Physical layer continued

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Transcript Physical layer continued

Physical layer continued
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
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.
Nothing more can be reconstructed, so no point of sending.
Detection
• 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
• 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
• 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
• 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=yx2. (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?
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, and
may use QAM-64.
– 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 – Synchronous Optical Network
– OC-1: 51.84Mbps
– OC-3: 3*51.84Mbps
– OC-9: 9*51.84Mbps
–…
• Used for backbone switching
Cellular Phone Networks
• User – base station – Telephone network
• FDMA – Frequency division multiplexing
– How to make sure that you are using this band,
not that band?
• TDMA – Time division multiplexing
• CDMA – Code division multiplexing
GSM – Global System for Mobile
Communications
• Second generation cell phone system (digital, first
generation analog).
• GSM-900 and GSM-1800 are most widely used
– GSM-900 uses 890 - 915 MHz to send information from the
Mobile Station to the Base Transceiver Station (uplink) and 935 960 MHz for the other direction (downlink).
• FDMA + TDMA
• Each user transmitting on a frequency and receiving on
another frequency.
• 124 pairs of 200 KHz channels. Each channel divided into 8
time slots for 8 users.
• Each user is has a chance to transmit every 4.615 ms. Each
time he can send 114 data bits – 24.7kbps.
CDMA
• Described in IS-95.
• A good analogy in the book – You have a
group of people in a room. TDMA means they
talk in turn. FDMA means that those who
wants to talk sit in different corners and can’t
hear other pair. CDMA means each pair talks
in a different language and other people’s
voice is noise to them.
CDMA
• The whole bandwidth is used by every user.
Meaning that they can send out symbols really
fast.
• The trick is to make what A sent appear as 0 to B.
– Because we have a fast symbol rate, for each data bit,
we send out, say, 8 bits, call the “small bits” chips.
– Given a bit, if 1, send out, say, -1,-1,-1,1,1,-1,1,1, and if
0, 1,1,1,-1,-1,1,-1,-1
– This is called the chip sequence.
– The key is that each station has a unique chip
sequence (language), and different languages are
orthogonal.
– Fig. 2-45.
Wireless LAN Physical Layer
• 802.11b,g in the 2.4G band and 802.11a in the 5G band.
People now consider 802.11 as the notion of MAC layer
protocol, while a, b, g, or n, are about physical layer.
• 802.11b. 1, 2, 5.5, 11Mbps.
– 1Mbps: BPSK modulation. 1 bit into 11 chips with Barker
sequence +1 +1 +1 −1 −1 −1 +1 −1 −1 +1 −1. Why spread
spectrum? Required by FCC but was later removed
– 2Mbps: QPSK.
– 5.5M and 11M: use some bits to select chip sequence and use
two bits for QPSK
• 802.11a. Up to 54Mbps. OFDM.
• 802.11g. Up to 54Mbps. OFDM.
OFDM (Orthogonal Frequency Division
Multiplexing)
• In wireless communications, in addition to the
bandwidth limit and additive noise, you also have
multipath fading!
• The faster your symbol rate is, the more badly you will
be affected by multipath fading.
• In effect, OFDM is like DSL: given a wideband channel,
divide it into many sub channels. Each sub-channel can
be modulated/demodulated independently. Because
each sub-channel is of a much smaller bandwidth,
multipath fading is much less severe.
• In implementation, use IFFT and FFT.
MIMO
• Used in 802.11n.
• t transmit antennas and r receive antennas.
With the knowledge of channel matrix, by preprocessing the data, equivalent to min{t,r}
channels.