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Old July 10th 07, 05:47 PM posted to rec.radio.cb
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First recorded activity by RadioBanter: Sep 2006
Posts: 298
Default Learn How VoiceMax Increases Your Transmission Range...

On Tue, 10 Jul 2007 05:28:56 +0100, " Peter"
wrote:

+++"james" wrote...
+++
+++ Let me be sure that I understand what you are saying.
+++
+++
+++If you are having trouble with my English, try these...
+++ (Full book titles, authors and ISBN numbers at bottom)
+++
+++
+++quote ref=Electronics For Engineers, page 11
+++ Amplification This essential process involves an increase
+++ in the amplitude or size of a signal without any change
+++ to the waveform.
+++/quote
+++
+++
+++quote ref=Electronics 2, page 114
+++ the input and output will have the same waveshape.
+++/quote
+++
+++
+++quote ref=Electronics Servicing Vol2, page 61
+++ we ideally require the output signal to be a faithful but
+++ magnified replica of the input signal.
+++quote
+++
+++
+++ If that is your position then that is utter bovine, canine, feline and
+++ any other *ine excrement you wish to use.
+++
+++Tell it to the authors of those books, but you had best consider
+++their qualifications and experience first.
+++
+++
+++ The only way the output waveform is equal to the input waveform is
+++ when the stage is at unity gain
+++
+++I have the choice of accepting the word of several well qualified
+++and experienced lecturers and engineers, or someone on a CB newsgroup.
+++Tough call... can I think about it?
+++
+++ period.
+++
+++ 1/f
+++
+++ A waveform of any continuous time varying signal is defined
+++ as a set of intantaineous points versus time that represent
+++ that signal.
+++
+++At the input of an amplifier, you have a single frequency signal...
+++ 10mV @ 1Khz
+++The waveform is...
+++ sin 2pi f t
+++
+++The signal at the output is amplified...
+++ 100mV @1KHz
+++The waveform should now be
+++ sin 2pi f t
+++
+++That is assuming a perfectly linear amplifier. We both
+++agree that perfection doesn't exists in this world.
+++Transistors are not perfectly linear but, with good design,
+++an amplifier can get pretty damn close.
+++

*********************

Partially correct in your formulae.

Consider this and reflect with your noted lecturers and writers.

let the input signal of an amplifier be:
f(t) = Ai * sin(2*PI*f*t)

Where Ai is the input signal amplitude.

Now let the scalar value of the transfer function of the amplifier be
a real number greater than one and represented by the constant K. This
becomes essentially a distorionless amplifier and does not consider
internal noise generated in the amplifier.

The output signal from the amplifier:
g(t) = Ai * K *sin(2*PI*f*t)

Now if you can prove that g(t) is equal to f(t) when K greater than
one I will be glad to nominate you for a Noble prize for mathematics.

The frequency component of the function, sin(2*PI*f*t),remains the
same in both equation. The amplitude does not. Therefore the two
functions are not equal. They will have similar waveshapes in that
they will be mathematical multiples of each other. To state that they
are the same is ludicrous. Now I can believe the mathematics that I
have been taught or accept your hypothesis.

Hmmm. Let me get back to you on that.



james
 
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