# What is the relationship between the bandwith on a wire and the frequency?

I am trying to learn networking (currently Link - Physical Layer); this is self-study.

I am very confused about one particular thing:

Suppose I want to send a data on the wire something like this:

01010101, where it will look some thing like this as a Signal:

``````__|‾‾|__|‾‾|__|‾‾|__|‾‾
``````

Well the data to be sent must be represented by a signal, and the signal in this situation is the "change in the voltage" on the link / wire (assume we are using cables, not wireless link).

So Fourier proved that with enough frequencies a signal can be represented pretty well.

Like: I still don't understanding the relationship between a signal on the wire, and the Frequencies.

The definition of frequency is: the number of occurrences of a repeating event per unit time. So what is repeating in the wire per unit time?

Also for example on a DSL line, for Frequency Division Multiplexing, because multiple users will be allocated less frequency, there will be less bandwidth per user on a given link / wire. What does it mean to allocate less frequency on a wire? Less repeating of what?

Are there many frequencies available on the wire? If there are ( lets say from 0 to 1 Mega Hertz ) can I represent the above using the range between 0 to 100 OR 100 to 200 OR 500 to 1000 ? Why do I have more bandwidth if I use more frequencies?

• Could you elaborate on what you would like answered that hasn't been answered by Mike Pennington and Malt? Both provided sufficiently in-depth answers to the OP. – Ryan Foley Jan 29 '14 at 16:20

### Modulation and symbols

the number of occurrences of a repeating event per unit time. So what is repeating in the wire per unit time?

The voltage patterns on the wire repeat.

In extremely simple communication systems, you might cycle the line's DC voltage above or below a threshold, as shown in your ASCII-art... `__|‾‾|__|‾‾|__|‾‾|__|‾‾`. Suppose your thresholds are +5v and -5vdc; modulating binary data through two DC voltages would only yield one bit per voltage level (each voltage transition is called a symbol in the industry).

DC voltage transitions are not the only way to represent data on the wire, as you mentioned, you can modulate the voltage of a signal on a given frequency, or shift between two frequencies to modulate data. This picture illustrates how the same `__|‾‾|__|‾‾|__|‾‾|__|‾‾` transitions are represented via Amplitude Modulation (AM) and Frequency Modulation (FM). More complex systems that are transmitted over longer distances use more complex modulation schemes, such as FDM or QPSK, to pack more data into a given bandwidth on the wire.

Generally speaking, you can modulate using combinations of:

### Bit rate and spectral efficiency

Are there many frequencies available on the wire? If there are (lets say from 0 to 1 Mega Hertz ) can I represent the above using the range between 0 to 100 OR 100 to 200 OR 500 to 1000 ? Why do I have more bandwidth if I use more frequencies?

Let's just consider an Frequency Modulation system, which has two states on the wire...

• The 0 symbol is represented by 1KHz
• The 1 symbol is represented by 2.5KHz

This modulation scheme requires 1.5KHz of bandwidth on the wire. However, that tells you nothing about the bit rate transmitted (which confusingly, is also known as 'bandwidth', but let's not use an overloaded term).

One reason that an FM system might space 0 and 1 symbols 1.5KHz apart is because there are limits to how well, how quickly, and how economically the modem can measure the frequency changes on the wire.

• How well the modem can measure frequency changes is one factor that drives how much bandwidth is required on the wire
• How quickly the modem can measure frequency (or other symbol) changes drives how high the modem's bit rate will be
• Economics play a big role, because you might be able to build a system that has extremely high spectral efficiency, but if nobody can afford it then it's not really a feasible solution.

As a general rule, you can build faster and cheaper modems if you have more bandwidth available to you.

### Edit: comment response

I have studied your response, but I am still confused about some things. I can only send 1 and 0s over a wire as far as I understand. So if 1.5 KHz is enough for this, why would I use more bandwidth?

I addressed the question in the last section, but let's continue with the FM modulation example. Real systems have to account for receiver sensitivity, and factors such as how well a band-pass filter can be implemented.

Suppose the 1.5KHz bandwidth available to the modem only yields 9600 baud, and that's not fast enough; however, you might build a 20KHz modem that is fast enough (maybe you need 56K baud).

Why is 20KHz better? Due to the realities and imperfect slopes on band-pass filters and other components, you may need that much bandwidth to implement the correct modulation and line code. Maybe with 20Khz, you could implement QAM scheme, which gave you 3 bits per symbol, resulting in a maximum bit rate of "9600*8", or 76.8 Kbaud (note: 2**3 = 8)

You're asking good questions, but it's very hard to explain this without getting into the guts of a real design. If you read some electronics books about receiver design, or take some electrical engineering courses this material is covered.

• Thank you very much for your detailed response. I have studied your response, but I am still confused about some things. I can only send 1 and 0s over a wire as far as I understand. So if 1.5 KHz is enough for this, why would I use more bandwidth? Why ( or how ) does it provide more bit rate? Because as far as I know, mode bandwidth on the wire = more bit rate / second. Does it mean I will also use for example 3.5 to 5 KHz for additional 1 and 0s in the same time? – Koray Tugay Jan 25 '14 at 19:23
• Hi, I updated my answer, perhaps that helps clarify – Mike Pennington Jan 25 '14 at 20:13
• When you change from one state (0) to another (1), you generate energy at various frequencies (spectra). How often you change state (modulation frequency) affects the bandwidth. Also, the faster you change state, the more energy you generate at higher frequencies. This adds to the bandwidth. – Ron Trunk Jan 25 '14 at 23:00
• @Ron, saying "faster you change state, the more energy you generate at higher frequencies." doesn't necessarily change the symbol rate (i.e. data bandwidth) within the signal. What we care about is information encoded on top of the signal; higher frequencies themselves don't inherently carry bits... if merely having higher frequencies was sufficient to increase the available bit rate, a microwave oven would be a fantastic communication tool. – Mike Pennington Jan 26 '14 at 0:19
• @MikePennington I'm well aware of that. I was trying to explain where the higher modulation frequency and therefore greater bandwidth come from. A higher symbol rate, and therefore a higher rate of change will generate more energy at higher frequencies and therefore increase (signal) bandwidth. – Ron Trunk Jan 26 '14 at 4:47

Mike offered an excellent answer but not exactly to what you were asking.

Bandwidth, by definition, is a range of frequencies, measured in Hz.

As you've said, the signal `__|‾‾|__|‾‾|__|‾‾|__|‾‾` can be broken down (using Fourier) into a bunch of frequencies. Let's say that we've broken it down, and saw that our signal is (mostly) made up of frequencies 1Mhz, 1.1Mhz,1.2Mhz,1.3Mhz... up to 2Mhz. That means that our signal has a bandwidth of 1Mhz.

Now, we want to send it through a channel, such as a copper wire, or an optical fiber. So first, let's talk a little bit about channels.

When talking about bandwidth in channels, we actually talk about passband bandwidth which describes the range of frequencies a channel can carry with little distortion. Say I have a channel that can only pass signals whose frequency is between f1 and f2. Its frequency response function (the channel's reaction to signals of different frequencies) might be something like this: The bandwidth of a channel depends on the physical properties of the channel, so a copper wire will have a different bandwidth from a wireless channel and from an optical fiber. Here, for example, is a table from wikipedia, specifying the bandwidths of different twisted pair cables.

If our example channel has a bandwidth of 1Mhz, then we can fairly easily use it to send a signal whose bandwidth is 1Mhz or less. Signals with a wider bandwidth will be distorted when passing through, possibly making them unintelligible.

Now let's get back to our example signal `__|‾‾|__|‾‾|__|‾‾|__|‾‾`. If we were to perform a Fourier analysis on it, we would discover that increasing the data rate (by making the bits shorter and closer to each other), increases the signal's bandwidth. The increase would be linear, so a two fold increase in the rate of bits, will mean a two fold increase in the bandwidth.

The exact relation between bit rate and bandwidth depends on the data being sent as well as the modulation used (such as NRZ, QAM, Manchseter, and others). The classic way in which people draw bits: `__|‾‾|__|‾‾|__|‾‾|__|‾‾` is what NRZ looks like, but other modulation techniques will encode zeroes and ones into different shapes, affecting their bandwidth.

Since the exact bandwidth of a binary signal depends on several factors, its useful to look at the theoretical upper bound for any data signal over a given channel. This upper bound is given by the Shannon–Hartley theorem: C is the channel capacity in bits per second;

B is the bandwidth of the channel in hertz (passband bandwidth in case of a modulated signal)

S is the average received signal power over the bandwidth (in case of a modulated signal, often denoted C, i.e. modulated carrier), measured in watts (or volts squared)

N is the average noise or interference power over the bandwidth, measured in watts (or volts squared)

S/N is the signal-to-noise ratio (SNR) or the carrier-to-noise ratio (CNR) of the communication signal to the Gaussian noise interference expressed as a linear power ratio (not as logarithmic decibels).

One important thing to note however, is that the Shannon-Hartley theorem assumes a specific type of noise - additive white Gaussian noise. The upper bound will be lower for other, more complex, types of noise.

• Also, on the receiving end, you have the Nyquist–Shannon sampling theorem that limits what can be detected – Remi Letourneau Feb 3 '14 at 21:56

Let me give the or practical, real-life network engineering answer. Here's the relationship bandwidth and frequency: Higher bandwidth, higher frequency. Done.

No, seriously, end of question and answer. You're done, move on to Layer 2.

I don't mean to be rude or smartass. Your question has delved way too far into the electrical engineering aspect of the Physical layer to be about what is known as network engineering. What you're asking is far more relevant to telecommunications, electrical engineering, or even computer science than network engineering in all but the strictest, most literal sense. It is also not relevant for anyone but extremely specialized personnel developing either the hardware or the protocols implemented by the hardware. I'd be quite surprised if most CCIE's could answer this question to the degree Mike Pennington did... and wouldn't be surprised at all if they didn't know enough to ask the original question with as much depth as you did!

Let me put it another way: If you're studying network engineering in the traditional sense, you have mastered Layer 1 far beyond (oh so far beyond) what is required, or even useful in a normal network engineering career. You're good, move on, there's far more to learn.

• On the one hand, it may be true that this isn't directly useful information day to day managing a wired network. On the other hand, I personally have NEVER come across a situation where I regretted knowing more about the fundamentals of how things work and many times been in situations where I wished I understood something better. In this particular case, this borders on many of the same principles that apply to RF, which is something I often do delve into on any given day as a network engineer. So a -1 from my perspective. – YLearn Aug 28 '14 at 19:33