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After reading this question I was wondering about another scenario where the signal is not a square wave but a pulse train with short ON time.

As far as I understand, if we have a 3 MHz pulse train signal with 50% duty cycle which is coupled by a 50Ohm coax cable to a scope, we can consider that the 7th harmonic of 3 MHz (21 MHz) is the highest frequency we need to be interested in maintaining. That has a wavelength of 14.28m. So if we exceed this length we need to use 50Ohm termination at scope input.

But how about if we have 3MHz pulse train with 2% duty cycle(ON time 6.6ns ), will the above logic apply to roughly determine the critical length to use 50Ohm termination?

user1999
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  • For 3 MHz, DuCy = 50% your pulses are 50% * 1/3MHz = 167 ns. At 3 MHz, DuCy = 0.2% your pulses are 0.2% * 1/3MHz = 667 ps (not 6.6 ns!). So your pulse is 167 ns / 667ps = 250 times faster (also: 50% / 0.2% = 250) so you will have a signal with a 250x higher bandwidth so your 14.28 m will be divided by 250 so a length of only 57 mm! – Bimpelrekkie Sep 29 '21 at 09:53
  • sorry it would be 2% in question – – user1999 Sep 29 '21 at 10:33
  • The answer Bimpelrekkie scales. – SteveSh Sep 29 '21 at 10:52
  • Take into account that 1 meter of coaxial cable lead to ... time of propagation is 5ns. If no matching, do you know what you would see ? – Antonio51 Sep 29 '21 at 11:02
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    You are right that that it's not the frequency of the square wave that governs the cable length to be concerned about, but wrong that it's some arbitrary Nth harmonic like the 5th. It's the RISETIME of the pulse that's important, if you want to transmit it without distortion or logic-crippling ringing on edges. – Neil_UK Sep 29 '21 at 12:41

1 Answers1

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enter image description here

Picture from here.

So, we are interested in finding \$a_n\$ and, the important part of the above formula is the \$\sin(n\pi d)\$ part because the \$n^{th}\$ harmonic and the duty cycle (d) will effectively make \$\sin(n\pi d)\$ equal to unity i.e. that part of the formula will be at a maximum value at some high harmonic.

In other words, the spectral content of a very low duty cycle pulse will peak at a much higher frequency than 21 MHz and, if you are to retain the pulse shape you need to consider that.

In simple terms, you need to regard the pulse as a series of pulses that look like a square wave. This will be a frequency equivalent to a time base of 13.333 ns (75 MHz). Then you consider harmonics of that: -

enter image description here

The above is the 3 MHz waveform (duty of 2%) and peak magnitude of 5 volts converted to harmonics by microcap.

Andy aka
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