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 B Hz, 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 if P(n=y-x1)>P(n=y-x2) output x1 else output x2 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 if p1 P(n=y-x1)> p2 P(n=y-x2) output x1 else output x2