[Elecraft] CW Bandwidth

[Jan Erik Holm]

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
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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