Physical Layer Continued

1. Physical Layer Continued

2. Review • Discussed how to design the communication scheme depending on the physical mediums – pulling voltage up and down for wired connections • Mentioned some important terms • Symbol: usually occupies a fixed length of time, carrying one or multiple bits. • Link speed: usually measured by bps (bits per second) • Discussed bandwidth and noise, and their influence on the maximum speed that can be achieved by the link. Roughly speaking, • Bandwidth determines how fast can we send symbols. For example, if the link does not allow the sudden change from 0v-5v, such a change will be smoothed out. So (roughly speaking) you cannot send symbols with durations shorter than the smoothing period. • Noise determines how close the constellations can be. For binary systems, the constellation lies in a line and is {-1, 1} where -1 is 0V and 1 is 5V, for example. This is called BPSK. The constellation can also be in a plane and take complex numbers when we send two voltages simultaneously, one representing the real component and the other representing the imaginary component, like {1+j, -1+j, -1-j, 1-j} , which is called QPSK. More points on the plane will result in something like 16QAM, 64QAM, 256QAM.

3. More on Bandwidth • Let’s dig a little deeper about bandwidth. • First Fact: Bandwidth is measured by Hz, so it is a frequency concept.

4. Frequency • Second Fact: Most of the signals we care about in communications can be represented by the summation of sine waves on different frequencies at certain magnitudes – Fourier transform. • So, when sending any signal waveform, think it as sending a whole bunch of sine waves all together. • http://www.facstaff.bucknell.edu/mastascu/elessonshtml/Freq/Freq4.html

5. Frequency • Third Fact: Most of the systems do NOT respond to all frequencies. What usually happens is that a system will respond only to a continuous range of frequencies. • For example, the human ear can pick up something like [0,20000]Hz. Our ears will not respond to a sound (a sine wave) produced by the dolphin on 120,000 Hz. • So, we say sine waves on frequency higher than 20000Hz are filtered out by our ears, and our ear is a low pass filter with a cutoff frequency of 20000Hz, or, the bandwidth of the human mouth to human ear link is no more than 20000Hz.

6. Bandwidth • So, for communication systems, the bandwidth is defined as the range of frequencies it passes. • The wider this range, i.e., the larger the bandwidth, the faster the signals are allowed to change, because there are higher frequency components in the signal.

7. Exercise • Suppose we are sending a signal cos(25000πt)cos(5000πt). It then passes through a Low Pass Filter with cut-off frequency of 12500Hz. What will be the signal coming out of the LPF? • For a signal like cos(2πft), the frequency is f, because it repeats every 1/f seconds.

8. Exercise • Suppose we are sending a signal cos(25000πt)cos(5000πt). It then passes through a Low Pass Filter with cut-off frequency of 12500Hz. What will be the signal coming out of the LPF? • Note that cos(25000πt)cos(5000πt)=1/2[cos(20000πt)+cos(30000πt)], so, while the frequency of the first sine wave is 10000Hz and the second is 15000Hz. So only the first one will stay.

9. 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.

10. More on Noise • In communication systems, we usually take a sample from the received waveform to determine what the transmitted symbol is. • With noise, the signal is added with noise. • For example, let’s say 0 is 0 volt 1 is 5 volts. When we send a `0’, the receiver could receive 0.6v, when we send `1’, the receiver could receive 4.2v. The receiver has to output a bit to the upper layer. So the problem is, given the received voltage, what bit should be output?

11. More on 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, 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?

12. 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)

13. Detection – An Example • Detection really depends on the noise. In the binary case when the transmitted signal is either 0 or 5V with equal probability, if we know that noise takes values -3V and 2V with probability 0.7 and 0.3, respectively. How would you design the detector?

14. Detection – An Example • Detection really depends on the noise. In the binary case when the transmitted signal is either 0 or 5V with equal probability, if we know that noise takes values -3V and 2V with probability 0.7 and 0.3, respectively. How would you design the detector? • First step, check the possible outcomes: • If send 0V, two possible outcomes: • -3V. Prob: 0.7. • 2V. Prob: 0.3. • If send 5V, two possible outcomes: • 2V. Prob: 0.7. • 7V. Prob: 0.3. • So, a total of only 3 possible outcomes.

15. Detection – An Example • Detection really depends on the noise. In the binary case when the transmitted signal is either 0 or 5V with equal probability, if we know that noise takes values -3V and 2V with probability 0.7 and 0.3, respectively. How would you design the detector? • First step, check the possible outcomes: • If send 0V, two possible outcomes: • -3V. Prob: 0.7. • 2V. Prob: 0.3. • If send 5V, two possible outcomes: • 2V. Prob: 0.7. • 7V. Prob: 0.3. • So, a total of only 3 possible outcomes. • Second step, check for each outcome, what should the output be. • No ambiguity when received -3V and 7V, 0 and 1, respectively. • What should we say when received 2v? From 0V, probability 0.5*0.3= 0.15. From 5V, 0.5*0.7=0.35. So will say when received 2V, the bit is 1. • What is the probability that we give the wrong detection result? When received 2V and was from bit 0 and we say it is from bit 1, so it is 0.15.

16. Maximum Likelihood Detection • There are two inputs, x1 and x2. Noise is n. What we receive is y. • If sent x1, we receive y=x1 + n. If sent x2, we receive y=x2+n. We don’t know • what was sent • how large n is. • The rule is: if P(Y=y|X=x1) > P(Y=y|X=x2), say sent x1, else say sent x2. • That is, compare the conditional probability.

17. Maximum Likelihood Detection • n follows some probability distribution known beforehand. • Note that • P(Y=y|X=x1) = P(n=y-x1) • P(Y=y|X=x2) = P(n=y-x2) • So the detection rule is ifP(n=y-x1)>P(n=y-x2) outputx1 else outputx2 • This will tell us for any received y whether to guess as x1 or as x2. That’s all.

18. Some Comments • When implemented in software, a BPSK detector (input signal from is a voltage level could be negative and the output is either bit 0 or bit 1 ) will be (for Gaussian noise) int BPSKDector(float sample) { return (sample > 0) ? 1 : 0; }

19. Beyond MLD – MAP (maximum a posteriori probability) • Wait, what if you know that x1 is more likely to be sent than x2? • The rule is: if P(X=x1|Y=y) > P(X=x2|Y=y), say sent x1, else say sent x2. • Let’s say that the probability that x1 is sent is p1 and x2 is p2, where p1!= p2. The detection rule should be changed to ifp1 P(n=y-x1)> p2 P(n=y-x2) outputx1 else output x2