coaxialelectromagnetismsignalwavewire
I know that cable TV use coaxial cable with alternating current frequencies varying from 5 MHz to 750 MHz.
Why/How a alternating current of, for example, 5 MHz , 10 MHz, 20 MHz, 100 MHz, 300 MHz, 500 MHz, 650 MHz, etc, do not adds ups or cancel each other, since all those current frequencies are passing through the wire at the same time?
I learnt from high school that all waves interfere with each other, and that is the principle of noise cancelling used in headphones.
the RF signals of each frequency are translated into voltage waves that are then superimposed onto one, thereby creating one electrical signal that is sent through the cable.
On the one hand, I think you've got it. On the other, I'm not sure you understand what's going on well enough to get it for the right reasons.
If we want to broadcast an rf signal through space, we have to emit that signal from an antenna. To get the signal from our generating equipment to the antenna we use a transmission line, typically a coaxial cable. The voltage signal on the transmission line excites the antenna, which generates EM waves that travel to the receiving antenna.
At the receiving antenna, the EM waves are delivered to another transmission line (again, typically coax) to get them to the receiving equipment.
Sending signals over coax is just like that, but without the antennas. The generating equipment is just connected to the receiving equipment by the cable. The same voltage signals in the cable that would have excited the antenna, travelled through space, been collected by the other antenna, etc, are just delivered straight to the coax that feeds the receiving equipment.
My point is there's no fundamental difference between the signals in a coax cable and "rf signals". Both are just voltages (or electric fields) varying with time.
And there's no need to "translate" rf signals into voltages. The RF signals already are (or started out as) voltages before they were broadcast as EM waves.
In fact you can also look at the signal in the coaxial cable as an EM wave, but it just happens to be travelling in a dielectric sandwiched between two conductors instead of travelling in free space.
Fortunately, someone more practical than me has fixed the cable, but it made me wonder: how do these cables work?
The very simplest and most basic of answers is there are two conductors and one conductor carries current in one direction while the other conductor carries current in the opposite direction.
But there is a lot more to the pair of wires that make up a cable to consider if you wanted to look into it in detail. For a coax: -
For all two-wire cables there are electric fields and magnetic fields set-up between the two conductors but the beauty about coaxial cables is that these fields, in a proper installation, do not extend outside the perimeter of the coax cable.
So, how does a signal get passed through a coaxial cable?
The signal's energy exists in the gap between the outer and inner conductor and it travels through the cable to the far-end (the load) as an electromagnetic wave. This EM wave carries the power of the signal and it carries the electric field and magnetic fields in a certain ratio. This ratio is known as the characteristic impedance of the cable.
There are also losses due to the resistance of the conductors and these can be significant. There are also losses in the dielectric (the material that seperates the inner and outer conductors) and at higher frequencies this loss can limit the use of a coax cable.
Giving a simple answer to the question is really problematic if all you might know is ohms law but if you are interested there are a lot of things you can look-up on google such as: -
All the above can contribute to signal reflections such as shown below: -
A signal travels from left to right along a perfect coax cable but that coax cable changes to a different characteristic impedance at a position shown by the vertical line. When the signal "hits" that point, some energy is reflected back up the cable whilst some energy continues down to the load.
This answer may be already more complex than you are currently able to cope with so I'll stop at this point.
Best Answer
... All waves on a potential field (say, the electric field in a coax wave guide, or the air) do superimpose, but they do not necessarily destructively interfere with each other.
Instead, model your signals \$s_1\$, \$s_2\$ and so on to be
$$s_n(t) = A_n(t)\cos(2\pi f_n t + \phi_n(t))$$
and \$A_n(t)\$ and \$\phi_n(t)\$ to carry the actual information. We call these the amplitude and phase of the signal, and they can change over time (not much info to get through if they couldn't!). \$f_n\$ is your carrier frequency, i.e. the 300, 500, whatever MHz in your question. In reality, you'll have to assume \$f_n\$ is so high that for a couple of oscillations of the cosine, the amplitude and phase stay pretty much constant.
Now, your over receive signal would be \$r(t)\$,
$$r(t) = \sum\limits_n s_n\text.$$
Then it's quite easy to show that you can re-extract the individual \$s_n\$ without any influence of the other \$s\$. You can, for example, multiply your \$r\$ with a cosine of the one carrier frequency you care about, i.e. with \$x(t) = \cos(2\pi f_n t)\$.
Get out your trigonometric identity table from high school! \$\cos(a)\cdot\cos(b) = \frac12 (\cos(a+b)+\cos(a-b))\$, and thus
\begin{align} s_n(t)\cdot x(t) &= A_n(t)\cos(2\pi f_n t + \phi_n(t))\cdot \cos(2\pi f_n t) \\ &= \frac{A_n(t)}2(\cos(2\pi f_n t + \phi_n(t) + 2\pi f_n t) + \cos(2\pi f_n t + \phi_n(t) - 2\pi f_n t))\\ &= \frac{A_n(t)}2(\cos(4\pi f_n t + \phi_n(t)) + \cos(\phi_n(t)))\\ &= \underbrace{\frac{A_n(t)}2\cos(4\pi f_n t + \phi_n(t))}_\text{at twice the original frequency!} + \underbrace{\frac{A_n(t)}2\cos(\phi_n(t))}_\text{around 0 Hz}\\ \end{align}
We can then just throw a low pass filter at this and get tha \$A_n\cos(\phi_n(t))\$ part alone – which contains all the information we put in there!
All the other frequencies \$f_m, m\ne n\$ don't end up around 0 Hz with this method (but at \$|f_n \pm f_m|\$); so, they'd be filtered away by the same low pass filter.
What we just did was prove that for cosine oscillations, a method exists that can eliminate the presence of all other oscillations. Cosines are hence orthogonal when projected onto cosines in this way.
This is a bit of a very short introduction into basic signal theory. If the math was too much for you, remember this:
Different carrier frequencies are independent. Just like your noise-cancelling headphone can't cancel a 100 Hz tone with a 312 Hz tone, a 50 MHz transmission can without any problem be separated from a 600 MHz transmission.
Example:
\begin{align} f_1&= 5\,\text{MHz}\\ f_2&= 10\,\text{MHz}\\ f_3&= 20\,\text{MHz}\\ f_4&= 100\,\text{MHz}\\ f_5&= 300\,\text{MHz}\\ f_6&= 500\,\text{MHz}\\ f_7&= 650\,\text{MHz} \end{align}
We want the signal at 300 MHz, so we mix with that:
\begin{align} r(t) &= \sum\limits_{n=1}^7 s_n(t)\\ &= \sum\limits_{n=1}^7 A_n(t)\cos(2\pi f_n t + \phi_n(t))\\ r(t)\cdot \cos(2\pi f_5t) &= \text{smth. at } f_1+f_5=305 \text{MHz and at } |f_1-f_5|= 295 \text{ MHz,}\\ &\phantom{=}\text{ at } f_2+f_5=310 \text{MHz and } |f_2-f_5|= 290 \text{ MHz,}\\ &\phantom{=}\text{ at } f_3+f_5=320 \text{, and } |f_3-f_5|= 280 \text{ MHz,}\\ &\phantom{=}\text{ at }f_4+f_5=400 \text{, and } |f_4-f_5|= 200 \text{ MHz,}\\ &\color{blue}{\phantom{=}\text{ at }f_5+f_5=600 \text{, and } |f_5-f_5|= 0 \text{ MHz,}}\\ &\phantom{=}\text{ at }f_6+f_5=800 \text{, and } |f_6-f_5|= 200 \text{ MHz,}\\ &\phantom{=}\text{ at }f_7+f_5=950 \text{, and } |f_7-f_5|= 350 \text{ MHz.}\\ \end{align}
If you now low pass filter this result with say, a filter that removes anything above 80 MHz, nothing survives but what was mixed down to 0 Hz. And that's coming from the frequency you cared about!