The truth about decibels

[Don Bowey]

On 7/5/05 12:28 PM, in article
8bBye.25924$d75.18306@fe1.news.blueyonder.co.uk, "Kevin Aylward"
<see_website@anasoft.co.uk> wrote:

John Larkin wrote:
On Mon, 04 Jul 2005 18:49:36 GMT, "Kevin Aylward"
see_website@anasoft.co.uk> wrote:

John Larkin wrote:
On Mon, 04 Jul 2005 05:52:49 GMT, "Kevin Aylward"
see_website@anasoft.co.uk> wrote:



Its this simple mate.

Zo = sqrt((R+jwL)/(G+jwC))

It is *not* Z0 = sqrt(L/C)

The derivation is here
http://www.sm.luth.se/~urban/master/Theory/4.html

R, L, C, G in per/M


Won't any unit length work?



Smart arse



No, really, the question was sincere.


Well, ok. I thought is was obvious that units are all in the wash. Sure,
I should have said per unit length, but ....

Anyway, meters are the only real lengths, you yanks need to catch up
with the rest of the known universe.

Some us have been working on catching up. I worked with a chap from north
North America who convinced a Standards working group to consider a
bit-error rate expressed in errors per fortnight.

Don

 

[Pooh Bear]

John Larkin wrote:

Are your road signs still in miles? They are in England.

Scotland, Wales and N.I. too indeed. But beware of kilometres over the border
from N.I. in the Republic. Not to mention km/h in place of mph.

I *have* seen some roadworks signs in metres here though.

You can still order a pint there, too.

I should damn well hope so ! ;-)

Graham

 

[Rich Grise]

On Sat, 02 Jul 2005 13:08:07 -0700, John Larkin wrote:

If you, say, apply 1 volt and graph current vs time, I'm not sure what
the exact expression for I(t) would be... does anybody know? I guess
you could spice it, using the lossy line model.

I(t) = V(t)/Zo. It's that simple. Your source provides a step to
V, and the line will draw current at V/Zo. It doesn't care what
the losses downstream are, albeit I could see a finite line that's
long enough that the losses would add up to the point that by the time
the reflection at the open end got back to you, it'd be attenuated to
negligibility.

But the impedance doesn't change over time, just because the
conductors have resistance!

Cheers!
Rich

 

[John Larkin]

On Mon, 18 Jul 2005 21:21:34 GMT, Rich Grise <richgrise@example.net>
wrote:

On Sat, 02 Jul 2005 13:08:07 -0700, John Larkin wrote:

If you, say, apply 1 volt and graph current vs time, I'm not sure what
the exact expression for I(t) would be... does anybody know? I guess
you could spice it, using the lossy line model.

I(t) = V(t)/Zo. It's that simple.

No it's not. If V(t) is just a step, say from 0 to +1 volt at t=0,
I(t) is initially 1/Zo but immediately begins to decline as line
series-loss resistance becomes visible (in microseconds for any real
transmission line), asymptotically approaching I=0, or a bit above
zero if there's any shunt leakage. In one second you're seeing
something like 100,000 miles of cable resistance.

As I said, I don't know the exact current decay curve for an ideal
series-lossy transmission line.

John

 

[keith]

On Mon, 18 Jul 2005 21:21:34 +0000, Rich Grise wrote:

On Sat, 02 Jul 2005 13:08:07 -0700, John Larkin wrote:

If you, say, apply 1 volt and graph current vs time, I'm not sure what
the exact expression for I(t) would be... does anybody know? I guess
you could spice it, using the lossy line model.

I(t) = V(t)/Zo. It's that simple. Your source provides a step to
V, and the line will draw current at V/Zo. It doesn't care what
the losses downstream are, albeit I could see a finite line that's
long enough that the losses would add up to the point that by the time
the reflection at the open end got back to you, it'd be attenuated to
negligibility.

Do note that the losses must be taken into account when calculating Z0.
A highly resistive line (such a on-chip wires) doesn't behave anthing
close to your ideal L/C model.

But the impedance doesn't change over time, just because the conductors
have resistance!

No. it's *different* because of the resistance.

--
Keith