Principles of Communication

Transcription

1 Principles of Communication The communication process: Sources of information, communication channels, modulation process, and communication networks Representation of signals and systems: Signals, Continuous Fourier transform, Sampling theorem, sequences, z-transform, convolution and correlation. Stochastic processes: Probability theory, random processes, power spectral density, Gaussian process. Modulation and encoding:

2 Basic modulation techniques and binary data transmission:am, FM, Pulse Modulation, PCM, DPCM, Delta Modulation Information theory: Information, entropy, source coding theorem, mutual information, channel coding theorem, channel capacity, rate-distortion theory. Error control coding: linear bloc codes, cyclic codes, convolution codes

3 Course Material 1. Text: Simon Haykin, Communication systems, 4th edition, John Wiley & Sons, Inc (2001) 2. References (a) B.P. Lathi, Modern Digital and Analog Communcations Systems, Oxford University Press (1998) (b) Alan V. Oppenheim and Ronald W. Schafer, Discrete-Time signal processing, Prentice-Hall of India (1989) (c) Andrew Tanenbaum, Computer Networks, 3rd edition, Prentice Hall(1998). (d) Simon Haykin, Digital Communication Systems, John Wiley & Sons, Inc.

4 *Duration:* 14 Weeks Course Schedule Week 1:* Source of information; communication channels, modulation process and Communication Networks Week 2-3:* Signals, Continuous Fourier transform, Sampling theorem Week 4-5:* sequences, z-transform, convolution, correlation Week 6:* Probability theory - basics of probability theory, random processes Week 7:* Power spectral density, Gaussian process Week 8:* Modulation: amplitude, phase and frequency Week 9:* Encoding of binary data, NRZ, NRZI, Manchester, 4B/5B

5 Week 10:* Characteristics of a link, half-duplex, full-duplex, Time division multiplexing, frequency division multiplexing Week 11:* Information, entropy, source coding theorem, mutual information Week 12:* channel coding theorem, channel capacity, rate-distortion theory Week 13:* Coding: linear block codes, cyclic codes, convolution codes Week 14:* Revision

6 Overview of the Course Target Audience: Computer Science Undergraduates who have not taken any course on Communication Communication between a source and a destination requires a channel. A signal (voice/video/facsimile) is transmitted on a channel: Basics of Signals and Systems This requires a basic understanding of signals Representation of signals Each signal transmitted is characterised by power. The power required by a signal is best understood by frequency characteristics or bandwidth of the signal: Representation of the signal in the frequency domain - Continuous Fourier transform

7 A signal trasmitted can be either analog or digital A signal is converted to a digital signal by first discretising the signal - Sampling theorem - Discrete-time Fourier transform Frequency domain interpretation of the signal is easier in terms of the Z-transform Signals are modified by Communication media, the communication media are characterised as Systems The output to input relationship is characterised by a Transfer Function Signal in communcation are characterised by Random variables Basics of Probability Random Variables and Random Processes Expectation, Autocorrelation, Autocovariance, Power Spectral Density

8 Analog Modulation Schemes AM, DSB-SC, SSB-SC, VSB-SC, SSB+C, VSB+C Frequency Division Muliplexing Power required in each of the above Digital Modulation Schemes PAM, PPM, PDM (just mention last two) Quantisation PCM, DPCM, DM Encoding of bits: NRZ, NRZI, Manchester Power required for each of the encoding schemes Information Theory Uncertainty, Entropy, Information Mutual information, Differential entropy Shannon s source and channel coding theorems

9 Shannon s information capacity theorem - Analysis of Gaussian channels Coding Repetition code Hamming codes Error detection codes: CRC

10 Analogy between Signal Spaces and Vector Spaces Consider two vectors V 1 and V 2 as shown in Fig. 1. If V 1 is to be represented in terms of V 2 where V e is the error. V 1 V 1 = C 12 V 2 + V e (1) C V 12 2 Figure 1: Representation in vector space V 2

11 The error is minimum when V 1 is projected perpendicularly onto V 2. In this case, C 12 is computed using dot product between V 1 and V 2. Component of V 1 along V 2 is = V 1.V 2 V 2 Similarly, component of V 2 along V 1 is = V 1.V 2 V 1 Using the above discussion, analogy can be drawn to signal spaces also. (2) (3)

12 Let f 1 (t) and f 2 (t) be two real signals. Approximation of f 1 (t) by f 2 (t) over a time interval t 1 < t < t 2 can be given by where f e (t) is the error function. f e (t) = f 1 (t) C 12 f 2 (t) (4) The goal is to find C 12 such that f e (t) is minimum over the interval considered. The energy of the error signal ε given by To find C 12, ε = 1 t 2 t 1 t2 t 1 [f 1 (t) C 12 f 2 (t)] 2 dt (5)

13 Solving the above equation we get C 12 = ε = 0 (6) C 12 t2 1 t 2 t 1 t 1 t2 1 t 2 t 1 f 1 (t).f 2 (t) dt t 1 f 2 2(t) dt The denominator is the energy of the signal f 2 (t). When f 1 (t) and f 2 (t) are orthogonal to each other C 12 = 0. Example: sinnω 0 t and sinmω 0 t be two signals where m and n are integers. When m n (7)

14 π ω 0 sinnω 0 t. sinmω 0 t dt = 0 (8) π Clearly sinnω 0 t and sinmω 0 t are orthogonal to each other.

15 Representation of Signals by a set of Mutually Orthogonal Real Functions Let g 1 (t), g 2 (t),..., g n (t) be n real functions that are orthogonal to each other over an interval t 1, t 2 : Let 1 t 2 t 1 1 t2 t 2 t 1 t 1 g i (t)g j (t)dt = 0, i j (1) t2 t 1 g j (t)g j (t)dt = K j (2) f(t) = C 1 g 1 (t) + C 2 g 2 (t) C n g n (t) (3)

16 n f(t) = C r g r (t) (4) To find C r, ε = When ε is expanded we have r=1 f e (t) = f ( t) 1 t 2 t 1 t2 n C r g r (t) (5) r=1 t 1 [f(t) n C r g r (t)] 2 dt (6) r=1 ε = ε =... = ε = 0 (7) C 1 C 2 C r

17 1 ε = t 2 t 1 t2 t 1 f ( t) 2f(t) Now all cross terms disappear 1 t 2 t 1 t2 n C r g r (t) + i=1 n C r g r (t) r=1 n C k g k (t) dt k=1 (8) t 1 C i g i (t)c j g j (t)dt = 0, i j (9) since g i (t) and g j (t) are orthogonal to each other. Solving the above equation we get

18 C j = t2 1 t 2 t 1 t 1 t2 1 t 2 t 1 t 1 f(t).g j (t) dt g 2 j(t) dt Analogy to Vector Spaces: Projection of f(t) along the signal g j (t) = C j (10)

19 Representation of Signals by a set of Mutually Orthogonal Complex Functions When the basis functions are complex. a E x = t2 represents the energy of a signal. t 1 x(t) 2 dt (11) Suppose g(t) is represented by the complex signal x(t) a u + v 2 = (u + v)(u + v ) = u 2 + v 2 + u v + uv

20 E e = = t2 Minimising the second term yields g(t) cx(t) 2 dt (12) t 1 t2 g(t) 2 dt 1 t2 2 g(t)x (t)dt t 1 Ex + (13) t 1 c E x 1 t2 2 g(t)x (t)dt Ex (14) c = 1 E x t2 t 1 t 1 g(t)x (t)dt (15) Thus the coefficients can be determined by projection g(t) along x (t).

21 Fourier Representation of continuous time signals Any periodic signal f(t) can be represented with a set of complex exponentials as shown below. f(t) = F 0 + F 1 e jω 0t + F 2 e j2ω 0t + + F n e jnω nt + (1) + F 1 e jω 0t + F 2 e j2ω 0t + F n e jnω nt + (2) The exponential terms are orthogonal to each other because + (e jnωt )(e jmωt ) dt = 0, m n The energy of these signals is unity since

22 + (e jnωt )(e jmωt ) dt = 1, m = n Representing a signal in terms of its exponential Fourier series components is called Fourier Analysis. The weights of the exponentials are calculated as F n = = 1 T t0 +T t 0 t0 +T t 0 t0 +T f(t).(e jnω 0t ) dt (e jnω 0 t).(e jnω 0 t) dt t 0 f(t).(e jnω 0t ) dt

23 Extending this representation to aperiodic signals: When T and ω 0 0, the sum becomes an integral and ω 0 becomes continuous. The resulting represention is termed as the Fourier Transform (F(ω)) and is given by F(ω) = + f(t)e jωt dt a The signal f(t) can recovered from F(ω) as a Analysis equation b Synthesis equation f(t) = + F(ω)e jωt dω b

24 Some Important Functions Delta function is a very important signal in signal analysis. It is defined as + δ(t) dt = 1

25 ε δ ( t ) 1/2ε ε Area under the curve is always 1 0 t Figure 1: The Dirac delta function The Dirac delta function is also called the Impulse function. This function can be represented as the limiting function of a number of sampling functions: 1. Gaussian Pulse δ(t) = lim 2 T 2 1 πt T 0 T e

26 2. Triangular Pulse δ(t) = lim T T 0 1 [ 1 t ], t T (3) T = 0, t > T (4) 3. Exponential Pulse 4. Sampling Function δ(t) = lim δ(t) = lim 1 T 0 2T e t T k π Sa(kt)dt = 1 k k π Sa(kt)

27 5. Sampling Square function δ(t) = lim k k π Sa2 (kt) The unit step function is another important function signal processing. It is defined by u(t) = 1, t > 0 = 1 2, t = 0 = 0, t < 0 The Fourier transform of the unit step can be found only in the limit. Some common Fourier transforms will be discussed.

28 Fourier Representation of continuous time signals Properties of Fourier Transform a Translation Shifting a signal in time domain introduces linear phase in the frequency domain. Proof: f(t) F (ω) f(t t 0 ) e jωt 0 F (ω) a F and F 1 correspond to the F orward and Inverse F ourier transforms

29 Put τ = t t 0 F (ω) = F (ω) = + + = e jωt 0 f(t t 0 )e jωt dt f(τ)e jω(τ+t 0 dt + f(τ)e jωτ dτ (1) = F (ω)e jωt 0 (2) Modulation A linear phase shift introduced in time domain signals results in a frequency domain. f(t) F (ω)

30 Proof: F (ω) = e jω 0t f(t) F (ω ω 0 ) = + + f(t)e jω 0t e jωt dt f(t)e j(ω ω 0)t dt (3) = F (ω ω 0 ) (4) Scaling Compression of a signal in the time domain results in an expansion in frequency domain and vice-versa.

31 Proof: Put τ = at If a > 0 f(t) F (ω) f(at) 1 a F (ω a ) F (ω) = F(f(at)) = + f(at)e jωt dt + = 1 a F (ω a ) f(τ)e j ω a τ dτ

32 If a < 0 F(f(at)) = = T herefore F(f(at)) = Duality Replace t with ω and ω with t in + 1 a F (ω a ) 1 a F (ω a ) f(t) F (ω) F (t) 2πf( ω) f(τ)e j ω a τ dτ

33 F (ω) = F (t) = Therefore + + f(t)e jωt dt f(ω)e jtω dω But the inverse Fourier transform of a given FT f(ω) is g(t) = 1 2π + f(ω)e jωt dω F (t) = 2πF 1 (f( ω))

34 or Example: F (t) 2πf( ω) δ(t) 1 1 2πδ( ω) = 2πδ(ω) b Convolution Convolution of two signals in the time domain results in multiplication of their Fourier transforms. f 1 (t) f 2 (t) F 1 (ω)f 2 (ω)

35 g(t) = f 1 (t) f 2 (t) = + f 1 (τ)f 2 (t τ)dτ Proof: F(g(t)) = = = f 1 (t) = F 1 (ω)f 2 (ω) f 1 (τ)f 2 (t τ)dτ e jωt dt + Multiplication Multiplication of two signas in the time domain results in convolution of their Fourier transforms f 2 (t)e jωt dtdτ f 1 (τ)f 2 (ω)e jωτ dτ

36 f 1 (t)f 2 (t) 1 2π F 1(ω) F 2 (ω) This can be easily proved using the Duality Property Differentiation in time d f(t) jωf (ω) dt Proof: f(t) = 1 2π Differentiation in Frequency + F (ω)e jωt dω Differentiating both sides w.r.t t yields the result.

37 ( jt) n f(t) dn F (ω) dω This follows from the duality property. Integration in time t f(t)dt 1 jω F (ω)

38 Some Example Continuous Fourier transforms F(δ(t)) F(δ(t)) = Given that + + Therefore F(δ(t)) = 1 + δ(t)e jωt dt f(t)δ(t) dt = f(0) = f(0) f(t)δ(t t 0 ) dt = f(t 0 ) + δ(t) dt

39 Linearity of the Fourier transform F(a 1 f 1 (t) + a 2 f 2 (t)) = a 1 F 1 (ω) + a 2 F 2 (ω) F(A), where A is constant Using the duality property and the linearity property of the Fourier transform F(A) = 2πAδ(ω) Fourier transform of e at, t > 0 (see Figure 1)

40 1 e at f(t) = e at, t > 0 F (ω) = = = t 0 1 a + jω +π/2 e (a+jω)t dt 0 0 1/a Phase Magnitude Figure 1: The exponential function and its Fourier transform ω π/2

41 Fourier transform of the unit step function The Fourier transform of the unit step function can be obtained only in the limit F(u(t)) = lim F(e at ) a 0 = 1 jω Fourier transform of e a t (see Figure 2)

42 1 e a t t 2/a F( ω) Figure 2: e a t and its Fourier transform Magnitude ω

43 f(t) = e at, t > 0 f(t) = 1, t = 0 F (ω) = = = e at, t < 0 = 0 e (a+jω)t dt 1 a + jω + 1 a jω Fourier transform of the rectangular function f(t) = A, T/2 t T/2 = 0, otherwise 0 e (a jω)t dt

44 F (ω) = + T T 2 2 Ae jωt dt = A e jωt/2 e +jωt/2 jω = AT sin ωt 2 ωt 2 = sinc( ωt 2 ) The rectangular function rect(t) and its Fourier transform F (ω) are shown in Figure 3

45 A T T 2 2 t 4π 2π 2π 2π T T T T ω Figure 3: rect(t) and its Fourier transform Fourier transform of the sinc function F( ω) Using the duality property, the Fourier transform of the sinc function can be determined (see Figure 4). AT

46 f(t) Aω π 4π 2π 2π 4π 6 π... ω ω ω ω ω ω t 2πA ω 2 2 ω 0 0 Figure 4: sinc(t) and its Fourier transform An important point is that a signal that is bandlimited is not time-limited while a signal that is time-limited is not bandlimited ω

47 Continuous Fourier transforms of Periodic Functions Fourier transform of e jnω 0t Using the frequency shifting property of the Fourier transform e jnω 0t Fourier transform of cos ω 0 t = 1.e jnω 0t F(e jnω 0t ) = F(1) shifted by ω 0 = 2πδ(ω nω 0 )

48 cos ω 0 t = ejω 0t + e jω 0t F(e jω 0t ) = F(1) shifted by ω 0 2 = 2πδ(ω ω 0 ) F(cos ω 0 t) = πδ(ω ω 0 ) + πδ(ω + ω 0 ) Fourier transform of a periodic function f(t) The periodic function is not absolutely summable. The Fourier transform can be represented by a Fourier series. The Fourier transform of the Fourier series representation of the periodic function (period T ) can be computed

49 f(t) = F(f(t)) = F( = = 2π n= F n e jnω0t, ω 0 = 2π T n= i= F n e jnω 0t ) F n F(e jnω 0t ) n= F n δ(ω nω 0 ) Note: The Fourier transform is made up of components at discrete frequencies. Fourier transform of a periodic function f(t) = n= δ(t nt ) (a periodic train of impulses)

50 f(t) = F n = 1 T n= F(f(t)) = 1 T F( = F n e jnω0t, ω 0 = 2π T n= e jnω 0t ) Note: A periodic train of impulses results in a Fourier i= = 2π 1 T = ω 0 F n F(e jnω 0t ) n= n= δ(ω nω 0 ) δ(ω nω 0 )

51 transform which is also a periodic train of impulses (see Figure 1). δ(t nt) δ( ω n ω 0 ) f(t) F( ω ) Figure 1: The periodic pulse train and its Fourier transform

52 Sampling Theorem and its Importance Sampling Theorem: A bandlimited signal can be reconstructed exactly if it is sampled at a rate atleast twice the maximum frequency component in it. Figure 1 shows a signal g(t) that is bandlimited.

53 ω The minimum required sampling rate f s = 2f m is called 0 G( ω) m m Figure 1: Spectrum of bandlimited signal g(t) The maximum frequency component of g(t) is f m. To recover the signal g(t) exactly from its samples it has to be sampled at a rate f s 2f m. ω ω

54 Nyquist rate. Proof: Let g(t) be a bandlimited signal whose bandwidth is f m (ω m = 2πf m ). (a) g(t) G( ω) ωm ωm Figure 2: (a) Original signal g(t) (b) Spectrum G(ω) δ T (t) is the sampling signal with f s = 1/T > 2f m. 0 (b)

55 T δ( t) (a) s δ(ω) Figure 3: (a) sampling signal δ T (t) (b) Spectrum δ T (ω) Let g s (t) be the sampled signal. Its Fourier Transform G s (ω) is given by ω (b)

56 F(g s (t)) = F [g(t)δ T (t)] [ = F = 1 2π G s (ω) = 1 T g(t) [ + n= G(ω) ω 0 + n= δ(t nt ) + n= ] δ(ω nω 0 ) G(ω) δ(ω nω 0 ) G s (ω) = F [g(t) + 2g(t) cos(ω 0 t) + 2g(t) cos(2ω 0 t) + ] G s (ω) = 1 T + n= G(ω nω 0 ) ]

57 g (t) s G ( ω) ω ω 0 ω m ω Figure 4: (a) sampled signal g s (t) (b) Spectrum G s (ω) If ω s = 2ω m, i.e., T = 1/2f m. Therefore, G s (ω) is given by G s (ω) = 1 T 1. Filter with a Gate function, H 2ωm (ω) of width 2ω m. + s n= To recover the original signal G(ω): m G(ω nω m ) s s

58 2. Scale it by T. G(ω) = T G s (ω)h 2ωm (ω). ω m 0 ω m H ( ) 2ω m ω Figure 5: Recovery of signal by filtering with a filter of width 2ω m Aliasing Aliasing is a phenomenon where the high frequency components of the sampled signal interfere with each other

59 because of inadequate sampling ω s < 2ω m. Interference of high frequency components ω ω 0 ω s m m Figure 6: Aliasing due to inadequate sampling Aliasing leads to distortion in recovered signal. This is the reason why sampling frequency should be atleast twice the bandwidth of the signal. ω s

60 Oversampling In practice signal are oversampled, where f s is significantly higher than Nyquist rate to avoid aliasing. ω s ω m 0 ω m ω Figure 7: Oversampled signal-avoids aliasing Problem: Define the frequency domain equivalent of the Sampling Theorem and prove it. s

61 Discrete-Time Signals and their Fourier Transforms Generation of Discrete-time signals Discrete time signals are obtained by sampling a continuous time signal. The continuous time signal is sampled with an impulse train with sampling period T which is usually taken greaterthan or equal to Nyquist Rate to avoid Aliasing. The Discrete-Time Fourier Transform (DTFT) of a discrete time signal g(nt )is represented by a G(e jω ) = + n= g(nt ).e jωnt a G(e jω ) represents the DTFT. It signifies the periodicity of the DTFT

62 In practice, it is assumed that signals are adequately sampled and hence T is dropped to yield: G(e jω ) = The inverse DTFT is given by: g(n) = 1 2π + n= +π π g(n).e jωn G(ω).e jωn dω

63 Some important Discrete-Time Signals Discrete time impulse or unit sample function Unit sample function is similar to impulse function in continuous time (see Figure 1. It is defined as follows δ(n) = 1, for n = 0 0, elsewhere

64 1 δ( n) Figure 1: The unit sample function Unit step function This is similar to unit step function in continuous time domain (see Figure 2) and is defined as follows n

65 3 2 1 u(n) = 1 u(n) 1, for n 0 0, elsewhere Figure 2: The unit step function n

66 Properties of the Discrete-time Fourier transform Time shift property a F(f(n n 0 )) F (e jω )e jωn 0 Modulation property F(f(n)e jω 0n ) F (e j(ω ω 0) ) Differentiation in the frequency domain F(nf(n)) df (ejω ) Convolution in the time domain F(f(n) g(n)) 1 2π F (ejω )G(e jω ) Prove that the forward and inverse DTFTs form a pair dω a A tutorial on this would be appropriate

67 F (e jω ) = + f(n) = 1 2π f(n) = 1 2π = f(n) = n= +π + l= + l= = f(n) π +π π f(n).e jωn F (ω).e jωn dω + l= +π f(l). 1 2π f(l).e jωl e jωn dω π f(l)δ(l n) )e j(ωl ωn) dω

68 Z-transforms Computation of the Z-transform for discrete-time signals: Enables analysis of the signal in the frequency domain. Z - Transform takes the form of a polynomial. Enables interpretation of the signal in terms of the roots of the polynomial. z 1 corresponds to a delay of one unit in the signal. The Z - Transform of a discrete time signal x[n] is defined as where z = r.e jω X(z) = + n= x[n].z n (1)

69 The discrete-time Fourier Transform (DTFT) is obtained by evaluating Z-Transform at z = e jω. or The DTFT is obtained by evaluating the Z-transform on the unit circle in the z-plane. The Z-transform converges if the sum in equation 1 converges

70 Region of Convergence(RoC) Region of Convergence for a discrete time signal x[n] is defined as a continuous region in z plane where the Z-Transform converges. In order to determine RoC, it is convenient to represent the Z-Transform as: a X(z) = P (z) Q(z) The roots of the equation P (z) = 0 correspond to the zeros of X(z) The roots of the equation Q(z) = 0 correspond to the poles of X(z) The RoC of the Z-transform depends on the convergence of the a Here we assume that the Z-transform is rational

71 polynomials P (z) and Q(z), Right-handed Z-Transform Let x[n] be causal signal given by x[n] = a n u[n] The Z - Transform of x[n] is given by

72 X(z) = = + n= + n= + x[n]z n a n u[n]z n The ROC is defined by az 1 < 1 or z > a. = = = = n=0 + n=0 a n z n (az 1 ) n 1 1 az 1 z z a

73 The RoC for x[n] is the entire region outside the circle z = ae jω as shown in Figure 1. Left-handed Z-Transform a z plane Figure 1: RoC(green region) for a causal signal RoC z > a

74 Let x[n] be an anti-causal signal given by y[n] = b n u[ n 1] The Z - Transform of y[n] is given by

75 Y (z) = = = = = = + n= + n= 1 n= + n=0 1 1 z b z y[n]z n b n u[ n 1]z n b n z n (b 1 z) n z b

76 Y (z) converges when b 1 z < 1 or z < b. The RoC for y[n] is the entire region inside the circle z = be jω as shown in Figure 2 z plane a RoC z < a Figure 2: RoC(green region) for an anti-causal signal

77 Two-sided Z-Transform Let y[n] be a two sided signal given by y[n] = a n u[n] b n u[ n 1] where, b > a The Z - Transform of y[n] is given by

78 Y (z) = = = = = = + n= + n= + n=0 + n=0 y[n]z n (a n u[n] b n u[ n 1])z n a n z n (az 1) n az 1. 1 z b z z a. z z b 1 n= + n=1 + 1 b n z n (b 1 z) n

79 Y (z) converges for b 1 z < 1 and az 1 < 1 or z < b and z > a. Hence, for the signal The ROC for y[n] is the intersection of the circle z = be jω and the circle z = ae jω as shown in Figure 3 a z plane RoC a < z < b Figure 3: RoC(pink region) for a two sided Z Transform b

80 Transfer function H(z) Consider the system shown in Figure 4. x[n] X(z) h[n] H(z) y[n] = x[n]*y[n] Y(z) = X(z)H(z) Figure 4: signal - system representation x[n] is the input and y[n] is the output h[n] is the impulse response of the system. Mathematically, this signal-system interaction can be represented as follows y[n] = x[n] h[n]

81 In frequency domain this relation can be written as or Y (z) = X(z).H(z) H(z) = Y (z) X(z) H(z) is called Transfer function of the given system. In the time domain if x[n] = δ[n] then y[n] = h[n], h[n] is called the impulse response of the system. Hence, we can say that h[n] H(z)

82 Some Examples: Z-transforms Delta function Z(δ[n]) = 1 Z(δ[n n 0 ]) = z n 0 Unit Step function x[n] = 1, n 0 x[n] = 0, otherwise X(z) = 1, z > 1 1 z 1 The Z-transform has a real pole at the z = 1.

83 Finite length sequence x[n] = 1, 0 n N x[n] = 0, otherwise X(z) = 1 z N 1 z 1 = z N 1 zn 1 z 1, z > 1 The roots of the numerator polynomial are given by: z = 0, Nzeros at the origin and the n th roots of unity: z = e j2πk N, k = 0, 1, 2,, N 1 (2)

84 Causal sequences x[n] = ( 1 3 )n u[n] ( 1 2 )n u[n 1] X(z) = z 1 3 z 1 1 1, z > 1 2 z 1 3 The Discrete time Fourier transform can be obtained by setting z = e jω Figure 5 shows the Discrete Fourier transform for the rectangular function.

85 1 1 N +N 4π 2π 2π 4π 2 2 Ν+1 Ν+1 Ν+1 Ν+1 Figure 5: Discrete Fourier transform for the rectangular function

86 Some Problems Find the Z-transform (assume causal sequences): 1. 1, a 1!, a2 2!, a3 3!, 2. 0, a, 0, a3 3! 3. 0, a, 0, a2 2! a5 a7, 0, 5!, 0, 7!, a4 a6, 0, 4!, 0, 6!, Hint: Observe that the series is similar to that of the exponential series.

87 Properties of the Z-transform 1. RoC is generally a disk on the z-plane. 0 r R z r L 2. Fourier Transform of x[n] converges when RoC includes the unit circle. 3. RoC does not contain any poles. 4. If x[n] is finite duration, RoC contains entire z - plane except for z = 0 and z =. 5. For a left handed sequence, RoC is bounded by z < min( a, b ). 6. For a right handed sequence, RoC is bounded by z > max( a ; b ).

88 Inverse Z-transform To determine the inverse Z-transform, it is necessary to know the RoC. RoC decides whether a given signal is causal (exists for positive time), anticausal (exists for negative time) or both causal and anticausal (exists forboth positive and negative time) Different approaches to compute the inverse Z-transform Long division method When Z-Transform is rational, i.e. it can be expressed as the ratio of two polynomials P (z) and Q(z) X(z) = P (z) Q(z) Then, inverse Z-transform can be obtained using long division: Divide P (z) by Q(z). Let this be:

89 X(z) = The method of adding inverse Z-transform is illustrated below. Let, i= a i z i (1) The coefficients of the RHS of equation (1) correspond to the time sequencei.e. the coefficients of the quotient of the long division gives the sequence. Partial Fraction method the Z-Transform is decomposed into partial fractions the inverse Z-transform of each fraction is obtained independently the inverse sequences are then added

90 X(z) = = where, = M b k.z k k=0, N a k.z k k=0 M (1 c k.z 1 ) k=1 N (1 d k.z 1 ) k=1 N k=1 A k (1 d k.z 1 ) M < N

91 For s multiple poles at z = d i X(z) = M N k=0 A k = (1 d k.z 1 )X(z) z=dk B r.z 1 + N k=1,k i A k s (1 d k.z 1 ) + m=1 C m (1 d i.z 1 ) m

92 Properties of the Z-Transform Linearity: a 1 x 1 [n] + a 2 x 2 [n] a 1 X 1 (z) + a 2 X 2 (z), RoC = R x1 R x2 Time Shifting Property: x[n n 0 ] z n 0 X(z), RoC = R x (except possible addition/deletion of z = 0 or z = ) Exponential Weighting: z0 n x[n] X(z0 1 0 R x The poles of the Z-transform are scaled by z 0

93 Linear Weighting nx(n) z dx(z), dz RoC = R x (except possible addition/deletion of z = 0 or z = ) Time Reversal x[ n] X(z 1 ), RoC = 1 R x Convolution x[n] y[n] X(z)Y (z), RoC = R x R y Multiplication x[n]w[n] 1 2πj X(v)w( z v )v 1 dv

94 Inverse Z-Transform Examples Using long division: Causal sequence 1 1 az 1, RoC = z > a = 1 + az 1 + az 2 + az 3 + IZT (1 + az 1 + a 2 z 2 + a 3 z 3 + ) = a n u[n] Using long division: Noncausal sequence 1, RoC = z < a 1 az 1 Here the IZT is computed as follows: This results in: 1 IZT ( 1 az ) = IZT ( z 1 a + z ) IZT ( a 1 z + a 2 z 2 + a 3 z 3 + ) = a n u[ n 1]

95 Inverse Z-transform - using Power series expansion X(z) = log(1 + az 1 ), z > a Using the Power Series expansion for log(1 + x), x < 1, we have ( 1) n+1 a n z n X(z) = n The IZT is given by x[n] = ( 1)n+1 a n, n 1 n = 0, n 0 1

96 Inverse Z-transform - Inspection method a n u[n] GivenX(z) = 1, z > a 1 az , z > 1 2 z 1 2 = x[n] = ( 1 2 )n u[n] Inverse Z-transform - Partial fraction method Example 1: All-Pole system X(z) = 1 (1 1 3 z 1 )(1 1 6 z 1 ), z > 1 3

97 Using partial fraction method, we have: X(z) = A A 2 3 z 1 1 1, 6 z 1 z > 1 3 A 1 = (1 1 3 z 1 )X(z) z= 1 A 2 = (1 1 6 z 1 )X(z) z= 1 A 1 = 2 A 2 = 1 x(n) = 2( 1 3 )n u[n] 1( 1 6 )n u[n] 3 6

98 Example 2: Pole-Zero system X(z) = 1 + 2z 1 + z z 1 + 1, z > 1 2z 2 (1 + z 1 ) 2 (1 1 2 z 1 )(1 z 1 ) = 2 + = z 1 (1 1 2 z 1 )(1 z 1 ) z 1 1 z 1 x[n] = 2δ[n] 9( 1 2 )n u[n] + 8u[n]

99 Example 3: Finite length sequences X(z) = z 2 (1 1 2 z 1 )(1 + z 1 )(1 z 1 ) = z 2 frac12z z 1 = δ[n + 2] 1 2 δ[n + 1] δ[n] + 1 δ[n 1] 2

100 1. Given X(z) = z Inverse Z-Transform Problem z 1 sequences for x[n]. z z 2 + z z 3, determine all the possible Hint: Remember that the RoC must be a continuous region

101 Basics of Probability Theory and Random Processes Basics of probability theory a Probability of an event E represented by P (E) and is given by P (E) = N E (1) N S where, N S is the number of times the experiment is performed and N E is number of times the event E occured. Equation 1 is only an approximation. For this to represent the exact probability N S. The above estimate is therefore referred to as Relative Probability Clearly, 0 P (E) 1. a required for understanding communication systems

102 Mutually Exclusive Events Let S be the sample space having N events E 1, E 2, E 3,, E N. Two events are said to be mutually exclusive or statistically N independent if A i A j = φ and A i = S for all i and j. Joint Probability Joint probability of two events A and B represented by P (A B) and is defined as the probability of the occurence of both the events A and B is given by Conditional Probability Conditional probability of two events A and B represented as i=1 P (A B) = N A B N S

103 P (A B) and defined as the probability of the occurence of event A after the occurence of B. Similarly, This implies, P (A B) = N A B N B P (B A) = N A B N B P (B A)P (A) = P (A B)P (B) = P (A B)

104 Chain Rule Let us consider a chain of events A 1, A 2, A 3,, A N which are dependent on each other. Then the probability of occurence of the sequence P (A N, A N 1, A N 2,, A 2, A 1 ) = P (A N A N 1, A N 2,, A 1 ). P (A N 1 A N 2, A N 3,, A 1 )..P (A 2 A 1 ).P (A 1 )

105 A A 5 4 Bayes Rule A 1 B A 3 A Figure 1: The partition space In the above figure, if A 1, A 2, A 3, A 4, A 5 partition the sample space S, then (A 1 B), (A 2 B), (A 3 B), (A 4 B), and (A 5 B) partition B. Therefore, 2

106 n P (B) = P (A i B) = In the above equation, P (A i B) is called posterior probability, i=1 n P (B A i ).P (A i ) i=1 In the example figure here, n = 5. P (A i B) = P (A i B) P (B) = P (B A i ).P (A i ) n P (B A i ).P (A i ) i=1

107 P (B A i ) is called likelihood, P (A i ) is called prior probability and n P (B A i ).P (A i ) is called evidence. i=1

108 Random Variables Random variable is a function whose domain is the sample space and whose range is the set of real numbers Probabilistic description of a random variable Cummulative Probability Distribution: It is represented as F X (x) and defined as F X (x) = P (X x) If x 1 < x 2, then F X (x 1 ) < F X (x 2 ) and 0 F X (x) 1. Probability Density Function: It is represented as f X (x) and defined as

109 This implies, f X (x) = df X(x) dx P (x 1 X x 2 ) = x2 x 1 f X (x) = df X(x) dx f X (x) dx

110 Random Process A random process is defined as the ensemble(collection) of time functions together with a probability rule (see Figure 2) Sample Space S 1 S 2 S n x (t) 1 x (t) n T +T x (t) 2 Figure 1: Random Processes and Random Variables

111 A random variable: x 1 (t) is an outcome of experiment 1 x 2 (t) is the outcome of experiment 2. x n (t) is the outcome of experiment n Each sample point in S is associated with a sample function x(t) X(t, s) is a random process is an ensemble of all time functions together with a probability rule X(t, s j ) is a realisation or sample function of the random process Probability rules assign probability to any meaningful event associated with an observation An observation is a sample function of the random process

112 {x 1 (t k ), x 2 (t k ),..., x n (t k )} = {X(t k, s 1 ), X(t k, s 2 ),..., X(t k, s n )} X(t k, s j ) constitutes a random variable. Outcome of an experiment mapped to a real number An oscillator with a frequency ω 0 with a tolerance of 1% The oscillator can take values between ω 0 (1 ± 0.01) Each realisation of the oscillator can take any value between (ω 0 )(0.99) to (ω 0 )(1.01) The frequency of the oscillator can thus be characterised by a random variable Stationary random process A random process is said to be stationary if its statistical characterization is independent of the observation interval over which the process was initiated. Mathematically,

113 F X(t1 +T ) X(t k +T ) = F X(t1 ) X(t k ) Mean, Correlation and Covariance Mean of a stationary random process is independent of the time of observation. µ X (t) = E[X(t)] = µ x Autocorrelation of a random process is given by: R X (t 1, t 2 ) = E[X(t 1 )X(t 2 )] = + + x 1.x 2 f X(t1 )X(t 2 )(x 1, x 2 ) dx 1 dx 2 For a stationary process the autocorrelation is dependent only on the time shiftand not on the time of observation.

114 Autocovariance of a stationary process is given by C X (t 1, t 2 ) = E[(X(t 1 µ x )(X(t 2 µ x )] Properties of Autocorrelation 1. R X (τ) = E[X(t + τ)x(t)] 2. R X (0) = E[X 2 (t)] 3. The autocorrelation function is an even function i.e, R X (τ) = R X ( τ). 4. The autocorrelation value is maximum for zero shift i.e, R X (τ) R X (0).

115 Proof: E[(X(t + τ) ± X(t)) 2 ] 0 = E[X 2 (t + τ)] + E[X 2 (t)] ± 2E[X(t + τ)x(t)] 0 = R X (0) + R X (0) + 2R X (τ) 0 = R X (0) R X (τ) R X (0)

116 A slowly varying random process A rapidly varying random process Figure 2: Autocorrelation function of a random process

117 Random Process: Some Examples A sinusoid with random phase Consider a sinusoidal signal with random phase, defined by X(t) = a sin(ω 0 t + Θ) where ω 0 and a are constants, and Θ is a random variable that is uniformly distributed over a range of 0 to 2π (see Figure 1) f Θ (θ) = 1 2π, 0 θ 2π = 0, elsewhere

118 Θ sin( ω t+θ) 0 Figure 1: A sinusoid with random phase This means that the random variable Θ is equally likely to have any value in the range 0 to 2π. The autocorrelation function of X(t) is

119 R X (t) = E[X(t + τ)x(t)] = E[sin(ω 0 t + ω 0 τ) + Θ) sin(ω 0 t + Θ)] = 1 2 E[sin(2ω 0t + ω 0 τ + 2Θ)] E[sin(ω 0τ)] = 1 2 2π 0 1 2π cos(4πf ct + ω 0 τ + 2Θ)] cos(ω 0 τ) dθ The first term intergrates to zero, and so we get R X (τ) = 1 2 cos(ω 0τ) The autocorrelation function is plotted in Figure 2.

120 cos( ω τ) 0 Figure 2: Autocorrelation function of a sinusoid with random phase Figure 3 shows the sample function x(t) of a random process 1 The autocorrelation function of a sinusoidal wave with random phase is another sinusoid at the same frequency in the τ - domain Random binary wave 2π ω 0

121 X(t) consisting of a random sequence of binary symbols 1 and t delay Figure 3: A random binary wave 1. The symbols 1 and 0 are represented by pulses of amplitude +1 and 1 volts, respectively and duration T seconds. 2. The starting time of the first pulse, t delay, is equally likely

122 to lie anywhere between zero and T seconds 3. t delay is the sample value of a uniformly distributed random variable T delay with a probability density function f Tdelay (t delay ) = 1 T, 0 t delay T = 0, elsewhere 4. In any time interval (n 1)T < t t delay < nt, where n is an interger, a 1 or a 0 is determined randomly (for example by tossing a coin: heads = 1, tails = 0 E[X(t)] = 0, for all t since 1 and 0 are equally likely. Autocorrelation function R X (t k, t l ) is given by E[X(t k )X(t l )], where X(t k ) and X(t l ) are random variables

123 Case 1: when t k t l > T. X(t k ) and X(t l ) occur in different pulse intervals and are therefore independent: E[X(t k )X(t l )] = E[X(t k )]E[X(t l )] = 0, for t k t l > T Case 2: when t k t l < T, with t k = 0 and t l < t k. X(t k ) and X(t l ) occur in the same pulse interval provided t delay satisfies the condition t delay < T t k t l. E[X(t k )X(t l ) t delay ] = 1, t delay < T t k t l = 0, elsewhere Averaging this result over all possible values of t delay, we get

124 E[X(t k )X(t l )] = = T tk t l 0 T tk t l 0 f Tdelay (t delay ) dt delay 1 T dt delay = (1 t k t l ), t k t l < T T The autocorrelation function is given by R X (τ) = (1 τ ), τ < T T = 0, τ > T This result is shown in Figure 4

125 T 1 Figure 4: Autocorrelation of a random binary wave T

126 Random Process: Some Examples Quadrature Modulation Process Given two random variables X 1 (t) and X 2 (t) X 1 (t) = X(t) cos(2ω 0 t + Θ) X 2 (t) = X(t) sin(2ω 0 t + Θ) where ω 0 is a constant, and Θ is a random variable that is uniformly distributed over a range of 0 to 2π, that is, f Θ (θ) = 1 2π, 0 θ 2π = 0, elsewhere The correlation function of X 1 (t) and X 2 (t) is

127 R 12 (τ) = E[X 1 (t)x 2 (t + τ)] = E[X(t) cos(ω 0 t + Θ)X(t τ) sin(2ω 0 (t τ) + Θ)] = 2π 0 1 2π X(t)X(t τ) cos(ω 0t + Θ) sin(ω 0 (t τ) + Θ) dθ = 1 2 R X(τ) sin(ω 0 τ)

128 Random Process: Time vs. Ensemble Averages Ensemble averages Difficult to generate a number of realisations of a random process = use time averages Mean Autocorrelation µ x (T ) = 1 2T +T T x(t) dt R x (τ, T ) = 1 2T +T T x(t)x(t + τ) dt

129 Ergodicity A random process is called ergodic if 1. it is ergodic in mean: lim µ x(t ) = µ X T + lim var[µ x(t )] = 0 T + 2. it is ergodic in autocorrelation: lim R x(τ, T ) = R X (τ) T + lim var[r x(τ, T )] = 0 T + where µ X and R X (τ) are the ensemble averages of the same random process.

130 Random Processes and Linear Shift Invariant Systems(LSI) The communication channel can be thought of as a system The signal that is transmitted through the channel is a realisation of the random process It is necessary to understand the behaviour of a signal that is input to a system. For analysis purposes it is assumed that a system is LSI Linear Shift Invariant(LSI) Systems Figure 1: An LSI system

131 In Figure 1, h[n] is an LSI system if it satisfies the following properties Linearity The system is called linear, if the following equation holds for all signals x 1 [n] and x 2 [n] and any a and b: x 1 [n] y 1 [n] x 2 [n] y 2 [n] = a.x 1 [n] + b.x 2 [n] a.y 1 [n] + b.y 2 [n] Shift Invariance The system is called Shift Invariant, if the following equation holds for any signal x[n] x[n] y[n] = x[n n 0 ] y[n n 0 ]

132 The assumption is that the output of the system is linear, in that if the input scaled, the output is scaled by the same factor. The system supports superposition When two signals are added in the time domain, the output is equal to the sum of the individual responses If the input to the system is delayed by n 0, the output is also delayed by n 0.

133 Random Process through a linear filter A random process X(t) is applied as input to a linear time-invariant filter of impulse response h(t), It produces a random process Y (t) at the filter output as shown in Figure 1 X(t) h(t) Figure 1: Transmission of a random process through a linear filter Difficult to describe the probability distribution of the output random process Y (t), even when the probability distribution of the input random process X(t) is completely specified for t +. Y(t)

134 Estimate characteristics like mean and autocorrelation of the output and try to analyse its behaviour. Mean The input to the above system X(t) is assumed stationary. The mean of the output random process Y (t) can be calculated m Y (t) = E[Y (t)] = E = = Z + Z +»Z + where H(0) is the zero frequency response of the system. h(τ)e[x(t τ)] dτ h(τ)m X (t τ) dτ Z + = m X h(τ) dτ = m X H(0) h(τ)x(t τ) dτ

135 Autocorrelation The autocorrelation function of the output random process Y (t). By definition, we have R Y (t, u) = E[Y (t)y (u)] where t and u denote the time instants at which the process is observed. We may therefore use the convolution integral to write R Y (t, u) = E =»Z + Z + Z + h(τ 1 )X(t τ 1 ) dτ 1 h(τ 2 )X(t τ 2 ) dτ 2 Z + h(τ 1 ) dτ 1 h(τ 2 )E [X(t τ 1 )X(t τ 2 )] dτ 2 When the input X(t) is a wide-stationary random process, The autocorrelation function of X(t) is only a function of the difference between the observation times t τ 1 and

136 u τ 2. Putting τ = t u, we get R Y (τ) = Z + Z + h(τ 1 )h(τ 2 )R X (τ τ 1 + τ 2 ) dτ 1 dτ 2 R Y (0) = E[Y 2 (t)] The mean square value of the output random process Y (t) is obtained by putting τ = 0 in the above equation. E[Y 2 (t)] = = = Z + Z + Putting τ = τ 2 τ 1 h(τ 1 )h(τ 2 )R X (τ 2 τ 1 ) dτ 1 dτ 2 1 Z + Z " + Z + # 2π H(ω) exp(jωτ 1 ) dω h(τ 2 )R X (τ 2 τ 1 ) dτ 1 dτ 2 1 Z + Z + Z + 2π H(ω) dω h(τ 2 ) dτ 2 R X (τ 2 τ 1 ) exp(j2ωτ 1 ) dτ 1

137 E[Y 2 (t)] = = 1 2π Z + H(ω) dω Z + h(τ 2 ) exp(jωτ 2 ) dτ 2 Z + R X (τ) exp( j2ωτ) dτ 1 Z + Z + Z + H(ω) dω 2π H (ω) dω R X (τ) exp( jωτ) dτ This is simply the Fourier Transform of the autocorrelation function R X (t) of the input random process X(t). Let this transform be denoted by S X (f). S X (ω) = + R X (τ) exp( jωτ) dτ S X (ω) is called the power spectral density or power spectrum of the wide-sense stationary random process X(t). E[Y 2 (t)] = 1 2π + H(ω) 2 S X (ω) df

138 The mean square value of the output of a stable linear time-invariant filter in response to a wide-sense stationary random process is equal to the integral over all frequencies of the power spectral density of the input random process multiplied by the squared magnitude of the transfer function of the filter.

139 Definition of Bandwidth Bandwidth is defined as a band containing all frequencies between upper cut-off and lower cut-off frequencies. (see Figure 1) f l Bandwidth Upper and lower cut-off (or 3dB) frequencies corresponds to the frequencies where the magnitude of signal s Fourier Transform is reduced to half (3dB less than) its maximum value. f u BW f b = f u f l Figure 1: Bandwidth of a signal 3 db

140 Importance of Bandwidth Bandwidth enables computation of the power required to transmit a signal. Signals that are band-limited are not time-limited Energy of a signal is defined as: E = + x(t) 2 dt Energy of a signal that is not time-limited can be computed using Parsevals Theorem a : + x(t) 2 dt = + X(ω) 2 dω a The power of a signal is the energy dissipated in a one ohm resistor

141 Multiple signals (of finite bandwidth) can be multiplexed on to a single channel. For all pulse signals, duration of the pulse and its bandwidth satisfies inverse relation between frequency and time. duration bandwidth = constant wider the pulse = smaller is the bandwidth required wider the pulse = Inter-Symbol-Interference is an issue ideally a pulse that is narrow and has a small bandwidth is required Noise equivalent Bandwidth: Let, a white noise signal with power N 0 /2 be fed to an arbitrary Low Pass Filter (LPF) wwith transfer function H(ω). Then, output noise power is given by

142 N out = 1 2π = N 0 2π For an ideal LPF, N H(0) 2 πb 0 2 π B H(ω) 2 dω H(ω) 2 dω Ideal LPF Practical LPF Figure 2: Ideal and Practical LPFs

143 = B = N out = N 0 BH 2 (0) + 1 2π 0 H(ω) 2 dω H 2 (0)

144 Modulation Modulation is a process that causes a shift in the range of frequencies in a signal. Signals that occupy the same range of frequencies can be separated Modulation helps in noise immunity, attentuation - depends on the physical medium Figure 1 shows the different kinds of analog modulation schemes that are available

145 Communication System Base band Modulation Carrier Modulation Carrier Modulation Amplitude (AM) Angle Frequency (FM) Figure 1: A broad view of communication system Phase (PM) Amplitude Modulation It is the process where, the amplitude of the carrier is varied proportional to that of the message signal. Amplitude Modulation with carrier Let m(t) be the base-band signal, m(t) M(ω) and c(t) be the carrier, c(t) = A c cos(ω c t). f c is chosen such that f c >> W, where W is the maximum frequency component

146 of m(t). The amplitude modulated signal is given by S(ω) s(t) = A c [1 + k a m(t)] cos(ω c t) = π A c 2 (δ(ω ω c) + δ(ω + ω c )) + k a A c (M(ω ω c ) + M(ω + ω c )) 2

147 m(t) s(t) t t f c 2f m M( ω) f f m A /2 S( ω) c 1/2 A/2 k a M(0) Figure 2: Amplitude modulation m f c ω ω

148 Figure 2 shows the spectrum of the Amplitude Modulated signal. k a is a constant called amplitude sensitivity. k a m(t) < 1 and it indicates percentage modulation. Modulation in AM: A product modulator is used for generating the modulated signal as shown in Figure 3.

149 m(t) A cos(2 f t) c Product Modulator π c + s(t) Figure 3: Modulation using product modulator Demodulation in AM: An envelope detector is used to get the demodulated signal (see Figure 4).

150 r + R v (t) Figure 4: Demodulation using Envelope detector The voltage v m (t) across the resistor R gives the message signal m(t) C m

151 Double Side Band - Suppressed Carrier (DSB-SC) Modulation In AM modulation, transmission of carrier consumes lot of power. Since, only the side bands contain the information about the message, carrier is suppressed. This results in a DSB-SC wave. A DSB-SC wave s(t) is given by s(t) = m(t)a c cos(ω c t) S(ω) = π A c 2 (M(ω ω c) + M(ω + ω c ))

152 s(t) t f c 2f m Figure 5: DSB-SC modulation S(f) 1/2 A c M(0) Modulation in DSB-SC: Here also product modulator is used as shown in Figure 3, but the carrier is not added. Figure 6 shows the spectrum of the DSB-SC signal. f c ω

153 2f c 0 LPF 1/2 A c M(0) cos( φ ) Figure 6: Spectrum of Demodulated DSB-SC signal Demodulation in DSB-SC: A coherent demodulator is used. The local oscillator present in the demodulator generates a carrier which has same frequency and phase(i.e. φ = 0 in Figure 7) a as that of the carrier in the modulated signal (see Figure 7) a Clearly the design of the demodulator for DSB-SC is more complex than that vanilla AM 2f c

154 s(t) Product Modulator Local Oscillator v(t) cos(2 π f t + φ ) c LPF Figure 7: Coherent detector v (t) o

155 v(t) = s(t). cos(ω c t + φ) = m(t)a c cos(ω c t) cos(ω c t + φ) = m(t) 2 A c [cos(2ω c t + φ) + cos(φ)] If, the demodulator (Figure 7) has constant phase, the original signal is reconstructed by passing v(t) through an LPF.

156 Single Side Band (SSB) Modulation In DSB-SC it is observed that there is symmetry in the bandstructure. So, even if one half is transmitted, the other half can be recovered at the received. By doing so, the bandwidth and power of transmission is reduced by half. Depending on which half of DSB-SC signal is transmitted, there are two types of SSB modulation 1. Lower Side Band (LSB) Modulation 2. Upper Side Band (USB) Modulation

157 ω ω c c ω c 2 π B M( ω ) 2 πb ω Baseband signal c ω c ω c DSBSC USB LSB Figure 1: SSB signals from orignal signal

158 Mathematical Analysis of SSB modulation M( ω) M ( ) 2 B 0 2 B + ω M ( ω ) 0 0 M ω ω ω ω ( + c ) M + ( c ) ω c M ( ω + ω + c ) ω c c M ( ω ω ) c Figure 2: Frequency analysis of SSB signals ω ω c c

159 From Figure 2 and the concept of the Hilbert Transform, So, Φ USB (ω) = M + (ω ω c ) + M (ω + ω c ) φ USB (t) = m + (t)e jω ct + m (t)e jω ct But, from complex representation of signals, m + (t) = m(t) + j ˆm(t) m (t) = m(t) j ˆm(t) φ USB (t) = m(t) cos(ω c t) ˆm(t) sin(ω c t)

160 Similarly, φ LSB (t) = m(t) cos(ω c t) + ˆm(t) sin(ω c t) Generation of SSB signals A SSB signal is represented by: φ SSB (t) = m(t) cos(ω c t) ± ˆm(t) sin(ω c t)

161 m(t) cos( t) π /2 ω c DSBSC π/2 DSBSC sin( ω t) + + Σ SSB signal Figure 3: Generation of SSB signals As shown in Figure 3, a DSB-SC modulator is used for SSB signal generation. Coherent Demodulation of SSB signals Φ SSB (t) is multiplied with cos(ω c t) and passed through low pass filter to get back the orignal signal. c

162 φ SSB (t) cos(ω c t) = 1 2 m(t) [1 + cos(2ω ct)] ± 1 2 ˆm(t) sin(2ω ct) = 1 2 m(t) cos(2ω ct) ± 1 2 ˆm(t) sin(2ω ct) M ( ω + ω M( ) c ) ω M ( ω ω c ) + 2ω 0 2ω c Figure 4: Demodulated SSB signal The demodulated signal is passed through an LPF to remove unwanted SSB terms. c

163 Vestigial Side Band (VSB) Modulation The following are the drawbacks of SSB signal generation: 1. Generation of an SSB signal is difficult. 2. Selective filtering is to be done to get the original signal back. 3. Phase shifter should be exactly tuned to To overcome these drawbacks, VSB modulation is used. It can viewed as a compromise between SSB and DSB-SC. Figure 5 shows all the three modulation schemes.

164 In VSB B ω c ν The transmission BW is BW v = B + v. where, v is the vestigial frequency band. The generation of VSB signal is shown in Figure 6 0 Φ (ω) VSB Figure 5: VSB Modulation 1. One sideband is not rejected fully. 2. One sideband is transmitted fully and a small part (vestige) of the other sideband is transmitted. ω c B

165 m(t) cos( ω t) c H ( ω ) Φ ( t ) Figure 6: Block Diagram - Generation of VSB signal Here, H i (ω) is a filter which shapes the other sideband. Φ V SB (ω) = [M(ω ω c ) + M(ω + ω c )].H i (ω) To recover the original signal from the VSB signal, the VSB signal is multiplied with cos(ω c t) and passed through an LPF such that original signal is recovered. i VSB

166 Φ ( t) VSB cos( ω t) c LPF H ( ω ) 0 Figure 7: Block Diagram - Demodulation of VSB signal From Figure 6 and Figure 7, the criterion to choose LPF is: M(ω) = [Φ V SB (ω + ω c ) + Φ V SB (ω ω c )].H 0 (ω) = [H i (ω + ω c ) + H i (ω ω c )].M(ω).H 0 (ω) = H 0 (ω) = 1 m(t) H i (ω + ω c ) + H i (ω ω c )

167 Appendix: The Hilbert Transform The Hilbert Transform on a signal changes its phase by ±90 0. The Hilbert transform of a signal g(t) is represented as ĝ(t). ĝ(t) = 1 π g(t) = 1 π + + g(τ) t τ dτ ĝ(τ) t τ dτ We, say g(t) and ĝ(t) constitute a Hilbert Transform pair. If we observe the above equations, it is evident that Hilbert transform is nothing but the convolution of g(t) with 1 πt. The Fourier Transform of ĝ(t) is computed from signum function sgn(t).

168 sgn(t) 2 jω = 1 jsgn(ω) πt Where, sgn(ω) = Since, ĝ(t) = g(t) 1 πt, 1, ω > 0 0, ω = 0 1, ω < 0 Ĝ(ω) = G(ω) jsgn(ω)

169 Properties of Hilbert Transform 1. g(t) and ĝ(t) have the same magnitude spectrum. 2. If ĝ(t) is HT of g(t) then HT of ĝ(t) is g(t). 3. g(t) and ĝ(t) are orthogonal over the entire interval to +. + g(t)ĝ(t) dt = 0 Complex representation of signals If g(t) is a real valued signal, then its complex representation g + (t) is given by

170 Therefore, g + (t) = g(t) + jĝ(t) G + (ω) = G(ω) + sgn(ω)g(ω) G + (ω) = 2G(ω), ω > 0 G(0), ω = 0 0, ω < 0 g + (t) is called pre-evelope and exists only for positive frequencies. For negative frequencies g (t) is defined as follows: g (t) = g(t) jĝ(t) G (ω) = G(ω) sgn(ω)g(ω)

171 Therefore, G (ω) = 2G(ω), ω < 0 G(0), ω = 0 0, ω > 0 Essentially the pre-envelope of a signal enables the suppression of one of the sidebands in signal transmission. The pre-envelope is used in the generation of the SSB-signal.

172 Angle Modulation In this type of modulation, the frequency or phase of carrier is varied in proportion to the amplitude of the modulating signal. A c c(t) Figure 1: An angle modulated signal If s(t) = A c cos(θ i (t)) is an angle modulated signal, then 1. Phase modulation: t

173 where ω c = 2πf c. 2. Frequency Modulation: θ i (t) = ω c t + k p m(t) ω i (t) = ω c + k f m(t) θ i (t) = t 0 = 2π ω i (t) dt t 0 f i (t) dt + t 0 k f m(t) dt Phase Modulation If m(t) = A m cos(2πf m t) is the message signal, then the phase modulated signal is given by

174 s(t) = A c cos(ω c t + k p m(t)) Here, k p is phase sensitivity or phase modulation index. Frequency Modulation If m(t) = A m cos(2πf m t) is the message signal, then the Frequency modulated signal is given by 2πf i (t) = ω c + k f A m cos(2πf m t) θ i (t) = ω c t + k fa m sin(2πf m t) 2πf m here, k f A m f is called frequency deviation ( f) and is called 2π f m modulation index (β). The Frequency modulated signal is given by

175 s(t) = A c cos(2πf c t + β sin(2πf m t)) Depending on how small β is FM is either Narrowband FM(β << 1) or Wideband FM(β 1). Narrow-Band FM (NBFM) In NBFM β << 1, therefor s(t) reduces as follows: s(t) = A c cos(2πf c t + β sin(2πf m t)) = A c cos(2πf c t) cos(β sin(2πf m t)) A c sin(2πf c t) sin(β sin(2πf m t)) Since, β is very small, the above equation reduces to s(t) = A c cos(2πf c t) A c β sin(2πf m t) sin(2πf c t)

176 The above equation is similar to AM. Hence, for NBFM the bandwidth is same as that of AM i.e., 2 message bandwidth(2 B). A NBFM signal is generated as shown in Figure??. m(t) Phase shifter DSB SC π/2 Asin( ω t) oscillator c A cos( ω t) c + NBFM signal Figure 2: Generation of NBFM signal

177 Wide-Band FM (WBFM) A WBFM signal has theoritically infinite bandwidth. Spectrum calculation of WBFM signal is a tedious process. For, practical applications however the Bandwidth of a WBFM signal is calculated as follows: Let m(t) be bandlimited to BHz and sampled adequately at 2BHz. If time period T = 1/2B is too small, the signal can be approximated by sequence of pulses as shown in Figure??

178 m p T Figure 3: Approximation of message signal If tone modulation is considered, and the peak amplitude of the sinusoid is m p, the minimum and maximum frequency deviations will be ω c k f m p and ω c + k f m p respectively. The spread of pulses in frequency domain will be 2π T = 4πB t

179 as shown in Figure?? ω k m π c f p 4 B ω + k m c f p Figure 4: Bandwidth calculation of WBFM signal Therefore, total BW is 2k f m p + 8πB and if frequency deviation is considered BW fm = 1 2π (2k f m p + 8πB) BW fm = 2( f + 2B)