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1 Why do we computer scientists need to know communication technologies? Professor R. C. T. Lee.

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Presentation on theme: "1 Why do we computer scientists need to know communication technologies? Professor R. C. T. Lee."— Presentation transcript:

1 1 Why do we computer scientists need to know communication technologies? Professor R. C. T. Lee

2 2 In these days, it is hard to imagine any computer which stands entirely alone. A computer is often connected to others, through communication technologies. Many CPU’s simply have built-in communication mechanisms.

3 3 Besides, many systems which were traditionally considered as communication systems are actually quite similar to computers. Examples: TV, mobile phone.

4 4 We can use our mobile phone to access internet. Is a mobile phone a computer? We can also insert a sim-card into a computer and the computer suddenly can access the base stations which were built for mobile phones? Is a computer a mobile phone?

5 5 But, unfortunately, almost all computer scientist students do not understand some very basic communication technologies. Question 1: How is a bit represented when it is transmitted in a wireless environment?

6 6 We often thought that a bit is represented by a pulse. (1,1,0,0,1)

7 7 But we cannot send a pulse in the wireless environment. Question 2: We often mix bits together. An example: In the ADSL system, 256 bits are bundled together.

8 8 If the bits are represented by pulses, how can they be mixed and later be distinguished. Question 3: We often mix bits by different users together. If the bits are represented by pulses, how can they be distinguished when they are received?

9 9 A very fundamental problem in communications is to study how digital information is represented, sent and later separated. Fundamental concept: There is no digital signal. Every digital bit is represented by an analog signal.

10 10 Consider the simplest case: There is only one user. We represent bit 1 by and represent bit 0 by. We further require that and are orthogonal.

11 11 That is, the inner product and for every i..

12 12 For the receiver, it either receives or. How do we know which function is received? We perform two inner products. Let the sent signal, which is received, be denoted by. We calculate and

13 13 Suppose the sent signal is, If, we conclude that we sent, which is 1(0). Conclusion:

14 14 There are millions of possible functions for and. We may let and We may also let and We may of course let and.

15 15 It can be easily seen that We can also prove, for instance, that if.

16 16 and iff If the sent bit is 1(0),

17 17 We should now always remember that every bit, 1 or 0, is represented by a cosine or a sine function. In the above, we assumed that every bit is sent alone. Can we send two bits together? Yes, we can.

18 18 Let us assume that we are going to send two bits: Bit 1 and Bit 2. Bit 1 can be 1 or 0 and Bit 2 can also be 1 or 0. We like to mix Bit 1 and Bit 2 together and send the mixed signal out. The important thing is that the receiver must be able to correctly determine what Bit 1 and Bit 2 are.

19 19 Let Bit 1(Bit 2) be represented by. The value of is determined by the value of Bit 1(Bit 2). We may let be 1(-1) if Bit i is 1(0) for Thus the sent signal is For instance. Suppose Bit 1 is 1 and Bit 2 is 0. Then the send signal is

20 20 The job of the receiver is to determine the values of for Note that To determine, we perform an inner product between and Let

21 21 Similarly, we have

22 22 For instance, let And Then, at any time,

23 23 For cases, (1)a 1 =1, a 2 =1 (2) a 1 =1, a 2 =-1 (3) a 1 =-1, a 2 =1 (4) a 1 =-1, a 2 =-1 This is called the QPSK system.

24 24 If we conclude that User 1 sends 1(0). Similarly, if we conclude that User 2 sends 1(0).

25 25 We can now see how two bits can be mixed and sent without any trouble. Essentially, we must understand that the digital signals are represented by analog signals which are orthogonal to each other. The receiver uses the property of orthogonality to separate the signals.

26 26 If we can mix 2 bits together, we can of course mix 256 bits together. In fact, the ADSL system uses this kind of scheme. This is why the ADSL system is a very fast system.

27 27 In the ADSL system, each bit i is represented by Thus, each signal is orthogonal to others due to the distinct frequencies. This is why the system is called Orthogonal Frequency Division Multiplexing (OFDM) system.

28 28 The coding of digital data by analog signals is often called digital modulation. The decoding of analog signals back to digital signals is called demodulation.

29 29 Let us go one step further by mixing bits from different people. Our trick is the following: A bit of 1 or 0 for User 1 is represented differently from a bit of 1 or 0 for User 2. Let us consider a simple case: two users.

30 30 Let the bit of 1 for User 1 be coded as and the bit of 0 for User 1 be coded as Let the bit of 1 for User 2 be coded as and the bit of 0 for User 2 be coded as We may say that User i sends where or

31 31 Since and are vectors, we use dot- product as the inner product. We therefore conclude that and are orthogonal to each other.

32 32 We may mix the bits of User 1 and User 2 and the mixed signal is therefore where or. The job of the receiver is to determine Again, this can be done by using the orthogonality of and

33 33 To find we calculate Thus can be found similarly.

34 34 Example: User 1 sends 1 and User 2 sends 0. User 1 sends 1. User 2 sends 0.

35 35 Thus we can see that we can even mix bits of two different users. The main trick is that we represent the bits of different users by vectors and make sure that they are orthogonal to each other. This can be easily extended to more than 2 users.

36 36 Let us now ask another interesting question. Our cables are often used to transmit digital data. We like our channel to transmit a large amount of bits per second. We are talking about high bit rate systems. Why do we often call these channels broadband systems?

37 37 By a broadband system, we mean it can transmit signals with a large range of frequencies. Why is a high bit system also a broadband system? This can be understood only through the knowledge of Fourier transform.

38 38 Fourier transform tells us that any signal contains a set of cosine functions with different frequencies.

39 39 A signal

40 40 The Discrete Fourier Transform Spectrum of the Signal in the Previous Slide f(t)=cos(2π(7)t)+3cos(2π(13)t).

41 41 A Music Signal Lasting 1 Second

42 42 A Discrete Fourier Transform Spectrum of the Music Signal in the Previous Slide.

43 43 Suppose that a bit is a rather wide one, as shown below: Then its Fourier transform spectrum is as follows: Not much frequencies are contained.

44 44 Suppose the system is a high bit rate system, the bit length is therefore very very short: Its Fourier transform is as follows: A large number of frequencies are contained.

45 45 Conclusion: In a low bit rate system, the bit length is very long and it actually contains a narrow band of frequencies. This can be understood by imagining the bit length to be so long that the signal becomes DC. In this case, the only frequency it contains is

46 46 On the contrary, in a high bit rate system, the bit length is very very short, it contains a wide range of frequencies. The system is consequently called a broadband system because for a wide range of frequencies, it must respond equally well.

47 47 By using Fourier transform, we can see that the frequency components in our human voice are roughly contained in 3k Hertz.

48 48 For a signal with frequency f, its wavelength can be found as follows: where f is the velocity of light?

49 49 If,.

50 50 It can also be proved that the length of an antenna is around. For human voice, this means that the wavelength is 50km. No antenna can be that long.

51 51 What can we do? Answer: By amplitude modulation.

52 52 Let be a signal. The amplitude modulation is defined as follows: where f c is the carrier frequency?

53 53 What is the Fourier transform of ?

54 54 Fourier transform tells us that every signal contains a bunch of cosine functions. Let us consider.

55 55 Thus every frequency f is lifted to.

56 56 The effect of amplitude modulation is to lift the baseband frequency to the carrier frequency level, a much higher one. Once the frequency becomes higher, its corresponding wavelength becomes smaller. An antenna is now possible.

57 57 After we receive, how can we take out of it? Answer: Multiply by.

58 58 Thus is recovered. We need a low-pass filter to get because is of low frequency.

59 59 We should all remember that digital signals are all sent as analog signals. Conceptually, the following is a typical one what we call a truncated cosine signal.

60 60 The Voice Hiding Problem The Artificial Music Pitch Generation Problem Analog modulation: We shall show that the voice hiding problem can be solved by analog modulation.

61 61 I want to hide a voice message into a music message, and the people who listen to the music can not detect. What can I do?

62 62 The frequencies of and are all low frequencies. If we make the frequencies of become high frequencies, people can not hear it. How do we make the frequencies higher? Approach: analog modulation:

63 63

64 64 Mixer

65 65 It is a signal of high frequency (We can not hear it.). It is a signal of low frequency (We can hear it.). High pass filter ( We still can not hear it. ).

66 66 If we want to extract the from, we multiply by.

67 67 The humans only can hear, because it is a low frequency signal.

68 68 We now show that the Fourier transform can be used to produce artificial music sounds.

69 69 What is the meaning of Do? Each music note corresponds to the vibration of the air in sinusoidal form with a certain frequency. Therefore, suppose one uses a signal generator to produce a cosine function with frequency 262Hz and let it go through a speaker, we shall hear a Middle C(Do). Cosine function of 262 Hz

70 70 Middle C of Piano

71 71 For the middle C range, let the frequency of the first Do be denoted as and the frequency of the next Do be. We then have.. We like to point out the following:

72 72 Let denote the frequency of the ith keyboard of a piano. Then.

73 73 For example, A4 (La) is the 49 th key from the left end of a piano (tuned to 440 Hz), and C4 (Do) is the 40 th key. These numbers can be used to find the frequency of C4: D4(Re) is the 42 th key, and the frequency of D4 is: CC#DD#EFF#GG#AA#B 262277294311330349370392415440466494 All frequencies of music tones of the middle C range

74 74 We can use a simple cosine function to represent a music tone.

75 75 Can we simulate the sound of a piano? The time domain function of middle C of piano is as following:

76 76 We transform the time domain function into the frequency domain function by using Discrete Fourier Transform(DFT).

77 77 Suppose we want to generate a simulated music note of any music instrument. We proceed as follows: 1.Obtain the time domain function of this music of note of the music instrument by asking someone to play it. 2.Perform a discrete Fourier transform of this function to obtain its frequency spectrum.

78 78 The Fourier transform will consist of roughly, say around 40,000 frequencies and their magnitudes. Among these frequencies, we pick frequencies with the largest magnitudes, and let us denote the frequencies as where and their magnitudes as,. We calculate for, and denote the function frequency as.

79 79 Example: For piano and middle C: we have the table as following: 10.26351262 20.70421.0038263 30.5051.0077264 40.33262.0038525 50.82.0077526 60.22552.0115527 70.30135.03451319 80.28237.07661854 90.26317.08051855 100.24027.08431856 Thus, we say that the music note Do for simulated piano has the frequency spectrum as described in the left table.

80 80 Similarly, we can generate frequency spectrum for other music notes for piano as well as for any other music instrument.

81 81 Having obtained the frequency spectrum of a music note, we use the Discrete Inverse Fourier Transform (DIFT) to transform the frequency domain function into time domain function, and now we can play a song by using the simulated music instrument. Amazing Grace played by simulated piano

82 82 We also simulated like violin or organ. Amazing Grace played by simulated violin Amazing Grace played by simulated organ

83 83 We can play other songs by the simulated instruments. Dreaming of home and mother (simulated violin) Auld lang syne (simulated organ and violin) 花非花 (simulated violin)

84 84 Conclusions You must have the analog feeling because digital signals are all analog signals. You should learn more about analog circuit design because this is where our country is exceedingly weak.

85 85 Thank you.


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