Mobile, copper and fiber comparison

Everybody knows some internet basics, but telecommunication technology and its limits across Europe and the West in general are often poorly understood.

People know that fiber internet is fast, but they don't know why to upgrade. They don't know why they can't get gigabit internet over their phone lines. They use mobile broadband but don't understand the technology behind it. Therefore I will present a fair comparison of these technologies.

Pictures in this page are provided by Wikipedia.

Communication channel capacity

Shannon-Hartley theorem works as the basis for our comparisons. It provides the maximum rate at which information can be transmitted over a communications channel in terms of bits per second:

C = B*log2(1+S/N)

C is the theoretical upper bound of the channel capacity in bits per second. This is known as a Shannon's limit. It tells us how much information a channel can carry if we know the bandwidth and signal-to-noise ratio of that channel.
B is the bandwidth of the channel in hertz. For example, the wavelengths from 1260nm to 1675nm are used in the fiber optic. This corresponds to the 187THz - 250THz frequency range. The bandwidth is the difference: 63THz.
S/N is the signal-to-noise (SNR) ratio. Both S and N are measured in Watts. The SNR is measured as amplitude ratio.

As a result of the SNR, to communicate successfully you must have a stronger signal than the noise around you.

This could be applied to anything from advertising to music, but let us take a practical example to lead us into telecommunication.

The signal-to-noise ratio practically means that to be able to communicate you must have a stronger signal than the noise around you.

This could be applied to about anything from advertising to music. But lets take a practical example to lead us into telecommunication.

You and your friend are talking with each other over a distance. You are at one end of a small park and your friend is at the other end. If it's quiet, you can hear each other even from quite far away, but if the wind shakes the leaves, you must either increase your volume or move closer to each other.

Shannon-Hartley theory describes how much information you could transmit with the speech, and how many more bits you get for the extra oomph. :). To do this you only need to know few values.

You need to know the bandwidth of your voice. A person can hear the sounds with frequencies ranging from 20Hz to 20kHz. We can approximate that the bandwidth is 20kHz.

Normal conversation tends to range between 40dB - 60dB when being listened from a meter distance. A calm room ambient noise is somewhere between 20dB - 30dB. A park might be anywhere between 20dB - 70dB.

The sound intensity drops by the distance. You are 10 meters afar from your friend. This means that the intensity of the sound drops by -20dB before it reaches your friend.

Everything being decibels mean that it is simple to calculate SNR in decibels:

Signal dB - Noise dB = SNR dB
Quiet park scenario:
60dB - 20dB - 20dB   = 20dB
Windy park scenario:
60dB - 20dB - 40dB   = 0dB

Next we need to convert the SNR from decibels to amplitude ratio.

no signal = 0    log2(1 + 0)  * 20kHz = 0kbps


0dB       = 1    log2(1 + 1)  * 20kHz = 20kbps
9.5dB     = 3    log2(1 + 3)  * 20kHz = 40kbps
17dB      = 7    log2(1 + 7)  * 20kHz = 60kbps
23.52dB   = 15   log2(1 + 15) * 20kHz = 80kbps
30dB      = 31   log2(1 + 31) * 20kHz = 100kbps

36dB      = 63   log2(1 + 63)   * 20kHz = 120kbps
42dB      = 127  log2(1 + 127)  * 20kHz = 140kbps
48dB      = 255  log2(1 + 255)  * 20kHz = 160kbps
54dB      = 511  log2(1 + 511)  * 20kHz = 180kbps
60dB      = 1023 log2(1 + 1023) * 20kHz = 200kbps

The base case means that if you do not speak then you cannot be heard.
An increase of noise means you must talk louder.
A louder voice can transfer more information.

If sound volume increases by ~10dB, you perceive it as twice as loud. Though to talk twice as loud requires ten times more power.

At somewhere between +15dB and +30dB SNR the capacity begins to increase linearly in respect to loudness in decibels. This means that after about 100kbps you will need ten times more power to provide each additional 30kbps.

Of course, if you yell you need more effort to control your voice. The above chart doesn't tell such a thing. The Shannon-Hartley theorem describes an upper bound and not the attainable rate, though it is a fine tool to compare different methods of communication.

We don't do anything by knowing how many bits we could shove down our ears, but there is an interesting human factor here. Children or foreigners require more SNR to understand what you say than what is sufficient for adults. Adults are okay with +4dB or +5dB. Children need +15dB SNR to comprehend what you say.

If you understood the example these ideas should be bit more intuitive for you now. You need lot of intuition to understand telecommunications.

Decibels

You may be confused that we subtracted decibels although in the equation the signal is divided by noise. It is ok because they are decibels. This is the convenience in the decibels and why they are used here. Observe:

SNR = S           / N
SNR = 10^(SdB/20) / 10^(NdB/20)
SNR = 10^(SdB/20 - NdB/20)
SNR = 10^((SdB-NdB)/20)
SNRdB = SdB - NdB

So in short, if we subtract two decibels, it means the same thing as dividing two amplitudes as linear quantities.

In telecommunications the SNR is usually represented as decibels while the S and N are represented as Decibel-milliwatts (dBm). 0dBm is 1mW. 30dBm is 1W. 40dBm is 10W.

So now we have some equipment to understand and compare telecommunication methods.

Copper wires and bandwidth

Let us start with the copper wires because this explains why we cannot make a faster ADSL. Copper wires cannot transmit high frequency signals over long distances.

ADSL2/2+ can easily download easily 20Mbps at 1km distance.
At a 2km distance it can download 15Mbps.
At a 3km distance it can download 10Mbps.
At a 4km distance it struggles to download 4Mbps.
At a 5km distance download falls below 1Mbps.

The chart under the link is adjusted for old wiring. Newer telephone cables may be about ten times better, but even then there is this problem: Telephone cables were designed for transmitting 5kHz signals.

ADSL annex a frequency plan

We have been extremely lucky that we can transmit 1MHz frequencies with ADSL. It has provided extra longevity for our now-obsolete telephone cables.

The total downstream bandwidth on ADSL in 966kHz. Typical output power of an ADSL modem is between 10dBm and 20dBm. Background noise level expected on telephone is -140dBm/Hz.

The noise is proportional to the bandwidth we use. This means that we multiply 966kHz to the -140dBm/Hz. Remember that is a decibel, so we get to add about 60dBm to the value and get -80dBm as our noise floor. So in ADSL the SNR is about 100dB.

The cable loss of a telephone cable is somewhere around 13dB/km. Lets use these values and check what the telephone capacity is according to Shannon-Hartley theorem:

1km log2(10^((100dB - 13dB)/10))*966kHz = 27Mbps
2km log2(10^((100dB - 26dB)/10))*966kHz = 23Mbps
3km log2(10^((100dB - 39dB)/10))*966kHz = 19Mbps
4km log2(10^((100dB - 52dB)/10))*966kHz = 15Mbps
5km log2(10^((100dB - 65dB)/10))*966kHz = 11Mbps
6km log2(10^((100dB - 78dB)/10))*966kHz = 7Mbps

The numbers are very different than in the earlier chart because this is an approximation. But they are very close to the downstream speeds charted.

What if we transmitted with 100W instead of 0.1W? That'd be 30dBm more. Lets try:

1km log2(10^((130dB - 13dB)/10))*966kHz = 37Mbps
2km log2(10^((130dB - 26dB)/10))*966kHz = 33Mbps
3km log2(10^((130dB - 39dB)/10))*966kHz = 29Mbps
4km log2(10^((130dB - 52dB)/10))*966kHz = 25Mbps
5km log2(10^((130dB - 65dB)/10))*966kHz = 20bps
6km log2(10^((130dB - 78dB)/10))*966kHz = 16Mbps

We would spend 1000 times more power for measly 2x improvement. How about increasing the frequency?

Increasing the frequency to 1GHz would increase the noise floor by 30dBm. Unfortunately the cable loss rises as well:

1km log2(10^((70dB - 20dB)/10))*1GHz = 16Gbps
2km log2(10^((70dB - 40dB)/10))*1GHz = 9Gbps
3km log2(10^((70dB - 60dB)/10))*1GHz = 3Gbps

This is an optimistic estimate because due to the skin effect different frequency components of the signal perceive different resistance in the wire. This causes the higher frequency signals to get dampened on much shorter distances than the low frequency signals.

In summary the copper wires have weakness in B and S over long distances.

Mobile communication and bandwidth

How does mobile connections compare? They are bandwidth(B) and noise(N) bounded. Even if more bandwidths were opened for access, we'd be bandwidth bounded.

The total bandwidth of 4G/LTE and 4G+/LTE in Finland is 130MHz. These frequencies are scattered around 800MHz, 1800MHz and 2600MHz. Wireless networks operate on microwaves. These waves have to propagate directly or reflect from flat surfaces such as lakes.

The typical base station has maximum 20W - 69W, or 43dBm - 48dBm, transmission power. This provides them an aerial bandwidth of about 1Gbps per sector. There tend to be 3 or 4 sectors at each mobile tower.

This means that in average Finnish town there are over 300 subscribers sharing the 1Gbps radio bandwidth. On average each subscriber has ~3Mbps of bandwidth, but they've been promised 100Mbps. Once about 10 or 20 subscribers use their whole bandwidth it means that the bandwidth has been congested.

The multiplexed bandwidths across the users introduce extra latency. 40ms latencies have been perceived. When the network is congested, the latency escapes as well.

Range of 4G is limited by the Free-space path loss. And as range increases, the signal strength decreases. This means you need more of the bandwidth to communicate with the distant subscribers, so anyone farther than 1km from a tower will consistently get considerably worse service. Especially if the farther reaching lower-frequency bands are consistently congested.

32.4 + 20*(log10(F MHz) + log10(D km))

32.4 + 20*log10( 800MHz) + 20*log10(0.5km) = 84dB
32.4 + 20*log10(1800MHz) + 20*log10(0.5km) = 91dB
32.4 + 20*log10(2600MHz) + 20*log10(0.5km) = 94dB

32.4 + 20*log10( 800MHz) + 20*log10(2km) = 96dB
32.4 + 20*log10(1800MHz) + 20*log10(2km) = 103dB
32.4 + 20*log10(2600MHz) + 20*log10(2km) = 106dB

32.4 + 20*log10( 800MHz) + 20*log10(5km) = 104dB
32.4 + 20*log10(1800MHz) + 20*log10(5km) = 111dB
32.4 + 20*log10(2600MHz) + 20*log10(5km) = 114dB

The signal at 5km is at least 1000 times weaker than around the tower just simply because of this. The noise floor of 4G goes somewhere around -100dBm. With +43dBm the shannon's equation gives something around 800Mbps to 10Gbps depending on how near you are the tower.

Increasing the bandwidth used by radio has two problems. First of all the bandwidth is limited, radio waves are the first 300GHz of the electromagnetic spectrum. The larger problem is that on every radio frequency there is an ambient noise of about -174dBm. The power you need to transmit increases linearly in respect to the bandwidth you use. This means to transmit 10 times as much, you need 10 times as much power.

For 1GHz bandwidth you got -84dBm of noise to defeat. For the whole 300GHz, although you couldn't use it all yourself, you will already get -60dBm noise floor. The easiest way to increase signal strength is to shorten the distance. To provide sufficient coverage and signal strength, the radio cell size should be shrunk a lot. For 5G networks, 100 meters wide cell sizes are recommended.

In summary. The radio networking is extremely limited and overprovisioned. Opening more frequencies wouldn't eliminate the fundamental problem in these systems.

Fiber optic communications and bandwidth

Copper and radio are heavily limited in all ways compared to fiber:

Signal strength must exceed the sensitivity of the receiver. There is about 10dB or 20dB power budget when designing fiber optic systems, meaning that much of signal strength is expected to be lost on the path from the transmitter to the receiver.
The communication medium is near-infrared light that cannot leak inside the cable from external sources. This means that the noise floor depends on the sensitivity of our receiver rather than the physics of the medium. The technology already present allows noise floors as low as -70dBm or even -90dBm for the whole bandwidth.
Power sent across fiber optic links range from -30dBm to 20dBm. This means that sufficient signal strength can be obtained with ease. It's not unheard to have a fiber optic network router consume less power than a lightbulb.
Fiber optic links have 63THz of bandwidth available to them. 1THz = 1 000GHz = 1 000 000 MHz.
+20dB signal strength and fiber bandwidth plugged in delivers a theoretical top-capacity of about 400Tbps on a single fiber.
The single mode fiber cable attenuates signal by 0.4dB/km. This allows for far distances across active devices. 30 kilometers of fiber attenuates signal only by -12dB.

Current fiber optic links use only a fraction of the theoretical bandwidth. Most consumer hardware cannot yet utilize the bandwidth that is already available on a single fiber. The technology is still simple. We are still just pulsing a laser to capture it in the other end and obtain tremendous transfer rates that way.

Because of the simplicity of fiber optic systems, the latency is low. 4ms and lower have been measured in practice.

In a Finnish fiber-to-the-home system a single cable carries all subscribers and each have an individual fiber reserved for them. This is by local standards but it makes sense elsewhere as well. Modern fiber cables may contain 192 individual fibers and the whole pack isn't thicker than your finger or cost more than few euros per meter.

In well-designed system each subscriber has their own fiber all the way to the operator's hall and the system can be affordably upgraded as the technology progresses.

One important question is, what sort of lifetime can you expect from fiber. The 30 years old early backbone fiber networks are still in use. It is expected that fiber cables last for about 50 to 100 years after installed.

In fiber optic networks the largest costs in establishing come from installing the cable. It can be installed to the ground, lakes or into aerial poles like the twisted pair copper cable. If installed into poles they require 100 or 150 meter pole span at maximum. Non-metallic cables are preferred everywhere because they don't need electrical grounding and aren't sensitive to lightning damage. Different types of installations require different type of cable packagings though.

Summary

In conclusion what you should catch from this article is that:

Long distance copper cables are outmoded form of telecommunication.
Mobile network capacity is physically bounded, it has been overprovisioned, upgrading it would require deploying fiber to attain small enough cell size.
Fiber optic networks deployed today will upgrade extremely well. Despite attaining high transfer rates it is still early days with the optic cables that are being deployed now.