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From wikipedia:

"A key characteristic of bandwidth is that any band of a given width can carry the same amount of information, regardless of where that band is located in the frequency spectrum. For example, a 3 kHz band can carry a telephone conversation whether that band is at baseband (as in a POTS telephone line) or modulated to some higher frequency."

(my emphasis)

From a previous post on this site:
Why is channel capacity a factor of bandwidth instead of frequency?

Question extract: "According to this formula [Shannon-Hartley], a fixed-frequency signal would have the same performance regardless of whether it's at high or low frequency..[]... For example say my bandwidth is 1Hz at a fixed frequency of 1Hz. Compare this with a bandwidth of 1Hz at a frequency of 2.4GHz. It's plainly obvious that I can cram way more bits into 2.4 x 109 cycles/second than I can with just 1/sec."

Extract from one answer: "Certainly you could". (note that the answer goes on to discuss signal modulation and energy, at which point I'm lost).

Question

I'm in way over my head on this topic, and the list of things I don't understand is embarrassingly long, despite having carefully read through the many related posts on this site.

However, the Wikipedia entry seems intuitively wrong. If we are using electro-magnetic waves to transmit 1s and 0s, then more waves per second must surely translate into a higher rate of 1s and 0s transmitted. So, to my simple mind, a 3 KHz band at 2.4 GHz must be able to carry more 1s and 0s per second than a 3 KHz band at say 4 KHz.

Is there any way of explaining this to someone who studied economics, rather than engineering or physics?

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  • $\begingroup$ Sorry, just to be clear. How do we reconcile the Wikipedia entry with the previous post on this topic? $\endgroup$ – Wish_I_was_an_engineer Feb 10 '17 at 18:11
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So, to my simple mind, a 3 KHz band at 2.4 GHz must be able to carry more 1s and 0s per second than a 3 KHz band at say 4 KHz.

Say you have a modulation scheme that provides 1 bit per second per hertz of bandwidth used.

If you use this modulation on a carrier at 2.4 GHz to send 3 kb/s, it will create a signal with a spectrum about 3 kHz wide. If you use it to send 10 kb/s, it will create a signal with a spectrum about 10 kHz wide. etc.

If you use this modulation on a carrier at 4 kHz to send 3 kb/s, it will create a signal with spectrum about 3 kHz wide. If you try to use it to send 10 kb/s, it will fail because you need to make a spectrum 10 kHz wide, but with a carrier at 4 kHz, this means the negative and positive spectra overlap in the frequency band from -1 to +1 kHz.

So if you have a higher-frequency carrier (or center frequency if your modulation scheme doesn't have a carrier), you can send higher data rates. But doing so creates a signal with wide bandwidth.

Edit

Here's an example of a simple modulation scheme: To send a one, send the carrier for 1 us. To send a zero, turn the carrier off for 1 us.

It doesn't matter if the carrier is 10 MHz or 10 GHz, or 350 THz, it takes one microsecond to send one bit, so the bit rate is 1 Mb/s.

And, whether the carrier is 10 MHz or 10 GHz, the spectrum of the modulated signal will have a peak with about 1 MHz spread. But the center frequency will be equal to the carrier frequency.

If you changed the modulation scheme to turn the carrier on or off for only 0.5 us for each bit, you'd get 2 Mb/s bit rate, and the spread of the spectrum around the carrier would be about 2 MHz instead of 1 MHz. Again, it wouldn't matter what the carrier frequency is.

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  • $\begingroup$ Thanks very much for that, which again I've thought about hard, and I'm still not quite getting it. I think a modulation scheme changes some characteristic of your electro-magnetic wave - amplitude, frequency, other - so it can represent 1s and 0s, i.e. binary digits. How then can a modulation scheme on is own determine the rate of bits per hertz of bandwidth used? Why is that rate of bits not a function of the carrier frequency, i.e. more bits at a higher frequency? Equally, why is it a function of the modulation scheme? $\endgroup$ – Wish_I_was_an_engineer Feb 11 '17 at 16:30
  • $\begingroup$ Many apologies, I know I am not being clear. Here is the nub of my question. This is my incorrect understanding. The modulation scheme decides how 1s and 0s are being represented by electro-magnetic waves. The frequency decides the rate of waves per second. Therefore, the frequency decides the bit-rate, no matter how wide the bandwidth is. What am I misunderstanding? $\endgroup$ – Wish_I_was_an_engineer Feb 11 '17 at 16:44
  • $\begingroup$ You're missing that applying the modulation changes the spectrum of the signal, spreading its bandwidth. And the amount that the bandwidth spreads depends on the bit rate. This spreading is the bandwidth we're talking about when we say "a certain bandwidth centered at 10 MHz can convey the same amount of data as the same bandwidth centered on 10 GHz." $\endgroup$ – The Photon Feb 11 '17 at 16:55
  • $\begingroup$ And that there is no one-to-one correspondence between cycles of the carrier and bits transmitted. $\endgroup$ – The Photon Feb 11 '17 at 16:57
  • $\begingroup$ Not looking for a one-on-one correpondance, but dumbfounded that there appears to be no correspondence. However, need some time to process your answer and the further answer below. Thanks v much for your effort, much appreciated $\endgroup$ – Wish_I_was_an_engineer Feb 12 '17 at 18:32
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Let's consider amplitude modulation. Consider a carrier at frequency $f_0$. Then the signal will look like $S(t) = A(t)\cos(2\pi f_0 t)$. When we perform amplitude modulation it means we control $A(t)$, the amplitude of the carrier, in such a way that $A(t)$ represents the signal we are trying to send. Let's consider a simple 1s and 0s scheme. Say if $A(t) = 10$ we consider the signal to be a 1 at time $t$ and if $A(t) = 5$ we consider the signal to be 0 at time $t$.

The question is now how fast can we switch the signal, $A(t)$ from 0 to 1. I think you're misconception is that we can switch $A(t)$ from zero to one within the period of the carrier, $T_0 = \frac{1}{f_0}$. Let's explore this for a minute.

Misconception Exploration

Let $A(t) = 0$ represent at 0 and $A(t)=10$ represent a 1. We could then send the signal $S(t) = 10 \cos(2\pi f_0t)$. Then at $t=0$ the signal would be a 1 and at $t = \frac{1}{4}T_0$ the signal would be a 0 and basically the signal would switch from 0 to 1 a couple of times per carrier period. However, if this was our method of sending signals then we couldn't send very interesting signals. Every signal would just be 010101010101.... The point is that what I have just describe here is NOT how signals are sent and does NOT represent how amplitude modulation (AM) works.

AM bandwidth

Let's return to the previous setup where $A(t) = 10$ is a 1 and $A(t) = 5$ is a 0. So if the bandwidth, $f_{BW}$ is not equal to the carrier frequency, $f_0$ what is it related to? Let's look into it. Like I said before $A(t)$ is going to carry our signal so we need to look into more detail at $A(t)$. Let's consider a signal $A(t)$ which is a step function. for $t=0$ to $t=10T_0$ we will set $A(t) = 10$ and for $t=10 T_0$ to $t=20 T_0$ we will let $A(t)=5$. What someone receiving this signal would then see is a cosine signal with amplitude of 10 for 10 periods and 5 for the next 10 periods. They would then interpret the signal as being 1 for the first 10 periods and 0 for the second 10 periods. It took $20T$ to send 2 bits then. We can then consider a bitrate $\frac{2 \text{bits}}{20 T} = \frac{1 \text{bit}}{10 T} = 0.1 f_0$. This bit rate is very closely related to the signal bandwidth. It will turn out that they're basically the same thing. I'll get to that in a little bit. But first, let's consider another scenario where we keep the signal high for 5 periods and then switch it low for 5 periods. In this case we were able to send 2 bits in only $10T_0$ so the bit rate is twice as fast, $0.2f_0$. We see that if we can switch the signal faster the bit rate goes up.

Bitrate vs. bandwidth

So far I've explained bitrate and we see that it has to do with how quickly we change $A(t)$, but I haven't related this to bandwidth at all. I think this is at the heart of your question. What is the relationship between bit rate and bandwidth. In fact, even more to the heart of your question is probably the question what is the bandwidth of a signal?

It turns out that we should think of the bandwidth as the inverse of how quickly we change the signal from one state to another state. So, for example, in the first example I gave above we were able to switch the signal from a 1 to a 0 after $10T_0$ so by the definition I just gave the bandwidth of that signal would be $f_{BW}\frac{1}{10T_0} = 0.1 f_0$. Ok, so I've basically just defined the bandwidth to be the same as the bitrate. The question then is why do we call it bandwidth and not bitrate?

Fourier transform interlude

The answer has to do with Fourier Transforms. If you're not familiar with Fourier transforms here is the quickest intro in the world. When we take the Fourier transform of a signal $S(t)$ we basically break the signal up into different 'components' at different frequencies and define a new function $\tilde{S}(f)$ which tells us how strong our signal is at different frequencies. So for example, if we take the fourier transform of a song with very strong bass (low frequency) then $\tilde{S}(f)$ will be high at low frequencies and lower at high frequencies whereas if we take the Fourier transform of a song with a lot of treble the fourier transform will be high at high $f$ and lower at low $f$.

Fourier transform and signal changing speed

In signal processing there is a very important relationship between frequency and time. Whenever a signal is high frequency that means it is changing very fast meaning the time for the signal to change is very short. Hammer that into head. High frequency means short switching time. Let's now think about $A(t)$. Remember $A(t)$ carries the information we are interested in. We want to take the Fourier transform of $A(t)$ to get $\tilde{A}(f)$, the frequency spectrum. Here is the heart of the answer to the heart of your question. It turns out (look up more about Fourier transform) that if your signal changes on a time scale of $T_a$ then the Fourier transform will have components up to $f_a = 1/T_a$. For example, in the first example up above $T_a = 10 T_0$ so $f_a = 0.1 f_0$. This is what we call the bandwidth of a signal. The bandwidth of a signal is the largest frequency in the fourier transform of a signal, $f_{BW} = f_a$. So we apply this to all the examples from before so that $f_{BW} = 0.1 f_0$ as claimed earlier. Consider the bass song and the treble song, we can see that the bandwidth of the bass song would be lower than the bandwidth of the treble song since the treble song needs to go up to higher frequencies.

Now we know what bandwidth is.

Carrier

But what about the carrier? I've totally stopped talking about the carrier, what happened? Ok, let's go back to that. Let's think JUST about the carrier, $C(t) = \cos(2\pi f_0 t)$. What does the Fourier transform look like? Well in time it only has a signal at 1 frequency so the fourier transform $\tilde{C}(f)$ is basically going to be a spike at $f=f_0$ and have no strength anywhere else. So let's recap our two fourier transforms. $\tilde{C}(f)$ has a spike at $f=f_0$ and $\tilde{A}(f)$ is non-zero from $f=0$ up until $f=f_{bw}$.

We now want to consider the Fourier transform of the actual signal $S(t) = A(t)C(t)$. It turns out* that when we look at the fourier transform of our carrier modulated with our signal it will look exactly like $\tilde{A}(f)$ but with the whole thing shifted up in frequency so that instead of being close to 0 with a width of $f_{bw}$ it will be close to $f_0$ with a width of $f_{bw}$. We say "the signal has been mixed up to the carrier frequency". The key point is that the width of the signal in Fourier space, the "band width" is still $f_{bw}$ even though it is now centered at the carrier frequency $f_0$.

Summary

To summarize the answer to your question:

1) The bitrate is how quickly we can change the signal $A(t)$.

2) How quickly a signal changes in time, $T_a$ is related to the inverse of the highest fourier frequency, $\frac{1}{f_a}$

3) We call this highest frequency $f_a$ the bandwidth, $f_{BW}$ since it represents the width (in frequency) of the Fourier spectrum.

4) Thus the bandwidth is equal to the bitrate.

5) When we modulate a carrier with a signal then the total signal $S(t)$ has the same bandwidth as the information signal, $A(t)$.

With the net result that if you want to change your modulated signal very quickly you need a high bandwidth, noting that this whole discussion does not depend on the carrier frequency, but rather the highest frequency in the Fourier spectrum of $A(t)$.

Caveat

I will note ONE final point. The signal bandwidth, $f_{BW}$ must be less than the carrier frequency $f_0$. So if we have a 10 MHz carrier it is not possible to have a 100 MHz bandwidth, whereas if we have a 1 GHz carrier it is possible to have a 100 MHz bandwidth. This means that for higher frequency signals we technically can cram in more information. HOWEVER, for a high frequency signal and a low frequency signal with the SAME bandwidth, you can cram in the same information rate. So a 1 GHz carrier with 1 MHz bandwidth and 10 MHz carrier with 1 MHz bandwidth would be able to transfer a signal at the same rate.

*Fourier Convolution Theorem

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  • $\begingroup$ Wow. Thanks very much for that. That will take me some time to process. Clearly, the issue is much more complex than I had appreciated, but with your and the previous answer i think i'm on the road to a better understanding. Once again, thanks v much $\endgroup$ – Wish_I_was_an_engineer Feb 12 '17 at 18:37

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