Digital Communica.ons Fredrik Rusek. Chapter 2 Proakis- Saleh

Transcription

1 Digital Communica.ons Fredrik Rusek Chapter 2 Proakis- Saleh

2 Schedule, tenta7ve Lectures Chapter 2. 2h Chapter 3. 2h Chapter 4. 4h Chapter 6. 2h Chapter 8. 2h Chapter 9. 2h Chapter 10. 2h Chapter 11. 2h Chapter 13. 2h Chapter 14. 2h Chapter 15. 2h Chapter 16. 2h 13 Lectures in total With one lecture/week, we finish on June 17 Problem: Easter break.?

3 Schedule, tenta7ve Examina7on is done via hand in assignements. One set per chapter in the book Must be handed in latest Friday, the week auer the lecture. Problems are discussed at seminars Thursdays the week auer the lecture. Reqs for passing degree 80% a[endance at lectures 80% a[endance at seminars Hand in all home- assignements Delivered one lecture each (I will give the rest)

4 Sec.on 2.1 Lowpass eqv of bandbass signal Objec.ve of this sec.on: To understand how to represent an arbitrary bandpass signal in terms of lowpass signals This is compactly done via the Hilbert transform

5 Chapter 1 Determinis.c and random signal analysis The unit step is defined as Its Fourier transform becomes

6 Sec.on 2.1 Lowpass eqv of bandbass signal Assume a real- valued bandpass signal x(t) and make defini7ons Define the pre- envelope of x(t) as

7 Sec.on 2.1 The func7on x (t) is the Hilbert transform of x(t) Define the lowpass equivalent (aka complex envelope) of x(t) as Small check point: is x l (t) real or complex valued?

8 Sec.on 2.1 Now take the inverse Fourier transform to get x l (t) (1) (2) Mul7plying with e j2π f 0 t on both sides (1) gives

9 Sec.on 2.1 With real- valued and, we can write In- phase quadrature Using the equa7on (2), we can iden7fy and

10 Sec.on 2.1 With real- valued and, we can write Any bandpass signal x(t) can be expressed in terms of two lowpass signals, the in- phase and the quadrature components In- phase quadrature Using the equa7on (2), we can iden7fy and

11 Sec.on 2.1 We can of course do a change into polar coordinates, so that we obtain Envelope phase

12 Modulator 1 and 2 Sec.on 2.1

13 Demodulator 1 Sec.on 2.1

14 Demodulator 2 Sec.on 2.1

15 Sec.on 2.1 Energy considera7ons Energy of passband signal is twice that of the lowpass signal. Why?

16 Home assignement. Prove: Sec.on 2.1 Interes7ng consequence (example 2.1-1) Passband signals are orthogonal, but the lowpass signals are not.

17 spectra of of the input input and the the output is is given by by spectra spectra ofthe the input and and the output output is given given by Y(f) = = X(f)H(f) Y(f) Y(f) = X(f)H(f) X(f)H(f) (2.1-28) (2.1-28) (2.1-28) Using Equation Equation 2.1-5, we we have Using Using Equation 2.1-5, 2.1-5, we have have Y1(f) = = 2Y(f + fo)u-j(f fo)u-j(f fo) Y1(f) Y1(f) = 2Y(f 2Y(f + fo)u-j(f + + fo) fo) = 2X(f 2X(f fo)h(f fo)u-j(f fo) = + fo)h(f + fo)u-j(f = 2X(f + fo)h(f + fo)u-j(f + + fo) fo) Sec.on == 1[2X(f [2X(f + + fo)u-j(f fo)u-j(f + + fo)] fo)] [2H(f [2H(f + +fo)u-j(f fo)u-j(f + + fo)] fo)] == lxi(f)hi(f) lxi(f)hi(f) (2.1-29) (2.1-29) where we the ff >> -- fo, where we have have used used the fact fact that that for for fo, which which isis the the range range of offrequencies frequencies of of Bandpass signal observed through binterest, andpass ifo)mpulse response == u_ interest, u:_ u:_ (f (f + + fo) u_ (f (f + + fo) fo) == 1.1.In In the thetime timedomain domainwe wehave have YI(t) YI(t) == 2XI(t) 2XI(t)**hi(t) hi(t) (2.1-30) (2.1-30) Equations Equations and and show show that that when when aa bandpass bandpass signal signal passes passes through through aa Impulse response of bandpass channel: bandpass system, the input-output relation between the lowpass equivalents bandpass system, the input-output relation between the lowpass equivalents isis very very similar similar to to the the relation relationbetween between the the bandpass bandpass signals, signals, the theonly only difference differencebeing beingthat thatfor for the the lowpass lowpass equivalents equivalents aafactor factor of of isis introduced. introduced. Channel output at bandpass: At baseband: In the 7me- domain: (c.f. energies at slide 15)

18 Sec.on 2.2: Signal space Inner product between signals: Induced norm: Assume finite energy signal and a set of orthonormal signals We can approximate with the error

19 Sec.on 2.2: Signal space Orthogonality principle states that the best approxima7on is found when the error is orthogonal to the func7ons in the expansion, i.e., Representa7on error: Why?

20 Sec.on 2.2 Complete basis: Gram- Schmidt orthogonaliza.on procedure: Given a set of finite energy signals, the G-S finds a set of Orthonormal basis functions for the signals

21 Example 2.2-3, G- S Find basis for this set of signals Typo in book. Should be 4

22 Example 2.2-3, G- S Find basis for this set of signals

23 Example 2.2-3, G- S Find basis for this set of signals

24 Example 2.2-3, G- S Find basis for this set of signals

25 Example 2.2-3, G- S Find basis for this set of signals This is equivalent to a 3x4 MIMO system. Hence, the DoF is 3 (aka pre- log factor), and signaling is very inefficient. Note that we did not break the Nyquist limit in any sense.

26 Pages Good stuff Read on your own

27 Let x be a N(0,1) variable Sec.on 2.3: Random variables Useful approxima.on: for large x Common nota.on in many books/papers Char- func.on

28 Sec.on 2.3: Chi- square random variable. Let Xi be zero- mean Gaussian r.v. With common variance, then Is a chi- square variable with n DoF. The pdf of X is managble to deal analy.cally with in most cases. Char func7on What it the variances are not the same in Xi?

29 Sec.on 2.3: Some facts: A chi- square measure something related to power, not amplitude With n=2, one gets an exponen7al random variable Special case of Gamma random variable Square of a chi- variable Variances must be equal, and Xi must be independent At a high n (DoF), X converges quickly to N(0,1) auer suitable scaling Non- central version exist (E(Xi)>0). Much harder to deal with. (Ricean)

30 Sec.on 2.3: Rayleigh distribu7on Square root of a chi- square with DoF=2 Models the amplitude of something. Important since the norm of x, where x is CN(0,1) distributed is Rayleigh distributed.

31 Sec.on 2.3: Ricean distribu7on Hard to deal with due to the Bessel func7on in its pdf Mean of X is given by a confluent hypergeometric func7on Standard misunderstanding: If a channel link, h, between A and B can be modeled as a known LOS component m plus a complex Gaussian r.v. v, then h is not Ricean distributed, h is complex Gaussian distributed. However, h is Ricean.

32 Sec.on 2.3: Ricean distribu7on The Rice- factor (or K- factor) is For K=0, we get a Rayleigh For large K, we get a Gaussian random variable

33 Rayleigh and Rice are commonly used to model the received signals in fading channels. Sec.on 2.3: However, the design parameters of Rice and Rayleigh are very limited, and ouen more general families of pdfs are needed. Rayleigh The most common extension is Nakagami fading One can play with the mass of the tail

34 Sec.on 2.3: Log- normal random variable Almost impossible to use for analysis Used to model shadow fading (houses, trees, trucks etc.) A simple, yet common channel model is to mul7ply one log- normal random variable with one Rayleigh variable.

35 Sec.on 2.3: Jointly normal variables. Assume two normal variables X and Y Assume that they are uncorrelated, i.e., E(XY)=0 What can be said about their independence?

36 Jointly normal variables. Sec.on 2.3: Assume two normal variables X and Y Assume that they are uncorrelated, i.e., E(XY)=0 What can be said about their independence? Not much! The word jointly has a precise meaning and should not be forgohen

37 Jointly normal variables. Sec.on 2.3: Assume two normal variables X and Y Assume that they are uncorrelated, i.e., E(XY)=0 What can be said about their independence? Not much! The word jointly has a precise meaning and should not be forgohen Let X be a normal variable, and let t be a discrete r.v. with P(t=1)=P(t=- 1)=1/2 Define Y=tX E(YX)=(tXX)=E(t)=0 X and Y are uncorrelated X = Y X and Y are dependent X and Y are indeed normal, but not jointly normal

38 Sec.on 2.4: The Chernov bound is very useful and is an extension of the Markov bound to nega.ve variables. Sec.on 2.6: Proper complex random variables loosely speaking means that the real and the imaginary parts are iden.cally distributed. In dig.com., proper complex channel inputs are close to always used

39 Sec.on 2.7: Random processes Mean, auto- correla7on, and cross- correla7on func7ons Wide- sense sta7onary (WSS): How about a sta.onary process?

40 Sec.on 2.7: Random processes Mean, auto- correla7on, and cross- correla7on func7ons Wiener- Khinchin Theorem. For a WSS, the power spectrum sa7sifes Define the cross spectral density Wide- sense sta7onary (WSS): X(t) through impulse response h(t) X(t) and Y(t) joint WSS:

41 Sec.on 2.7: Random processes Mean, auto- correla7on, and cross- correla7on func7ons Wiener- Khinchin Theorem. For a WSS, the power spectrum sa7sifes Home assignement: Read (e.g. Define On Wikipedia) the cross spectral about density the Bussgang Theorem and compare with Wide- sense sta7onary (WSS): X(t) through impulse response h(t) X(t) and Y(t) joint WSS:

42 White Gaussian processes Sec.on 2.7 A process is Gaussian, if all possible subsets of samples taken from X(t) are jointly Gaussian A process is called white if its power spectral density is constant Power is infinite, and therefore only models physical phenomenon Thermal noise N(t) can be modeled as Typo in book. kshould be 1.38

43 Sec.on 2.7 Cyclo- sta.onary processes Wiener- Khinchin theorem must be modified A cyclo- sta7onary process is most common in dig.com, e.g., linear modula7on is cyclo- sta7onary

44 Sec.on 2.8 The sampling theorem Assume a complex signal x(t). Let W1 and W2 be the smallest and the largest frequencies in the signal. Define W = W1 - W2 The signal x(t) can be perfectly reconstructed from its samples if the rate is at least 2W Implica.on: The signal is completely specified from 4W real samples per second. 2W complex per sec = 4W real per sec Real signals: 2W real samples per second.

45 Sec.on 2.8 The sampling theorem The sampling theorem was proved by Shannon, and not by Nyquist Hand in problem Prove

46 Sec.on 2.8 Karhunen- Loeve expansion Pure sampling of the signal x(t) is of course not the only way to reach a series expansion of x(t). One can reach an expansion where the coefficients are uncorrelated through the K- L framwork. This is, in some sense, the op7mal set of basis func7ons, as it compresses the energy of the signal into a few basis func7ons (since it avoids the redundancy of having correlated variabels)

47 Sec.on 2.8, Example Karhunen- Loeve expansion of White Gaussian noise Any orthonormal basis works as K- L basis for white Gaussian noise. This is a bad result, as one would have hoped (in general) that most of the energy would have concentrated to a few basis func.ons

48 Sec.on 2.9: bandpass random processes Assume a WSS bandpass random process X(t) We can define its in- phase, quadrature, and low- pass equivalent signals as Long, but straighqorward, manipula7ons gives In- phase and quadrature components are jointly WSS with equal power spectrums

49 Sec.on 2.9: bandpass random processes Take Fourier transform to get power spectrums:

50 Sec.on 2.9: bandpass random processes We can also reach (by using results for complex vectors that I did not cover at the lecture) From slide 8

51 Sec.on 2.9: bandpass random processes We can also reach (by using results for complex vectors that I did not cover at the lecture) Take Fourier transforms use results at last slide

52 Example Important!

53 Hand in problems 2.2, 2.9, 2.11, 2.22, 2.38, 2.43, 2.44 Read about the Bussgang Theorem at e.g., Wikipedia