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communication system Chapter 2

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communication system Chapter 2

1. 1. Communication System Ass. Prof. Ibrar Ullah BSc (Electrical Engineering) UET Peshawar MSc (Communication & Electronics Engineering) UET Peshawar PhD (In Progress) Electronics Engineering (Specialization in Wireless Communication) MAJU Islamabad E-Mail: ibrar@cecos.edu.pk Ph: 03339051548 (0830 to 1300 hrs) 1
2. 2. Chapter # 2 2
3. 3. Chapter-2 • • • • • • • • • Signals and systems Size of signal Classification of signals Signal operations The unit impulse function Correlation Orthogonal signals Trigonometric Fourier Series Exponential Fourier Series 3
4. 4. Signals and systems • A signal is a any time-varying quantity of information or data. • Here a signal is represented by a function g(t) of the independent time variable t. Only one-dimensional signals are considered here. • Signals are processed by systems. • "A system is composed of regularly interacting or interrelating groups of activities/parts which, when taken together, form a new whole." (from Wikipedia) • Here a system is an entity that processes an input signal g(t) to produce a new output signal h(t). 4
5. 5. Size of Signals The size of any entity is a number that indicates the largeness or strength of that entity • Energy The energy Eg of a signal g(t) can be calculated by the formula For complex valued signal g(t) it can be written as The energy is finite only, if 5
6. 6. Size of Signals (cont…) • Power The power Pg of a signal g(t) can be calculated by the formula For complex valued signal g(t) it can be written as The power represents the time average(mean) of the signal amplitude squared. It is finite only if the signal is periodic or has statistical regularity. 6
7. 7. Size of Signals (cont…) • Examples for signals with finite energy (a) and finite power (b): • • Remark: The terms energy and power are not used in their conventional sense as electrical energy or power, but only as a measure for the signal size. 7
8. 8. Example 2.1 Page: 17 • Determine the suitable measures of the signals given below: • The signal (a) → 0 as t → ∞ Therefore, the suitable measure for this signal is its energy Eg given by ∞ 0 ∞ −∞ −1 0 Eg = ∫ g 2 (t )dt = ∫ (2) 2 dt + ∫ 4e −t dt = 4 + 4 = 8 • 8
9. 9. Example 2.1 Page: 17 (Cont.) The signal in the fig. Below does not --- to 0 as t  ∞ . However it is periodic, therefore its power exits. 9
10. 10. Example 2.2 page-18 (a) 10
11. 11. Example 2.2 (cont…) Remarks: A sinosoid of amplitude C has power of frequency and phase . regardless of its 11
12. 12. Example 2.2 (cont…) 12
13. 13. Example 2.2 (cont…) And rms value We can extent this result to a sum of any number of sinusoids with distinct frequencies. 13
14. 14. Example 2.2 (cont…) Recall that Therefore The rms value is 14
15. 15. Classification of signals 1) 2) 3) 4) 5) 6) Continuous-time and discrete-time signals Analog and digital signals Periodic and aperiodic signals Energy and power signals Deterministic and random signals Causal vs. Non-causal signals 15
16. 16. Classification of signals Continuous time (CT) and discrete time (DT) signals CT signals take on real or complex values as a function of an independent variable that ranges over the real numbers and are denoted as x(t). DT signals take on real or complex values as a function of an independent variable that ranges over the integers and are denoted as x[n]. Note the use of parentheses and square brackets to distinguish between CT and DT signals. 16
17. 17. Classification of signals Analog continuous time signal x(t) Analog discrete time signal x[n] 17
18. 18. Classification of signals Digital continuous time signal Digital discrete time signal 18
19. 19. Classification of signals periodic and aperiodic signals Examples: 19
20. 20. Classification of signals 20
21. 21. Classification of signals 21
22. 22. Classification of signals Energy and Power Signals 22
23. 23. Classification of signals Remarks: • A signal with finite energy has zero power. • A signal can be either energy signal or power signal, not both. • A signal can be neither energy nor power e.g. ramp signal 23
24. 24. Classification of signals Deterministic and Random signals • A signal g(t) is called deterministic, if it is completely known and can be described mathematically • A signal g(t) is called random, if it can be described only by terms of probabilistic description, such as – distribution – mean value (The average or expected value) – squared mean value (The expected value of the squared error) – standard deviation (The square root of the variance) 24
25. 25. Classification of signals Causal vs. Non-causal signals A causal signal is zero for t < 0 and an non-causal signal is zero for t > 0 or A causal signal is any signal that is zero prior to time zero. Thus, if x(n) denotes the signal amplitude at time (sample) n, the signal x is said to be causal if x(n)=0 for all n< 0 25
26. 26. Classification of signals Right- and left-sided signals A right-sided signal is zero for t < T and a left-sided signal is zero for t > T where T can be positive or negative. 26
27. 27. Classification of signals  Even signals xe(t) and odd signals xo(t) are defined as xe(t) = xe(−t) and xo(−t) = −xo(t).  If the signal is even, it is composed of cosine waves. If the signal is odd, it is composed out of sine waves. If the signal is neither even nor odd, it is composed of both sine and cosine waves. 27
28. 28. Signal operations Time Shifting 28
29. 29. Signal operations (cont'd) Time-Scaling 29
30. 30. Signal operations (cont'd) Time- Inversion/ Time-Reversal 30
31. 31. Signal operations (cont'd) Example: 2.4 For the dignal g(t), shown in the fig. Below , sketch g(-t) 31
32. 32. Unit Impilse Function 32
33. 33. Unit Impilse Function 33
34. 34. Unit Step Function 34
35. 35. Signals and Vectors • • • • • • Analogy between Signals and Vectors --A vector can be represented as a sum of its components --A Signal can also be represented as a sum of its components Component of a vector: A vector is represented by bold-face type Specified by its magnitude and its direction. • E.g • • • Vector x of magnitude | x | and Vector g of magnitude | g | Let the component of vector g along x be cx Geometrically this component is the projection of g on x The component can be obtained by drawing a perpendicular from the tip of g on x and expressed as g = cx + e 35
36. 36. Component of a Vector • There are infinite ways to express g in terms of x • g is represented in terms of x plus another vector which is called the error vector e • If we approximate g by cx 36
37. 37. Component of a Vector (cont..) The error in this approximation is the vector e e = g - cx The error in the approximation in both cases for last figure are 37
38. 38. Component of a Vector (cont..) • We can mathematically define the component of vector g along x • We take dot product (inner or scalar) of two vectors g and x as: g.x = | g || x | cos θ • The length of vector by definition is |x|² = x.x • The length of component of g along x is | g | cos θ Multiply both sides by | x | c| x | = c | x | ² = | g | | x | cos θ = g.x 38
39. 39. Component of a Vector (cont..) Consider the first figure again and expression for c • Let g and x are perpendicular (orthogonal) • g has a zero component along x gives c = 0 • From equation g and x are orthogonal if the inner (scalar or dot) product of two vectors is zero i.e. g.x = 0 39
40. 40. Component of a Signal • Vector component and orthogonality can be extended to signals • Consider approximating a real signal g(t) in terms of another real signal x(t) The error e(t) in the approximation is given by 40
41. 41. Component of a Signal • As energy is one possible measure of signal size. • To minimize the effect of error signal we need to minimize its size-----which is its energy over the interval This is definite integral with This is definite integral with dummy variable t t dummy variable Hence function of cc(not t) Hence function of (not t) For some choice of c the energy is minimum Necessary condition Necessary condition 41
42. 42. Component of a Signal 42
43. 43. Component of a Signal Recall equation for two vectors •• Remarkable similarity between behavior of Remarkable similarity between behavior of vectors and signals. Area under the product of two vectors and signals. Area under the product of two signals corresponds to the dot product of two signals corresponds to the dot product of two vectors vectors ••The energy of the signal is the inner product of The energy of the signal is the inner product of signal with itself and corresponds to the vector signal with itself and corresponds to the vector length squared (which is the inner product of the length squared (which is the inner product of the vector with itself) vector with itself) 43
44. 44. Component of a Signal Consider the signal equation again: t2 1 c= ∫ g (t ) x(t )dt E x t1 • Signal g(t) contains a component cx(t) • cx(t) is the projection of g(t) on x(t) • If cx(t) = 0 ⇒ c = 0 ⇒ signal g(t) and x(t) are orthogonal over the interval [ t1 , t 2 ] 44
45. 45. Example 2.5 Component of a Signal (cont..) For the square signal g(t), find the component of g(t) of the form sint or in other words approximate g(t) in terms of sint g (t ) ≅ c sin t 0 ≤ t ≥ 2π 45
46. 46. Example 2.5 (cont…) x(t ) = sin t and From equation for signals c= ⇒ 1 c= π t2 1 ∫ g (t ) x(t )dt E x t1 π 2π  4 1 ∫ g (t ) sin tdt = π ∫ sin tdt + π − sin tdt  = π ∫ o 0  2π g (t ) ≅ 4 sin t π 46
47. 47. Orthogonality in complex signals For complex functions of t over an interval g (t ) ≅ cx(t ) t2 2 Ex = ∫ x(t ) dt t1 Coefficient c and the error in this case is e(t ) = g (t ) − cx(t ) t2 Ee = ∫ g (t ) − cx(t ) t1 2 47
48. 48. Orthogonality in complex signals 2 t2 Ee = ∫ g (t ) − cx(t ) t1 We know that: ( ) u + v = ( u + v ) u + v = u v + u v + uv 2 t2 2 Ee = ∫ g (t )dt − t1 1 Ex t2 ∗ ∗ 2 2 2 ∗ 1 ∫ g (t ) x (t )dt + c Ex − Ex t1 ∗ ∗ 2 t2 g (t ) x ∗ (t )dt ∫ t1 48
49. 49. Orthogonality in complex signals t 1 2 c= g (t )x ∗ (t )dt Ex ∫ t1 So, two complex functions are orthogonal over an interval, if t2 ∫ x (t ) x (t )dt = 0 1 ∗ 2 t1 or t2 x1∗ (t ) x2 (t )dt = 0 ∫ t1 49
50. 50. Energy of the sum of orthogonal signals • Sum of the two orthogonal vectors is equal to the sum of the lengths of the squared of two vectors. z = x+y then 2 2 z = x + y 2 • Sum of the energy of two orthogonal signals is equal to the sum of the energy of the two signals. If x(t) and y(t) are orthogonal signals over the interval, [ t1 ,t 2 ] and if z(t) = x(t)+ y(t) then Ez = Ex + E y 50
51. 51. Correlation Consider vectors again: • Two vectors g and x are similar if g has a large component along x OR • If c has a large value, then the two vectors will be similar c could be considered the quantitative measure of similarity between g and x But such a measure could be defective. The But such a measure could be defective. The amount of similarity should be independent of the amount of similarity should be independent of the lengths of g and x lengths of g and x 51
52. 52. Correlation Doubling g should not change the similarity between g and x Doubling g doubles the value of c Doubling g doubles the value of c Doubling x halves the value of c Doubling x halves the value of c ⇒ However: However: c is faulty measure for similarity • Similarity between the vectors is indicated by angle between the vectors. • The smaller the angle , the largest is the similarity, and vice versa • Thus, a suitable measure would be c = cos θ , given by n g .x cn = cos θ = g x Independent of the lengths of g and x 52
53. 53. Correlation g .x cn = cos θ = g x This similarity measure cn is known as correlation co-efficient. The magnitude of cn is never greater than unity − 1 ≤ cn ≥ 1 •Same arguments for defining a similarity index (correlation co-efficient) for signals • consider signals over the entire time interval • normalize c by normalizing the two signals to have unit ∞ 1 energies. c = g (t ) x(t )dt n Eg Ex ∫ −∞ 53
54. 54. Correlation consider g (t ) = kx(t ) If k is positive then: cn = 1 Related signals-------Best friends Related signals-------Best friends Negative then: c n = −1 Dissimilarity worst enemies Dissimilarity worst enemies If g(t) and x(t) are orthogonal then cn = 0 Unrelated signals-------Strangers Unrelated signals-------Strangers 54
55. 55. Example 2.6 Find the correlation co-efficient cn between the pulse x(t) and the pulses g i (t ) =, i = 1,2,3,4,5,6 5 5 0 0 E x = ∫ x 2 (t )dt = ∫ dt = 5 cn = 1 Eg Ex ∞ ∫ g (t ) x(t )dt −∞ Similarly E g1 = 5 5 1 ⇒ cn = ∫ dt = 1 5× 5 0 Maximum possible similarity Maximum possible similarity 55
56. 56. Example 2.6 (cont…) 5 5 0 cn = E g 2 =1.25 0 E x = ∫ x 2 (t )dt = ∫ dt = 5 1 Eg Ex ∞ ∫ g (t ) x(t )dt −∞ 5 1 ⇒ cn = ∫ (0.5)dt = 1 1.25 × 5 0 Maximum possible similarity……independent of amplitude Maximum possible similarity……independent of amplitude 56
57. 57. Example 2.6 (cont…) 5 5 0 0 E x = ∫ x 2 (t )dt = ∫ dt = 5 Similarly cn = 1 Eg Ex ∞ ∫ g (t ) x(t )dt −∞ E g1 = 5 5 1 ⇒ cn = ∫ (1)(−1)dt = −1 5× 5 0 57
58. 58. Example 2.6(cont…) 5 5 0 0 E x = ∫ x 2 (t )dt = ∫ dt = 5 T E = ∫ (e 2 − at 0 Here 1 a= 5 T ) dt = ∫ e E g 4 = 2.1617 0 T =5 − 2 at 1 dt = (1 − e − 2 aT ) 2a 5 −t 5 1 cn = ∫ e dt = 0.961 5 × 2.1617 0 Reaching Maximum similarity Reaching Maximum similarity 58
59. 59. Orthogonal Signal Space 59
60. 60. Orthogonal Signal Space 60
61. 61. Trigonometric Fourier series Consider a signal set: {1, cos wot , cos 2wot........ cos nwot ,.... sin wot , sin 2wot.... sin nwot ,....} •A sinusoid function with frequency nwo is called the nth harmonic of the sinusoid of frequency w o when n is an integer. • A sinusoid of frequency wo is called the fundamental •This set is orthogonal over any interval of duration because: 0  cos nw o t cos mw o tdt =  T ∫ To  o 2  n≠ m n= m≠ o wo n≠m 0  sin nw o t sin mw o tdt =  T ∫ To  o 2  To = 2π n=m≠o 61
62. 62. Trigonometric Fourier series and ∫ sin nw o for all n and m t cos mw o tdt = 0 To The trigonometric set is a complete set. Each signal g(t) can be described by a trigonometric Fourier series over the interval To : g ( t ) = a o + a 1 cos w o t + a 2 cos 2 w o t + ... or t 1 ≤ t ≤ t 1 + To + b1 sin w o t + b 2 sin 2 w o t + ... ∞ g ( t ) = a o + ∑ a n cos nw o t + b n sin nw o t t 1 ≤ t ≤ t 1 + To n =1 wn = 2π To 62
63. 63. Trigonometric Fourier series We determine the Fourier co-efficient Cn = ∫ t 1 +T o t1 ∫ t1 2 an = To 2 bn = To as: g ( t ) cos nw o tdt t 1 +T o 1 a0 = To ao ,an ,b n cos 2 nw o tdt t 1 +T o ∫ g ( t )dt t1 t 1 +T o ∫ g ( t ) cos nw o tdt t1 t 1 +T o ∫ g ( t ) sin nw t1 o tdt n = 1, 2 , 3 ,...... n = 1, 2 , 3 ,...... 63
64. 64. Compact Trigonometric Fourier series Consider trigonometric Fourier series g ( t ) = a o + a 1 cos w o t + a 2 cos 2 w o t + ... t 1 ≤ t ≤ t 1 + To + b1 sin w o t + b 2 sin 2 w o t + ... It contains sine and cosine terms of the same frequency. We can represents the above equation in a single term of the same frequency using the trigonometry identity a n cos nw o t + b n sin nw o t = C n cos( nw o t + θ n ) Cn = 2 2 a n + bn  − bn θ n = tan −1   a  n Co = ao     64
65. 65. Compact Trigonometric Fourier series ∞ g ( t ) = C 0 + ∑ C n cos( nw o t + θ n ) t 1 ≤ t ≤ t 1 + To n =1 65
66. 66. Example 2.7 Find the compact trigonometric Fourier series for the following function 66
67. 67. Example 2.7 Solution: We are required to represent g(t) by the trigonometric Fourier series over the interval 0 ≤ t ≤ π and To = π wo = 2π = 2 rad sec To Trigonometric form of Fourier series: n =1 a0 ?, an ?, bn ? π g (t ) = ao + ∑ an cos 2nt + bn sin 2nt 0 ≤≤ t ∞ 67
68. 68. Example 2.7 π 1 −t 2 a0 = ∫ e dt = 0.50 π 0 C o = ao π 2 −t 2   a n = ∫ e 2 cos 2 ntdt = 0.504   2 π 0  1 + 16 n  2 2 Cn = an + bn π 2 −t 2  8n  bn = ∫ e sin 2ntdt = 0.504 2  π 0  1 + 16n  Compact Fourier series is given by n =1 π g (t ) = C0 + ∑ Cn cos(nwot + θ n ) 0 ≤≤ t ∞ 68
69. 69. Example 2.7 Co = ao = 0.504 Cn = a + b = 0.504 2 n 2 n 64n 2 2 + = 0.504( ) 2 2 2 (1 + 16n ) 1 + 16n 4 (1 + 16n ) 2 2  −b  θ n = tan −1  n  = tan −1 ( − 4n ) = − tan 4n  a   n  + 0.084 cos(6t − 85.24o ) + 0.063 cos(8t − 86.42o ) + ....... π = 0.504 + 0.244 cos(2t − 75.96o ) + 1.25 cos(4t − 82.87 o ) 0 ≤≤ t ) π ( 2 cos 2nt − tan −1 4n 1 + 16n 2 n =1 ⇒ g (t ) = 0.504 + 0.504∑ 0 ≤≤ t ∞ 69
70. 70. Example 2.7 n 0 1 2 3 4 5 6 7 Cn 0.504 0.244 0.125 0.084 0.063 0.054 0.042 0.063 Өn 0 -75.96 -82.87 -85.24 -86.42 -87.14 -87.61 -87.95 Amplitudes and phases for first seven harmonics 70
71. 71. Periodicity of the trigonometric Fourier series The co-efficient of the of the Fourier series are calculated for the interval [ t1 , t1 + To ] ∞ φ ( t ) = C o + ∑ C n cos( nw o t + θ n ) for all t n =1 ∞ φ ( t + T 0 ) = C o + ∑ C n [cos( nw o ( t + T 0 ) + θ n ] n =1 ∞ = C + ∑ C cos( nwt + 2 n π + θ ) o n o n n =1 ∞ = C + ∑ C cos( nwt + θ ) o n o n n =1 for all t = φ (t ) 71
72. 72. Periodicity of the trigonometric Fourier series ao = 1 To ∫g (t )dt To an = 2 ∫g (t ) cosnw otdt To To n= 1,2,3,…… bn = 2 ∫g (t ) sinnw otdt To To n= 1,2,3,…… ∫ Means integration over any interval of To Means integration over any interval of To To 72
73. 73. Fourier Spectrum Consider the compact Fourier series ∞ g (t ) = C0 + ∑ Cn cos(nwot + θ n ) n =1 This equation can represents a periodic signal g(t) of frequencies: 0(dc), wo ,2wo ,3wo ,....., nwo Amplitudes: C0 , C1 , C2 , C3,......,Cn Phases: 0, θ1 , θ 2 ,θ 3 ,.....θ n 73
74. 74. Fourier Spectrum Frequency domain description of φ ( t ) cn vs w (Amplitude spectrum) θ vs w (phase spectrum) Time domain description of φ(t ) 74
75. 75. Fourier Spectrum Consider the compact Fourier series ∞ g (t ) = C0 + ∑ Cn cos(nwot + θ n ) n =1 This equation can represents a periodic signal g(t) of frequencies: 0(dc), wo ,2wo ,3wo ,....., nwo Amplitudes: C0 , C1 , C2 , C3,......,Cn Phases: 0, θ1 , θ 2 ,θ 3 ,.....θ n 75
76. 76. Fourier Spectrum Frequency domain description of φ ( t ) cn vs w (Amplitude spectrum) θ vs w (phase spectrum) Time domain description of φ(t ) 76
77. 77. Example 2.8 Find the compact Fourier series for the periodic square wave w(t) shown in figure and sketch amplitude and phase spectrum Fourier series: ∞ w ( t ) = a o + ∑ a n cos nw o t + b n sin nw o t W(t)=1 only over (-To/4, To/4) W(t)=1 only over (-To/4, To/4) and and n =1 1 a0 = To t 1 +T o ∫ g ( t )dt t1 1 ⇒ a0 = To To ∫ To 4 4 1 dt = 2 w(t)=0 over the remaining w(t)=0 over the remaining segment segment 77
78. 78. Example 2.8 2 an = To To 4 ∫ cos nw o tdt = −T o 4  0   2 =  nπ −2  nπ  2 bn = To To ∫ To 2  nπ  sin   nπ 2   n − even n = 1 , 5 , 9 , 13... n = 3 , 7 , 11 , 15... 4 4 sin ntdt = 0 ⇒ bn = 0 All the sine terms are zero All the sine terms are zero 78
79. 79. Example 2.8 w(t ) = 1 2 1 1 1  +  cos w o t − cos 3 w o t + cos 5 w o t − cos 7 w o t + ....  2 π 3 5 7  The series is already in compact form as there are no sine terms The series is already in compact form as there are no sine terms Except the alternating harmonics have negative amplitudes Except the alternating harmonics have negative amplitudes The negative sign can be accommodated by aaphase of π radians as The negative sign can be accommodated by phase of radians as − cos x = cos( x − π ) Series can be expressed as: w(t ) = 1 2 1 1 1 1  +  cos w o t + cos( 3 w o t − π ) + cos 5 w o t + cos( 7 w o t − π ) + cos 9 w o t + ....  2 π 3 5 7 9  79
80. 80. Example 2.8 1 Co = 2 0 −π θn = Cn 0  = 2  nπ  n − even n − odd for all n≠ 3,5,7,11,15,….. for all n = 3,5,7,11,15,….. We could plot amplitude and phase We could plot amplitude and phase spectra using these values…. spectra using these values…. In this special case ififwe allow Cnnto In this special case we allow C to take negative values we do not need aa take negative values we do not need phase of − π to account for sign. phase of to account for sign. Means all phases are zero, so only Means all phases are zero, so only amplitude spectrum is enough amplitude spectrum is enough 80
81. 81. Example 2.8 Consider figure w o ( t ) = 2 ( w ( t ) − 0 .5 ) w(t ) = 4 1 1 1  cos w o t − cos 3 w o t + cos 5 w o t − cos 7 w o t + ....   π 3 5 7  81
82. 82. Exponential Fourier series 82
83. 83. Exponential Fourier series 83
84. 84. Example Consider example 2.7 again, calculate exponential Fourier series wo = 2π = 2 rad sec To To = π ∞ ϕ (t ) = D n e j 2 nt ∑ n = −∞ 1 Dn = To π π 1 −t 2 − j 2 nt − j 2 nt ∫To ϕ ( t )e dt = π ∫ e e dt 0 1 = ∫e π 0 −( 1 +2 n ) t 2 dt = 84
85. 85. Example and 0.504 = 1+ j 4 n ∞ 1 ϕ ( t ) = 0.504 ∑ e j 2 nt n = −∞ 1 + j 4 n 1 1 1   1+ e j 2t + e j 4t + e j 6 t + ...   1+ j 4 1+ j 8 1 + j 12  = 0.504  1 1  1  e− j 2t + e−j 4t + e − j 6 t + ...   1− j 4 1− j 8 1 + j 12   Dnnare complex D are complex Dnnand D-n are conjugates D and D-n are conjugates 85
86. 86. Example 1 D n = D −n = C n 2 < D n = θ n and thus D n = D n e jθ n < D −n = − θ n and D − n = D n e − jθ n D o = 0.504 o 0.504 ⇒ 0.122 e − j 75.96 1+ j 4 o 0.504 = ⇒ 0.122 e − j 75.96 1− j 4 D1 = D −1 < D 1 = −75.96 o < D −1 = 75.96 o 86
87. 87. Example o 0.504 ⇒ 0.625 e − j 82.87 1+ j 8 0.504 − j 82.87 o = ⇒ 0.625 e 1− j 8 D2 = < D 1 = −82.87 o D −2 < D −1 = 82.87 o And so on…. 87
88. 88. Exponential Fourier Spectra 88