Driver to drive?

Hi Ken,

You could change to using sound: 25KHz is high enough that most people
can't hear it unless it is quite loud.


Please don't do that. If can be quite a torture for animals such as
Click to expand...
dogs. Even if it's out in the woods we should mind foxes and others.

Remember those "toneless" whistles? Blow gently into them and the dog
obeys some command such as 'come' if trained. Blow hard and you get
growls and hisses.

Regards, Joerg

http://www.analogconsultants.com
 
In article <mJS9d.27027$QJ3.16630@newssvr21.news.prodigy.com>,
Joerg <notthisjoergsch@removethispacbell.net> wrote:
Hi Ken,

You could change to using sound: 25KHz is high enough that most people
can't hear it unless it is quite loud.


Please don't do that. If can be quite a torture for animals such as
dogs. Even if it's out in the woods we should mind foxes and others.

Remember those "toneless" whistles? Blow gently into them and the dog
obeys some command such as 'come' if trained. Blow hard and you get
growls and hisses.
Click to expand...
Good point.

A 50 foot twisted pair would be the best answer. It would be very
directional.

--
--
kensmith@rahul.net forging knowledge
 
In article <ck8pf402cgo@drn.newsguy.com>,
Winfield Hill <Winfield_member@newsguy.com> writes:
John S. Dyson wrote...

... the very lousy traitor against America, John Kerry. ...
... If John Kerry did become president, it JUST MIGHT be time
for even those who are non-religious to pray for some divinity
to deliver us from John Kerry, who would be (at least partially)
elected based upon 'evil and destructive hatred.'

You are making a veiled suggestion that should John Kerry
be elected, he should be killed?

Of course not -- however, your comment does show that that
Click to expand...
killing is in your mind.

John
 
Dyson is off his meds again, someone get the straight jacket. When he is
helpless we can show him pictures of Clinton and watch him squirm.
 
Don Pearce wrote:
On Sat, 09 Oct 2004 11:59:21 GMT, "Kevin Aylward"
salesEXTRACT@anasoft.co.uk> wrote:

maTheMatic wrote:
Hi,all
I often see the following transmission line model:

------Zrl---------Zrl----------
|
Zin Ygc Ygc ......
|
--------------------------------

Where Zrl=R+jwL ,Ygc=G+jwC.
so for infinite transmission line, we have

------Zrl---------------
|
Zin Ygc Zin
|
------------------------

so There is the equation :

Zin/Ygc
Zin = Zrl + ---------------
Zin + 1/Ygc

solve this quadratic equation, we get

Zrl + sqrt( (Zrl)^2 + 4(Zrl/Ygc))
Zin = ---------------------------------
2


But we ALL KNOW the transmission line characteristic impedance is

Zin= sqrt(Zrl/Ygc)

what am I missing? why doesn't this model work, any insight
explaniation are appreciated.

The characteristic impedance of a transmission line is *only* the
value of the lines input impedance when terminated by its
characteristic impedance!

For other loads, the input impedance wil *not* be given only by its
x/Length spec.

This simply isn't so.
Click to expand...
Ho hummm....


The impedance of the transmission line does not
change in response to the load on the end of it.
Click to expand...
I see you are doing your best to read this as incorrectly as you can.

I wonder why this would be so.

What does change is
the impedance of the entire *network* as seen by the source.
Click to expand...
Oh dear...

I'll give you a hint quote: "...is *only* the value of the lines input
impedance..."

What do you suggest "the lines input impedance" actually means. Hint:
its not the impedance of the line on its own.

This will
be the load impedance rotated and transformed by the line
characteristics. To understand this in a reasonably intuitive manner,
get hold of a book on Smith charts.
Click to expand...
You have obviously got some chip on your shoulder here.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
 
In article <41675DA0.6090709@nospam.com>,
Fred Bloggs <nospam@nospam.com> wrote:
[....]
groups have pulled the plug on Halliburton stocks and a few other
businesses. And this is not just a "liberal" slant- it is integrity-
Click to expand...

It could just be that they see investments in companies that deal with
murderers as extra risky. It could be very bad press that your machine
was used to grind up Hitler's victims. Your sales would likely fall as
would the stock. Anyone who invests, looking for a safe bet, would not
want that added risk.

There are perfectly good "green eyeshade" reasons for not murdering people
or poluting a lot or laying off a large number of workers all at once.

--
--
kensmith@rahul.net forging knowledge
 
In article <ck7mus$21bn$1@news.iquest.net>,
John S. Dyson <toor@iquest.net> wrote:
[...]
paying an emotional price for their support of the very lousy traitor
against America, John Kerry.
Click to expand...
Please define how you can consider Kerry a traitor and not Bush.

This should be interesting, since the Bush crowd has outted a CIA agent
during "a time of war" and it has been proven that the charge of treason
against Kerry is just spewing from the nut case right.


--
--
kensmith@rahul.net forging knowledge
 
In article <20041009111220.22449.00000566@mb-m06.aol.com>,
Rolavine <rolavine@aol.com> wrote:
Dyson is off his meds again, someone get the straight jacket. When he is
helpless we can show him pictures of Clinton and watch him squirm.
Click to expand...
Kerry's health care proposal will lower the cost of his drugs enough that
he won't have to choose between taking Prozac and Viagra.
--
--
kensmith@rahul.net forging knowledge
 
To summarize: is a MS-FF one FF, or two FFs?

http://focus.ti.com/lit/ds/symlink/sn74ls74a.pdf
Click to expand...
That's a edge-triggered D flip flop. Does it have anything to do with
my question?


Wouter van Ooijen

-- ------------------------------------
http://www.voti.nl
Webshop for PICs and other electronics
http://www.voti.nl/hvu
Teacher electronics and informatics
 
On 9 Oct 2004 02:55:23 -0700, maTheMatic wrote:

Hi,all
I often see the following transmission line model:

------Zrl---------Zrl----------
| |
Zin Ygc Ygc ......
| |
--------------------------------

Where Zrl=R+jwL ,Ygc=G+jwC.
so for infinite transmission line, we have

------Zrl---------------
| |
Zin Ygc Zin
| |
------------------------

so There is the equation :

Zin/Ygc
Zin = Zrl + ---------------
Zin + 1/Ygc

solve this quadratic equation, we get

Zrl + sqrt( (Zrl)^2 + 4(Zrl/Ygc))
Zin = ---------------------------------
2


But we ALL KNOW the transmission line characteristic impedance is

Zin= sqrt(Zrl/Ygc)

what am I missing? why doesn't this model work, any insight
explaniation are appreciated.
Click to expand...
I think it does work, but it's a better model than the simpler one you've
grown accustomed to using.

Here's what I get:

Zin = jwL + 1/(jwC + Go)

where Go = 1/Zo. If I let w -> 0, then

Zin = 1/Go = Zo.

At low frequencies, such that jwC << Go, 1/(jwC + Go) ~ (1/Go)(1 - jwC/Go)

Since Go = 1/Zo = sqrt(L/C), this becomes 1/Go - jwL. Substituting this
into the equation for Zin,

Zin = jwL + 1/Go - jwL = 1/Go = Zo.

The model is valid as long as jwC << Go. You'll probably find that a real
transmission line has the same limitations (any many more).

-- Mike --
 
On 09 Oct 2004 15:12:20 GMT, rolavine@aol.com (Rolavine) wrote:

Dyson is off his meds again, someone get the straight jacket. When he is
helpless we can show him pictures of Clinton and watch him squirm.
Click to expand...
---
Poor sonofabitch is helpless _now_.

Just throw him a crumb, anything at all, and watch him bounce around
in his cage.

--
John Fields
 
S. hashem Aref wrote...
Is there anyone how can design a simple circuit for Avalanch
photodiode bias.
I want to use it instead of PIN photodiode and I just need to
measure changing of light. I don't need high responsivity of APD.

Specification of APD:
Sensitivity@1310: .86 A/W
Vr=5v
Click to expand...
Read the "Avalanche Photodiodes: A User's Guide" article at this
site, and consider the issue of bias voltage control (figure 2).
http://optoelectronics.perkinelmer.com/Service/whitepapers.html


--
Thanks,
- Win

(email: use hill_at_rowland-dotties-org for now)
 
John S. Dyson wrote...
Winfield Hill writes:
John S. Dyson wrote...

... the very lousy traitor against America, John Kerry. ...
... If John Kerry did become president, it JUST MIGHT be time
for even those who are non-religious to pray for some divinity
to deliver us from John Kerry, who would be (at least partially)
elected based upon 'evil and destructive hatred.'

You are making a veiled suggestion that should John Kerry
be elected, he should be killed?

Of course not -- however, your comment does show that that
killing is in your mind.
Click to expand...
Don't be silly. But please explain exactly what you mean by
divinity delivering us from John Kerry, should he win?


--
Thanks,
- Win

(email: use hill_at_rowland-dotties-org for now)
 
On Sat, 09 Oct 2004 15:20:57 GMT, "Kevin Aylward"
<salesEXTRACT@anasoft.co.uk> wrote:

Don Pearce wrote:
On Sat, 09 Oct 2004 11:59:21 GMT, "Kevin Aylward"
salesEXTRACT@anasoft.co.uk> wrote:

maTheMatic wrote:
Hi,all
I often see the following transmission line model:

------Zrl---------Zrl----------
|
Zin Ygc Ygc ......
|
--------------------------------

Where Zrl=R+jwL ,Ygc=G+jwC.
so for infinite transmission line, we have

------Zrl---------------
|
Zin Ygc Zin
|
------------------------

so There is the equation :

Zin/Ygc
Zin = Zrl + ---------------
Zin + 1/Ygc

solve this quadratic equation, we get

Zrl + sqrt( (Zrl)^2 + 4(Zrl/Ygc))
Zin = ---------------------------------
2


But we ALL KNOW the transmission line characteristic impedance is

Zin= sqrt(Zrl/Ygc)

what am I missing? why doesn't this model work, any insight
explaniation are appreciated.

The characteristic impedance of a transmission line is *only* the
value of the lines input impedance when terminated by its
characteristic impedance!

For other loads, the input impedance wil *not* be given only by its
x/Length spec.

This simply isn't so.

Ho hummm....


The impedance of the transmission line does not
change in response to the load on the end of it.

I see you are doing your best to read this as incorrectly as you can.

I wonder why this would be so.

What does change is
the impedance of the entire *network* as seen by the source.

Oh dear...

I'll give you a hint quote: "...is *only* the value of the lines input
impedance..."

What do you suggest "the lines input impedance" actually means. Hint:
its not the impedance of the line on its own.

This will
be the load impedance rotated and transformed by the line
characteristics. To understand this in a reasonably intuitive manner,
get hold of a book on Smith charts.

You have obviously got some chip on your shoulder here.

Kevin Aylward
Click to expand...

The OP is asking about the impedance of a transmission line - not the
impedance of some random impedance hanging on the end of the line. No
chip needed - you simply weren't following what the thread is about.

d
Pearce Consulting
http://www.pearce.uk.com
 
And in case you reached the conclusion that laptops are safer because they
use Li-ion batteries,

http://www.tmcnet.com/usubmit/2004/May/1040980.htm
 
On Friday 08 October 2004 06:57 pm, ChrisGibboGibson did deign to grace us
with the following:

Rich Grise wrote:

[snip]

Heh. Yeah. Every year 400,000 people are killed by smoking. And 2,500,000
are killed by something else.

So NON-SMOKERS ARE OVER SIX TIMES AS LIKELY TO DIE!!!!!!!!!!!!!!!!


That's like 15% of fatal road accidents involve a drunk driver. So surely
it's the sober drivers we should be banning from driving ?
Click to expand...
Well, the sober drivers should be held to at least as high a standard
of driving expertise as the drunk ones are, if you want the "accident" rate
to drop.

"Oh, little Timmy was killed by a drunk driver" is just a way of placing
blame: "Oh, there was a drunk on the road, so that relieves little Timmy of
responsibility to operate his vehicle safely." A drunk driver is just
another road hazard.

Thanks,
Rich
 
Winfield Hill wrote:

John S. Dyson wrote...

Winfield Hill writes:

John S. Dyson wrote...

... the very lousy traitor against America, John Kerry. ...
... If John Kerry did become president, it JUST MIGHT be time
for even those who are non-religious to pray for some divinity
to deliver us from John Kerry, who would be (at least partially)
elected based upon 'evil and destructive hatred.'

You are making a veiled suggestion that should John Kerry
be elected, he should be killed?

Of course not -- however, your comment does show that that
killing is in your mind.


Don't be silly. But please explain exactly what you mean by
divinity delivering us from John Kerry, should he win?
Click to expand...
God does the killing.

--
Dirk

The Consensus:-
The political party for the new millenium
http://www.theconsensus.org
 
On Saturday 09 October 2004 05:08 am, Dirk Bruere at Neopax did deign to
grace us with the following:

Scott Stephens wrote:

Dirk Bruere at Neopax wrote:

Rich Grise wrote:

On Friday 08 October 2004 01:26 am, Scott Stephens did deign to grace
us with the following:


The best thing to do is, if they assholes are worth tolerating, is to
find someone weaker and torment the shit out of them. Give as good as
you get.



...

But in the end, its best not to associate with assholes.




Good Idea.

Plonk.


How mean. And I was going to nominate Rich to be the Duck-Bill Platypus
of S.E.D.

http://www.wunderland.com/WTS/Ginohn/cetera/hankisms.html


Perhaps its better to face the true God having ignorantly believed in a
false one, than face truth having believed in nothing, shirking
responsibility for risking being wrong in search for the truth.

And which Hank would that be, and do we have to kiss Karl's ass as well?

Another great lesson in politics we can learn in this "mean season". Get
your way by criticizing those in power because they aren't perfect. You
can always find some fault to blame, if you're not responsible for doing
any better.

We don't ask for perfect, just for not total moronic fuckup.

I'd even take a mere moronic fuckup. There's intentional death and
Click to expand...
destruction going on, and GWB seems to be the figurehead. I'm really
saddened that doublethink has got so bad that people can convince
themselves that long-distance murder has anything to do with "defense."

Thanks,
Rich
 
On 6 Oct 2004 19:53:02 -0700, bokiteam@ms21.hinet.net (boki) wrote:

Dear All,
I have a portable device that has a 5-V to 0-V pulse output(
without constant frequency, but below 10Hz range), I want this signal
could be detected by PC RS232 port.

I try many connection type, the RS232 still can't detect the device
pulse.
Click to expand...
The RS-232 data input is designed to receive "characters" sent in a
specific format - it is not intended to detect random pulses.
Do we have to design RS-232 handshake circuits with device output to
make the RS232 detect that pulse?
Click to expand...
You can use a handshake input line to detect your pulse - but you will
have to read the UART status register to see it.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca
 
On Sat, 09 Oct 2004 15:37:45 GMT, wouter@voti.nl (Wouter van Ooijen
(www.voti.nl)) wrote:

To summarize: is a MS-FF one FF, or two FFs?

http://focus.ti.com/lit/ds/symlink/sn74ls74a.pdf

That's a edge-triggered D flip flop. Does it have anything to do with
my question?
Click to expand...
Geez, ya don't have to get snotty about it...

On page 1, read the description and take a look at the logic diagram
on the lower right hand corner of the page and you'll find that it's
not _really_ an edge-triggered flip-flop, its inputs are all voltage
sensitive and cause triggering to occur at particular _voltage_
levels. It's specified as an edge-triggered flip-flop because if the
timing constraints placed on the input signal transitions are adhered
to, the outputs will be guaranteed to occur as predicted.

To go on: referring to the logic diagram, notice that when the rest of
the input conditions are right, the state of the D input will be
entered into the lower flip-flop, (the master) and then, if
everything's cool when the clock goes high, the state of the master
will be entered into the second flip-flop, the slave. So, if your
question was, "is a MS-FF one FF, or two FFs?" , the answer is: "Two
FF's."

OK?

--
John Fields
 
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