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I'm quite new to transmission line theory, so I decided to follow an introduction course online.

I am struggling with understanding the solution to this question.

Here is what the course is giving as an example: enter image description here

Here we see that the source has a operating frequency of f=1/(100 ns), so I don't understand where the 30 ns comes from. Also, I assume that the source in question has only 1 cycle, since we see that the yellow scope goes to 0 eventually.

I tried solving it using a bounce diagram, but I can't seem to find the correct answer. Can someone help me?

Here is my bounce diagram solution:

enter image description here

ocrdu
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Dimitar Zhekov
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3 Answers3

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I don't understand where the 30 ns comes from. Can someone help me?

A typical 50 Ω transmission-line has an inductance per metre of 250 nH and, a capacitance per metre of 100 pF. The characteristic impedance relates to the "per metre" L and C values thus: -

$$Z_0 = \sqrt{\dfrac{L/m}{C/m}} = \sqrt{\dfrac{L}{C}} = \sqrt{\dfrac{250}{0.1}} = 50\text{ }\Omega\hspace{1cm}\text{(just a confirmation)}$$

The speed at which a wave travels in that typical cable is this: -

$$\dfrac{1}{\sqrt{LC}} = \dfrac{10^{9}}{\sqrt{25}} = 200,000,000\text{ m/s}$$

Or, about two thirds the speed of light.

So, if the cable is 6 metres long, a wave takes \$\frac{6}{200,000,000}\$ seconds to reach the end.

That equals 30 ns.

Of course, you can choose a coax cable that is slightly fast or slower and this depends entirely on L and C per metre. I chose the values above knowing that they gave the stated velocity of propagation and Z0. Call it experience.

Little footnote

The speed of light in free-space is defined by the permeability and permittivity of a vacuum. These are also terms measured in inductance per metre and capacitance per metre. Hence, for a vacuum, the speed of light is: -

$$\dfrac{1}{\sqrt{\mu_0 \cdot \epsilon_0}} = \dfrac{1}{\sqrt{4\times \pi \times 10^{-7}\hspace{0.5cm}\times \hspace{0.5cm}8.85418782\times 10^{-12}}} = 299,792,458 \text{ m/s}$$

Or, approximately 300,000,000 m/s. Quote from wiki: -

The speed of light in vacuum, commonly denoted c, is a universal physical constant that is important in many areas of physics. The speed of light c is exactly equal to 299,792,458 metres per second (approximately 300,000 kilometres per second; 186,000 miles per second; 671 million miles per hour)

Andy aka
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  • But the cable is not 30 ns delay for 1 length rather two which is not your fault. – Tony Stewart EE75 Sep 08 '22 at 15:18
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    It says in the question (and I quote) that "the green signal rises 30 ns later". I can't attribute any other meaning than the output responding 30 ns after the input (yellow) rises. – Andy aka Sep 08 '22 at 15:34
  • Yes, I see that. Do you see anything wrong with the plot at all. The pulse shows <3 divisions at 50 ns /div or ~140 ns wide and the schematic reads 100 ns input and the output must also be 100 ns wide . Which is wrong? Or is that incorrect constructive interference? (wrong does not affect pulse width at output)) What's wrong with that picture? – Tony Stewart EE75 Sep 08 '22 at 17:10
  • When you are able to respond I'll tell you. – Andy aka Sep 10 '22 at 12:38
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It takes 30ns for the pulse to travel through the cable from input to output.

Yellow is input, green is output. It is explained in the text below the image.

Justme
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  • Yes I can read that as well. I do not understand why it is 30ns... IF you look at the calculation they made lambda= vp*T lambda = (2*10^8)*100ns lambda = 20m – Dimitar Zhekov Sep 07 '22 at 14:01
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    Because that is the amount of time the signal travels through the cable they used in the measurement. If the speed of signal in the cable is known, then you know the cable length, because you know the time it took to travel. – Justme Sep 07 '22 at 14:02
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    No, they used 30ns which is 6m, the time is not 100ns so cable is not 20m. – Justme Sep 07 '22 at 14:09
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    Just so I understand that, the 30ns is arbitrary... meaning it cannot be deduced by the information given in the circuit digram? – Dimitar Zhekov Sep 07 '22 at 14:24
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    30ns is the measured time it takes for the signal to travel. It just happens to be 30ns for this length and type of cable. – Justme Sep 07 '22 at 14:27
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    Okay thanks, so I need to take it as at "face value" – Dimitar Zhekov Sep 07 '22 at 14:33
  • It is no different from walking between point A and B in a city. If you don't know the distance or the speed, the only thing you can measure is how long a time it took. By measuring the time and estimating what typically the speed is, you can approximate the distance. – Justme Sep 08 '22 at 06:12
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As vp = 2*10^8 m/s ... or 5 ns/m ... cable is RG58 or similar.

Here is a view of your transmission line (Td = 30 ns --> length= 6 m) in the "ideal" case, and in a real configuration with an "equivalent" input impedance of the scope (1 Meg // 15 pF).

I did not use a step voltage with a "rise & fall" time of some ns (generally 3 to 5 ns) for a classic "pulse" generator. This changes slightly the waveforms.
Made with microcap v12, see "Download" web page.

enter image description here

Antonio51
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