Telamon wrote:
In article .com,
Denny wrote:

Snip

plonk


Anyone who would do that deserves a

Bigger SNIP

and a

Monster PLONK

tom
K0TAR

Telamon wrote:
In article .com,
Denny wrote:

Snip

plonk


Do you only post here to plonk people? I guess that could be fun.... ;^)

- 73 de Mike KB3EIa -

wrote:


Let's see how many more times you are added to his killfile.

Beware lest thou enrage the gods of plonkification, Jim.

- 73 de Mike KB3EIA -

On Jun 21, 1:08 pm, Michael Coslo wrote:
Telamon wrote:
In article .com,
Denny wrote:

Snip

plonk

Do you only post here to plonk people? I guess that could be fun.... ;^)

- 73 de Mike KB3EIa -

Thats about it. Don't mind him. He was abused as a child. He's
plonked
me twice in the last 5 years or so, and I'm sure this post will bring
me
up to three.. Which begs the answer... If I was plonked before, how
can I be replonked, unless I was unplonked..
I noticed he had to plonk Denny twice.. I guess his plonking
device is about as effective as viagra on a 98 year old..
I call him Telaprick... And yes, he has worked hard in the past
to deserve the new name I have bestowed upon him. I quit posting to
the shortwave group about a year ago, cuz I got tired of dealing
with a couple of brain dead morons that think they own the whole
show over there. And if you post anything that is even slightly
technical, they will whine and complain that it's too much for
the average shortwave listener to handle, or comprehend.
So I decided, $%&* em if they can't take a joke.
The only thing that goes "plonk" around here are the turds
that fall into the toilet each morning after I've had my first cup
of go juice. Only a whiny bitchette would go around constantly
"plonking" people. It adds more clutter than the posts he is
plonking..
MK

In article ,
Michael Coslo wrote:

Telamon wrote:
In article .com,
Denny wrote:

Snip

plonk


Do you only post here to plonk people? I guess that could be fun.... ;^)

Nope. I just figure that if a person will entertain a Troll then they
are not worth reading.

Plonk

--
Telamon
Ventura, California

In article ,
Tom Ring wrote:

Telamon wrote:
In article .com,
Denny wrote:

Snip

plonk


Anyone who would do that deserves a

Bigger SNIP

and a

Monster PLONK

Good for you.

Plonk

--
Telamon
Ventura, California

On Jun 20, 1:01 pm, (Mark Zenier) wrote:
In article . com,

Radium wrote:
On Jun 18, 11:05 am, (Mark Zenier) wrote:

1. What is the energy of a photon at 150 kHz?

6.2 X 10^-10 eV

You didn't answer the other questions. How much power does a good radio
need to get a listenable signal?

Here's some numbers

input impedance 50 ohms
Noise Figure 5 db
Bandwidth 6 kHz
signal to noise ratio 40 dB

(This would be what some of the newsgroups listeners here would want if
they had one of their pretty damn good radios listening to a broadcast
station. Really picky ones would probably want a 8-15 kHz bandwidth
with a 60 dB s/N ratio).

There are equations out there that will give you how much power
you need for this signal...

I guess the dynamic range needs to be at least 140 dB [to match human
ear's loudness perception] and bandwidth at least 40 KHz [to match
human's pitch perception].

I am not sure about the input impedance or noise figure.

In article .com,
Radium wrote:
On Jun 20, 1:01 pm, (Mark Zenier) wrote:
In article . com,

Radium wrote:
On Jun 18, 11:05 am, (Mark Zenier) wrote:

1. What is the energy of a photon at 150 kHz?

6.2 X 10^-10 eV

You didn't answer the other questions. How much power does a good radio
need to get a listenable signal?

Here's some numbers

input impedance 50 ohms
Noise Figure 5 db
Bandwidth 6 kHz
signal to noise ratio 40 dB

(This would be what some of the newsgroups listeners here would want if
they had one of their pretty damn good radios listening to a broadcast
station. Really picky ones would probably want a 8-15 kHz bandwidth
with a 60 dB s/N ratio).

There are equations out there that will give you how much power
you need for this signal...

I guess the dynamic range needs to be at least 140 dB [to match human
ear's loudness perception] and bandwidth at least 40 KHz [to match
human's pitch perception].

I am not sure about the input impedance or noise figure.


We're not talking about human perceptions, we're talking about a
radio.

Anyway, how much power does a radio signal to need to be picked up?

To start with, there is electronics noise. This comes from temperature
of the circuit, and how wide the frequency response. The equation is
kTB, where k is Boltzmann's Constant, T it the absolute temperature
in Kelvin, and B is the bandwidth in Hertz. This is called the thermal
noise power.

Next, there's a fudge factor, called either the Noise Factor or the
same thing stated in decibels, the Noise Figure. It's determined by
the quality of the transistors (or other active amplifying element)
and the topology of the circuit. It specifies how much more noise is
added in the circuit. The best way to get it is to measure it.
(5db, about a factor of 3, would be for a pretty good shortwave radio).

Multiply thermal noise power by the noise factor, and you find the
"noise floor", or "minimum detectable signal" for a circuit.

But that doesn't do you much good, because all you can do with that is,
maybe, tell if there is a signal there, or perhaps not. So you need
more power in the signal to overcome the noise. For reasonable audio,
a factor of 10,000 for power is in the ball park. (This is usually
expressed as decibels, ie. 40 dB).

So here's a cut and paste from a unix program that does calulations with
stated dimensions, along with some comments.

[mzenier@localhost mzenier]$ units --verbose
1948 units, 71 prefixes, 28 functions

You have: boltzmann * 290 kelvin * 6 kilohertz * 3 * 10000

Boltzmann's constant * room temperature * bandwidth * Noise Factor * 40 dB

You want: watts
boltzmann * 290 kelvin * 6 kilohertz * 3 * 10000
= 7.2069944e-13 watts

This is how much power is needed to receive the signal with a signal
to noise ratio of 40 dB in a shortwave radio.

You have: 150 kilohertz h

frequency * Planck's Constant

You want: joules
150 kilohertz h = 9.9391031e-29 joules

Energy per photon at 150 kHz

You have: (boltzmann * 290 kelvin * 6 kilohertz * 3 * 10000)/(150 kilohertz h)
You want: hertz
(boltzmann * 290 kelvin * 6 kilohertz * 3 * 10000)/(150 kilohertz h)
= 7.2511517e+15 hertz

Divide the power needed by the energy in each photon, and you get the
number of Photons per second at 150 kHz to give a 40 dB signal to noise
ratio in a pretty good quality receiver with a 6 kilohertz bandwidth.

7.2511517e+15 is a lot more than 20,000.

(And a good demonstration why they don't use the math for quantum physics
for radio frequency calculations.)

Mark Zenier
Googleproofaddress(account:mzenier provider:eskimo domain:com)