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)