[CW] Speed vs Bandwidth
David J. Ring, Jr.
[email protected]
Sun, 14 Mar 2004 21:04:18 -0500
Ken,
Thanks for the message - which I include below.
You're right! But there is more!
First if we just key a carrier off/on with no shaping, we get a square
wave - which has lots of harmonics and this makes the signal wider.
But ALSO the speed of the interuptions (keying morse speed) makes the signal
wider. That's why pulsed radar is so wide.
I couldn't find the formula for bandwidth related to morse keying speed, but
I remembered that Kai KE4PT had written a book about digital modes
(including CW) and had some information about bandwidth vs. code speed. I
wrote him and asked if I could post his answer on this reflector. Kai is in
Orlando, FL attending the IEEE802 but he emailed me with the needed
information and formula.
>From Kai, KE4PT:
Lets look at a 'DIT' sequence at 25 WPM. That will produce an ON-OFF
sequence of pulses at a 20 Hz rate - the occupied BW will be a sin(x)/x with
20 Hz between the first nulls. Filter this a little and we can call it 20 Hz
BW for 25 WPM, or 0.4 Hz per WPM.
The bandwdith is proportional to the keying rate.. BW=(WPM)x(0.4) Hz for a
well filtered CW signal. Is that what you wanted?
Now most people listen in a 3-500 Hz BW or something like that no matter
what the keying speed. Such a BW could easily accomodate CW up to hundreds
of WPM!! But the actual OCCUPIED bandwidth is still WPM)x(0.4) Hz . Maybe
that is where the confusion is!
73
Kai
I think Kai made in error in stating that the occupied bandwidth is 0.4 Hz
per WPM because if I divide 20 Hz by 25 wpm, I get 0.8 wpm.
60 wpm produces 25 dots and 25 inter-element spaces (50 total code elements)
in sixty seconds. Each dot or each inter-element space occupies 1/50 of a
second or 20 ms.
The "period" of the pulse at 60 wpm is equal to 20 ms, therefore the
frequency of the pulse with is the inverse of the period is 50 Hz. The
frequency of the pulse would be 0.834 times the speed in WPM, so that at 25
wpm the pulse would be 20.8 Hz.
This agrees with Kai's figure, so I believe Kai was right about the
bandwidth for 25 wpm, but made a error in stating the conversion factor by a
factor of two. BW=(WPM)x(0.4) Hz . I believe it should be times 0.8.
This would make 25 wpm have a 20 Hz bandwidth which is what he said.
If Kai made an error by understating the ratio by a factor of two, and
giving me the correct bandwidth for 25 wpm (as I believe) then the figures
for various bandwidths of CW at 0.8 Hz per WPM would be:
8 Hz -------- 10 wpm
24 Hz -------- 30 wpm
48 Hz -------- 60 wpm
I don't think this answers the whole question because there is a difference
between it seems there is a difference between occupied bandwidth and
"necessary" bandwidth. I find the "
Looking futher into the ITU and FCC I find the formual for "necessary"
bandwidth which takes into account the type of circuit - fading or
non-fading and I find the formula:
Morse (CW) - Typical Bandwidth
Formula Bn = BK
where:
Bn = necessary bandwidth (Hz)
B = telegraph speed (Bauds)
K = (a factor that depends on type of emission and level of allowable
distortion) = 3 in a non-fading circuit, and 5 is in a fading circuit.
By definition 60 wpm is the word PARIS repeated SIXTY times in one minute.
If we count the number of code elements in the word PARIS we arrive at this
figure for "signaling elements".
PARIS
P A R
I S
1(I)3(I)3(I)1 L 1(I)3 L 1(I)3(I)1 L 1(I)1 L
1(I)1(I)1 W
Where 1 is a DOT, 3 is a DASH, (I) is an inter-element space (equal to one
DOT in length) which occurs between every dot or dash, and L is a
inter-letter space which is equal in time to three DOTS, and W is a word
space which is equal to 7 DOTS in length.
P XXXXXXXXXXX 11 code length elements.
A XXXXX 5 code length elements.
R XXXXXXX 7 code length elements.
I XXX 3 code length elements.
S XXXXX 5 code length elements.
Units taken up by dots and dashes and inter-element spaces: 31
There are FOUR LETTER SPACES which are 3 units each:
Letter spaces 4 X 3 units each = 12
There is ONE WORD SPACE which is 7 units:
Word space 1 X 7 units each = 7
We add them all up and we get
TOTAL --------------------------------------------- 50 code elements
The signaling rate per second defined as BAUD, thus for 60 wpm we will send
ONE complete word PARIS in one second (one word per second) which is
composed of 50 code elements. 50 code elements in one second is 50 baud.
Thus the signaling rate for 60 wpm Morse (Baud) is 50 code elements/second
or 50 baud.
Also since a dot followed by the inter-element spacing occupies 2 elements,
it follows that if we send a string of dots, we will send 50 code elements
in one second, or 25 dots followed by 25 inter-element spaces.
We can manipulate these numbers to make formulas for code speed.
Code Speed (WPM) = 2.4 times the number of dots per second.
(Check on this multiply 25 dots/second of 60 wpm morse by 2.4).
Dots per second = Code Speed (WPM) divided by 2.4
At 60 wpm the dot length is 20 ms and the intercharacter space is 20 ms.
Baud rate = number of code elements per second (definition)
Baud rate = Code Speed (WPM) times 0.8334
Baud rate = 1.2 times the number of dots per second
(Check on this, 30 wpm times 0.8334 should equal half the 60 wpm baud rate,
or 25 baud.)
Now pluging in different baud rates in the above equation, we get the
bandwidth for different morse code rates:
Bn = BK
where:
Bn = necessary bandwidth (Hz)
B = telegraph speed (Bauds)
K = (a factor that depends on type of emission and level of allowable
distortion) = 3 in a non-fading circuit, and 5 is in a fading circuit.
NOTICE that without the K factor, that Bn (necessary band
WPM Bauds Bandwidth (non-fading circuit) Bandwidth
(fading circuit)
10 8.34 25 Hz
41.67 Hz
20 16.67 50 Hz
83.37 Hz
30 25.00 75 Hz
125.00 Hz
40 33.34 100 Hz
166.67 Hz
50 41.67 125 Hz
208.34 Hz
60 50.00 150 Hz
250.00 Hz
Just to make things both simple and easy (notice that 60 wpm returns the 50
baud speed), lets compute 120 wpm - which is just twice 60 wpm figures.
120 100.0 300 Hz
500.00 Hz
I have a ICOM IC-756PRO with digital i.f. filters and if I use 200 Hz and
the sender is sending 50 wpm or so, I find it difficult to copy, so I open
the bandwidth.
I hope this is helpful to you and answers your questions better.
73
David J. Ring, Jr., N1EA
----- Original Message -----
From: "Ken Brown" <[email protected]>
To: "David J. Ring, Jr." <[email protected]>
Sent: Sunday, March 14, 2004 5:45 PM
Subject: Re: [CW] Speed vs Bandwidth
> Hi David,
>
> I believe what you have said below is not entirely correct in every
> detail. Here is the way I understand it: The bandwidth of a keyed CW
> signal is dependant on the slope of the keying envelope. If the
> transition from signal off to full power signal on was an infinite steep
> slope, that is to say the time from zero power output to full power
> output is zero, then infinite bandwidth would be occupied by the signal
> during the transistion. With this infinitely steep slope the transition
> would take zero time. So there would be a key click that lasts for zero
> time and occupies infinite bandwidth. Of course there are many factors
> which prevent this from ever being acheived as well a lot of good
> reasons that it would not be desireable. The finite bandwidth of RF
> amplifiers, and the antenna system are some of the factors that won't
> let us do this. The fact that there is no good reason to spread our
> signal out over a wider than necessary bandwidth is one of the reasons
> it is not desireable.
>
> So we are limited to a finite rise time of our keying envelope. When we
> key the signal on, the RF output takes some amount of time to transition
> from zero power output to full power output. The slope of this
> transition determines the bandwidth the signal will occupy. When we key
> the signal on there will be a key click that lasts as long as the
> transition time (from zero output to full output) and occupies the
> bandwidth determined by the slope of the envelope. After the transition
> is finished and the output is a constant full power, the bandwidth
> occupied becomes zero.
>
> If the slope of the envelope remains constant the bandwidth will remain
> constant. If you send at a slow speed with this transmitter it will
> occupy the same bandwidth as if you send at a high speed. The key clicks
> will be more frequent at a higher speed, but the bandwidth they occupy
> will be the same. The shorter the envelope rise time is the higher
> maximum keying speed we will be able to use, and the higher bandwidth
> the signal will occupy. If we set up the transmitter to be capable of
> 100 WPM keying speed and only send at 5 WPM we will still be using a
> bandwidth that would be occupied if we were going 100 WPM. We would only
> be occupying the full bandwidth during the keying transistions, and
> during the constant full power output of dits and dahs the bandwidth is
> zero.
>
> If I'm wrong about this, I would like to know what is right. I want to
> understand.
>
> Thanks Ken N6KB
>
>
>
> >If you take a signal and key it at a rate if 5 dots per second and
compare
> >it with a signal keyed at a rate of 800 dots per second, you would notice
a
> >large change in bandwidth.
> >
> >
> >As the frequency of the modulation goes up, the width of the sidebands
goes
> >up. A 60 wpm signal is keyed at 25 dots/second, a 6 wpm signal is keyed
at
> >2.5 dots/second. Since the frequency by which the carrier is changed
> >(modulated) the bandwidth is changed.
> >
> >On SSB (or AM or FM etc...) bandwidth doesn't increase when you speak
> >faster, but if you increase the audio frequencies (modulation
frequencies)
> >you will change the bandwidth, just like on CW when you change the
> >modulation frequency. However having a wider bandwidth does mean you can
> >convey more "intelligence" in a second - even if that intelligence is
more
> >fidelity or multiplexing signals.
> >
> >When you key a CW signal faster, you increase the pulse rate at which the
> >signal is modulated, thus increasing the bandwidth. The higher the
"Pulse
> >Repetition Rate" the wider the signal.
> >
> >
> >
>
>
>