks,
> chris
>
>
> --- In si-list@xxxxxxxxxxxxxxx, steve weir wrote:
> > matthias, in the time domain we would solve the differential
> equations for=20
> > the network, or more likely using a computer program we would
solve =
> the=20
> > difference equations over a series of discrete time steps. Now
in
> either=20
> > case we could express impedance as dv/dt / di/dt. But I don't
know =
> how=20
> > useful it would be towards either visualizing behavior, or
solving the =
>
> > equations. Let's take the trapezoidal wave for instance. An =
> effective=20
> > impedance is pretty easy to come by on each: the rising, and
falling=20
> > portions of the waveform from the capacitance expression C =3D
> i/dv/dt, Z =3D=20
> > dv/dt / di/dt =3D 1/(dv/dt * C ). The flat portions are
troublesome
> as are=20
> > the vertices, since dv/dt theoretically goes to zero and the
> impedance from=20
> > the formula jumps to an infinite value. Intuition should tell us
> that this=20
> > is wrong, as
>
> coupling capacitors routinely pass high frequency pulses.
> >=20
> > In the frequency domain, we have this nailed. We don't have=20
> > discontinuities at the vertices. The vertices and flat portions
=
> follow=20
> > curves formed by the frequency components, and rather than a flat
> section=20
> > containing DC and no HF, quite the opposite is true: the
flatter we
> want=20
> > the pulse tops to be, the higher the frequency content
required. This =
>
> > aligns with our intuition. But when we transform the
representation
> back=20
> > to the time domain, those piecewise linear segments are now
curved
> solving=20
> > the discontinuities at the vertices and eliminating the flat
slopes
> with=20
> > theoretically infinite Z between the edges.
> >=20
> > So if someone wanted to look only at the rising and falling
edges, an=20
> > impedance in the time domain is reasonable, and possibly even
> useful. But=20
> > it really gets awkward when dealing with the whole waveform
unless
> we first=20
> > perform frequency limiting operations, most easily performed in
the=20
> > frequency domain.
> >=20
> > I am not an expert on algorithms, so I really can't say from an
error=20
> > analysis and computational efficiency standpoint what is really
the
> best=20
> > way to perform a transient analysis. But in my naivete, I would
be=20
> > inclined to transform everything into the frequency domain,
compute =
> the=20
> > solution and transform back. In my feeble mind, this would avoid
> some of=20
> > the discontinuity and convergence problems in SPICE and more
closely=20
> > follows nature. But since people a whole lot better at math
than I
> have=20
> > worked long and hard on those algorithms, I suspect either the=20
> > computational overhead, or error build-up of my naive approach
would =
> be=20
> > unacceptably high. Maybe what this world needs is a five cent,
256 =
> bit=20
> > floating point, matrix solver!
> >=20
> > Steve.
> >=20
> > At 10:13 PM 1/26/2005 +0100, Matthias Bergmann wrote:
> > >
> > >Hello, I don`t understand why impedance should be limited to =
> Frequency
> > >domain. What impedance are we speaking about ? For example the
> > >characteristicimpedance Z of a transmission line also exists in
> time domain.
> > >If you look along a transmission line, v(t) / i(t) have got
> singularities
> > >(undefined, infinite), these are called short and open ?!?!?
> Furthermore
> > >mostof the simulation programs use the time domain because it
permits
> > >non-linearities. I don`t know how what happens when your
impulse is
> > >trapezoidal, but if it was a rectangular and your load is a
> capacitance, you
> > >are answer would look like an exponential function, with your
> reflection
> > >co-efficient as initial value. Regards, Matthias Bergmann P.S.:
> Yes, use
> > >SPICE or ADS ! _m |---------+---------------------------------->
> > >
> > >
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> > >
> > >-
list@xxxx>
> > >-LIST] Re: capacitor impedance in time
> >
>
>domain         
0;&#=
> 160;
> >
>
>60I
;=
> 60;
> > >          |
> >
> > >-----------------------------------------------------------
------=
> ----
> > >--  -----------------------------------------| >I could be
> wrong >but
> > >tome >impedance is a concept strongly related to Frequency
domain.
> >>It is
> > >meaningful just in that domain. Absolutely. If you define
impedance =
> as
> > >voltage/current, then you run into great difficulties if you
try to
> do it in
> > >the time domain.  In general, with any complex impedance,
> v(t)/i(t) has
> > >singularities (undefined, infinite). I consider impedance =3D
> v(s)/i(s) or
> > >v(f)/i(f), which makes it a strictly frequency domain parameter.
> Regards,
> > >Andy
> ------------------------------------------------------------------
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