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K3 transmitt filter

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MontyS

K3 transmitt filter

> Just so someone says it, a CW signal has no width, it's carrier only. The K3 does a fine job of generating it with a 2.7 or 2.8khz filt

> Monty K2DLJ
>
>
>>> Kind of like narrow filters for CW.
>>>
>>> I am still new to this hobby and it was just an idea that I had. Maybe when I finish going thru the extra study guide over the holidays I will know the theory if this was a bad random thought.
>>>
>>> Thanks
>>> Don
>>> KD8NNU
>>>
>

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N5GE

Re: K3 transmitt filter

Theoretically that is correct, but the true width of a CW carrier is determined
by the purity of the signal. I doubt there are many TX's that can transmit a
perfectly pure signal in any mode.

73,

Tom Childers
Radio Amateur N5GE
Licensed since 1976
QCWA Member 35102
ARRL Life Member

On Sat, 4 Dec 2010 12:16:35 -0500, Monty Shultes <[hidden email]> wrote:

>> Just so someone says it, a CW signal has no width, it's carrier only. The K3 does a fine job of generating it with a 2.7 or 2.8khz filt
>
>> Monty K2DLJ
>>
>>
>>>> Kind of like narrow filters for CW.
>>>>
>>>> I am still new to this hobby and it was just an idea that I had. Maybe when I finish going thru the extra study guide over the holidays I will know the theory if this was a bad random thought.
>>>>
>>>> Thanks
>>>> Don
>>>> KD8NNU
>>>>
>>
>
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Amateur Radio Operator N5GE

KK7P

CW Bandwidth (was: K3 transmitt filter)

Look at the Tx spectrum plots of radios tested by ARRL.
CW signals defintely occupy bandwidth! If they didn't,
we could do CW contests in a 10 Hz slice of each band and
accommodate every Amateur on earth with no QRM! And have
essentially 10 Hz left over!

In fact, a CW signal might have no width if you never
keyed it. The "C" in CW. But in fact we turn the carrier
on and off, and this creates sidebands. If the signal
on/off keying is properly shaped, these sidebands will
occupy little spectrum. And, if you think about it, the
total occupied spectrum width will be very nearly the same
no matter your keying speed up to the limit allowed by the
"waveshaping." If the waveshaping, for example, is a
raised cosine and you key it at a rate such that an
element just allows the complete waveshaping to occur, you
will have a 100% modulated AM signal.

So, if the waveshaping were, say, a raised cosine of 5
msec, and you keyed on for 5 msec, then off for 5 msec,
then on... you'd be generating a AM signal 100% modulated
with a 100 Hz tone. You occupied bandwidth would then be
200 Hz. Thus, to a first approximation, 5 msec
waveshaping will result in a signal at least 200 Hz wide.
As you reduce the keying speed (increase the length of
the elements, or the "dwell time" for carrier on and
carrier off, the energy in the sidebands will be reduced
and the energy in the carrier will be increased. But the
signal will still "occupy" 200 Hz of spectrum.

You can get narrow spectrum at these lower keying rates by
doing more complex waveshaping, but in the extreme (look
at a sinc function) you'd need to modulate the carrier
BEFORE the key is closed and for a similar amount of time
AFTER the key is opened. This translates to latency,
since the DSP can't look ahead in time to see what you
will be doing in the future (if it could, I'd apply THAT
signal processing algorithm to the stock market!).

We're restricted to filtering that attempts to minimize
occupied bandwidth while operating with absolute minimum
delays/latencies. We CAN apply this sort of waveshaping
to other digital data modes -- see, for example, James
Miller, G3RUH, classic article on "The Shape of Bits to
Come" <
URL:http://www.amsat.org/amsat/articles/g3ruh/108.html >
and think about CW rather than PSK while reading it.

We'll never get to zero bandwidth if we want to convey
information. I suppose if Shannon hadn't written his
paper, we'd be able to communicate with zero bandwidth,
but he did and now we can't... :-)

Enjoy!

73,

Lyle KK7P

> ... a CW signal has no width,
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P.B. Christensen

Re: CW Bandwidth (was: K3 transmitt filter)

> 'And, if you think about it, the total occupied spectrum width will be
> very nearly the same
no matter your keying speed up to the limit allowed by the "waveshaping." '

Lyle,

I think you've just hit upon the core of the argument often seen between two
camps (See eHam article archives for examples):

Camp 1 says CW bandwidth is a function of only the shape of keyed envelope,
rise/fall time, etc. They vary their keyers between 5 and 60 WPM and of
course keying bandwidth stays the same because any increase in speed at
these rates is masked by the larger waveform shape component. There's a
time component to measuring bandwidth too.

Camp 2 (usually the Fourier experts), emphatically state that it is only the
keying rate that determines bandwidth. Who is correct?

Both are correct.

At normal CW keying speeds for the ear, it's the waveform shape component
that determines bandwidth. But as the keying speed increases, something has
to give. When? Probably as we near the reciprocal of the time/fall time.
So, assuming that the rise/fall is symmetrical at 5 msec, that comes to a
switch rate of about 200 times per second. If 25 Hz equates to roughly
60WPM (from W8XR's paper) , then as an approximate gauge, let's call 200 Hz
equal to approximately 480 WPM. If we wanted to send faster, then the
slope of the waveform will have to start becoming more vertical in order to
keep the shape (e.g., Gaussian) of the waveform constant. As the rise/fall
becomes more vertical, it must consume more bandwidth, even if it follows a
Gaussian curve at say 1 GHz. So, faster keying produces additional
bandwidth. And so do changes to the rise/fall and keyed envelope shape.

I've heard some folks say that a raised cosine does not consume bandwidth.
Any amplitude-moving continuous wave will produce some bandwidth. What's
probably important in the raised cosine context is that by making the on/off
transitions follow the slope of a sinusoidal curve, the keying bandwidth is
minimized, but not eliminated. If the keying were allowed to completely
follow the curve without stopping (e.g., the flat top portion of the CW
waveform), then as you noted we have a continuous wave and no bandwidth.

Paul, W9AC

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Jan Erik Holm

Re: CW Bandwidth

For a CW signal with clicks at the break side (like the majority)
the faster you key the stronger and wider will the click bandwith
be. This can clearly be heard every day on the ham bands.
So this calls for Camp 2.

However I do not agree that it´s either 1 or 2, in practice it works
differently. I have explained before in a previous mail.

/James SM2EKM
------------------
On 2010-12-04 21:57, Paul Christensen wrote:

>> 'And, if you think about it, the total occupied spectrum width will be
>> very nearly the same
> no matter your keying speed up to the limit allowed by the "waveshaping." '
>
> Lyle,
>
> I think you've just hit upon the core of the argument often seen between two
> camps (See eHam article archives for examples):
>
> Camp 1 says CW bandwidth is a function of only the shape of keyed envelope,
> rise/fall time, etc. They vary their keyers between 5 and 60 WPM and of
> course keying bandwidth stays the same because any increase in speed at
> these rates is masked by the larger waveform shape component. There's a
> time component to measuring bandwidth too.
>
> Camp 2 (usually the Fourier experts), emphatically state that it is only the
> keying rate that determines bandwidth. Who is correct?
>
> Both are correct.
>
> At normal CW keying speeds for the ear, it's the waveform shape component
> that determines bandwidth. But as the keying speed increases, something has
> to give. When? Probably as we near the reciprocal of the time/fall time.
> So, assuming that the rise/fall is symmetrical at 5 msec, that comes to a
> switch rate of about 200 times per second. If 25 Hz equates to roughly
> 60WPM (from W8XR's paper) , then as an approximate gauge, let's call 200 Hz
> equal to approximately 480 WPM. If we wanted to send faster, then the
> slope of the waveform will have to start becoming more vertical in order to
> keep the shape (e.g., Gaussian) of the waveform constant. As the rise/fall
> becomes more vertical, it must consume more bandwidth, even if it follows a
> Gaussian curve at say 1 GHz. So, faster keying produces additional
> bandwidth. And so do changes to the rise/fall and keyed envelope shape.
>
> I've heard some folks say that a raised cosine does not consume bandwidth.
> Any amplitude-moving continuous wave will produce some bandwidth. What's
> probably important in the raised cosine context is that by making the on/off
> transitions follow the slope of a sinusoidal curve, the keying bandwidth is
> minimized, but not eliminated. If the keying were allowed to completely
> follow the curve without stopping (e.g., the flat top portion of the CW
> waveform), then as you noted we have a continuous wave and no bandwidth.
>
> Paul, W9AC

S Sacco

Re: K3 transmitt filter

In reply to this post by N5GE

More on this here: http://www.w8ji.com/occupied_bw_of_cw.htm

On Sat, Dec 4, 2010 at 2:26 PM, Amateur Radio Operator N5GE
<[hidden email]> wrote:
> Theoretically that is correct, but the true width of a CW carrier is determined
> by the purity of the signal.
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