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Author Topic: Pulse width of a CW signal?  (Read 9020 times)
QRPNEW
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Posts: 51




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« on: March 16, 2012, 02:48:43 AM »

Can anyone tell me the pulse width of a CW signal at 50WPM  for dots and dashes in micro seconds?

Thanks



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N3QE
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Posts: 2196




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« Reply #1 on: March 16, 2012, 03:28:03 AM »

Standard definition of WPM in the handbook, is the number of times that the word "PARIS" can be sent in 1 minute. PARIS has 50 "bits" in it (dit is two bits, one for each on and off; dash is 4 bits 3 on and 1 off; intercharacter space is 2 bits if you don't count the off at the end of the previous character; end of word space is 7 bits).

So 50 WPM = 2500 bits in one minute. That works out to 41.7 bits a second which means that each bit is 0.024 seconds.

So a dit at 50 WPM is 0.024 seconds on and 0.024 seconds off
And a dash at 50 WPM is 0.072 seconds on and 0.024 seconds off

For typical keying, the nominal bandwidth is going to be bits per second times 3 or 5. So most of the energy of a 50WPM signal will be in 120 to 200. Depending on keying shape there are lobes way way out from 200Hz though.
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PA0BLAH
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« Reply #2 on: March 16, 2012, 10:29:43 AM »

QRONEW Do you really know what you are asking?

Exactly, as explained, for PLAIN LANGUAGE (I just corrected Palin language)
1 WPM is 50 signalelements per minute. So 50 wpm is 5/6 times 50 bits per second.

However when you are transmitting random codegroups, each 5 characters long, you lose the advantage of Morse coding, which is that the characters most present in American English language such as E and T, obtain in random text  the same probability of occurrence.

So your question is not OK, and with the information I provide you here, you can answer your own question.

Try it;  will be  very rewarding.

When every question is answered by somebody else, you forget asking how somebody else acquired the knowledge.
The primary source is just thinking, not asking somebody else.
« Last Edit: March 16, 2012, 10:52:39 AM by PA0BLAH » Logged
QRPNEW
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Posts: 51




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« Reply #3 on: March 16, 2012, 01:47:46 PM »

Thanks, I got it.

Standard definition of WPM in the handbook, is the number of times that the word "PARIS" can be sent in 1 minute. PARIS has 50 "bits" in it (dit is two bits, one for each on and off; dash is 4 bits 3 on and 1 off; intercharacter space is 2 bits if you don't count the off at the end of the previous character; end of word space is 7 bits).

So 50 WPM = 2500 bits in one minute. That works out to 41.7 bits a second which means that each bit is 0.024 seconds.

So a dit at 50 WPM is 0.024 seconds on and 0.024 seconds off
And a dash at 50 WPM is 0.072 seconds on and 0.024 seconds off

For typical keying, the nominal bandwidth is going to be bits per second times 3 or 5. So most of the energy of a 50WPM signal will be in 120 to 200. Depending on keying shape there are lobes way way out from 200Hz though.
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QRPNEW
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Posts: 51




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« Reply #4 on: March 16, 2012, 01:56:11 PM »

Yeah, I know what I am asking. Do you understand the question?

If you inject your CW signal into a dummy load and a oscilloscope. You then  run a string of dahs at 50wpm, assuming you have a digital oscilloscope, you should be be able to get a reading of the duty cycle and width of the pulse?
If you vary the CW speed say go slower the pulse width or pulse on time increases. I am not interested in the duty cycle just the pulse width versus speed.
So what is the pulse width in microseconds? Thats the simple  question.

This is how radar works, by controlling the width and pulse repetition rate which increase the resolution etc etc

Thanks OM!

QRONEW Do you really know what you are asking?

Exactly, as explained, for PLAIN LANGUAGE (I just corrected Palin language)
1 WPM is 50 signalelements per minute. So 50 wpm is 5/6 times 50 bits per second.

However when you are transmitting random codegroups, each 5 characters long, you lose the advantage of Morse coding, which is that the characters most present in American English language such as E and T, obtain in random text  the same probability of occurrence.

So your question is not OK, and with the information I provide you here, you can answer your own question.

Try it;  will be  very rewarding.

When every question is answered by somebody else, you forget asking how somebody else acquired the knowledge.
The primary source is just thinking, not asking somebody else.
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PA0BLAH
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Posts: 0




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« Reply #5 on: March 16, 2012, 03:19:57 PM »

Yeah, I know what I am asking. Do you understand the question?

If you inject your CW signal into a dummy load and a oscilloscope. You then  run a string of dahs at 50wpm, assuming you have a digital oscilloscope, you should be be able to get a reading of the duty cycle and width of the pulse?
If you vary the CW speed say go slower the pulse width or pulse on time increases. I am not interested in the duty cycle just the pulse width versus speed.
So what is the pulse width in microseconds? Thats the simple  question.

This is how radar works, by controlling the width and pulse repetition rate which increase the resolution etc etc

Thanks OM!

QRPNEW I am pretty sure, you did not get the point.

With PARIS standard,  the bitrate in bits per second is 5/6 wpm. So when you transmit 50 wpm your dots are 24 millisecond= 24000 microsecond.

However when you don't sent plain text but code groups of random characters A..Z 0..9 grouped in 5 characters separated by a wordspace, then you get with the same number of bits per second a lower number of wpm, in this case the words are codegroeps of 5 characters. The difference is significant. One word than doesn't take 50 bits but 68 bits on the average. Hence with the same bitrate your wpm is factor .73 lower.

And when you will a throughput of 50 codegroups of 5 characters per second your bit is only 17.5 ms .

With a bitrate of 50 wpm PARIS standard your throughput is not 50 codegroups of 5 characters but only nearly 37.

Did you get that point already in the previous posting or right now?

When you start transmitting you are on-off keying a carrier, that is Amplitude modulation, and in order to get reasonable leading and trailing edges, you need to transmit or receive at least up to the fifth harmonic, so the required bandwidth of the signal is 10 times (2 sidebands) the half bitrate. So your 50 wpm PARIS standard requires 208 Hz bandwidth.
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VA7CPC
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« Reply #6 on: March 16, 2012, 11:19:23 PM »

If you can lay hands on a copy of "QST" magazine with a transceiver test, you'll find an oscilloscope trace of a "string of dits" CW signal.  There's a time scale.  They test at 60 wpm.

A dah (the on-time) is three times as long as a dit.

       Charles

PS -- If you can't find a QST, let me know.
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N3QE
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« Reply #7 on: March 17, 2012, 08:19:38 AM »

The primary source is just thinking, not asking somebody else.

You expect someone to figure out what the "standard word" for WPM measurement is, just by thinking? That's like saying that anyone could define ab initio the volt, or the conversion factor from calories into joules, just by thinking.

I've been doing CW for almost half a century now, I have advanced degrees in physics and information theory, and yet I still open the ARRL handbook whenever I want the conversion factor from WPM to baud, or when I've forgotten the "standard word" is that defines the WPM measurement :-).

If like me you don't know the conversion factor by heart, that's fine, but I see no need to lambaste someone on how bad their question is, when in fact they asked a very good question.

Tim.
« Last Edit: March 17, 2012, 08:28:29 AM by N3QE » Logged
PA0BLAH
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Posts: 0




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« Reply #8 on: March 17, 2012, 11:17:57 AM »

The primary source is just thinking, not asking somebody else.

You expect someone to figure out what the "standard word" for WPM measurement is, just by thinking? T

No, but when you know, and that is generally known, that PARIS is the standard word, and you know the Morse code, you can find out without your degrees, what the pulse width is as function of the wpm (in the subject 50 wpm).

AND you can find out using  that pulsewidth , what the thoughput is with random characters, dependent on enclosing the figures 0..9 and also on punctuation.

So, when the questioner get his answer, he has to realise that his 50 wpm gives pulsewidths dependent on the character distribution. I doubted that he realises that.

Not the slightest intention here, to humuliate or whatever. Just wondering why a group of people always ask everything they want to know, and not trying to find out themselve based on the wellknown PARIS in this case.

Best thing should be to answer that PARIS is the standard word, which means that the number of times PARIS can be sent in one minute, is wpm. Rest is deductable knowledge.
« Last Edit: March 17, 2012, 11:21:35 AM by PA0BLAH » Logged
HA7AP
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Posts: 19




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« Reply #9 on: March 25, 2012, 09:09:48 AM »

Can anyone tell me the pulse width of a CW signal at 50WPM  for dots and dashes in micro seconds?

Thanks




[/quote

e.g at 60 WPM dot lenght 1200/60=20msecs. If this is what you need?

73 Imi HA7AP
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W0WCA
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« Reply #10 on: March 26, 2012, 10:09:23 AM »

I sure hope that I can remember to never, ever post anything on this forum!  You are just too quarlsome!
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PA0WV
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Posts: 133




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« Reply #11 on: March 26, 2012, 10:13:28 AM »

I sure hope that I can remember to never, ever post anything on this forum!  You are just too quarlsome!
Extreme short memory? Why did you post this anyway?
« Last Edit: March 26, 2012, 10:15:04 AM by PA0WV » Logged

Using an appliance without CW is just CB
LB3KB
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Posts: 227


WWW

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« Reply #12 on: March 26, 2012, 03:36:40 PM »

the advantage of Morse coding, which is that the characters most present in American English language such as E and T

What advantage are you talking about here ?

I realize that the effective WPM will be lower for random letters than for English text, but I don't see the connection between Morse codes and the most used characters in American English.
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M0LEP
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Posts: 209




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« Reply #13 on: March 27, 2012, 04:20:45 AM »

I don't see the connection between Morse codes and the most used characters in American English.

Here are a couple of random relevant web links about letter frequency in written English:
on oxforddictionaries.com
on wikipedia.org
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N3QE
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Posts: 2196




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« Reply #14 on: March 27, 2012, 09:00:18 AM »

the advantage of Morse coding, which is that the characters most present in American English language such as E and T

What advantage are you talking about here ?

I realize that the effective WPM will be lower for random letters than for English text, but I don't see the connection between Morse codes and the most used characters in American English.

PA0BLAH might imply there's a perfect correlation that does not require any thing other than the application of information theory to figure out.

In fact, it's not a perfect correlation but the many commonly used letters are the shortest and the least commonly used letters are the longest in most Morse type codes (specifically both American Morse and International Morse).

Although the science and mathematics of information theory may not have been formalized until Bell Labs inthe 1920's, certainly in the 19th century the frequency of letters in the English language was well known and in fact of great practical use in say the design and implementation of moveable type printing systems (which actually predate the 19th century) and Morse code. http://en.wikipedia.org/wiki/Letter_frequency

5 of the top 6 Most commonly used letters in American English are, E T A I and N. These are the shortest in International Morse, requiring two dit or dah  elements or less with no more than one dash.

The 4 least commonly used letters in American English are J X Q Z. These are the longest in International Morse, all requiring at 4 dit or dah elements, at least two of witch are dashes.

The one letter that obviously breaks the frequency rule that someone might impose as obvious, is the letter O. It is three dahs, taking quite a long time to send, yet is 4th most common letter in English.

Tim.
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