Decreasing Channel Bandwidth by Using CW Key Clicks

Decreasing Channel Bandwidth by Using CW Key Clicks
The text has previously mentioned that it’s important for the
transmitter not to turn on or off too quickly. For example, C56 in
the Driver Amplifier is used to gradually turn off the transmitter:

In this lecture, we will see that by gradually turning off (and on)
the transmitter, much less bandwidth is required for each CW
“channel.” (A channel is the contiguous frequency spectrum
needed for clear communications.) To understand this, first consider transmitter pulses with no rise or fall time:

This waveform is really the time domain product of the high
frequency carrier
V_{c}( t) = 2cos(?t) V
and a pulse train of frequency f_{k}. From equation (B.22) with V_{m} = 1, the Fourier series expansion of this keying waveform is

Multiplying these last two equations and simplifying gives:

Now, let’s compute the average power contained in each
frequency harmonic. Using P = |V|^{2} / (2R) , normalizing to R = 1? and defining f = f_{c} ± nf_{k} where f_{c} is the carrier frequency:

A plot of this spectrum is shown in .
The FCC requires that for QRP transmitters (those 5 W and
less), the spurious radiation must be ? 30 dB below the carrier
(dBc).
However, for keying transmitters such as the NorCal 40A, there
are many spurious components. While most, or all, of them may
be reduced 30 dB from the carrier, the sum of these may cause a
problem for another person’s receiver.
Consequently, for keying-type “sidebands” a more appropriate
transmitter specification is the channel bandwidth required so
that the average power contained in all frequency components
outside of the channel is 30 dB below the carrier.
This quantity can be easily computed. From (12.31), the total
average power (computed in the frequency domain) is:

where n is an odd and positive integer. The extra factor of 2
accounts for the average power in both the upper and lower
sidebands. Simplifying gives

Imagine that harmonics up to and including n are needed for the
channel. Then p, which we’ll define as the total average power
in HO harmonics relative to the carrier for either the upper or
lower sideband, is

There is no additional factor of 2 here since we’re looking only
at one sideband. If n is large, we can evaluate p with the approximation:

The factor of one-half is present in this expression since we have
only odd harmonics in the keying waveform [see (12.30)].
From (12.35), we find that

This result allows us to approximately compute the number of
harmonics needed for one sideband of a keying waveform so
that the average power contained outside the channel relative to
the carrier is p. For example, imagine we wish to determine the number of keying harmonics (i.e., the channel width) required so that the total average power transmitted outside this channel is 30 dB smaller than that transmitted within the channel. Then,
p = ?30 dBc = 0.001? n =101
using (12.36). If we next assume a 10-Hz keying rhythm, then
the bandwidth needed for this communication channel is
BW=2.101 harmonics .10 Hz /harmo c?2kHz.
This is a pretty large bandwidth and it’s needed if we require
that the keying waveform rise and fall instantaneously. (For
comparison 2 kHz is the BW needed for a SSB voice channel.) Decreasing Channel Bandwidth for CW
The BW required for a CW channel can be greatly reduced
(?10x) by introducing a rise and fall time to the transmitter
keying pulses:

This waveform is the product of carrier and keying waveforms.
However, the Fourier series expansion of this waveform is more
complicated than the one we considered earlier (? = 0).
Your text has a clever method for computing the average power
in the harmonics using the RC network in

From this circuit

where ? = RC and n is the keying-harmonic number.
Your text uses this frequency response as well as the carrier
waveform to solve for the average power outside the channel
relative to the carrier to be

For p = -30 dBc as before

Some commercial transmitters use ? = 3 ms. Then with f_{k} = 10 Hz this gives n ?10 harmonics
Hence,
BW =2.10 harmonics 10 Hz 200/harmonic=200 Hz
This is the BW per channel needed for CW communications
when the keying waveform has rise and fall times equal to 3 ms.
This BW is 10x smaller than without the rise and fall
characteristic. A huge improvement. Key Clicks
The IF Filter BW in the NorCal 40A is approximately 400 Hz.
Consequently, a 200-Hz CW channel can easily pass through
without significant distortion. A roughly 200-Hz channel is common for CW communications. Operators will space themselves a few hundred Hz more than this from other CW “QSO’s” to avoid interference. However, if there is a transmitter turning on and/or off too quickly, operators on nearby frequencies will hear clicking sounds that will interfere with their QSO. These are called key clicks. In Prob. 30, you will measure the output signal produced by your NorCal 40A when it is transmitting. This output will have the smoothed keying waveform shown earlier in this lecture:

Previously in this course, we have characterized the rise and fall
times of waveforms in terms of the time t2. This works only for
exponential waveforms. In the case of non-exponential waveforms, such as the keying waveform above, it is customary to use different definitions of rise and fall times:
• The rise time in terms of t_{10} ?90 : the time it takes themodulated waveform to go from 10% to 90% of its final value, and
• The fall time in terms of t_{90} ?10 : the time it takes the modulated waveform to go from 90% to 10% of its initialvalue.