Envelope-to: dave@picks.force9.co.uk
Delivery-date: Wed, 01 Feb 2006 19:27:03 +0000
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From: "Andy Talbot" <actalbot@southsurf.com>
To: rsgb_lf_group@blacksheep.org
References: <000e01c6273f$7b8710f0$018cf8d4@standalone> <001501c6274a$04bb8860$67b0fea9@lark>
Date: Wed, 1 Feb 2006 19:25:45 -0000
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Subject: LF: Re: Re: Can't see the wood for the trees
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Think I've found a way of getting the answer without having to back to first
FFT principles - it had to be possible.

Take a DC voltage of level = 1.0, In a unit resistance the  power also = 1.
When chopped by a 2% duty cycle pulse train, this will have a mean DC
VOLTAGE of 0.02
For any signal so chopped, its  mean POWER will also be 0.02
Now, a voltage of 0.02  NORMALLY means a power of 0.02^2  = 0.0004 so
something is going on.
The solution is that the extra power (0.02 - 0.0004) is going into the
sidebands / frequency comb.

So, an easy way to calculate the absolute level of any comb component is to
first find out the zero frequency term, which is the 0.02  VOLTAGE in this
case, then apply sin(x)/x to any other terms as needed and multiply the
result by the zero frequency value to obtain the absolute level.

So, using the 5MHz sounder sequence, DC term (carrier when at RF) = 0.02
voltage = -34db on the CW power.  At 1kHz away, X = 1kHz * 500us pulse width
= 0.5, so   further attenuation = sin (pi.0.5)/pi.0.5 =  0.64 = -4dB.
So sideband level at 1kHz away from 5.29MHz = -39db on the CW tone.

Which is very close to what I'm seeing  - QED

Seems a rather convoluted calcualtion, but it got there!

Andy  G4JNT


----- Original Message -----
From: "Alan Melia" <alan.melia@btinternet.com>
To: <rsgb_lf_group@blacksheep.org>
Sent: Wednesday, February 01, 2006 4:09 PM
Subject: LF: Re: Can't see the wood for the trees


> Hi Andy I think what you are looking for is given on page 22 of my copy of
> "Radio Engineers Handbook" by Terman I am sure you have access to a copy
if
> not I could scan
>
> Cheers de Alan G3NYK
>
> ----- Original Message -----
> From: Andy <actalbot@southsurf.com>
> To: <rsgb_lf_group@blacksheep.org>
> Sent: 01 February 2006 14:51
> Subject: LF: Can't see the wood for the trees
>
>
> > Can someone help with what should be obvious.
> >
> > I have a train of constant width pulses at a fixed repetition rate.   In
> the
> > frequency domain these appear as a spectral comb with spacing at the
> > repetition rate, whose amplitude  follows a SIN(X) / X shape depending
on
> > the pulse width, ie. the first null occuring at at 1/width and so on.
> >
> > What I'm getting tied up in knots trying to calculate is :
> >
> > What is the absolute amplitude (power) of just one individual tooth of
the
> > comb at any particular spacing.
> >
> > Assume the pulse waveform has, say,  1mW or 0dBm mean amplitude, and
> > consists of 500us pulses at 40Hz PRI.   The duty cycle is 0.02, so the
> > individual pulse power would have to be 50mW or 17dBm to get this mean.
> > But what is the amplitude of the component at, say, 1kHz, or 1040Hz, or
> > 10kHz ??
> >
> > It must be obvious, but I keep feeling the urge to integrate SIN(X) / X
> > which is not funny and way beyond my maths capabilities!!
> >
> > The figures given above are those for the 5MHz beacon sounder sequence.
> > The amplitude trace on the monitoring software during the sounder
sequence
> > is measuring just one line of the comb ( F = 0,  the carrier) , and
> appears
> > to suggest this is about 30 - 35dB down on the CW part.  That is -17dB
> from
> > the peak/mean  as above, but where does the other 13 - 18dB come from?
> >
> > Tearing hair out
> >
> > Andy  G4JNT
> > www.scrbg.org/g4jnt/
> >
> >
> >
>
>
>
>
>
> --
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2006-01-30
>
>


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