TV communication ???

Discussion in 'Wireless & RF Design' started by Mathematics!, Jul 23, 2008.

1. Mathematics! Thread Starter Senior Member

Jul 21, 2008
1,022
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OK, let's say you have cable TV.
The TV signal's come thru the coaxial cable in mpeg 2 format.
But what I don't get is how can all the channels come into the same cable all at the same time won't their be constructive or destructive interference??? I just don't get how one can seperate the signals from each other.

And my other problem is how the computer knows what channel data we want. If their is a tuner in each TV then how does it filter out the correct
data for the TV.
Note I know the basic's of LC circuit's so spare nothing.

What I am trying to understand is how the tv can seperate the electromagnetic waves for each channel and how the tuner (if any)
works to filter the data.

Last edited: Jul 23, 2008
2. Mathematics! Thread Starter Senior Member

Jul 21, 2008
1,022
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change computer to TV in last post

3. roddefig Active Member

Apr 29, 2008
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The channels are all located at different frequencies. Each channel has a fixed bandwidth so if we space them out far enough they will not interfere with each other. The tuner chooses the channel by mixing the incoming TV signal with a generated sinusoid at the frequency of the selected channel. The selected channel is translated to a set frequency where it can be filtered, thus we only need to have one filter but we can tune many different channels.

4. Mathematics! Thread Starter Senior Member

Jul 21, 2008
1,022
4
YOU SAID
"Each channel has a fixed bandwidth so if we space them out far enough they will not interfere with each other."

I am unclear what bandwidth really is?
I thought bandwidth was the difference in the highest and lowest frequency waves being transmitted.

If this is true then we have w * f = c (where w = wavelength, f = frequency , c = speed of light. )

I know that c is the speed of light in a vaccum but it is approximate.
My point is that if we know the different frequencies then we know
the wavelength of the waves.
But what difference in wavelength do they use so that the waves when super imposed will create mininal interference.

I am finding it kind of hard to see that these channels other then the highest channel won't get lost. When adding up all the wave height's
(using superimposed principle from physics)

Basically if you are transmitting different waves at the same time you add their height's to get the resulting frequency. But then how do you go from the resulting wave back to a specific wave. Mathematically I don't see how this is possible???

Last edited: Jul 23, 2008

Apr 28, 2004
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6. Mathematics! Thread Starter Senior Member

Jul 21, 2008
1,022
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it doesn't go thru an example or how they do it.
It just talks about devices and the kind's of ...etc etc etc.

I was hoping for an example. I am looking into filter theory right now.
Maybe this contains the key.
Anyway So we have this one resulting wave obtained from adding all the wave's together. How can we filter for a given channel once we have the resulting wave.
Does the tuner just have a wave that when add to the resulting wave gives back the channel wave???

I have posted this question all over the place but nobody seems to know the whole picture.

Mathematically the only way to get back the channel wave from the resulting wave is to subtract a wave (or another words add a negative wave)

7. m4yh3m AAC Fanatic!

Apr 28, 2004
186
42
The reason why it's not adding up is you are trying to use Frequency Division Multiplexing -- which is used for analog signals. Cable is now digital and uses Time Division Multiplexing.

Last edited: Jul 23, 2008
8. Mathematics! Thread Starter Senior Member

Jul 21, 2008
1,022
4
YOU SAID
"Cable is now digital and uses Time Division Multiplexing."

But the only difference in digital and anolog signals is that one leads to sign waves where as the other leads to step waves
__ __
_| |__| digital

The only difference between anolog signal's and digital signals is that
digital is less prown to interference.

I will look up Time Division Multiplexing but I would think this just leads me to them talking about the different stuff that make's it up not the actual HOW OR WHY?

9. m4yh3m AAC Fanatic!

Apr 28, 2004
186
42
You are focusing on the wrong aspects. It has little to do with the shape of the signal. It deals with splitting the signal up by frequencies or times. Again, if you want to understand how it works, then you will need to look up Time Division Multiplexing. If you want a deeper understanding, look up how a T1 works. None of the materials I have been exposed to took us down to a component level "This chip performs such-and-such logic functions on so-and-so signals". You will have to get info for a specific multiplexer and find out what kind of chip does the MUXing. Those pages explain concisely how TDM works, and say similar material to whats in my telecommunications encyclopedia and Telecommunications for Managers book.

Apr 28, 2004
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11. Mathematics! Thread Starter Senior Member

Jul 21, 2008
1,022
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Time Division Multiplexing is the process of dividing up one communication time slot into smaller time slots

I am confused I thought the waves get sent all at once but have different frequencies that some how don't interfere with each other that much?

Now I am think it works the way threads do in programming.
Each channel gets a very small timeslice to send data and then it keeps switching channels giving you the illusion of them sending it all at once?

This would make more sense to my sense their is no good way to extract
channel waves from a resulting wave.

They do say in the artical a time slot for a T1 is 1/8,000th of a second.
Do they mean each channel gets 1/8000 of a second to send data before the channel data changes to the next channel data.

Or is this the total time it takes to go thur all 24 channels.

But now where does bandwidth come into play???

I am going in circles now.

12. m4yh3m AAC Fanatic!

Apr 28, 2004
186
42
well, a T1 can carry 24 circuits of 64,000 bps (24 x 60,000 = 1.536 Mbps, the extra 8,000 bps are used for signaling). It can carry one or two digitized voice signals depending on whether it uses PCM or ADPCM. Then it gets increased as you go up in T-Carrier System numbers...

T2 = 6.312 Mbps
T3 = 44.736 Mbps
T4 = 274.176 Mbps

and remember bps is bits per second, not bytes per second (Bps).

Be aware that European carrier systems exist as well. They have 32 64kbps channels for 2.048 Mbps capacity.

As for time slots, I don't know, but it reads as if each channel gets 1/8k

13. Mathematics! Thread Starter Senior Member

Jul 21, 2008
1,022
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So It's Not Senting The Waves At The Same Time.

"it Is Timesliceing It For Each Channel."

Yes Or No ???

Because I Heard From Different People Here And Their That It Is Send All At The Same Time.

Who Is Correct???

14. AlexR Well-Known Member

Jan 16, 2008
735
55
The TV signal for each channel is first digitised i.e. turned into a stream of binary numbers that describe the original image's luminance and color. Then the digitised signal is put through the mpeg encoder to get rid of redundant data and reduce the amount of data that needs to be transmitted. No waveforms are sent over the cable, just a stream of numbers in binary format that describe the original picture.
All the channel digital streams are then time division multiplexed onto the cable with each channel allocated to its own timeslot.

15. roddefig Active Member

Apr 29, 2008
149
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TDMA is a method of allowing different transmissions to use a single medium (cable/frequency/etc.) but it is not used in the transmission of analog television signals.

Bandwidth is the range of frequencies occupied by a signal. For example, our (fictitious) TV signal has a 1 MHz bandwidth and is transmitted at 70 MHz (the carrier frequency). It occupies the band from 69.5 MHz to 70.5 MHz (70.5 - 69.5 = 1 MHz, assuming our TV signal is centered at the carrier frequency, which it should be). We can transmit a second TV signal with the same bandwidth at 71 MHz. Using the same math it will occupy the band from 70.5 MHz to 71.5 MHz and will not interfere with our first signal (in theory, in reality we would want some more space between the two). In communications the distance between two signals is called the "channel spacing."

Edit after reading the other responses:

Digital cable still transmits "channels" at different frequencies, but each of these physical channels is composed of several television channels transmitted using TDMA.

Last edited: Jul 24, 2008
16. roddefig Active Member

Apr 29, 2008
149
0
Nothing is lost...the resulting function is still made up of two cosines. While it may not be obvious looking at the picture on an oscilloscope, the signal is definitely still made up of the two sinusoids.

Example: We have two sinusoids at different frequencies (going off my previous post, 70 and 71 MHz). They are represented as follows

$f(t) = \cos \left ( 2 \pi \cdot 70 \cdot 10^6 \cdot t\right ) + \cos \left (2 \pi \cdot 71 \cdot 10^6 \cdot t \right)$

At the receiver, we wish to tune the 71 MHz signal. Thus, we generate a corresponding carrier signal at 71 MHz

$g(t) = \cos \left (2 \pi \cdot 71 \cdot 10^6 \cdot t \right)$

When we "mix" two signals together we multiply them


\begin{align}
g(t)\cdot f(t) &= \cos \left (2 \pi \cdot 71 \cdot 10^6 \cdot t \right) \cdot \left [\cos \left ( 2 \pi \cdot 70 \cdot 10^6 \cdot t \right ) + \cos \left (2 \pi \cdot 71 \cdot 10^6 \cdot t \right) \right ] \\
&= \cos \left (2 \pi \cdot 71 \cdot 10^6 \cdot t \right) \cdot \cos \left ( 2 \pi \cdot 70 \cdot 10^6 \cdot t \right ) + \cos \left (2 \pi \cdot 71 \cdot 10^6 \cdot t \right) \cdot \cos \left (2 \pi \cdot 71 \cdot 10^6 \cdot t \right)
\end{align}

Using the trig identity for a product of cosines


\begin{align}
g(t) \cdot f(t) &= \frac{\cos \left ( \left (2\pi \cdot 71 \cdot 10^6 - 2\pi \cdot 70 \cdot 10^6 \right ) \cdot t \right ) + \cos \left ( \left (2\pi \cdot 71 \cdot 10^6 + 2\pi \cdot 70 \cdot 10^6 \right ) \cdot t \right )}{2} + \frac{\cos \left ( \left (2\pi \cdot 71 \cdot 10^6 - 2\pi \cdot 71 \cdot 10^6 \right ) \cdot t \right ) + \cos \left ( \left (2\pi \cdot 71 \cdot 10^6 + 2\pi \cdot 71 \cdot 10^6 \right ) \cdot t \right )}{2} \\
&= \frac{1}{2}\, \left [\cos \left (2\pi \cdot 1 \cdot 10^6 \cdot t \right ) + \cos \left (2\pi \cdot 141 \cdot 10^6 \cdot t \right ) + \cos \left (0 \right ) + \cos \left (2\pi \cdot 142 \cdot 10^6 \cdot t \right ) \right ]
\end{align}

So, we are left with a signal at 0 MHz (baseband), 1 MHz, 141 MHz, and 142 MHz. We now use our filter to filter out all the other signals except for the one located at baseband. This is sort of a canned example, however, this is how early radio worked with Morse Code, when the presence and absence of the signal was used to indicate a dot or a dash.

Last edited: Jul 24, 2008
17. loosewire AAC Fanatic!

Apr 25, 2008
1,584
438
You have a lot freq.-a.m. -T.V.-short wave-police-the short answer is
a tuned curcuit that shorted all freq.to ground,except the freq.your

LOOSEWIRE- HANGING ON BY A THREAD
copy right-ALL RIGHTS DESERVED

18. Mathematics! Thread Starter Senior Member

Jul 21, 2008
1,022
4
Wait in your example the f(t) is the resulting wave.
Then why do you multiple g(t)???

QUOTE: When we "mix" two signals together we multiply them???

I thought you add them? (super imposed princple)

And why did you have to go thru the math.
When we had f(t) which was the sum of all the waves.

You could have just said use the filter on f(t).

So I guess i don't know how a filter can filter a given wave from a
resulting wave.

Do you call carrier wave what I call resulting wave?

Either way I get the simple math but I don't see how it makes it easier for the filter to tune into the correct channel.

And I guess I don't see why bandwidth matters so much.
What is the point of spacing the channels out more with more bandwidth
have to do with anything. You still have to extract the correct channel from the resulting wave. Witch I can't see being easier with more bandwidth.

Their are different types of multplexing out their so what does the TV broadcasting companies using frequency division multplexing or time division multplexing.

I get time multiplexing because it is the equivalent of threads in programming where each channel takes a turn. And know to channels go at the same time.
But with frequency multiplexing the all are going at the same time.
Seperation is still to me impossible.
Sorry if I don't get it but thanks for the help.
I would love to know if given a resulting wave if their is some mathematical formula to decompose it back into it's component waves?

The answer to this probably NO. But we can use Fouriers series to approximate it I guess. But all this seems impossible for a filter to compute
having to deal with speed's close to light.

19. m4yh3m AAC Fanatic!

Apr 28, 2004
186
42
I'll answer the few questions I can... keep in mind, you keep asking the same questions and I keep repeating the same answers:

In the USA, almost all cable companies now use digital cable. Look up how digital cable works. You keep bringing up waves. In digital cable, the content is what's important, not the signal. I get the feeling you're still stuck in sine waves, not square waves. If you want to know how the sine wave aspects of transmission works, look up how radio broadcasting works. That seems to coincide more with what you're constantly questioning about. Digital cable is no different than streaming video on the internet. Your digital cable box has a MAC address that talks to the cable company for getting channels. Thats why you can get VOIP, internet, and digital cable all on the same line. They use the same technology that your computer uses.

20. Mathematics! Thread Starter Senior Member

Jul 21, 2008
1,022
4
OK , yes I did want to know about the wave stuff but I will forget it for now.
But with cable we are going thru the same line with computer's , TV, and some times even phone.

Then what is stopping me from watching TV on my computer without a TV tuner card. At least the public channels!

I could just run a wire from the coaxial cable to the audio port of my computer. And write a program to translate it into the correct pixel's.

I guess I would need a tuner to listen in on a channel specific channel.
But even still could I get a mixed picture of a ton of different channels?
Maybe the input voltage's for the audio port and the coaxial cable are way off.

You said the coaxial cable has it's waves as square wave's (i.e digital waves)

But I thought they where still using anolog signals until the end of 2009.
Maybe they are just talking about air wave signals being changed to digital? Because if they had to change it for the cable line then we would need a DAC for are TV's.

Was Coaxial Cable always transmitting digital signal's ??
"Digital Cable" was it ever "Anolog Cable"