Chapter 6: Applications of Fourier Representation Houshou Chen

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1 Chapter 6: Applications of Fourier Representation Houshou Chen Dept. of Electrical Engineering, National Chung Hsing University

2 H.S. Chen Chapter6: Applications of Fourier Representation 1 Applications of FS, DTFS, FT, and DTFT In the previous chapters, we developed the Fourier representations of four distinct classes of signals. 1. FS for periodic continuous-time signals. 2. DTFS for periodic discrete-time signals. 3. FT for aperiodic continuous-time signals. 4. DTFT for aperiodic discrete-time signals. In this chapter, we consider mixed signals such as 1. periodic and aperiodic signals 2. continuous- and discrete-time signals

3 H.S. Chen Chapter6: Applications of Fourier Representation 2 If we apply a periodic signals to a stable LTI system, the convolution operation involves a mixing of aperiodic impulse response and periodic input. A system that samples continuous-time signals involves both continuous- and discrete-time signals. In order to use Fourier methods to analyze such interactions, we must build bridges between the Fourier representation of different classes of signals.

4 H.S. Chen Chapter6: Applications of Fourier Representation 3 We can develop FT and DTFT representations of continuous- and discrete-time periodic signals, respectively. FT can also be used to analyze problems involving mixtures of continuous- and discrete-time signals. FT and DFTT are most commonly used for analysis applications. The DTFS is the primary representation used for computational applications. We will consider the sampling theorem and FFT in this chapter. A thorough understanding of the relationship between the four Fourier representations is a critical step in using Fourier methods to solve problems involving signals and systems.

5 H.S. Chen Chapter6: Applications of Fourier Representation 4 Strictly speaking, neither the FT nor the DTFT converges for periodic signals. However, by incorporating impulse into the FT and DTFT, we may develop FT and DTFT representation for periodic signals. We may use them and the properties of FT and DTFT to analyze problems involving mixtures of periodic and aperiodic signals. We will consider the convolutional and multiplication of aperiodic and periodic signals in time domain and see what happens in frequency domain.

6 H.S. Chen Chapter6: Applications of Fourier Representation 5 Relating FT and FS Given a continuous-time periodic signal x(t) with FS representation x(t) = k= X[k]e jkw 0t. Recall the following FT pair (with impulse in frequency domain) e jkw 0t FT 2πδ(w kw 0 ) We thus have the FT for x(t) as follows. x(t) = k= X[k]e jkw 0t FT X(jw) = 2π k= X[k]δ(w kw 0 )

7 H.S. Chen Chapter6: Applications of Fourier Representation 6 Thus, the FT of a periodic signal is a series of impulses spaced by the fundamental frequency w 0. The kth impulse has strength 2πX[k], where X[k] is the kth FS coefficient.

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9 H.S. Chen Chapter6: Applications of Fourier Representation 8 Find the FT representation of x(t) = cos w 0 t

10 H.S. Chen Chapter6: Applications of Fourier Representation 9 The FT of a unit impulse train p(t) = n= δ(t nt) Since p(t) is periodic with fundamental frequency w 0 = 2π/T and the FS coefficients are given by p[k] = 1/T T/2 T/2 δ(t)e jkw 0t dt = 1/T. Therefore the FT of p(t) is given by P(jw) P(jw) = 2π T k= δ(w kw 0 ). Hence, the FT of p(t) is also an impulse train; that is, an impulse train is its own FT.

11 H.S. Chen Chapter6: Applications of Fourier Representation 10 The spacing between the impulses in the frequency domain is inversely related to the spacing between the impulses in the time domain.

12 H.S. Chen Chapter6: Applications of Fourier Representation 11

13 H.S. Chen Chapter6: Applications of Fourier Representation 12 Relating DTFT and DTFS Given a discrete-time periodic signal x[n] with DTFS representation x[n] = N 1 k=0 X[k]e jkw 0n. As in the FS case, the key observation is that the inverse DTFT of a frequency-shifted impulse is a discrete-time complex sinusoid. The DTFT is a 2π-periodic function of frequency, so we may express for w [ π, π] and kw 0 [ π, π] e jkw 0n DTFT δ(w kw 0 )

14 H.S. Chen Chapter6: Applications of Fourier Representation 13 Or as the following DTFT pair (with impulse in frequency domain) e jkw 0n DTFT m= 2πδ(w kw 0 m2π)

15 H.S. Chen Chapter6: Applications of Fourier Representation 14 We thus have the DTFT for x[n] as follows. x[n] = N 1 k=0 X[k]e jkw 0n N 1 DTFT X(e jw ) = 2π Or equivalently as follows. k=0 X[k] m= δ(w kw 0 m2π) x[n] = N 1 k=0 X[k]e jkw 0n DTFT X(e jw ) = 2π m= X[k]δ(w kw 0 )

16 H.S. Chen Chapter6: Applications of Fourier Representation 15 Thus, the DTFT of a periodic signal is a series of impulses spaced by the fundamental frequency w 0. The kth impulse has strength 2πX[k], where X[k] is the kth DTFS coefficient.

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18 H.S. Chen Chapter6: Applications of Fourier Representation 17 Convolution of periodic and aperiodic signals Use the fact that convolution in the time domain corresponds to multiplication in the frequency domain. That is, y(t) = x(t) h(t) FT Y (jw) = X(jw)H(jw). If the input x(t) is periodic with period T, then x(t) FT X(jw) = 2π k= where X[k] are the FS coefficients of x(t). X[k]δ(w kw 0 ), We substitute this representation into the convolution property to obtain y(t) = x(t) h(t) FT Y (jw) = 2π k= H(jkw 0 )X[k]δ(w kw 0 )

19 H.S. Chen Chapter6: Applications of Fourier Representation 18 The form of Y (jw) implies that y(t) corresponds to a periodic signal with the same period T as x(t) Indeed, The strength of the kth impulse in X(jw) is adjusted by the value of H(jw) evaluated at the frequency at which it is located, or H(jkw 0 ), to yield an impulse in Y (jw) at w = kw 0. The results show that the periodic extension in time domain corresponds to the discrete operation in frequency domain.

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21 H.S. Chen Chapter6: Applications of Fourier Representation 20 Example: Let the input signal applied to an LTI system with impulse response h(t) = 1/(πt) sin(πt) be the periodic square wave. Find the output of the system. The frequency response of h(t) can be shown to be low pass filter h(t) FT 1, w π H(jw) = 0, w > π The FT of square wave can be found by the FS coefficients as follows. x(t) FT X(jw) = k= 2 sin(kπ/2) δ(w kπ/2) k

22 H.S. Chen Chapter6: Applications of Fourier Representation 21 There are five terms of X(jw) inside [ π, π], i.e., k = 2, 1, 0, 1,, 2, 2 sin( 2π/2) sin( 0π/0) 0 δ(w ( 2)π/2) + 2 sin( 1π/2) δ(w ( 1)π/2) ( 1) δ(w 0π/2)+ 2 sin(1π/2) 1 = 2δ(w + π/2) + πδ(w) + 2δ(w π/2) δ(w 1π/2)+ 2 sin(2π/2) δ(w 2π/2) 2 Finally, Y (jw) is obtained by the fact that H(jw) acts as a low-pass filter, passing the harmonics at π/2, 0, and π/2, while suppressing all others. Y (jw) = πδ(w) + 2δ(w π/2) + 2δ(w + pi/2) Taking the inverse FT of Y (jw) gives the output. Thus y(t) = 1/2 + 2/π cos(π/2t)

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24 H.S. Chen Chapter6: Applications of Fourier Representation 23 Multiplication of periodic and aperiodic signals Recall the multiplication property of the FT, represented as y(t) = g(t)x(t) FT Y (jw) = 1 G(jw) X(jw). 2π If the signal x(t) is periodic with period T, then x(t) FT X(jw) = 2π Therefore, the FT of y(t) is k= y(t) = g(t)x(t) FT Y (jw) = G(jw) X[k]δ(w kw 0 ). k= X[k]δ(w kw 0 ). Finally, by the sifting property of the impulse function, the convolution of any function with a shifted impulse results in a shifted version of the

25 H.S. Chen Chapter6: Applications of Fourier Representation 24 original funciton, i.e., y(t) = g(t)x(t) FT Y (jw) = k= X[k]G(j(w kw 0 )). Multiplication of g(t) with the periodic function x(t) gives an FT consisting of a weighted sum of shifted version of G(jw). As expected, the form of Y (jw) corresponds to the FT of a continuous-time aperiodic signal, since the product of periodic and aperiodic signals is aperiodic.

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27 H.S. Chen Chapter6: Applications of Fourier Representation 26 Let x(t) be the continuous-time impulse train. x(t) = n= δ(t nt) Remark: By introducing the delta function in time domain, we can represent a discrete-time function as a continuous-time function. For example, the discrete-time periodic signal x[n] = 1, for all n corresponds to continuous-time periodic x(t) as above. The FT of x(t) is also a periodic impulse train in frequency domain X(jw) = 2π T k= δ(w kw 0 ).

28 H.S. Chen Chapter6: Applications of Fourier Representation 27 Now y(t) = g(t)x(t) is the sampled version of g(t) and the FT of y(t) is y(t) = g(t)x(t) FT Y (jw) = k= 2π T G(j(w kw 0)). We see that Y (jw) is the periodic extension of G(jw). The corresponded result is called the sampling theorem and we will discuss now. The results show that the discrete operation in time domain corresponds to the periodic extension in frequency domain

29 H.S. Chen Chapter6: Applications of Fourier Representation 28 Relating the FT to the DTFT We can derive an FT representation of discrete-time signals by incorporating impulses into the description of the signals in the time domain. Therefore, the FT is a powerful tool for analyzing problems involving mixtures of discrete- and continuous-time signals. Combine the results of the relationship between FT and FS, also FT and DTFT, the FT can be used for the four classes of signals.

30 H.S. Chen Chapter6: Applications of Fourier Representation 29 Consider the DTFT of an arbitrary discrete-time signals x[n]: X(e jω ) = n= x[n]e jωn. We seek an FT pair x δ (t) FT X δ (jw) that corresponds to the DTFT pair x[n] DTFT X(e jω ).

31 H.S. Chen Chapter6: Applications of Fourier Representation 30 Now, let Ω = wt s, where x[n] = x δ (nt s ). I.e., x[n] is equal to the samples of x(t) taken at intervals of T s. By this substitution, Ω = wt s, we transform X(e jω ) of Ω into X δ (jw) of w X δ (jw) = X(e jω ) Ω=wTs = x[n]e jwtsn. n= Taking the inverse FT of X δ (jw) and use the following fact δ(t nt s ) DTFT e jwt sn, we obtain the continuous-time x δ (t) x δ (t) = n= x[n]δ(t nt s ).

32 H.S. Chen Chapter6: Applications of Fourier Representation 31 Hence, we have x[n] DTFT X(e jω ) = n= x[n]e jωn and x δ (t) FT X δ (jw) = n= x[n]e jwt sn

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34 H.S. Chen Chapter6: Applications of Fourier Representation 33 Sampling Theorem : We use the FT representation of discrete-time signals to analyze the effects of uniformly sampling a signal. The sampling operation generates a discrete-time signal from a continuous-time signal. We will see the relationship between the DTFT of the sampled signals and the FT of the continuous-time signal. Sampling of continuous-time signals is often performed in order to manipulate the signal on a computer or microprocessor.

35 H.S. Chen Chapter6: Applications of Fourier Representation 34 A signal x(t) with X(jw) as follows X(jw ) - w b w b w is called band-limited signal and can be exactly reconstructed from its samples {x(nt S )} n= provided that the sampling frequency ω s = 2π T S 2ω B, 2ω B : Nyquist sampling rate.

36 H.S. Chen Chapter6: Applications of Fourier Representation 35 Ideal Sampling : X(t) Xs(t) H(jw)? Xr(t) = X(t) P(t) First, we multiply x(t) by the impulse train P(t) = n= δ(t nt s ) (period T s ) x s (t) = p(t)x(t) X(t) Xs(t) -2Ts -Ts 0 Ts 2Ts

37 H.S. Chen Chapter6: Applications of Fourier Representation 36

38 H.S. Chen Chapter6: Applications of Fourier Representation 37 Now, x s (t) = x(t)p(t) = x(t) δ(t nt s ) n= = x(t)δ(t nt s ) n = n= x(nt s )δ(t nt s ) Next, we want to find X s (jw) p(t) = δ(t nt s ) = C k e jkw st n= where C k = 1 T s <T s > = 1 T s Ts 0 n= P(t)e jkw st dt δ(t)e jkw st dt = 1 T s

39 H.S. Chen Chapter6: Applications of Fourier Representation 38 Since x s (t) = x(t)p(t) p(t) = P(jw) = n= n= 1 T s e jkw st 2π T s δ(w kw s ) X s (jw) = 1 X(jw) p(jw) 2π = 1 2π X(jw) = 1 T s k= = 1 T s k= k= 2π T s δ(w kw s ) X(jw) δ(w kw s ) X(j(w kw s ))

40 H.S. Chen Chapter6: Applications of Fourier Representation 39 The sampling theorem says that the discrete operation in time domain x s (t) = x(t)p(t) corresponds to the periodic extension in frequency domain X s (jw) = 1 T s X(j(w kw s )) k= X(jw) Xs(jw) 1 1/Ts -W B W B W B Ws

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42 H.S. Chen Chapter6: Applications of Fourier Representation 41 We can recover X(jw) from X s (jw) if and only if w s w B w B w s 2w B To recover x(t), we multiply X s (jw) by H(jw) Ts -W B W B w X r (jw) = X s (jw) H(jw)

43 H.S. Chen Chapter6: Applications of Fourier Representation 42 Since H(jw) = T s rect( w 2w B ) h(t) = T s 1 2π 2w Bsinc( 2w B 2 t) = T s w B π sinc(w Bt) = 2w B sinc(w B t) w s x r (t) = x s (t) h(t) = [ = n= n= x(nt s )δ(t nt s )] [ 2w B w s sinc(w B t)] x(nt s ) 2w B w s sinc(w B (t nt S )) = x(t) iff w s 2w B

44 H.S. Chen Chapter6: Applications of Fourier Representation 43 FFT (Fast Fourier Transform) The role of DTFS as a computational tool is greatly enhanced by the availability of efficient algorithms for evaluating the forward and inverse DTFS. We call these algorithms fast Fourier transform (FFT) algorithm. FFT use the divide and conquer principle by dividing the DTFS into a series of lower order DTFS and using the symmetry and periodicity properties of the complex sinusoid e jk2πn/n. The total computations of FFT is substantially less than the original DTFS.

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46 H.S. Chen Chapter6: Applications of Fourier Representation 45 The computation of x[n] from X[k] or the computation of X[k] from x[n] requires N 2 complex multiplications and N N complex additions. Assume N is a power of 2, we can thus split x[n], 0 n N 1, into even and odd indexed signals, i.e., x[2n] and x[2n + 1], 0 n N/2 1.

47 H.S. Chen Chapter6: Applications of Fourier Representation 46 (1) Decimation in time F k = N 1 n=0 2πnk j f n e N DFT if N is a power of 2 e.q. {f n } = {f 0, f 1, f 2, f 3, f 4, f 5, f 6, f 7 } N = 8 Define g n = f 2n (even-number samples) h n = f 2n+1 (odd-number samples) n = 0, 1,, N 2 1

48 H.S. Chen Chapter6: Applications of Fourier Representation 47 N F k = = = { N 1 n=0 N 2 1 n=0 N 2 1 n=0 2πnk j f n e N = DFTN {f n } 2π(2n)k j f 2n e N + 2πnk j g n e N 2 1 n=0 N N/2 } + { 2 1 n=0 2π(2n+1)k j f 2n+1 e N 2πnk j h n e N/2 }e j 2πk N 2πk j = DFT N/2 {g n } + e N DFTN/2 {h n } Define W N = e j 2π N W k N 2πk = e j N NF k = [ N 2 NF k = [ N 2 G k N 2 G k ] + W k N[N 2 ] + W k N[N 2 H k ] 0 k N 2 1 H k N 2 ] N 2 k N 1

49 H.S. Chen Chapter6: Applications of Fourier Representation 48 In summary, we have the following = NF k = [ N 2 G k ] + W k N [ N 2 NF k = [ N 2 G k N 2 H k ] 0 k N 2 1 ] + W k N [ N 2 H k N 2 ] N 2 k N 1 This indicates that F[k] and F[k + N/2], 0 k N/2 1, are a weighted combination of G[k] and H[k] This structure is called a butterfly structure.

50 H.S. Chen Chapter6: Applications of Fourier Representation 49 For example : N=8 k = 1 8F 1 = 4 G 1 + W 1 8 ( 4 H 1 ) k = 5 8F 5 = 4 G 1 + W 5 8 ( 4 H 1 ) G 0 G 1 1 F 0 F 1 G 2 G 3 W 8 1 F 2 F 3 This structure is call Butterfly =2 complex adds +1 complex multiplication H 0 1 F 4 H 1 H 2 H 3 W 8 1 =W 8 4 W 8 1 = -W 8 1 F 5 F 6 F 7

51 H.S. Chen Chapter6: Applications of Fourier Representation 50 We can further simplify the results by exploiting the following fact. For k N/2, we have W k N = W N 2 N W k N 2 N = W k N 2 N W N 2 N = e j 2π N N 2 = e jπ = 1

52 H.S. Chen Chapter6: Applications of Fourier Representation 51 For example : 8-point FFT

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58 H.S. Chen Chapter6: Applications of Fourier Representation 57 Operation Counted in FFT algorithm/dft algorithm Total operations in FFT ( 2 M = N) =(M sections) (N/2 butterfly/section) (3 operations/butterfly) = 3 2 NM = 3 2 N log 2N

59 H.S. Chen Chapter6: Applications of Fourier Representation 58 Decimation in Frequency F k = N 1 n=0 2πnk j f n e N if N is a power of 2,eg. N=8 Def ine {f n } = {f 0, f 1, f 2, f 3, f 4, f 5, f 6, f 7 } g n = f n ( first half samples ) n = 0, 1,, N 2 1 h n = f n+ N 2 ( second half sample )

60 H.S. Chen Chapter6: Applications of Fourier Representation 59 F k = = = N 2 1 n=0 N 2 1 n=0 N 2 1 2πnk j f n e N 2πnk j g n e N N n=0 N n=0 (g n + h n e jπk ) e n=0 f n+ N 2 2πnk j h n e N 2πnk j N e j 2π(n+ N )k 2 N e jπk N=8 {F k } = {F 0, F 1, F 2, F 3, F 4, F 5, F 6, F 7 } Define 2 sequences {R k } = {F 0, F 2, F 4, F 6 }even samples {S k } = {F 1, F 3, F 5, F 7 }odd samples

61 H.S. Chen Chapter6: Applications of Fourier Representation 60 even number group k = 2k F 2k = = N 2 1 (g n + h n e } jπ 2k {{} n=0 =1 N 2 1 (g n + h n )e n=0 j 2πnk )e j 2πn(2k ) N N/2 N 2 point DFT NF 2k = DFT N 2 {g n + h n } k = 0, 1,, N 2 1

62 H.S. Chen Chapter6: Applications of Fourier Representation 61 odd number group k = 2k + 1 F 2k +1 = = N 2 1 (g n + h n e jπ(2k +1) }{{} n=0 = 1 N 2 1 [(g n h n )e n=0 NF 2k+1 = DFT N 2 2πn j N )e j 2πn(2k +1) N ]e j 2πnk N/2 2πn j {(g n h n )e N } = NF 2k = DFT N 2 NF 2k+1 = DFT N 2 = DFT N 2 {(g n h n )W n N } k = 0, 1,, N 2 1 [g n + h }{{ n ] k = 0, 1,, N } 2 1 g n [(g n h n )WN n ] k = 0, 1,, N }{{} 2 1 h n

63 H.S. Chen Chapter6: Applications of Fourier Representation 62 We already reduce N-point DFT to N 2 -point DFT. we can repeat this process until we get one-point DFT if N is a power of 2. N=2 M f 0 g 0 g 0 f 1 g 1 1 g 1 g n + h n =g n fl f 2 g 2 1 g 2 (g n - h n n )W N = h n fl f 3 g 3 g 3 f 4 f 5 h 0 h 1-1W 1 h 0 h 1 Butterfly = 2 complex adds n +1 complex multiplication N ) (W f 6 h 2 h 2 f 7 h 3 h 3

64 H.S. Chen Chapter6: Applications of Fourier Representation 63 Example:

65 H.S. Chen Chapter6: Applications of Fourier Representation 64 Total operations in FFT (2 M = N) =( M sections) ( N 2 butterfly/section) (3 operations/butterfly) = 3 2 NM = 3 2 N log 2N Operations counted in FFT algorithm: total operations in DFT N 2 ( X k = N 1 n=0 x 2πnk j N ne, k = 0, 1,, N 1) total operations in FFT N log 2 N Improvement Ratio= DFT FFT = N log 2 N