Re: AM radio receiver - design


"MooseFET" <kensmith@xxxxxxxxx> wrote in message
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On Feb 9, 1:15 pm, "christofire" <christof...@xxxxxxxxxxxxxx> wrote:
"MooseFET" <kensm...@xxxxxxxxx> wrote in message

news:775c12df-094a-499d-bfc6-eefab7c1eb21@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Feb 8, 8:34 am, "christofire" <christof...@xxxxxxxxxxxxxx> wrote:

"MooseFET" <kensm...@xxxxxxxxx> wrote in message

- - - snip - - -

I can't explain why Outlook Express isn't inserting '>' characters at the
beginnings of lines, but I will precede each of my comments by @@ below:

- - - -

I think I was completely clear in the first posting, but that you
jumped to assuming something because you were not thinking about the
OP's case. The OP is making a radio for the standard AM band.

@@ I had never lost sight of that.

Keep that point foremost in your mind as you reread what I wrote back
there
and read this explanation below. They are two different cuts at
trying to explain the same ideas.

Disregarding signal strength, the signal in a normal AM radio is
ideally:

Y = f(t) * sin(wt) + sin(wt) (1)

Where:

f(t) is the program material that is centered around DC. It contains
frequencies from 20Hz to roughly 7KHz. The peak amplitude is always
less than one.

sin(wt) is the carrier. It appears twice because this could also have
been written as:

Y = (1 + f(t)) * sin(wt)

The version (1) of the equation is the better because it breaks the
carrier and the sidebands apart and makes it far more clear what the
signal is:

What you actually receive will be:

Y = fn1(t)*cos(wt) + fn2(t)*sin(wt) + f(t)*sin(wt)
+ sin(wt)

The fn1 and fn2 are two parts of the noise. Any function (noise or
signal) in a band that does not include DC can be broken into two
parts in this way.

@@ This does not represent noise in a band of frequencies, as is
encountered
in AM reception. Simply multiplying a function by sin and cos waves works
fine for a single vector but does not account for the different Fourier
components of the noise. There are many well-respected text books that
provide the correct way to resolve noise into amplitude and phase
components, and this always involves integrals.

Since you know single side band methods, I won't
bother to explain how. Just remember to remind yourself that this is
not SSB radio before reading the next bit.

Now that we have what the receiver actually has as a signal. The next
step in the argument is to assume that we have a PLL that is locked
onto the carrier and that its bandwidth is much less than 20Hz so that
as far as we are concerned, it is a constant frequency sine wave. We
can multiply each term in the equation by sin(wt)

fn1(t)*cos(wt) * sin(wt) no base band result

@@ Only true for a single vector, not for a frequency band of noise.

Yes, it does apply to the noise in this case because we only care
about the base band result and not the high frequency results. If we
cared about the signals out of the demodulator at high frequencies, it
would be quite a different matter.

Also note my comment about my being unwilling to write the infinite
series needed for a full treatment.

- - - -

@@ Both irrelevant. You don't have to believe what I say but you owe it to
yourself to at least investigate the possibility that you have made an
error, and look up the correct way to resolve noise components in a band.

I believe what is stated in a number of well-respected books, and is
demonstrated in a small proportion, that there is no difference in the S/N
provided by an envelope detector or a synchronus detector in the conditions
we've been discussing, and you disagree. Your analysis hasn't convinced me
that you're right.

Let's end the argument here - I don't suppose many others are interested.

Chris


.

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