Physical explanation of bandwidth (analog)

[manulal]

Hello friends,

I am finding it a bit difficult to comprehend what the author of a famous textbook has written about TV signal band width. Here is the excerpt from his book. Could any of you please explain what exactly he means? I don’t get why the required bandwidth is around 5.5 MHz.



"Another example of a nonperiodic composite signal is
the signal received by an old-fashioned analog blackand-
white TV. A TV screen is made up of pixels. If we
assume a resolution of 525 × 700, we have 367,500
pixels per screen. If we scan the screen 30 times per
second, this is 367,500 × 30 = 11,025,000 pixels per
second. The worst-case scenario is alternating black and
white pixels. We can send 2 pixels per cycle. Therefore,
we need 11,025,000 / 2 = 5,512,500 cycles per second, or
Hz. The bandwidth needed is 5.5125 MHz."

 

[DiracRules]

I think it's easier if you think of it in terms of cycles (single waveform).

If we scan the screen 30 times per
second, this is 367,500 × 30 = 11,025,000 pixels per
second.

This means you should have a 11,025,000 cycles per second, that is 11.025 MHz.

The worst-case scenario is alternating black and
white pixels. We can send 2 pixels per cycle. Therefore,
we need 11,025,000 / 2 = 5,512,500 cycles per second, or
Hz. The bandwidth needed is 5.5125 MHz."

If we send within a single waveform two signals (2 pixels per cycle), we have a 2 sign per cycle: this halves the number of required cycles to 5,512,500 per second, hence we have 5.5125 MHz.

I'm quite confident about this explanation, despite having studied only digital transmission. The principle here is quite similar. If you are interested in seeing how a single waveform can send two or more "bits" (here, pixels), you can read something about PAM modulation http://en.wikipedia.org/wiki/Pulse-amplitude_modulation"

 

[yungman]

I think it takes two pixels to make one cycle because the worst case is one high and one low and this is consider one cycle. so for 11.0125 mega pixels, the 5.512MHz should be the correct one.

 

[Dickfore]

Suppose you want to represent an analog waveform with N 'values' per second. In general, this sequence has to be sampled N times per second, so the sampling frequency is $f_{s} = 1/N \; \mathrm{Hz}$. But, according to the sampling theorem, the sampling frequency has to be at least twice the highest frequency component (bandwidth). Therefore:
$$f_{s} = 2 f_{b} \Rightarrow f_{b} = \frac{1}{2 \, N} \, \mathrm{Hz}$$

 

[yungman]

I don't think this is sampling. For digital recording, you need twice the pixel speed to sample all the information. For example if you want to digitize a 1MHz sine wave, you need a sample rate of 2MHz to capture all the information. I think this is call Nyquest or something! It been 30 years and I forgot most of it already.but when you display, the fastest speed is only from alternate 1 and 0 which is half the bit rate. They are not related.

I worked 3 years for LeCroy that produce Digital scope designing transient recorders in the early 80s, I know.

 

[DiracRules]

yungman is right, it does not involve sampling, as it is an analog transmission. The trick that halves the required bandwidth is carrying two pixels in each waveform.

In digitalization, since sampling in time means obtaining a periodic waveform in frequencies, to avoid aliasing you need to sample with a frequency that it is at least double than the highest frequency of the signal. This is the Nyquist-Shannon sampling theorem http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem" .

However, this is not the case, since manulal said that the signal is analog.

 

[Dickfore]

yungman is right, it does not involve sampling, as it is an analog transmission. The trick that halves the required bandwidth is carrying two pixels in each waveform.

In digitalization, since sampling in time means obtaining a periodic waveform in frequencies, to avoid aliasing you need to sample with a frequency that it is at least double than the highest frequency of the signal. This is the Nyquist-Shannon sampling theorem http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem" .

However, this is not the case, since manulal said that the signal is analog.

It doesn't matter if the signal is analog or not.

 

[manulal]

Thank you very much to all for your replies.

I think I have made a mistake of not phrasing my question correctly and not stating what exactly the question is for which I seek the answer.

I am a mechanical engineer and I am trying to grasp the basics of electrical engineering on my own at my own pace.

When I reached AC circuits and its applications, I stumbled upon the above quoted paragraph and this doubt started bothering me since then.

I clearly understood the logic behind 11,025,000 pixels per second and since a cycle can have two pixels 5,512,500 cycles per second or 5.5125 MHz.

So, if I have understood correctly any signal with a frequency exceeding 5.5125 or let us say 6 MHz can theoretically contain all the information required for the TV transmission in the worst case scenario.

Now, let me mention an example.

North American channel 2 occupies the spectrum from 54 MHz to 60 MHz. A bandwidth of 6 MHz.

My question is, provided that any frequency more than 6 MHz can contain the all information in the TV transmission, why can’t Channel 2 use a particular frequency above 6MHz, say 57 MHz, instead of using the spectrum from 54 to 60 MHz?

I am sure that the reason for my doubt is my ignorance about electronic communication fundamentals. But this doubt is pestering me and I would be grateful to anyone kind enough to clear it.

Could you please also suggest some good books on this subject which covers the subject from basics?

 

[yungman]

Thank you very much to all for your replies.


North American channel 2 occupies the spectrum from 54 MHz to 60 MHz. A bandwidth of 6 MHz.

My question is, provided that any frequency more than 6 MHz can contain the all information in the TV transmission, why can’t Channel 2 use a particular frequency above 6MHz, say 57 MHz, instead of using the spectrum from 54 to 60 MHz?

Could you please also suggest some good books on this subject which covers the subject from basics?

You should re-phrase "the BW is more than 6MHz".

I don't understand your second question:

why can’t Channel 2 use a particular frequency above 6MHz, say 57 MHz, instead of using the spectrum from 54 to 60 MHz?[End quote]

If the Ch2 center at 57MHz, and BW is 6MHz, then the spectrum of CH2 is 57+/-3MHz which is 54 to 60 MHz! that is if I guess your question correctly.

 

[manulal]

You should re-phrase "the BW is more than 6MHz".

I don't understand your second question:

why can’t Channel 2 use a particular frequency above 6MHz, say 57 MHz, instead of using the spectrum from 54 to 60 MHz?[End quote]

If the Ch2 center at 57MHz, and BW is 6MHz, then the spectrum of CH2 is 57+/-3MHz which is 54 to 60 MHz! that is if I guess your question correctly.

Thanks again yungman.

My question is why should channel 2 use the spectrum from 54 to 60 Mhz and not simply (and only) 57 MHz. as per the excerpt I pasted in the beginning, any signal with more than 6 million cycles per second should be able to contain the complete information required to refresh the picture 30 times a second. 57 MHz is far more than 6 MHz. So why can’t CH 2 use only 57 MHz? This is what my question was.

 

[DiracRules]

@Dickfore: Shannon-Nyquist theorem applies to digital signals, or better to system of transmission that implies the digitalization of the signal, since the sampling frequency required to digitalize the signal is responsible for the possible aliasing.
If you are dealing only with analog circuits that do not sample anything, you can't have aliasing, so the shannon-nyquist theorem does not apply.

My question is why should channel 2 use the spectrum from 54 to 60 Mhz and not simply (and only) 57 MHz. as per the excerpt I pasted in the beginning, any signal with more than 6 million cycles per second should be able to contain the complete information required to refresh the picture 30 times a second. 57 MHz is far more than 6 MHz. So why can’t CH 2 use only 57 MHz? This is what my question was.

I'm not sure if the two questions -the one related to the TV signal and the one related to channel 2- are related.
However, channel 2 can't use only 57 MHz because the signal usually is composed of several frequencies, so when you say that its bandwidth is 6MHz you are saying that the distance (in term of frequency) between the lowest and the highest frequency is 6MHz.
Even if the signal has one frequency, the interference and the disturb along the path from source to receiver broadens the band of the signal transmitted.

Thus, channel 2 must use 6MHz to transmit its signal, but it has no limitation about the central frequency: it could be 20-26MHz as well as 30-36MHz.

However, I am sure there are other reason for the band being around 57MHz, but either I didn't study them, or I can't remember :D

 

[Dickfore]

@Dickfore: Shannon-Nyquist theorem applies to digital signals, or better to system of transmission that implies the digitalization of the signal, since the sampling frequency required to digitalize the signal is responsible for the possible aliasing.
If you are dealing only with analog circuits that do not sample anything, you can't have aliasing, so the shannon-nyquist theorem does not apply.

And what kind of signal is the one representing $N$ 'black' or 'white' pixels per second?

 

[yungman]

You should re-phrase "the BW is more than 6MHz".

Thanks again yungman.

My question is why should channel 2 use the spectrum from 54 to 60 Mhz and not simply (and only) 57 MHz. as per the excerpt I pasted in the beginning, any signal with more than 6 million cycles per second should be able to contain the complete information required to refresh the picture 30 times a second. 57 MHz is far more than 6 MHz. So why can’t CH 2 use only 57 MHz? This is what my question was.

I think you and confused between BW(bandwidth) and frequency. 6MHz is the BW, not the frequency of the CH2. For FM(frequency modulation), the frequency vary a little when modulated with the tv signal. If CH2 is 57MHz AND we know the BW has to be 6MHz, when the tv signal modulate ( add into the center frequency of 57MHz of CH2), the resulting signal will vary between 54MHz to 60MHz.

I am going to explain in a very simple way which is not FM modulation, but the idea is similar:

Say your tv signal is from 0hz to 6MHz, so your middle frequency is 3MHz. Let say we make the center frequency at 3MHz, so if the tv signal is 0Hz, then it become -3MHz. If the signal is 6MHz, then it become +3MHz. All I am doing is putting an offset into this to change from (0 to 6MHz) to (-3MHz to +3MHz).

Now if you add the signal to the 57MHz, then the resulting signal will vary between 57+/-3MHz. That is the reason the CH2 cover between 54 to 60MHz.


Remember the above is not FM modulation. You are not EE, you just want to know why, so I give you the reason why. FM signal is the same, you modulate a 6MHz BW signal to a carrier frequency, you make the resulting frequency vary +/-3MHz from the center frequency. This is true for AM or any other modulation.

 

[DiracRules]

And what kind of signal is the one representing $N$ 'black' or 'white' pixels per second?

Well, I'm not a technician, I think an oversimplified example could be
$-|\sin(t/2)|$ if the pixel is white
$|\sin(t/2)|$ if the pixel is black

or something like this.
This doesn't need any sampling of the signal. The waveform source generate an analog sine and then you just make it positive or negative (i'm not EE, so i don't know actually how).

It's just that Shannon theorem applies to digital signals, so if you are dealing with analog signals it means nothing applying it.

 

[Dickfore]

Well, I'm not a technician, I think an oversimplified example could be
$-|\sin(t/2)|$ if the pixel is white
$|\sin(t/2)|$ if the pixel is black

or something like this.

And what is the value of $t$?

 

[DiracRules]

t is a continuous variable

 

[Dickfore]

So, how does one distunguish between, let's say, the 1st and the 3rd white pixel?

 

[AlephZero]

So, how does one distunguish between, let's say, the 1st and the 3rd white pixel?

In an analog TV system, there aren't any real "pixels". The electron beam just scans across the screen and its brightness varies as it goes along, following the continuous analog signal.

(To keep things simple, let's ignore practical details like how the TV set "knows" where each line of the picture starts, etc.)

However, since there are a fixed number of vertical lines (525 for the US system) and the proportions of the screen are 4 wide to 3 high, it wouldn't make much sense for one horizontal line to display more than about 525 x 4/3 = 700 separate "pixels", because there is no sense having a higher resolution of the picture horizontally than vertically. That's where the idea of the "525 x 700 = 367500 pixels in the screen image" comes from.

 

[Dickfore]

In an analog TV system, there aren't any real "pixels". The electron beam just scans across the screen and its brightness varies as it goes along, following the continuous analog signal.

(To keep things simple, let's ignore practical details like how the TV set "knows" where each line of the picture starts, etc.)

However, since there are a fixed number of vertical lines (525 for the US system) and the proportions of the screen are 4 wide to 3 high, it wouldn't make much sense for one horizontal line to display more than about 525 x 4/3 = 700 separate "pixels", because there is no sense having a higher resolution of the picture horizontally than vertically. That's where the idea of the "525 x 700 = 367500 pixels in the screen image" comes from.

This does not answer my question. What you are calculating has been done in the first post.

 

[yungman]

For pure analog tv, there is really no pixels to talk about, it is the frequency components we are talking about. You don't distinguish between pixels. You really care about the transient response of the tv reciever circuits being able to distinguish the transient ( as small edge detail ) on the screen.

It is not the pixels that matter, it's the ability of the tv to resolve the highest frequency component of the signal that is important. That is where the bandwidth come it. From the original example, the BW is 6MHz, so the tv circuit has to have the BW of 6MHz to resolve the most detail signal of the channel.

 

[Dickfore]

For pure analog tv, there is really no pixels to talk about, it is the frequency components we are talking about. You don't distinguish between pixels. You really care about the transient response of the tv reciever circuits being able to distinguish the transient ( as small edge detail ) on the screen.

It is not the pixels that matter, it's the ability of the tv to resolve the highest frequency component of the signal that is important. That is where the bandwidth come it. From the original example, the BW is 6MHz, so the tv circuit has to have the BW of 6MHz to resolve the most detail signal of the channel.

Exactly, and the point is how we had calculated that bandwidth.

 

[yungman]

Exactly, and the point is how we had calculated that bandwidth.

I think the BW is given by the spec of the CH2. It is really not up to me and you or the tv. In a way, it is the standard set by FCC to limit the side band. The wider the BW, the wider the side band. You can make the picture even nicer but making the BW at 8MHz, the problem is then CH2 will cover from 53 to 61MHz, then FCC can only fit one channel in every 8MHz spacing and therefore less channel can be fitted into a given band.

It is all a standard set by FCC and everybody follow. You don't exactly calculate it, you are given it!

 

[Zryn]

http://en.wikipedia.org/wiki/Frequency_allocation"

Government regulating bodies give you a center frequency and a bandwidth, keeping in mind there is also usually a small separation between each Channel. Its the companies job to fit as much information into their allotted bandwidth as their money and technology allows for.

 

[yungman]

Yep, you are given the BW restriction and you are not allow to go over it. That is the reason people come up with different method of modulation to incorporate more info in the given BW...like the QAM 128 that you put a lot more info than just the frequency number by I Q modulation. It is done by constellation in a 2D plane and by location, you can get like 128 different info per each bit( or whatever you call it). All these to conform to the given BW.