Representing a Signal

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1 The Fourier Series

2 Representing a Signal The convolution method for finding the response of a system to an excitation takes advantage of the linearity and timeinvariance of the system and represents the excitation as a linear combination of impulses and the response as a linear combination of impulse responses The Fourier series represents a signal as a linear combination of complex sinusoids 5/10/04 M. J. Roberts - All Rights Reserved 2

3 Linearity and Superposition If an excitation can be expressed as a sum of complex sinusoids the response can be expressed as the sum of responses to complex sinusoids. 5/10/04 M. J. Roberts - All Rights Reserved 3

4 Real and Complex Sinusoids cos( x)= e jx + e 2 jx sin( x)= e jx e j2 jx 5/10/04 M. J. Roberts - All Rights Reserved 4

5 Jean Baptiste Joseph Fourier 3/21/1768-5/16/1830 5/10/04 M. J. Roberts - All Rights Reserved 5

6 Continuous- Time Fourier Series Concept 5/10/04 M. J. Roberts - All Rights Reserved 6

7 Continuous- Time Fourier Series Concept 5/10/04 M. J. Roberts - All Rights Reserved 7

8 Continuous- Time Fourier Series Concept 5/10/04 M. J. Roberts - All Rights Reserved 8

9 CT Fourier Series Definition The Fourier series representation, x F t, of a signal, x(t), over a time, t 0 < t < t 0 + T F, is x F k = () 2π ( F ) ()= t X[ k] e j kf t where X[k] is the harmonic function, k is the harmonic number and ff = 1/ TF (pp ). The harmonic function can be found from the signal as t0 TF 1 + j2 kf t F X[ k]= x() te π ( ) T F t 0 The signal and its harmonic function form a Fourier series FS pair indicated by the notation, x t X k. dt () [ ] 5/10/04 M. J. Roberts - All Rights Reserved 9

10 CTFS of a Real Function It can be shown (pp ) that the continuous-time Fourier series (CTFS) harmonic function of any real-valued function, x(t), has the property that X[ k]= X * k [ ] One implication of this fact is that, for real-valued functions, the magnitude of the harmonic function is an even function and the phase is an odd function. 5/10/04 M. J. Roberts - All Rights Reserved 10

11 The Trigonometric CTFS The fact that, for a real-valued function, x(t), X[ k]= X * k [ ] also leads to the definition of an alternate form of the CTFS, the so-called trigonometric form. k= 1 xf()= t Xc[ 0]+ Xc[ k] cos 2π( kff) t Xs k sin 2π( kff )t where { ( )} ( )+ [ ] 0 F 2 Xc[ k]= x() t cos( 2π( kff ) t) dt T F t + T t 0 0 F 2 Xs[ k]= x() t sin( 2π( kff ) t) dt T F t + T t 0 5/10/04 M. J. Roberts - All Rights Reserved 11

12 The Trigonometric CTFS Since both the complex and trigonometric forms of the CTFS represent a signal, there must be relationships between the harmonic functions. Those relationships are Xc[ 0]= X[ 0] Xs [ 0]= 0 * Xc[ k]= X[ k]+ X [ k] * Xs k j X k X k ( ) [ ]= [ ] [ ], k = 123,,, K X[ 0]= Xc[ 0] X X c X[ [ k] j s[ k] k]= 2 * Xc X X[ ]= X [ [ k]+ j s[ k] ]= k k 2, k = 123,,, K 5/10/04 M. J. Roberts - All Rights Reserved 12

13 Periodicity of the CTFS It can be shown (pg. 218) that the CTFS representation, x F t of a function, x(t), is periodic with fundamental period, T F. Therefore, if x(t) is also periodic with fundamental period, T 0 and if T F is an integer multiple of T 0 then the two functions are equal for all t, not just in the interval, t 0 < t < t 0 + T F. () 5/10/04 M. J. Roberts - All Rights Reserved 13

14 CTFS Example #1 ()= ( ) Let a signal be defined by x t 2 cos 400πt and let T F = 5mswhich is the same as T 0 t0 + TF t0 + TF 2 2 Xc[ k]= x() t cos( 2π( kff ) t) dt Xs [ k]= x() t sin( 2π ( kff ) t) dt T T F t 0 F t 0 5/10/04 M. J. Roberts - All Rights Reserved 14

15 CTFS Example #1 5/10/04 M. J. Roberts - All Rights Reserved 15

16 CTFS Example #2 ()= ( ) Let a signal be defined by x t 2 cos 400πt and let T F = 10 ms which is T 2 0 5/10/04 M. J. Roberts - All Rights Reserved 16

17 CTFS Example #2 5/10/04 M. J. Roberts - All Rights Reserved 17

18 CTFS Example #3 Let x()= t cos( 20πt)+ sin( 30πt) and let T F = 200 ms 5/10/04 M. J. Roberts - All Rights Reserved 18

19 CTFS Example #3 5/10/04 M. J. Roberts - All Rights Reserved 19

20 CTFS Example #3 5/10/04 M. J. Roberts - All Rights Reserved 20

21 CTFS Example #3 5/10/04 M. J. Roberts - All Rights Reserved 21

22 Linearity of the CTFS These relations hold only if the harmonic functions, X, of all the component functions, x, are based on the same representation time. 5/10/04 M. J. Roberts - All Rights Reserved 22

23 CTFS Example #4 Let the signal be a 50% duty-cycle square wave with an amplitude of one and a fundamental period, T 0 = 1 ()= ( ) () x t rect 2t comb t 5/10/04 M. J. Roberts - All Rights Reserved 23

24 CTFS Example #4 5/10/04 M. J. Roberts - All Rights Reserved 24

25 CTFS Example #4 5/10/04 M. J. Roberts - All Rights Reserved 25

26 CTFS Example #4 5/10/04 M. J. Roberts - All Rights Reserved 26

27 CTFS Example #4 A graph of the magnitude and phase of the harmonic function as a function of harmonic number is a good way of illustrating it. 5/10/04 M. J. Roberts - All Rights Reserved 27

28 CTFS Example #5 ()= ( ) Let x t 2 cos 400πt and let T F = 75. ms which is 1.5 periods of this signal. 5/10/04 M. J. Roberts - All Rights Reserved 28

29 CTFS Example #5 5/10/04 M. J. Roberts - All Rights Reserved 29

30 CTFS Example #5 5/10/04 M. J. Roberts - All Rights Reserved 30

31 CTFS Example #5 The CTFS representation of this cosine is the signal below, which is an odd function, and the discontinuities make the representation have significant higher harmonic content. This is a very inelegant representation. 5/10/04 M. J. Roberts - All Rights Reserved 31

32 CTFS of Even and Odd Functions For an even function The complex CTFS harmonic function, X k, is purely real The sine harmonic function, k, is zero For an odd function The complex CTFS harmonic function, X k, is purely imaginary The cosine harmonic function, X c k, is zero X s [ ] [ ] [ ] [ ] 5/10/04 M. J. Roberts - All Rights Reserved 32

33 CTFS Example #6 This signal has no known functional description but it can still be represented by a CTFS. 5/10/04 M. J. Roberts - All Rights Reserved 33

34 CTFS Example #6 5/10/04 M. J. Roberts - All Rights Reserved 34

35 CTFS Example #6 5/10/04 M. J. Roberts - All Rights Reserved 35

36 CTFS Example #6 5/10/04 M. J. Roberts - All Rights Reserved 36

37 CTFS Properties Let a signal, x(t), have a fundamental period, 0 and let a signal, y(t), have a fundamental period, T 0 y. Let the CTFS harmonic functions, each using the fundamental period as the representation time, T F, be X[k] and Y[k]. In the properties which follow the two fundamental periods are the same unless otherwise stated. T x Linearity ()+ () [ ]+ [ ] FS αx t βy t αx k βy k 5/10/04 M. J. Roberts - All Rights Reserved 37

38 CTFS Properties Time Shifting FS j2π ( kf0) t0 x( t t ) e X k 0 [ ] FS j( kω ) t x( t t ) e 0 0 X k 0 [ ] 5/10/04 M. J. Roberts - All Rights Reserved 38

39 CTFS Properties Frequency Shifting (Harmonic Number Shifting) j e 2 π ( k0f0) t FS () t k k x X[ 0] x X[ 0] ( ) () j k0ω 0 t FS e t k k A shift in frequency (harmonic number) corresponds to multiplication of the time function by a complex exponential. Time Reversal x( ) FS t X [ k] 5/10/04 M. J. Roberts - All Rights Reserved 39

40 Time Scaling CTFS Properties Let z()= t x ( at), a> 0 T0 x Case 1. TF = = a Z[ k]= X[ k] T 0z for z(t) Case 2. T F = T 0 for z(t) If a is an integer, Z[ k]= x k X k, a an integer a 0, otherwise 5/10/04 M. J. Roberts - All Rights Reserved 40

41 CTFS Properties Time Scaling (continued) 5/10/04 M. J. Roberts - All Rights Reserved 41

42 CTFS Properties Change of Representation Time With T = T 0, x() t FS X [ k ] F F x With T = mt0, x t FS () X [ k ] X m [ k]= x k X k, an integer m m 0, otherwise (m is any positive integer) m 5/10/04 M. J. Roberts - All Rights Reserved 42

43 CTFS Properties Change of Representation Time 5/10/04 M. J. Roberts - All Rights Reserved 43

44 CTFS Properties Time Differentiation FS d FS ( x() t ) j2π kf0 X k dt d FS ( x() t ) j( kω ) X[ k] dt 0 ( ) [ ] FS 5/10/04 M. J. Roberts - All Rights Reserved 44

45 CTFS Properties Time Integration Case 1. X0 [ ]= 0 t t FS x( λ) dλ FS x( λ) dλ j [ ] ( ) X k π kf 2 0 [ ] ( ) X k jkω 0 Case 1 Case 2 Case 2. X[ 0] 0 t ( ) x λ dλ is not periodic 5/10/04 M. J. Roberts - All Rights Reserved 45

46 CTFS Properties Multiplication-Convolution Duality () () FS x t y t X k Y k [ ] [ ] (The harmonic functions, X[k] and Y[k], must be based on the same representation period,.) T F () () [ ] [ ] FS x t y t T X k Y k The symbol,, indicates periodic convolution. Periodic convolution is defined mathematically by x t y t x t y t () ()= ( ) ( ) x t y t x τ y t τ dτ T 0 () ()= () () where x ap t is any single period of ap 0 () x t () 5/10/04 M. J. Roberts - All Rights Reserved 46

47 CTFS Properties 5/10/04 M. J. Roberts - All Rights Reserved 47

48 CTFS Properties Conjugation x FS () t X * * [ k] Parseval s Theorem 1 T 0 T x() t dt = X[ k] k = The average power of a periodic signal is the sum of the average powers in its harmonic components. 5/10/04 M. J. Roberts - All Rights Reserved 48

49 Convergence of the CTFS For continuous signals, convergence is exact at every point. Partial CTFS Sums x N N 2 ( 0 ) ()= t X[ k] e π k= N j kf t A Continuous Signal 5/10/04 M. J. Roberts - All Rights Reserved 49

50 Convergence of the CTFS Partial CTFS Sums For discontinuous signals, convergence is exact at every point of continuity. Discontinuous Signal 5/10/04 M. J. Roberts - All Rights Reserved 50

51 Convergence of the CTFS The Gibbs Phenomenon 5/10/04 M. J. Roberts - All Rights Reserved 51

52 Discrete- Time Fourier Series Concept 5/10/04 M. J. Roberts - All Rights Reserved 52

53 The Discrete-Time Fourier Series The formal derivation of the discrete-time Fourier series (DTFS) is on pages The results are x F 2π ( F ) [ n]= X[ k] e k= N F j kf n n0 + NF 1 1 j2 kf n F n= n F X[ k]= x[ ne ] π ( ) N 0 where N F is the representation time, FF = 1, and the notation, N k = N F means a summation over any range of consecutive k s exactly in length. N F F 5/10/04 M. J. Roberts - All Rights Reserved 53

54 The Discrete-Time Fourier Series Notice that in x F 2π ( F ) [ n]= X[ k] e k= N F j kf n the summation is over exactly one period, a finite summation. This is because of the periodicity of the complex sinusoid, e j 2π kf F n ( ) in harmonic number, k. That is, if k is increased by any integer multiple of the complex sinusoid does not change. N F e ( ) (( + ) ) = e j2π kf n j2π k mn F n F F F (m an integer) This occurs because discrete time, n, is always an integer. 5/10/04 M. J. Roberts - All Rights Reserved 54

55 The Discrete-Time Fourier Series In the very common case in which the representation time is taken as the fundamental period,, the DTFS is N 0 j kf n x[ n]= 2π( ) X[ k] e X k 0 FS 1 x N k= N 0 or in terms of radian frequency [ ]= [ ] 0 n= j k n x[ n]= ( Ω ) X[ k] e X k 0 FS 1 x N k= N 0 2π where Ω 0 = 2πF 0 = N0 N N 0 ne [ ]= [ ] 0 n= 0 ne ( ) j2π kf n 0 j( kω ) n 0 5/10/04 M. J. Roberts - All Rights Reserved 55

56 DTFS Example 5/10/04 M. J. Roberts - All Rights Reserved 56

57 DTFS Properties Let a signal, x[n], have a fundamental period, N 0 x, and let a signal, y[n], have a fundamental period, N 0 y. Let the DTFS harmonic functions, each using the fundamental period as the representation time, N F, be X[k] and Y[k]. In the properties to follow the two fundamental periods are the same unless otherwise stated. Linearity ()+ () [ ]+ [ ] FS αx t βy t αx k βy k 5/10/04 M. J. Roberts - All Rights Reserved 57

58 Time Shifting DTFS Properties FS j2π ( kf0) n0 x[ n n0 ] e X k FS j( kω n x[ n n ] e 0 ) 0 0 X[ k] [ ] 5/10/04 M. J. Roberts - All Rights Reserved 58

59 DTFS Properties Frequency Shifting (Harmonic Number Shifting) j e 2 π ( k0f0) n FS [ n] k k x X[ 0] x X[ 0] j( k0ω0) n FS e [ n] k k Conjugation x FS [ n] X * * [ k] Time Reversal x FS [ n] X [ k] 5/10/04 M. J. Roberts - All Rights Reserved 59

60 Time Scaling Let z n x an, a DTFS Properties [ ]= [ ] > 0 If a is not an integer, some values of z[n] are undefined and no DTFS can be found. If a is an integer (other than 1) then z[n] is a decimated version of x[n] with some values missing and there cannot be a unique relationship between their harmonic functions. However, if then z[ n]= n x n, an integer m m 0, otherwise 1 Z[ k]= X[ m k ], N mn F = 0 5/10/04 M. J. Roberts - All Rights Reserved 60

61 DTFS Properties 5/10/04 M. J. Roberts - All Rights Reserved 61

62 DTFS Properties Change of Representation Time With N = N 0, x n FS [ ] X F x [ k ] With N = qn 0, x n FS [ ] X X q k F x (q is any positive integer) [ k ] k k X, [ ]= q an integer q 0, otherwise q 5/10/04 M. J. Roberts - All Rights Reserved 62

63 DTFS Properties 5/10/04 M. J. Roberts - All Rights Reserved 63

64 DTFS Properties Accumulation n m= n m= [ ] FS X k x[ m] j 1 e FS X k x[ m] j k 1 e π ( kf ) 2 0 [ ] Ω 0 ( ),, k 0 k 0 Parseval s Theorem 1 N 0 x[ n] = X[ k] n N k N 2 2 = = 0 0 5/10/04 M. J. Roberts - All Rights Reserved 64

65 DTFS Properties Multiplication- Convolution Duality [ ] [ ] q= N [ ] FS x n y n Y[ k] X[ k]= Y[ q] X k q [ ] [ ] FS x n y n N Y k X k [ ] [ ] 0 0 First Backward Difference FS j2π ( kf0 ) x[ n] x[ n 1] 1 e X k FS j( kω 0 ) x[ n] x[ n 1] 1 e X k ( ) [ ] ( ) [ ] 5/10/04 M. J. Roberts - All Rights Reserved 65

66 DTFS Properties FS FS FS 5/10/04 M. J. Roberts - All Rights Reserved 66

67 Convergence of the DTFS The DTFS converges exactly with a finite number of terms. It does not have a Gibbs phenomenon in the same sense that the CTFS does 5/10/04 M. J. Roberts - All Rights Reserved 67

68 LTI Systems with Periodic Excitation The differential equation describing an RC lowpass filter is RC v t v t v t ()+ ()= () out out in () If the excitation, v in t, is periodic it can be expressed as a CTFS, j2 kf0 t v t V k e π in k = ()= [ ] ( ) The equation for the kth harmonic alone is RC v t v t v t V k e out, k out, k in, k in in 2 ( 0 ) ()+ ()= ()= [ ] j π kf t 5/10/04 M. J. Roberts - All Rights Reserved 68

69 LTI Systems with Periodic Excitation If the excitation is periodic, the response is also, with the same fundamental period. Therefore the response can be expressed as a CTFS also. v out, k 2 ( 0 ) ()= t V [ k] e π out j kf t Then the equation for the kth harmonic becomes j kf t j kf t j kπf RCV k e V k e V k e 2 0 out [ ] + [ ] = [ ] ( ) ( ) ( ) 2π 0 2π 0 j2π kf0 t out in Notice that what was once a differential equation is now an algebraic equation. 5/10/04 M. J. Roberts - All Rights Reserved 69

70 LTI Systems with Periodic Excitation Solving the kth-harmonic equation, V out [ k]= [ ] Vin k j2kπf RC+ 1 0 Then the response can be written as v out j2π kf t ()= t V [ k] e = k = out ( ) k = [ ] V k e j2kπf RC+ 1 ( ) in j2π kf t /10/04 M. J. Roberts - All Rights Reserved 70

71 LTI Systems with Periodic Excitation Vout k The ratio,, is the frequency response of the system. V k in [ ] [ ] 5/10/04 M. J. Roberts - All Rights Reserved 71

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