Fourier Series and Fourier Transforms

Transcription

1 Fourier Series and Fourier Transforms Houshou Chen Dept. of Electrical Engineering, National Chung Hsing University

2 H.S. Chen Fourier Series and Fourier Transforms 1 Why Fourier 1. eigenfunction e jwt 2. x(t) periodic x(t) = k= c ke jkw 0t 3. e k 1w 0 t and e k 2w 0 t (k 1 k 1 ) are orthogonal over [0, T] c k = 1 T T 0 x(t)e jkw 0t dt

3 H.S. Chen Fourier Series and Fourier Transforms 2 Representing signals as superpositions of complex sinusoids not only leads to a useful expression for the system output, but also provides an insightful characterization of signals and systems. We represent the signal as the linear sum of the sinusoids. The weight associated with a sinusoid of a given frequency represents the contribution of that sinusoids to the overall signal. The study of signals and systems using sinusoidal representations is termed Fourier analysis after Joseph Fourier ( ).

4 H.S. Chen Fourier Series and Fourier Transforms 3 There are four distinct Fourier representations, each applicable to a different class of signals, determined by the periodicity properties of the signal and whether the signal is discrete or continuous in time. Fourier series (FS) for periodic continuous-time signals Discrete time Fourier series (DTFS) for periodic discrete-time signals Fourier transform (FT) for aperiodic continuous-time signals Discrete time Fourier transform (DTFT) for aperiodic discrete-time signals

5 H.S. Chen Fourier Series and Fourier Transforms 4

6 H.S. Chen Fourier Series and Fourier Transforms 5 Fourier Series Consider representing a periodic signal as a weighted superposition of complex sinusoids. Since the weighted superposition must have the same period as the signal, each sinusoid in the superposition must have the same period as the signal. This implies that the frequency of each sinusoid must be an integer multiple of the signal s fundamental frequency.

7 H.S. Chen Fourier Series and Fourier Transforms 6 If x(t) is a continuous-time signal with fundamental period T, then we seek to represent x(t) by the FS: x(t) = k A[k]e jkw 0t, where w 0 = 2π T.

8 H.S. Chen Fourier Series and Fourier Transforms 7 Fourier Transform In contrast to the case of the periodic signal, there are no restrictions on the period of the sinusoids used to represent aperiodic signals. Hence, the Fourier transform representations employ complex sinusoids having a continuum of frequencies. The signal is represented as a weighted integral of complex sinusoids where the variable of integration is the sinusoid s frequency.

9 H.S. Chen Fourier Series and Fourier Transforms 8 Continuous-time sinusoids with distinct frequencies are distinct, so the FT involves frequencies from to : x(t) = 1 2π X(jw)e jwt dw Here, X(jw) 2π is the weight at frequency w.

10 H.S. Chen Fourier Series and Fourier Transforms 9 FS We will do the following step by step 1. The derivation of FS 2. Properties of FS 3. Some examples of FS

11 H.S. Chen Fourier Series and Fourier Transforms 10 Continuous-time periodic signals are represented by the Fourier series (FS). We write the FS of a signal x(t) with fundamental period T (fundamental frequency w 0 = 2π T ) as where x(t) = X[k] = 1 T k= T 0 X[k]e jkw 0t x(t)e jkw 0t dt are the FS coefficients of the signal x(t).

12 H.S. Chen Fourier Series and Fourier Transforms 11 We say that x(t) and X[k] are an FS pair and denote this relationship as x(t) FS;w 0 X[k] In some problem it is advantageous to represent the signal in the time domain as x(t), while in others the FS coefficients X[k] offer a more convenient description. The FS coefficients are known as a frequency domain representation of x(t) because X[k] is the coefficient associated with complex sinusoid at frequency kw 0.

13 H.S. Chen Fourier Series and Fourier Transforms 12 The reason that x(t) and X[k] = c k with the following relationship x(t) = c k e jkw 0t and X[k] = c k = 1 T k= T 0 x(t)e jkw 0t dt The reason is that any two signals e k 1w 0 t e k 2w 0 t at distinct frequencies k 1 w 0 and k 2 w 0 are orthogonal over [0, T]. The derivation is as follows.

14 H.S. Chen Fourier Series and Fourier Transforms 13 since = = = = T 0 T 0 e jk 1w 0 t e jk 2w 0 t dt e j(k 1 k 2 )w 0 t dt 1 j(k 1 k 2 )w 0 e j(k 1 k 2 )w 0 t T 0,if k 1 k 2 T 0 1dt,if k 1 = k 2 1 j(k 1 k 2 )w 0 (e j(k 1 k 2 )2π 1),if k 1 k 2 T,if k 1 = k 2 0,if k 1 k 2 T,if k 1 = k 2

15 H.S. Chen Fourier Series and Fourier Transforms 14 I.e., any two signals e k 1w 0 t e k 2w 0 t at distinct frequencies k 1 w 0 and k 1 w 0 are orthogonal over [0, T], or in general over any period [t 0, t 0 + T 0 ] We said that {e jkw 0t } k= are orthogonal basis for x(t) whose fundamental period is T 0 Apply this fact to x(t), we have x(t) = k= c k e jkw 0t

16 H.S. Chen Fourier Series and Fourier Transforms 15 (x(t), e jmw 0t ) = = = T0 0 T0 0 k= x(t)e jmw0t dt c k e jkw0t e jmw0t dt k= c k T0 0 e j(k m)w 0t dt = T 0 c m c m = 1 T 0 T0 0 x(t)e jmw 0t dt for any m 0 Z I.e., we have the Fourier series pair for a periodic signal x(t) of fundamental period T 0.

17 H.S. Chen Fourier Series and Fourier Transforms 16 x(t) = k= c ke jkw 0t c k = 1 T 0 (x(t), e jkw 0t ) = 1 T 0 T 0 x(t)e jkw 0t dt

18 H.S. Chen Fourier Series and Fourier Transforms 17 Recall: V = v 1,..., v n n v V v = a i v i i=1 if {v 1,..., v n } is an orthogonal basis i.e. (v i, v j ) = A,i = j 0,i j

19 H.S. Chen Fourier Series and Fourier Transforms 18 then n (v, v j ) = ( a i v i, v j ) = i=1 n a i (v i, v j ) = Aa j i=1 i.e. a j = 1 A (v, v j) for j Therefore the coefficient a j of an orthogonal basis can be obtained from the inner product between v and v j. v = n i=1 a iv i a i = 1 A (v, v i)

20 H.S. Chen Fourier Series and Fourier Transforms 19 For the space of periodic signals with period T 0, we have V = e jkw 0t k= w 0 = 2π T 0 Now e jkw 0t k= is an orthogonal basis since (e jk 1w 0 t, e jk 2w 0 t ) = x(t) = k= c ke jkw 0t T 0,k 1 = k 2 0,k 1 k 2 c k = 1 T 0 (x(t), e jkw 0t ) = 1 T 0 T 0 x(t)e jkw 0t dt as we expect.

21 H.S. Chen Fourier Series and Fourier Transforms 20 If x(t) is real, we can use {cos(kw 0 t), sin(kw 0 t)} k=0 orthogonal basis for x(t). as the Original basis: Since {e jkw 0t } k= e jkw 0 t +e jkw 0 t 2 = cos(kw 0 t) e jkw 0 t e jkw 0 t 2j = sin(kw 0 t) we can replace {e jkw 0t, e jkw 0t } k=1 by {cos(kw 0t), sin(kw 0 t)} k=1

22 H.S. Chen Fourier Series and Fourier Transforms 21 Now if x(t) is real, we have c k = 1 T0 x(t)e jkw0t dt T 0 0 c k = 1 T0 x(t)e j( k)w0t dt T 0 0 = 1 T0 x(t)e jkw0t dt T 0 0 = ( 1 T0 x(t)e jkw0t dt) T 0 0 = c k = c k e j c k i.e. c k = c k e j c k = c k e j c k c k = c k and c k = c k

23 H.S. Chen Fourier Series and Fourier Transforms 22 x(t) = k= = c 0 + = c 0 + = c 0 + = c 0 + c k e jkw 0t 1 k= c k e jkw 0t + c k e jkw0t + k=1 c k e jkw 0t k=1 c k e jkw 0t k=1 (c ke jkw0t ) + k=1 c k e jkw 0t k=1 [(c k e jkw0t ) + c k e jkw0t ] k=1

24 H.S. Chen Fourier Series and Fourier Transforms 23 = c 0 + = c 0 + = c 0 + 2Re{c k e jkw0t } k=1 2Re{(Re{c k } + jim{c k })(cos(kw 0 t) + jsin(kw 0 t))} k=1 2(Re{c k }cos(kw 0 t) Im{c k }sinkw 0 t) k=1

25 H.S. Chen Fourier Series and Fourier Transforms 24 x(t) = c 0 + A k cos(kw 0 t) + B k sin(kw 0 t) k=1 where A k = 2Re{c k } = 2Re{ 1 T0 x(t)e jkw0t dt} T 0 = 2 T 0 T0 0 0 x(t)cos(kw 0 t)dt B k = 2Im{c k } = 2Im{ 1 T0 x(t)e jkw0t dt} T 0 = 2 T 0 T0 0 0 x(t)sin(kw 0 t)dt

26 H.S. Chen Fourier Series and Fourier Transforms 25 Properties of Fourier Series 1.linearity x(t) FS;w 0 a k y(t) FS;w 0 b k Ax(t) + By(t) FS;w 0 Aa k + Bb k c(t) = k= c k e jkw 0t

27 H.S. Chen Fourier Series and Fourier Transforms 26 and c(t) = Ax(t) + By(t) = A k a k e jkw 0t + B k b k e jkw 0t = k (Aa k + Bb k )e jkw 0t c k = Aa k + Bb k

28 H.S. Chen Fourier Series and Fourier Transforms time shifting x(t) a k x(t t 0 ) a k e jkw 0t 0 }{{} b k b k = 1 T = 1 T = 1 T = ( 1 T T T 0 T t0 t 0 T = a k e jkw 0t 0 x(t t 0 )e jkw 0t dt x(t t 0 )e jkw 0t dt x(τ)e jkw 0(τ+t 0 ) dτ x(τ)e jkw 0τ dτ)e jkw 0t 0

29 H.S. Chen Fourier Series and Fourier Transforms frequency shifting x(t) a k b k = 1 T x(t)e jmw0t b k = a k M x(t)e jmw0t e jkw0t dt = 1 x(t)e j(k M)w0t dt T T T = a k M

30 H.S. Chen Fourier Series and Fourier Transforms conjugation x(t) a k x (t) b k = a k b k = 1 T T x (t)e jkw 0t dt = ( 1 T T x(t)e j( k)w 0t dt) = a k in particular if x(t) is real x(t) = x (t) therefore a k = a k

31 H.S. Chen Fourier Series and Fourier Transforms multiplication x(t) a k y(t) b k x(t)y(t) a k b k = h k i.e. h k = a l b k l l= x(t) = a k e jkw 0t k= y(t) = b l e jlw 0t l=

32 H.S. Chen Fourier Series and Fourier Transforms 31 Finally, we have x(t)y(t) = a k b l e j(k+l)w 0t k= l= where m = k + l. = ( a m l b l )e jmw 0t m= l= = h m e jmw 0t m=

33 H.S. Chen Fourier Series and Fourier Transforms convolution proof: z(t) = x(t) y(t) = z k = 1 T 0 = 1 T 0 = 1 T 0 = T 0 1 T 0 z(t) = x(t) y(t) z k = T 0 a k b k t T 0 t T 0 τ T 0 = T 0 a k b k τ T 0 z(t)e jkw 0t dt τ T 0 τ T 0 x(τ) x(τ)y(t τ)dτ x(τ)y(t τ)dτe jkw 0t dt t T 0 y(t τ)e jkw 0t dtdτ x(τ)e jkw 0τ dτ 1 T 0 λ T 0 y(λ)e jkw 0λ dλ

34 H.S. Chen Fourier Series and Fourier Transforms Parseval s theorem for F.S. P = 1 T 0 T0 0 x(t) 2 dt = k= c k 2 I.e., to find the average power P of x(t), we can either calculate in time domain or in frequency domain.

35 H.S. Chen Fourier Series and Fourier Transforms 34 proof: 1 T0 T 0 0 = 1 T 0 T 0 = 1 T 0 x(t) 2 dt c k e jkw 0t c me jmw0t dt k= k= m= = 1 c k c T kt 0 0 = k= k= c k 2 c k c m m= T 0 e j(k m)w 0t dt

36 H.S. Chen Fourier Series and Fourier Transforms 35 For real signal x(t), we have c k = c k P = c c k 2 k=1 also we have A k = 2Re{c k } B k = 2Im{c k } therefore c k 2 = Re 2 {c k } + Im 2 {c k } = A2 k 4 + B2 k 4

37 H.S. Chen Fourier Series and Fourier Transforms 36 Finally for real signal, we have another form: P = c k=1 A 2 k 2 + B2 k 2

38 H.S. Chen Fourier Series and Fourier Transforms 37 FT We will do the following step by step 1. The derivation of FT 2. Some examples of FT 3. Properties of FT

39 H.S. Chen Fourier Series and Fourier Transforms 38 Fourier Transform: from F.S. to F.T. f(t) FT F(jw) F(jw) = F {f(t)} and f(t) = F 1 {f(jw)} F(jw) = f(t)e jwt dt f(t) = 1 2π F(jw)ejwt dw

40 H.S. Chen Fourier Series and Fourier Transforms 39 Fourier transform pair for some common signals Example 1: f(t) = 1, T 1 t T 1 0, otherwise

41 H.S. Chen Fourier Series and Fourier Transforms 40 F(jw) = = = = T1 f(t)e jwt dt 1 e jwt dt T 1 1 jw e jwt T 1 T 1 1 jt +jw (ejt 1w e 1 w ) = 2T 1 e jt1w e jt1w wt 1 2j sin(t 1 w) = 2T 1 wt 1 = 2T 1 sinc(t 1 w) therefore F(jw) is the envelop function of T 0 C k

42 H.S. Chen Fourier Series and Fourier Transforms 41 i.e. rect( t w ) Wsinc(wW 2 )

43 H.S. Chen Fourier Series and Fourier Transforms 42 Example 2: x(t) = e at u(t) (a > 0) = X(jw) = = = = = 0 x(t)e jwt dt e at u(t)e jwt dt e (jw+a)t dt 1 (jw + a) e (jw+a)t 0 1 (ifa > 0) jw + a

44 H.S. Chen Fourier Series and Fourier Transforms 43 Example 3: x(t) = e a t (a > 0)

45 H.S. Chen Fourier Series and Fourier Transforms 44 = X(jw) = = = = = = = 0 0 x(t)e jwt dt e a t e jwt dt e at e jwt dt + e (a jw)t dt a jw e(a jw)t a jw + 1 jw + a 2a a 2 + w 2 if a > 0 0 e at e jwt dt e (a+jw)t dt 1 jw + a if a > 0

46 H.S. Chen Fourier Series and Fourier Transforms 45 Example 4: x(t) = δ(t) = X(jw) = = x(t)e jwt dt δ(t)e jwt dt = e jwt t=0 = 1 i.e. δ(t) 1 similarly if x(t) = δ(t t 0 ) = X(jw) = δ(t t 0 )e jwt dt = e jwt t=t0 = e jwt 0 δ(t t 0 ) 1 e jwt 0

47 H.S. Chen Fourier Series and Fourier Transforms 46 Example 5: X(jw) = 2πδ(w) i.e. = x(t) = 1 2π = 1 2π = e jwt w=0 = 1 1 2πδ(w) similarly if X(jw) = 2πδ(w w 0 ) = x(t) = 1 2π = e jwt w=w0 = e jw 0t X(jw)e jwt dw 2πδ(w)e jwt dw 2πδ(w w 0 )e jwt dw

48 H.S. Chen Fourier Series and Fourier Transforms 47 i.e. 1 e jw 0t 2πδ(w w 0 ) (frequency shifting) ) With this, we can represent the periodic signal by FT: x(t) = k= C k e jkw 0t = X(jw) = F{x(t)} = F{ C k e jkw0t } k= = C k F{e jkw0t } = k= k= 2πC k δ(w kw 0 )

49 H.S. Chen Fourier Series and Fourier Transforms 48 i.e.

50 H.S. Chen Fourier Series and Fourier Transforms 49 Example 6: x(t) = cosw 0 t = 1 2 ejw 0t e jw 0t = X(jw) = 1 2 2πδ(w w 0) πδ(w + w 0) = πδ(w w 0 ) + πδ(w + w 0 ) x(t) = sinw 0 t = 1 2j ejw 0t 1 2j e jw 0t = X(jw) = 1 2j 2πδ(w w 0) 1 2j 2πδ(w + w 0) = jπδ(w + w 0 ) jπδ(w w 0 )

51 H.S. Chen Fourier Series and Fourier Transforms 50 In conclusion x(t) X(jw) 1 2πδ(w) e jw 0t 2πδ(w w 0 ) δ(t) 1 δ(t t 0 ) e jwt 0 e at u(t) e a t C k e jkw 0t k= k= 1 jw + a (a > 0) 2σ a 2 + w 2 (a > 0) 2πC k δ(w kw 0 )

52 H.S. Chen Fourier Series and Fourier Transforms 51 cosw 0 t πδ(w + w 0 ) + πδ(w w 0 ) sinw 0 t π j δ(w w 0) π j δ(w + w 0) u(t) πδ(w) + 1 jw sgn(t) 2 jw

53 H.S. Chen Fourier Series and Fourier Transforms 52 Properties of FT x(t) X(jw) y(t) Y (jw) ax(t) + by(t) ax(jw) + by (jw) x (t) X ( jw) x(at) 1 a X(w a ) x(t t 0 ) X(jw)e jwt 0 x(t)e jw 0t X(j(w w 0 )) X(t) 2πx( w)

54 H.S. Chen Fourier Series and Fourier Transforms linearity x(t) X(jw) y(t) Y (jw) ax(t) + by(t) ax(jw) + by (jw)

55 H.S. Chen Fourier Series and Fourier Transforms conjugation x(t) X(jw) x (t) X ( jw) x (t)e jwt dt = ( x(t)e ( jwt) ) = X ( jw) if x(t) = x (t) real = X ( jw) = X(jw) Re{X( jw)} = Re{X(jw)} Im{X( jw)} = Im{X(jw)} = X( jw) = X(jw) X( jw) = X( jw)

56 H.S. Chen Fourier Series and Fourier Transforms Time scaling Interpretation: f(αt) F 1 α F(w α ) If α > 1, we obtain that compression in the time domain corresponds to expansion in the frequency domain. If 0 < α < 1, we obtain that compression in the time domain corresponds to expansion in the frequency domain.

57 H.S. Chen Fourier Series and Fourier Transforms Time shifting f(t t 0 ) F e jwt 0 F(w)

58 H.S. Chen Fourier Series and Fourier Transforms 57 Example: Recall the Fourier transforms of δ(t) and δ(t t 0 ). Proof: F[f(t t 0 )] = = = e jwt 0 F(w) f(t t 0 )e jwt dt f(σ)e jw(σ+t 0) dσ

59 H.S. Chen Fourier Series and Fourier Transforms 58 Combining Time Scaling and Time shifting : f(at + t 0 ) F 1 a F(w a )e+jt 0w/a This is obtained by (i) shifting f(t) by t 0 and (ii) scaling the result of (i) by a.

60 H.S. Chen Fourier Series and Fourier Transforms Frequency shifting e jw 0t f(t) F F(w w 0 ) for any real w 0. Example: Recall the Fourier transforms of 1 and e jw 0t, and u(t) and cos(w 0 t)u(t). Also,examine the modulation property below.

61 H.S. Chen Fourier Series and Fourier Transforms 60 Proof: F[e jw0t f(t)] = = = F(w w 0 ) e jw 0t f(t)e jwt dt f(t)e j(w w 0)t dt

62 H.S. Chen Fourier Series and Fourier Transforms Duality F( t) 2πf(w) F(t) 2πf( w)

63 H.S. Chen Fourier Series and Fourier Transforms 62 Proof: f(t) = 1 2π 2πf( t) = 2πf( w) = F(w)e jwt dw F(w)e jwt dw F(t)e jwt dt = F[F(t)] where the last line is obtained by interchanging t with w.

64 H.S. Chen Fourier Series and Fourier Transforms Convolution If all the involved Fourier transforms exist, then x(t) h(t) F X(w)H(w)

65 H.S. Chen Fourier Series and Fourier Transforms 64 Proof: F[x(t) h(t)] = = = = H(w) [ h(t τ)x(τ)dτ]e jwt dt x(τ)[ x(τ)[e jwt H(w)]dτ = H(w)X(w) h(t τ)e jwt dt]dτ x(τ)e jwt dτ

66 H.S. Chen Fourier Series and Fourier Transforms Modulation If all the involved Fourier transforms exist, then x(t)m(t) F 1 X(w) M(w) 2π

67 H.S. Chen Fourier Series and Fourier Transforms 66 Proof: F[x(t)m(t)] = = 1 2π = 1 2π x(t)[ 1 2π M(w )[ = 1 M(w) X(w) 2π M(w )e jw t dw ]e jwt dt M(w )[X(w w )]dw x(t)e jwt e +jw t dt]dw Example: This result can be used to calculate F[cos(w 0 t)u(t)].

68 H.S. Chen Fourier Series and Fourier Transforms Time Differentiation If f(t) is continuous and if f(t) and f (t) = df(t) dt are both absolutely integrable (and thus satisfy the Dirichlet conditions),then df(t) dt F jwf(w)

69 H.S. Chen Fourier Series and Fourier Transforms 68 Proof: f (t)e jwt dt = f(t)e jwt + jw f(t)e jwt dt The key step now is to observe that since f(t) is assumed to be absolutely integrable, then f(t) 0 as t ± and thus the first term on the RHS evaluates to zero. The second term is simply jwf(w)

70 H.S. Chen Fourier Series and Fourier Transforms Parseval s Theorem Parseval s Theorem f(t) 2 dt = 1 2π F(w) 2 dw

71 H.S. Chen Fourier Series and Fourier Transforms 70 Proof: f(t) 2 dt = = = 1 2π = 1 2π = 1 2π f(t)f (t)dt f(t)[ 1 2π F (w)[ F (w)e jwt dw]dt F (w)f(w)dw F(w) 2 dw f(t)e jwt dt]dw