# Tag Info

1

It depends on the circumstances. If the Z0 line is long, or can be any length like a piece of coaxial cable connecting two pieces of test gear, or is required to cover a wide band of frequencies, then each bit of gear should be matched to the cable. If the line is very short and narrowband, like the track on a board connecting an RF chip to an antenna, then ...

1

The equation you have is for a quarter wave transformer. The red line in your formulas are wrong. When at pi/2 (90°) the first formula reduces to this: - $$Z_{IN}=Z_{LINE}\cdot\dfrac{j\cdot Z_{LINE} \tan(90)}{j\cdot Z_{L} \tan(90)}$$ And clearly the TAN(90°) parts and the "j" parts cancel leaving you with this: - $$Z_{IN}=\dfrac{Z_{LINE}^2}{Z_{L}} ... 1 First, $\tan(\pi/2)$ equals to $\infty$, not to zero. Second, the math shown in your question is absolutely incorrect. Because you can't just do the following:$$ \mathrm{ A+j\ B\ C\\ =C \ (\frac{A}{C}+j\ B) }  when C is $\infty$ since $\frac{\infty}{\infty}$ is indefinite. Look at the calculation in your question again. The author cancels out \\$\...

2

First, the speed at which a signal propagates is irrelevant to bandwidth. Fiber optics, which have the highest bandwidths propagate signals at 2/3 the speed of light, but a 100 year old AM radio, cable of sending only a few thousand bits per second sends those bits at the speed of light. If you send a signal over a radio, it will get there first, then after ...

6

because fiber optics use the speed of light rather than the speed of electrons A complete nonsense. Both methods propagate the signal at the speed of light in the particular media used. This can be assumed roughly to be 2/3 of the speed of light in vacuum in either case. Both methods involve moving the electrons in the media somewhat (way below the speed of ...

0

It is not only about the fibre itself, but also about the light source used to transmit the data and the light detector used to receive data. Any part of the system can and will affect the data rate, so the fiber may not be the limitation. Like with any other data transfer system, as long as the transmission medium itself does not distort the sharp ...

4

The question and the slide deck are purple monkey dishwasher. ‘Digital’ transmission, if anything, is less efficient than analog in terms of spectrum usage for conveying an uncompressed baseband signal. Compare for example sending 3kHz audio vs. 64kbps digital for telephony. Clearly, the baseband audio is more efficient. On the other hand, careful encoding ...

7

As a communications engineer (and someone dealing with a lot more digital signal processing than is good), this slides ... weak to wrong. Advantages of Digital transmission We have to be very clear here: when we say "data", we typically mean things that are already digital to begin with. Now, of course, some kind of digital data has an analog ...

3

This is a very interesting question, and I particularly enjoyed reading the comments and see the difference between an academic's attempt to "simplify" a networking concept versus the accurate interpretation of the slide's words by practitioners. So here is the situation - what the slide is trying to say, clearly not very successfully, is that when ...

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