# bandwidth etc.

Discussion in 'Mathematics and Physics' started by PG1995, Jul 23, 2013.

1. ### PG1995Active Member

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Hi

Why does a copper wire has relatively low bandwidth? Perhaps, it has something to do inductive and capacitive effects in copper wires; for instance, in a coaxial cable electromagnetic effects are cancelled which means less capacitive and inductive effects. But what about microwaves? They only need free space to travel which means capacitive and inductive effects. What effects limit their bandwidth. The same goes for optical fiber because it also has upper bandwidth cap. Please help me.

Regards
PG

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2. ### JimBWell-Known Member

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Your attachment appears to be something cut from a PowerPoint presentation.
Taken out of context it is misleading at best, and could be considered wrong in some ways.

A coax cable is copper wire. Mechanically different from plain wire, but non the less it is copper.

Let us say we want to transfer data across a distance of 10km.

We could just hang two copper wires from a series of poles, just like the old telephone and telegraph.
The wires would behave like a low-pass filter and high frequencies would be attenuated. If we tried to send data as a series of voltage levels, the transitions between the voltage levels would be slow, our data rate (bandwidth) would be low.
We could try using a modem with a high carrier frequency, this would likely be more successfull but the losses due to radiation from the line would attenuate the signal.

If we used a twisted pair cable, we could probably getup to, or maybe exceed the 1Mhz bandwidth stated. There would be less direct radiation from the cable.

A coax cable would be better, but then there are many different types of coax cable with different losses at high frequencies.
Again the coax behaves like a low pass filter, but depending on the construction of the cable the losses can be low enough that we can send carriers of several Ghz along that cable.
There will be minimal radiation from the cable, so we will not annoy our RF neighbours.
The bandwidth we could achieve would depend on the carrier frequencies and the modulation type we used.
It should certainly be better than 100Mhz.

Microwave, what is meant by microwave in this context?
One definition would be RF at greater than 1Ghz, we could send that down a coax cable.
Or does it mean a microwave radio link?

If we wanted to use a microwave radio link to span our 10km, we may have an RF system with a carrier frequency of 5Ghz.
Does that mean that we have 5Ghz bandwidth for our data? NO WAY!
We may have a channel 50 or 100Mhz wide at 5Ghz, not the whole 5Ghz, other users of the RF spectrum want to use the other 4.9Ghz.

So in terms of usable capacity, a coax cable may have more usable bandwidth than the microwave link.

But 10km of cable is going to cost a LOT of money.
Not just for the cable itself, but for installation as well. Digging a trench to bury the cable, or installing the poles to carry it above ground.
A couple of towers and some radio kit is starting to look cheap in comparison.

Then we come to optical fibre.
Sending light along a strand of glass (usually).
Just because the light frequency is in the TeraHertz region, it does not mean that the laser diodes that send the light can be modulated at Thz rates.

So, as I said, context is very important when considering this problem of bandwidth.

JimB
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3. ### dougy83Well-Known Member

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The inductance and capacitance of cables will affect the bandwidth through attenuation, filtering and consequent reduction in SNR. Parallel copper wires are especially vulnerable to external interference as one wire will likely be more affected than the other. Twisted pair will be less affected as both wires will receive the same interference signal (more or less), which will be removed as common mode noise from the differential signal. Bad termination will also increase interference caused by reflections.

The microwave transmission will have bandwidth limited by the channel width, which will be further reduced by attenuation from signal divergence, multipath propagation, atmospheric conditions and interference from other RF sources.

Fibre optic transmission doesn't suffer from external electromagnetic interference, but does suffer from attenuation.
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4. ### PG1995Active Member

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Thank you, Jim, Doug.

I won't say that I understand everything at this stage because I have just started my communication course.

Could you please also help me with this query. Thanks.

Regards
PG

PS:
Sorry. I forgot to include this. Yes, you are right. It was taken from this presentation.

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5. ### steveBWell-Known MemberMost Helpful Member

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PG,

One point to consider is that the frequency of the carrier is a big part of the potential bandwidth. Optical fiber uses light with frequency in the THz range. Microwave is in the GHz range. Radio is in the kHz-MHz range. Audio is in the 20-20kHz range. Hence, speaking has lower bandwidth and less information than radio which is less than microwave which is then less than optical frequencies.

On your query above, we can say the human voice does not use the entire 20 Hz to 20 kHz range. Even a symphony orchestra does not use this entire range. It is well known from telephone communications research that the human voice is much more limited in bandwidth. And, even when speech is filtered more than this, it is still quite understandable.

Why is the carrier frequency important? Because we modulate the carrier with information and typically the information frequency (which is the baseband signal, by the way) is a fraction of the carrier frequency. For example, you can't speak a microwave frequencies without becoming a microwave transmitter. You can't modulate a microwave carrier at optical frequencies because then you would have an optical (light) source. So you can think of a carrier having a channel bandwidth used for the information.

Why does the information frequency and the carrier frequency need this relation? Well, think of your Fourier transform analysis theory. The act of modulation makes the frequency content of the carrier broader than the narrow carrier itself. A pure carrier has a Fourier transform with impulse spikes at the carrier frequency (the negative frequency too). But a modulated carrier takes up the full channel bandwidth from F-f to F+f, where F is the carrier frequency and f is the information frequency bandwidth.

The transmission medium itself does not fully define the bandwidth limit of the method. You must also consider the technology used to implement the full system. So, optical frequencies might be in the 100s of THz range, but every device used will fall short of this. Optical fiber is transparent with it's lowest loss in the 1500-1600 nm wavelength range (bandwidth 12.5 THz). A communications laser might only have a 1 GHz modulation bandwidth, or an external modulator might get to 50 GHz. Electronic circuits limits will limit attainment of the 50 GHz. Photodiodes and receiver electronics might be limited to 10 GHz. etc. .... Multiple lasers can be used to increase the bandwidth utilization by a factor of 100 or so, but frequency stabilization and multiplexing and demultiplexing devices limit this. So, of this 12.5 GHz, you might be able to use a few THz. You can debate some of these numbers, but the point should be clear. The technology might prevent you from fully utilizing the transmission medium bandwidth.

Why is copper wire itself a low bandwidth medium? Because a copper wire is more like an antenna than a transmission cable. As the wire gets longer and the frequency gets higher, the energy radiates away, rather than being guided along the desired path.

A twisted wire is more like a transmission medium since the twisting acts like a shielding. But, this is limited because at higher frequency the twists are not fast enough and the length between the twists act like antennas. Coax is better because the shielding is better, and energy can't get lost until even higher frequencies. Optical fiber is quite different, and we have fortunate physics that makes the silica glass have low loss and low dispersion.

Dispersion is another issue for bandwidth we haven't mentioned too much, but you can imagine that if the different frequency components of the signal each travel at different speeds, then over a long distance, the signal will be distorted and unrecognizable.
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6. ### JimBWell-Known Member

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For voice communication, most information is carried in the range 300 to 3000hz (ish).
So for a telephone system, or a radio communication system, there is advantage in limiting the frequency range to 3 or 3.5kHz.

An AM (double sideband with carrier) radio system will occupy a bandwidth of twice the highest modulating frequency.
So if the highest modulating frequency is 20kHz, the transmission needs 40kHz.
If we restrict the highest modulating frequency to 3kHz the transmitter bandwidth is only 7kHz. As a result, more transmissions can occupy a given amount of RF spectrum.

This bandwidth reduction can be taken further by using SSB* (Single Side Band Suppressed Carrier), where the required bandwidth is the same as the highest modulating frquency.

* At this point we expect a rant from AudioGuru telling us that SSB sounds like quacking ducks and is muffled and difficult to understand with no high frequencies.

JimB
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7. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

I dont think anyone mentioned distance yet, so i'll do that now.

We have different mechanisms coming into play as the distance the signal travels increases. In copper wires we have resistance for one thing, in radio we have a decrease with the square of the distance, with fiber optics we have a decrease due to the imperfect optical properties of the fiber.
As the distance increases we have more and more of the imperfect medium between us and where we want to send the signal, so it gets worse.

Way back when i had DSL for my internet, the phone company said that in order to use that i had to have copper wires and had to be within a certain distance from the station. Since i am within about 3 city blocks (or something like that) i was able to get DSL and i could get a data rate of 3Mb/s (three megabits per second). That was over copper wires that are normal for telephone. How they got such a high rate i dont know as i never looked into it. But if i was too far away i could not get this.
Now with fiber however i get 10 times that bandwidth, 30Mb/s and it can go higher too.

Also, a while back i was building a switch hub for my USB devices. What we see commonly for sale is a USB hub where you can plug several USB devices into one USB computer port. But what we dont see too often is a hub with switches so you can turn the USB devices off without having to unplug them.
I did find one, but was disappointed to find that it ONLY turned off the +5v power it did not break the data connections which does not help isolate it from the computer port when turned off. I wanted one that would also break the data connections, so i built one. Now to do this required wiring switches into the hub so that it could actually disconnect the USB device completely, and wiring those switches took a certain amount of wire lengths. Using around 1 inch or maybe 1.5 inch of wire per lead, it seemed to work ok but going beyond that it did not want to work right at all where it was not recognized by the computer port software. So in that case the lead length had to be held to some maximum value or else the whole thing just did not work.

As a side note, transmission lines can act as resonant bandpass filters too where the main losses come from resistances in the wires not from any capacitance or inductance, even though there is capacitance and inductance just like any transmission line remember that capacitors and inductors do not dissipate power only their conductive materials do.

As another side note, because twisted pairs wires are so close together they receive external signals nearly the same in both wires so they main advantage is that the external signals appear as common mode signals.

As yet another side note, a coax cable can be made with inner wire with much smaller diameter than a twisted pair and this along with inner insulation thickness keeps capacitance down as low as possible.
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8. ### steveBWell-Known MemberMost Helpful Member

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Maybe I didn't stress it, but I did mention it. Certainly, distance as a key factor with regards to loss and dispersion of the transmission medium. Both effects accumulate and worsen as distance increases. Within limits, the loss and dispersion can be offset by amplifiers and dispersion compensators, but distance is a key factor, and both loss and dispersion will limit the bandwidth more as distance increases. In a digital communications system, loss reduces the energy per pulse and dispersion spreads the pulses so that they overlap. Too much overlap will mix the pulse energies and increase error rate beyond acceptable values and too little pulse energy will fall below the receiver sensitivity, - again increasing error rate unacceptably.
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9. ### PG1995Active Member

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Thank you very much, Steve, Jim, MrAl, for your help, time and patience. @MrAl: I was very happy to see your reply. My special thanks.

I will use Steve's post to make some important follow-on queries. The replies to queries here will also help me to understand Jim's reply here. I have tried my best to state my queries but still it's quite possible that I will sound like a raving lunatic. But I hope you will be bear with me! Thanks.

Q1:
I interpret that you are saying that maximum frequency supported by a medium plays a big part in determining potential bandwidth. I hope I have it correct. Here one should be careful that bandwidth relates to maximum data supported by Shannon's formula. For example, a telephone line normally has a bandwidth of 3000Hz (300 to 3300Hz). Does this mean if frequency is increased to, say, 10kHz then the signal will be affected in such a way communication will be no longer possible? Perhaps, such a high frequency will result in too much losses?

Another factor which is also important to consider is distance traveled by a signal. For example, in telephone line the more distance a signal has to travel the more there will be loss which means deterioration of the signal.

So, it means most of the information of human voice is carries in frequency range 300 to 3300Hz range and if other frequencies are skipped even then voice is understandable. Normal human frequency range is 20Hz to 20000Hz.

So, I believe you are saying that carrier frequency should always be higher than the information frequency (e.g. for voice information frequency is 300Hz-3300Hz).

Q2:
Mostly a carrier signal is a pure sinusoidal signal. Fourier series and Fourier transform represent any given signal using sinusoidal signals. If a signal is periodic then Fourier transform looks like this.

You said, "But a modulated carrier takes up the full channel bandwidth from F-f to F+f, where F is the carrier frequency and f is the information frequency bandwidth". For clarification purposes, I would say that f is center frequency. For instance, for a voice signal the center frequency is (300+3300)/2=1800Hz.

I think that's not a right place to ask this. Anyway, suppose we have sine signal of frequency 10 rad/sec. Fourier transform of this signal will give us one spectral line at 10 rad/sec and another at -10 rad/sec. Do I have it correct? Here, -10 rad/sec is negative frequency, right? What is this negative frequency and what is 'physical' interpretation of it?

Q3:
In frequency modulation, amplitude of information signal modifies or modulates the frequency of carrier signal which means higher the amplitude of information signal at some instant the higher the change in carrier signal's frequency at that instant. There is going to be certain upper and lower limit on the amplitude. Let me elaborate on it. Consider a microphone in which resistance changes in response to pressure gradient. Let's say when pressure gradient is 50μPa the amplitude of the modified or modulated current flowing through microphone is 10μA and when pressure gradient is -50μPa the amplitude of the current is -10μA. We can say that even when the pressure gradient is 100μPa the current flowing through microphone will be 10μA because that's the upper limit on current that can pass through microphone.

Again I'm referring to your comment, "But a modulated carrier takes up the full channel bandwidth from F-f to F+f, where F is the carrier frequency and f is the information frequency bandwidth". Once again, I will use telephone line example where the line has a bandwidth of 3000Hz (300 to 3300Hz) and center frequency, f, 1800Hz. I don't understand how you get "the full channel bandwidth from F-f to F+f". From microphone example above the change in frequency of carrier signal is dependent on pressure gradient and not on frequency of words spoken into microphone. Although a microphone, at least carbon microphone, uses DC current, for the sake argument assume that it uses carrier sinusoidal current with frequency 60Hz. Suppose when pressure gradient is 50μPa or more the change in frequency is 20Hz and when the pressure gradient is -50μPa the change is -20Hz, and this gives F-x=60-20=40Hz and F+x=60+20=80Hz. In this calculation you see no role of f which is 1800Hz.

To me, what you say makes sense if we superimpose voice signal which varies between 300Hz and 3300Hz onto some some carrier with frequency, F, of , say, 5000Hz. Then, I think after decomposition we will be able to see what you say, that is channel bandwidth from 3200Hz to 6800Hz. Perhaps, this example might help you to see what I'm trying to say.

This part is relevant to Q1 above.

Regards
PG

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10. ### steveBWell-Known MemberMost Helpful Member

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There is a lot of information and questioning there. It's too hard to hit every point, but I will try to convey the important responses to what you are asking. For the most part, you are getting things right, or at least right enough considering that you are just entering the course now and you are not expected to have a full grasp on this yet. I won't comment on the questions you ask and you have interpreted an answer correctly.

One key point that you hit on is that audio signals, such as that we deal with in an audio amplifier, speakers, microphones and primitive telephone systems use the baseband audio signal directly without modulation. So, when you talk about 300-3300 Hz for a baseband signal, there is indeed no carrier to consider. Or, you can say that the carrier frequency is zero, if you like to visualize it that way. However, when you want to multiplex many voice signals, you would likely modulate different carriers (channels) with the different audio signals, and in this way use frequency division multiplexing of many channels. Old style telephone, analog cable TV, or air transmitted television and radio are examples of having channel frequencies. Because of linearity and superposition (which is often just an approximation) those channels can coexist without interference.

Negative frequency can be interpreted differently in different situations. For now, I would say that you can think of the negative and positive frequency as representing the same thing. When we use complex exponentials, we have exp(jwt) and exp(-jwt) which combine to make the real cosine function. (remember Euler's relation). We are biased to think of positive frequency as a more tangible thing, but really negative and positive are equally valid. Ultimately, the sign relates to rotation (clockwise is negative and counter-clockwise is positive) whether it's a real rotation (like with motors) or an imagined rotations (like with complex vectors in system theory). These negative frequencies can actually manifest themselves in very interesting ways when you get into modulation methods. So, keep an eye out for this. For example, you didn't question the F+f and the F-f frequency band. Doesn't this look like double the bandwidth needed? Later you study upper and lower sidebands, and these are a manifestation of negative frequencies sliding up into the positive frequency domain. Don't expect this kind of thing to makes sense right away. It takes time to absorb it and make sense of it.
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11. ### PG1995Active Member

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modulation

Hi

Suppose we want to transmit baseband voice signal which has bandwidth of 3kHz (300-3300Hz) using a carrier of 700kHz and assigned bandwidth is 500Hz, i.e. we have frequency range from 650-750kHz at our disposal and 700 kHz is center frequency, fc, of carrier signal. Q1: Do I have the information correct up to this point?

Do I need to choose center frequency for baseband signal too so that the signal has equal range of frequencies on both sides of the center frequency? For example, the center frequency for voice signal would be (300+3300)/2=1800Hz. Q2: Do I have it correct?

Assuming what I say above is correct then it would mean that when baseband signal has frequency 3300Hz, the corresponding frequency of carrier signal is 750kHz, right? Q3: What would be the corresponding frequency of the carrier if baseband signal has frequency of 2500Hz or 800Hz?

Thank you for the help.

Regards
PG
12. ### JimBWell-Known Member

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No, you are totally confused and messed up in your head.

I had a look at your questions based on some wikipeedia article about full duplex, half duplex etc, and decided that based on the confusing information in the article it was too difficult to formulate a reply.
Something which adds to the confusion is that practice and terminology are different between radio systems and line based systems, even though the fundamental principles are the same.

So, lets try and clear some confusion.
I am talking about a radio system.

If we want to use a carrier frequency of 700kHz and modulate with an audio signal (300 to 3000kHz), the modulation scheme is AM (Amplitude Modulation).
Assuming that there is no distortion, the modulated signal will occupy 697 to 703 kHz, ie the occupied bandwidth is 6kHz.

Q2 - No

Q3 - The modulation frequency does not change the carrier frequency.
(For the pedantic, yes it could be argued that in FM systems the modulation does change change the carrier frequency, but lets keep it simple for now).

JimB
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13. ### steveBWell-Known MemberMost Helpful Member

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PG,

How did you come up with those numbers? There seems to be no logic to what you presented.

Why would only 500 Hz be assigned to a channel that needed to carry 3300 Hz information? And, why would you then have plus and minus 50 kHz? Was this a typo? Did you mean 50 kHz bandwidth?

Modulation is basically a multiplication operation for the carrier and baseband signals. Think of your math for system theory. Multiplication in the time domain becomes convolution in the frequency domain. This is why the frequencies add to give the upper and lower side bands. Your spectrum will be a lower sideband of 696700-699700 Hz and a carrier spike at 700 kHz and an upper sideband of 700300-703300 Hz.
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14. ### PG1995Active Member

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Thank you, Jim, Steve.

I'm really sorry. I had it wrong from the very start because I had in mind that information signal's frequency modifies the frequency of the carrier. That's totally wrong although I had myself noted in Q3 in post #9 that in frequency modulation, amplitude of information signal modifies or modulates the frequency of carrier signal which means higher the amplitude of information signal at some instant the higher the change in carrier signal's frequency at that instant. But to tell you the truth now I seem to be more confused! I will start by rephrasing the queries in my previous post. But I think you should go through the Q3 in post #9 because it shows how I view frequency modulation in general and perhaps some of the confusion stems from there.

Now coming back to my previous post.

Suppose we want to transmit baseband voice signal which has bandwidth of 3kHz (300-3300Hz) using a carrier of 700kHz. Here, we can say 700kHz is center frequency, fc, of the carrier, and (300+3300)/2=1800Hz=1.8kHz is fc of the baseband signal. The bandwidth of the carrier is going to be (700-1.8)kHz to (700+1.8)kHz or 698.2kHz to 701.8kHz with 700kHz as the center frequency. Q1: Do I have the information correct up to this point?

I believe this is the part where my confusion lies. In frequency modulation amplitude of the information signal modifies the frequency of a carrier signal. It means that even when the two information signals are different but have the same amplitude then they would modulate the carrier the same way. Let me put it differently. If you say the letter 'k' and 'q' in such a way that they both create same pressure gradient (or, amplitude) then both of them modifies the carrier's frequency the same way, and hence two different kinds of signals (letter 'k' and 'q') get mixed up. Q2: Is this possible? If not, then why not? Why can't two different signals give rise to same amplitude?

I understand that convolution in time domain becomes multiplication in frequency domain, i.e. convolution of f(t) and g(t) becomes F(s)*G(s) in frequency domain, and convolution in frequency domain becomes multiplication in time domain, i.e. convolution of F(s) and G(s) becomes f(t)*g(t) in time domain. But I don't see the point where you say "This is why the frequencies add to give the upper and lower side bands". Q3: What are you trying to point out? Could you please clarify it?

This is how I see it. Suppose we have a information signal f(t)=cos(2∏ft) and carrier signal g(t)=cos(2*∏*50*t). In frequency modulation the factor "50" of the carrier is modified by amplitude of the information signal as shown here.

Now I turn to amplitude modulation. In amplitude modulation, the amplitude of the carrier is modified by the amplitude of an information signal and the frequency of the carrier remains constant. This means that in case of amplitude modulation, we don't need bandwidth because frequency of the carrier always remains constant, i.e. if it's 500kHz then it's always going to remain at this value and hence no upper and lower frequency limits and therefore no bandwidth. Q4: Where do I have it wrong?

Thank you very much for the help and your time.

Regards
PG

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15. ### JimBWell-Known Member

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A carrier, amplitude modulated by a sine wave, has a frequncy spectum of:

The Carrier (fc)

The Upper Side Frequency (fc + fm)

The Lower Side Frequency (fc - fm)

And so the bandwidth occupied by the AM signal is 2fm.

If you had access to a signal generator and a spectrum analyser or a very selective receiver, you could easily prove it to yourself.

JimB
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16. ### steveBWell-Known MemberMost Helpful Member

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PG,

One of the problems I see here is that you are not using enough math to sort out the issues. There are differences between types of modulation. AM is different than FM and there are several ways to achieve both AM and FM. To see exactly what happens, you need to do it out mathematically for the case at hand. There are similarities in all cases, and were trying to give you the overall information in preparation for your more detailed studies, but you are not going to make sense of all this without using the math and without being patient and going through the course step by step.

I'll try to address your above questions in more detail later today. I'll start introducing a little bit of the math, but not too much for now.

One thing I want to ask is why you keep bringing up the center frequency of the baseband signal. Why do we care about this and why do you try to use this to determine bandwidth after modulation? Why is 1800 Hz any more special than the other frequencies? Is there a basis for this you found in a reference?
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17. ### PG1995Active Member

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Thank you, Jim, Steve.

Exactly true. But there are reasons for not using enough math. Yes, I have learned to be patient but I believe this is the fundamental confusion and if it's not cleared then it will beget more confusion

Because I think you need to partition baseband frequency range at some middle point so that we have equal range (or, spread) of frequencies around center frequency, 700 kHz, of the carrier after modulation. For instance, if you don't partition baseband frequency range, 300-3300 Hz, at 1800 Hz, then how would you know that how the spectrum is spread around fc of the carrier? And if you don't partition the baseband frequency range and go on to use it as it is then bandwidth of the carrier becomes (700-3)kHz to (700+3) kHz or 697kHz-703kHz; (3300-300)=3kHz. Is this number, 697kHz-703kHz, for the bandwidth correct in your opinion?

Please help me when you have time. Thank you.

Regards
PG
Last edited: Aug 2, 2013
18. ### steveBWell-Known MemberMost Helpful Member

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OK,

We'll try to clear up the confusion, as it is important to have a good intuitive understanding as a basis before diving into the math and the details. I'll do most of this later today, with just a little bit now at the bottom of this post.

I would tend to say the 1800 Hz is not a critical frequency to think about. There is nothing wrong with identifying the center, but don't use this as the basis for the bandwidth. We've tried to point out that the 697-703 kHz is the final bandwidth (assuming you have 300-3000 Hz baseband). Perhaps what bothers you is that this is twice the baseband bandwidth. But, that's the way it works out, which we'll see in the math soon.

Before looking at the math, consider that the baseband signal can be viewed as a spectrum of -3300 to -300 Hz and +300 to +3300 Hz. Then, when you modulate, these frequencies get slided up to the carrier frequency. The details of the final spectrum depend on the exact modulation scheme and implementation, but lets look at simple multiplication AM modulation. Keep in mind that a real system might not use this exact multiplication method, but it is instructive to look at this very simple case.

So, we have a carrier at frequency ωc and a signal at frequency ωs, which are simple cosine functions, and we multiply them as follows.

Now recall the trig identities as follows ...

Now add these two identities together to get ...

Applying this result means that the modulated signal is as follows ...

Here you see the signal sidebands are created by adding the carrier frequency to the signal frequency and by subtracting the signal frequency from the carrier frequency, but the carrier gets cancelled out. However, not all modulation schemes will cancel the carrier.

Now, what if you look at this in the frequency domain? You will have a signal with two impulses, one at -ωs and one at +ωs, and then a carrier with two impulses, one at -ωc and one at +ωc. Convolution of impulses is easy and you will get 4 frequency spikes at -ωc-ωs, -ωc+ws, +ωc-ωs, and +ωc+ωs.

Now, what if your baseband signal is not a simple impulse function with one frequency, but is instead a 300-3300 Hz band of frequencies? Calculating this in the time domain is hard to do, but convolution in the frequency band is easy to do. The carrier impulse function will generate 4 separate bands as follows ...

BAND1 -ωc-2∏3300 t0 -ωc-2∏300
BAND2 -ωc+2∏300 t0 -ωc+2∏3300
BAND3 +ωc-2∏3300 t0 +ωc-2∏300
BAND4 +ωc+2∏300 t0 +ωc+2∏3300

Hence, the full signal band spectrum is maintained, but just shifted. The plus and minus frequencies are just the same thing, so we only look at the positive frequencies. Hence, we conclude that two bands (upper and lower sidebands) are generated by modulation of this type.

I would recommend that you not proceed until you fully understand this most basic step. It forms the basis of everything else, and as you get to more complicated methods, the math gets more difficult and in some cases becomes intractable. But, whether or not we can follow the math, the effect of modulation is always the same. The baseband information is shifted up in frequency to one or more bands. Demodulation will shift it back down again.

Hopefully you see why the math is important. Personally, I would not be able to understand this without using math as the tool to make things clear. Although, the math can eventually get too complex to rely on and we use experiment or number crunching to get answers, the simple math cases are prototypical models that gives us a good feel for what is happening.
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19. ### JimBWell-Known Member

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Steve

Nice explaination, but I think that you have complicated things a bit further by generating a Double Side Band Suppressed Carrier (DSBSC), rather than good old AM (Double Side Band with Full Carrier).

If you run through a similar process but starting with the expression:

v = A[1 + M.cos(ωm.t)]cos(ωc.t)
where A = unmodulated carrier amplitude and M = modulation depth

you end up with a carrier and the two side frequencies.

I fully agree with you that worrying about the centre frequency of the baseband is just a distraction and does not contribute to understanding the modulation process.

JimB
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20. ### steveBWell-Known MemberMost Helpful Member

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JimB

Thanks, that's a good point. I would also recommend to PG to trace through the math for that, as you said.

Steve