3

If I am using a channel only for morse code (CW), with one frequency and interruptions for dots and dashes, is the necessary bandwidth of the channel twice the audio frequency of the tone used for CW?

Ken - Enough about Monica
  • 181
  • 1
  • 8
  • CW uses a local beat frequency oscillator to generate the audio frequency, so that doesn't really affect the bandwidth. I'm not sure technically the minimum bandwidth required, but it won't be related to the audio frequency you hear. – PeterJ Jul 21 '14 at 15:59
  • The bandwidth is, as, always determined by the modulation, in this case the transmitter keying, rather than the carrier frequency. The ARRL handbook is a good practical reference - if I recall, there's an article on the bandwidth implications of "hard" vs "soft" transmitter pulse shaping. Crystal receive filters a few hundred HZ wide were not uncommon before the DSP era granted continuous variability. – Chris Stratton Jul 21 '14 at 16:06
  • Is the "channel" a piece of wire, or is it an RF radio wave channel that you will modulate at the audio frequency? – George Herold Jul 21 '14 at 16:28
  • That channel is RF – Ken - Enough about Monica Jul 21 '14 at 18:51
  • 2
    You only need twice the maximum audio frequency if you are transmitting audio and you want to recover sine waves at the maximum audio frequency. Nyquist-Shannon theorem. None of that applies in this case. – user207421 Jul 21 '14 at 23:01
  • See https://ham.stackexchange.com/questions/1412/bandwidth-of-a-cw-signal for a more complete answer. – hotpaw2 Oct 13 '17 at 04:16

2 Answers2

4

If I pushed my doorbell button I'd hear a tone coming from the thing that makes the sound but that sound is not coming from the doorbell switch - it's on-off dc coming from the doorbell switch (because it takes two batteries to power it). The tone is created inside the wall thing - it recognizes the doorbell switch and activates an oscillator that sounds.

Similarly CW morse code works with a carrier that is turned on and off by the switch - the carrier is not the tone you hear but the carrier is (or can be) used to gate a tone generator at the receiver - the carrier can also be used to create the tone more directly by mixing it (at the receiver) with a frequency of similar frequency - this creates a difference frequency that can be audible (even though both received carrier and local oscillator are not).

The bandwidth of the channel therefore only needs to be a few hertz - enough to be able to recognize a dot from a dash at high transmission speed. However, if your morse code modulator is a bit crap it may generate a much wider bandwidth that can interfere with other stuff tuned close by.

Andy aka
  • 456,226
  • 28
  • 367
  • 807
  • Almost correct, except the sufficiently over 2X the dot rate is more than a few Hz for “high speed” (QRQ) CW. – hotpaw2 Oct 13 '17 at 04:11
  • @hotpaw2 thanks for the down vote and thanks for not really explaining what your point is or was. – Andy aka Oct 13 '17 at 07:46
3

Yes. The rules of aliasing still apply. "Interruptions", by the way, impact the frequency content of the tone. The resulting spectrum is the frequency of the tone CONVOLVED in frequency space with the windowing pulses (windowing multiplication in time maps to convolution in frequency), so your frequency content is fairly higher than that of your tone.

As Chris points out, the bandwidth, as opposed to the spectrum, is a function of the pulses (though that bandwidth will be displaced by the tone frequency, the carrier frequency).

The transform of a pulse, though, is going to have frequency content that is mainly at the frequency of the pulse, but there is energy at high frequency at the fast transitions (see http://faculty.kfupm.edu.sa/EE/muqaibel/Courses/EE207%20Signals%20and%20Systems/notes/FourierS/Figure%203_11%20Spectra%20for%20periodic%20pulse%20train.jpg) -- it takes the form of sin(pi*x)/(pi*x), and has "side lobes" that can go on for quite some time in frequency.

Scott Seidman
  • 29,939
  • 4
  • 44
  • 110
  • No, the bandwidth is determined entirely by the modulation (the keying), NOT the carrier. – Chris Stratton Jul 21 '14 at 15:58
  • Why is the frequency content higher than the tone, when the tone is, maybe 1500 hz, and the interruptions happen maybe 4 Hz? – Ken - Enough about Monica Jul 21 '14 at 16:00
  • 1
    @ChrisStratton that is exactly what Scott is saying, it's just a more mathematically complete version. A sinusoid of finite length has finite BW, a Sinusoid that has been running from -'ve infinity to +'ve infinity has infinitesimal BW. – placeholder Jul 21 '14 at 16:11
  • No, this answer is fundamentally mistaken as it includes the frequency of the tone, ie, the carrier. But the channel bandwidth includes only the variation not the base. The conclusion stated in the last sentence is dead wrong. – Chris Stratton Jul 21 '14 at 16:12
  • @ChrisStratton, doesn't make a difference which is the carrier, your spectrum of f(t)g(t) is F(s)*G(s). It's a transform pair! See http://www.mechmat.ethz.ch/Lectures/tables.pdf, or any other good table of pairs. – Scott Seidman Jul 21 '14 at 16:13
  • 2
    The question doesn't ask about the spectrum, it asks about the bandwidth. And the answer to that stated here is dead wrong, as the carrier frequency is irrelevant to the bandwidth. What people who've never communicated via CW forget is that the tone you hear is simply the RF carrier mixed to an audible IF. – Chris Stratton Jul 21 '14 at 16:14
  • @ChrisStratton -- are you sure about this?? Perhaps we have a misunderstanding, but I'm trying to directly describe the properties of the multiplication transform pair. If my language is wrong, perhaps you can help me fix it (but, f(t)g(t) transforms to F(s)*G(s), no doubt about that). – Scott Seidman Jul 21 '14 at 16:16
  • Step back from the misapplied math and think about it practically. If I key the transmitter once per second, what is the bandwidth? If I key it ten times per second, what is bandwidth? (Obviously it depends on the shape of the keying transients too, but in general). Now, if I move from the 40 meter band to the 15 meter band, does the bandwidth change? Obviously not, as the carrier frequency is irrelevant to the bandwidth. – Chris Stratton Jul 21 '14 at 16:18
  • @Chris, right you are about bandwidth vs spectrum, but the question relates to ALIASING as well, to some extent. I see your issue now, and I'll edit – Scott Seidman Jul 21 '14 at 16:19
  • No, it is not about aliasing at all. Aliasing and imaging only come into it in certain receiver implementations, but are private to that receiver design and not an aspect of the signal. – Chris Stratton Jul 21 '14 at 16:20
  • @ChrisStratton -- "is the necessary bandwidth of the channel twice the audio frequency of the tone" push me right to aliasing concerns, as that's the only thing it can refer to! – Scott Seidman Jul 21 '14 at 16:27
  • 1
    No, that phrase is most likely prompted by the behavior of a trivial double sideband AM modulator. But such is not normally used for CW. When a keyed audio tone does get fed into an audio channel, it is called MCW (modulated CW) and generally only used for secondary purposes on a voice service - code practice, automated ID sequences on repeaters, etc as it is extremely wasteful of spectrum to modulate audio modulation that way and take up an entire voice channel, rather than to modulate the carrier. – Chris Stratton Jul 21 '14 at 16:32
  • I understand there will be frequency content due to the sinc function. How much matters to reconstruct this single frequency signal (or is that another question)? – Ken - Enough about Monica Jul 21 '14 at 19:30