Physical layer continued Limited Bandwidth Bandwidth is limited

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 2 B – if B is 4 KHz, you send/receive 8 K symbols per second – the baud rate is 8 K per second. • Why? If a signal is band-limited by BHz, by taking 2 B 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, x 1 and x 2. Noise is n. What you receive is y. • If I sent x 1, you receive y=x 1 + n. If I sent x 2, you receive y=x 2+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 L 1 > L 2, you say I sent x 1. Else you say I sent x 2. • How to compute L 1 and L 2?

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=x 1 if I sent x 1 and y=x 2 if I sent x 2. Of course x 1 != x 2. Will you make mistake in this case? What is the likelihood of y=x 2 if I sent x 1? • 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 x 1 being sent is the likelihood that n=y-x 1. Similarly, the likelihood of x 2 being sent is the likelihood that n=y-x 2. (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-x 1 and n=y-x 2. So the detection rule is if p(n=y-x 1)/p(n=y-x 2) > 1, output x 1, else output x 2. • That’s all! • Wait, what if you know that x 1 is more likely to be sent than x 2?

Wired Communication – Telephone Company • Dial-up – 56 kbps • DSL – Digital Subscriber Line – ADSL: Asymmetric DSL, different upload and download bandwidth – Available bandwidth is about 1. 1 MHz, 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 100 kbps upstream and 768 kbps 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 6 MHz, 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. 84 Mbps – OC-3: 3*51. 84 Mbps – OC-9: 9*51. 84 Mbps –… • 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. 7 kbps.

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, and if 0, 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. 11 b, g in the 2. 4 G band 802. 11 a in the 5 G band. People now consider 802. 11 as the notion of MAC layer protocol, while a, b, g, or n, are about physical layer. • 802. 11 b. 1, 2, 5. 5, 11 Mbps. – 1 Mbps: BPSK modulation. 1 bit into 11 chips with Barker sequence +1 +1 +1 − 1 − 1 +1 − 1. Why spread spectrum? Required by FCC but was later removed – 2 Mbps: QPSK. – 5. 5 M and 11 M: use some bits to select chip sequence and use two bits for QPSK • 802. 11 a. Up to 54 Mbps. OFDM. • 802. 11 g. Up to 54 Mbps. 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. 11 n. • t transmit antennas and r receive antennas. With the knowledge of channel matrix, by preprocessing the data, equivalent to min{t, r} channels.