Physical Layer – How bits are sent

Goal

• Physical layer design goal: send out bits as fast as possible with acceptable low error ratio

• Goal of this lecture: get to know the basics of physical layer design, the constraints, the solutions, should be able to solve some simple problems

Some simple schemes

• There is a wire between A and B. If A wants to send a bit `1’, he connects the wire to the positive end of a battery. Otherwise he disconnects it from the battery.

• Or A can hold a radio, if `1’, he sends at frequency f1 and if `0’ he sends at frequency f2.

• Or there is an optical fiber between A and B and if `1’ A lit up a light and if `0’ A does nothing.

Ethernet

• So, why not simply do that? We can let a electronic switch to do the switching between 0s and 1s.

• In fact, yes we can.• This is what is done in 10Mbs Ethernet.

Ethernet Physical Layer

• Get bits from upper layer, do Manchester encoding, then pull the voltage up and down

• The only thing that is added other than simply sending 1 and 0 is the Manchester encoding – why do we need it? What could happen if you miss the sample time?

• How do you know the beginning of a frame?

Ethernet

• 10 M Ethernet is probably the simplest case – computers all connect to a bus of limited length (<1km) and the speed is not required to be very high – so it can afford to be simple

• In a sense, it does not have to deal with the two things that can make life a lot more complicated– Bandwidth– Noise

Bandwidth and noise

• Bandwidth basically means how fast your signal can change or how fast can you send out symbols.– Symbol is something you send out to represent

bit(s)

• Noise means that although you sent 1 to me, I may receive something like 1+x, where x is the noise added by the media.

Ideal case

• If the bandwidth is infinite and absolutely no noise, how fast can you send/receive data?

Bandwidth

• If the media is of infinite bandwidth but with some noise, how fast can you send/receive data? Assuming that your device is fast enough.

Noise

• If there is absolutely no noise but the bandwidth is limited, how fast can you send/receive data? Assuming that your device is fine enough to tell the slightest differences of signal voltage.

Shannon’s Theorem

• C=B*log(1+S/N)– C is the capacity of the channel, B is the

bandwidth of the channel, S is power of the signal and N is the power of the noise

– Channel capacity means how many bits you can send out per second reliably

Shannon’s Theorem

• There is actually a very simple way to understand Shannon’s theorem– B means how fast can you send out symbols– S/N determines how many bits each symbol

carries – why there is a log?

Limited bandwidth

• The trouble is, we live in a world with limited bandwidth and some noise.

• Noise is easy to model. • In mathematical languages, any signal can be

viewed as the sum of a series of sine waves on different frequencies

• You pass the signal to a channel that can pass frequencies up to B, all sine waves on frequencies higher than B will be lost – you will receive a distorted signal

Nyquist Theorem

• If the bandwidth is limited to B, in the ideal case when there is no noise, how fast can you send/receive symbols? – Note that the channel capacity is infinity because each

symbol can carry infinite number of bits• Nyquist Theorem says that it only makes sense for you

to send/receive symbols at a speed of 2B – if B is 4KHz, you send/receive 8K symbols per second – the baud rate is 8K per second.

• Why? If a signal is band-limited by BHz, by taking 2B samples per second, you can completely reconstruct it.

Limited Bandwidth

• Bandwidth is limited because of many reasons– The wire itself, if too long, is a capacitor and slows

down voltage transition– In wireless transmissions, the whole spectrum

shared by many communication parties and each can have only a limited chunk of it

Modulation

• Even in the simplest 10Mbs Ethernet case, the signals you send out is different from the 0-1 waveform.

• Modulation is to convert the 0-1 waveform to the actual signal to be sent out.

• In most cases the signals are derived from sine waves.– Given the 0-1 waveform, you do some tricks to the

sine wave at some frequency f, and send it out– This sine wave is called the carrier

Different modulation techniques

• Amplitude modulation• Frequency modulation• Phase modulation• In digital communications, phase modulation

is used

Demodulation

• The problem is, given a (0,180) phase modulated signal, how to determine whether the bit is 0 or 1?

• This is demodulation.• Assume that you know the beginning of a

binary symbol.

Demodulation

• You multiply the received signal with a sine wave and do an integral. If the result is positive, it was 1 and if it was negative, it was 0.

• If no noise, things are simple!

Demodulation• With noise, it’s all about guessing, because you

don’t know what the noise is when this symbol is sent as noise is random.

• You may know some statistics of the noise, based on which you make your best guess.

• For example, let’s say 0 is 0 volt 1 is 5 volts. Suppose you know that very rarely the noise exceeds 2.5 volts. If you received a 2.2 volts, you would guess it to be 0 or 1? What is the chance that you got it right/wrong?

Maximum Likelihood Detection

• Detection – given a received signal, determine which of the possible original signals was sent. There are finite number of possible original signals (2 for the binary case – 0 or 1)

• Compute a likelihood value for every possible input, choose the one with largest likelihood – maximum likelihood detection

Maximum Likelihood Detection

• There are two inputs, x1 and x2. Noise is n. What you receive is y.

• If I sent x1, you receive y=x1 + n. If I sent x2, you receive y=x2+n. You don’t know what I sent and how large n is.

• You compute the likelihood of receiving y if I sent xi, Li (i=1,2). If L1 > L2, you say I sent x1. Else you say I sent x2.

• How to compute L1 and L2?

Maximum Likelihood Detection• If n=0 always, y=x1 if I sent x1 and y=x2 if I sent x2. Of course x1 !=

x2. Will you make mistake in this case? What is the likelihood of y=x2 if I sent x1?

• If n is not always zero, we assume n follows some probability distribution. If it is Gaussian, the channel is called AWGN.

• Given y, the likelihood of x1 being sent is the likelihood that n=y-x1. Similarly, the likelihood of x2 being sent is the likelihood that n=y-x2. (likelihood is derived from probability, but likelihood could be taking some values that probability cannot take depending on how you define likelihood)

• So what you are doing is to compare the likelihood of n=y-x1 and n=y-x2. So the detection rule is if p(n=y-x1)/p(n=y-x2) > 1, output x1, else output x2.

• That’s all!• Wait, what if you know that x1 is more likely to be sent than x2?

I,Q channels• In modern communications, you have two carriers on the

same frequency but their phases differ by 90 degrees. So one is sine and the other is cosine. You can apply two symbol streams (called the baseband signals) to them separately, and add the two together and send out.

• The receiver won’t be confused because what he does is to multiply the received signal with two locally generated sine waves on the same frequency as the sender side, one sine and one cosine. Then he passes each of the two multiplied signals to a device that only allows signals on frequency no higher than B to pass (assuming the highest frequency of your signal is B). Why?

QAM

• With two carriers, you have two dimensions. So the signal will appear on a plane. 16-QAM, 64-QAM, 256-QAM

Wired Communication – Telephone Company

• Dial-up – 56kbps• DSL – Digital Subscriber Line– ADSL: Asymmetric DSL, different upload and

download bandwidth– Available bandwidth is about 1.1MHz, divided into

256 channels, one for voice, some unused or for control, the rest divided among upstream and downstream data. My DSL at Pittsburgh was 100kbps upstream and 768kbps downstream

– How ADSL is set up. Fig. 2-29. The ADSL modem is 250 QAM modems operating at different frequencies. The actual QAM depends on the noise.

Wired Communications – The Cable TV Company

• Cable frequency allocation. Fig. 2-48.– Downstream channel bandwidth is 6MHz. If using

QAM-64, how much of a speed we can get? – Upstream channel is worse so use QAM-4.– Upstream – stations contend for access (MAC

layer issue, will be discusses later)– Downstream – no contention, from the head end

to user– Shared medium, so some security is needed

Wired communication – Optical Backbone

• SONET

Wireless communications

• FDMA – Frequency division multiplexing• TDMA – Time division multiplexing• CDMA – Code division multiplexing