2 Fourier Transforms and Sampling

Transcription

1 2 Fourier ransforms and Sampling 2.1 he Fourier ransform he Fourier ransform is an integral operator that transforms a continuous function into a continuous function H(ω) =F t ω [h(t)] := h(t)e iωt dt (1) he inverse Fourier ransform recovers the original function h(t) =F ω t [H(ω)] := 1 H(ω)e iωt dω (2) he Fourier Series If h(t) is periodic with period (i.e. h(t + k) =h(t) k N ) then it can be expanded h(t) = a a n cos(nω 0 t)+ b n sin(nω 0 t) (3) n=1 where ω +0 =. he coefficients {a i,b i } are obtained by integration and invoking the trigonometric relation /2 /2 /2 /2 ( 2nπt cos ( 2nπt cos ) cos ) sin ( 2mπt ( 2mπt n=1 ) dt = /2 2 δ nm = /2 ) dt = 0 whereδ nm = ( 2nπt sin { 1 n = m 0 n m ) sin ( 2mπt ) dt to give a n = 2 /2 /2 h(t) cos (nω 0 t)dt, b n = 2 /2 /2 h(t)sin(nω 0 t)dt (4) 3

2 If we define complex coefficients :- c n = 1 2 (a n ib n ) = 1 c n = 1 2 (a n +ib n ) = 1 then we reconsider the Fourier Series as with c n given (for all integer n) h(t) = /2 /2 /2 /2 n= /2 h(t)e inω0t dt n > 0 h(t)e inω0t dt n > 0 c n e inω0t (5) c n = 1 h(t)e inω0t dt. (6) /2 hus the Fourier Series expansion transforms a continuous function to a descrete series. he relation Eq. 6 is also known as the Discrete ime Fourier ransform, and Eq. 5 as the Inverse Discrete ime Fourier ransform Discrete Fourier ransform (DF) he DF is a transform from a discrete series to a discrete series. GivenalistoflengthN : {h 0,h 1,...h N 1 } then we can derive another list : {H 0,H 1,...H N 1 } by the following relations H m = h k = N 1 k=0 N 1 k=0 h k e imk N (7) H k e imk N (8) Relationship Between ransforms Clear the transform relations (Eq. 1,Eq. 2 ), (Eq. 5,Eq. 6), (Eq. 7,Eq. 8) are related, but the precise relationships contains some subtleties. Often the Fourier Series is used to define the Fourier ransform as a limit from Eq. 6 c n = /2 /2 h(t)e inω0t dt lim From Eq. 5 and Eq. 9 and using ω 0 = 0gives h(t) = 1 c n e inω0t ω 0 lim n= 4 h(t)e inω0t dt =: H(nw 0 ) (9) 1 H(ω)e iωt dω (10)

3 his depends on accepting that a sum in a limit becomes an integral which is mathematically a bit flakey. Furthermore it does not lead easily to an interpretation of the DF. It is preferable to use the F as the starting point for all other forms. 5

4 2.1.4 Fourier ransforms Involving δ-functions o find the F of periodic functions consider cos(t), sin(t). [ 1 ( F [cos(ω 0 t)] = F e iω 0t +e iω0t)] 2 = π [δ(ω ω 0 )+δ(ω + ω 0 )] (11) [Note that the result is purely real] [ 1 ( F [sin(ω 0 t)] = F e iω 0t e iω0t)] 2i =iπ [δ(ω ω 0 ) δ(ω + ω 0 )] (12) [Note that the result is purely imaginary] 6

5 Fourier ransform of a Pulse rain An extremely important function in sampling theory is a pulse train of Comb function Comb (t) := δ(t n ) (13) n= consisting of an infinite sequence of equidistant δ-functions. We will prove the following : Comb (t) F F 1 ω 0Comb ω0 (ω) (14) It suffices to show that the inverse transform is true (since Fourier ransforms are symmetric). First F 1 [Comb ω0 (ω)] = 1 δ(ω nω 0 )e iωt dt = 1 e inω0t (15) n= n= We need to show that the sum on the right in Eq. 15 is equivalent to a sequence of δ-functions in the time domain. Consider the partial sum k N (t) = 1 N n= N e inω0t = ei(n+1)ω0t e inω0t (e iω0t 1) = sin ((N +1/2)ω 0t) sin ( ω 0t) 2 his function is known as the Fourier Series Kernel with period ω 0 =. 7

6 We need to prove that in the interval [ /2,/2] k N (t) tends to δ(t) since then lim k N (t) =Comb (t). N Consider k N (t) = sin ((N +1/2)ω 0t) sin ( ω 0t) = sin ((N +1/2)ω 0t) t 2 We know, from the definitiion of a δ-function that and, since sin ((N +1/2)ω 0 t) lim = π N t δ(t) t is bounded in [ /2,/2], we have sin( ω 0 t 2 ) lim N k N (t) = π t sin ( ω 0t 2 t sin ( ω 0t 2 )δ(t) = π t sin ( ω 0t )δ(t) = δ(t) 2 t=0 as required. In the final step, note that from the definition of the δ-function that f(t)δ(t) = t f(0)δ(t) and sin( ω 0 t 2 ) = πsinc(ω whic has value 0t/2) π at t =0 Inversion Formula Proof Using properties of the δ-function we can now prove the inversion formula F 1 [F [h(t)]] = 1 e iωt e iωt h(t )dt dt = 1 e iω(t t ) h(t )dt dt as required. = = h(t) h(t )δ(t t )dt ) 2.2 Convolution Next to the inversion formula Eq. 2, the convolution theorem is the most important tool in Fourier Analysis he Convolution Integral given two functions f(t),g(t), their convolution produces a third function : h(t) = f(t )g(t t )dt := f(t) g(t) (16) 8

7 It is trivially seen by substitution of variables that convolution is commutative, i.e. f(t) g(t) =g(t) f(t) Interpretation of he Convolution Integral he convolution of f(t) byg(t)) is often interpreted as a weighted averaging of f(t) byafilter g(t) where the extent of the average is the support of g(t) he Convolution heorem We state the convolution theorem as he Fourier ransform of the convolution of two functions equals the product of the Fourier ransform of the functions individually F [f(t) g(t)] = F [f(t)] F [g(t)] (17) he proof is straightforward : F [f(t) g(t)] = (substitute z = t τ) = = e iωt f(τ)g(t τ)dtdτ e iω(z+τ) f(τ)g(z)dτdz e iωτ f(τ)dτ e iωz g(z)dz = F [f(t)] F [g(t)] he complementray form of Eq. 17 is the Frequency Convolution heorem 9

8 he Fourier ransform of the product of two functions equals ( 1 ) times the convolution of the Fourier ransform of the functions individually F [f(t)g(t)] = 1 F [f(t)] F[g(t)] (18) 10

9 2.2.4 Parseval s heorem there is a relation between the area under the function curve in the temporal and Fourier domains: h(t) 2 dt = 1 H(ω) 2 dω (19) First we require a result regarding conjugation If F (ω) =F [f(t)] then F ( ω) =F [ f(t) ] where a is the complex conjugate of a. 11

10 Proof From this F (ω) = F (ω) = F ( ω) = (Re [f(t)]+iim [f(t)]) e iωt dt (Re [f(t)] i Im [f(t)]) e iωt dt (Re [f(t)] i Im [f(t)]) e iωt dt }{{} F [ h(t) 2] = F [ h(t)h(t) ] = 1 ( ) H(ω) (H)( ω) (set ω =0) h(t) 2 dt = 1 H(y)H( (ω y))dy = 1 f(t) H(y) 2 dy ω=0 2.3 Fourier Series and Sampled Waveforms In the technical literature, Fourier Series is usually developed independently of the Fourier Integral. We already saw (Eq. 9, Eq. 10 ) that the Fourier Integral could be developed by taking the limit of the Fourier Series. Now we derive it in the other direction. We want to show that if h(t + ) = h(t) i.e. h(t) is periodic, period then h(t) = n= /2 c n e inω0t where ω 0 = (same as Eq. 5 ) where c n = 1 h(t)e inω0t dt (same as Eq. 6 ). /2 It follows from Eq. 5 and the properties of δ-functions (viz. δ(t) =F 1 [1] = 1 eiωt dω ) that the F of a periodic function is a sequence of equidistant δ-functions: H(ω) =F [h(t)] = c n δ(ω ω 0 n) (20) n= 12

11 [Note : at this point we say nothing about whether or not H(ω) is a finite or infinite series.] Firstly. it is simple to prove the inverse, i.e. that the inverse F of a sequence of equidistant δ-functions is a periodic function : [ ] F 1 A n δ(ω nω 0 ) = 1 A n e inω0t =: φ(t) (21) n= n= ( ) and since e inω0 t+ ω 0 =e inω0t we must have φ(t + ω 0 )=φ(t). oprove20wemakeuseofthe Fourier ransform pair Eq. 14. Consider the function { h(t) t </2 h 0 (t) = = h(t)rect /2 (t) 0 t >/2 which is h(t) masked ot zero except in a single period ( /2,/2). he original, periodic, function is therefore h(t) = h 0 (t + n )=h 0 (t) Comb (t) n= hen we have but F [h(t)] = F [h(t) Comb (t)] = F [h 0 (t)] ω 0 Comb ω0 (ω) H 0 (ω) = h 0 (t)e iωt dt = /2 /2 h(t)e iωt dt therefore H(ω) =H(ω 0 ) π H 0 (nω 0 ) δ(ω nω 0 )= δ(ω nω 0 ) n= n= and since c n from Eq. 5 is equal to H0(nω0) we have proved Eq. 20. he coefficients c n of the Fourier Series expansion of a periodic function h(t) are equal to 1/ times the value of the Fourier ransform of h 0 (t) sampled at the frequencies ω = nω 0 = n Example riangular wave a n = 1 /2 n= /2 1 2 n =0 2 h(t) cos(nω 0 t)dt = n 2 π n even,n>0 2 0 n odd and H 0 (ω) = 8sin2 (ω 4 ) ω 2 hus a n = H 0 (ω) ω= n 13

12 2.3.1 Waveform Sampling he fourier Series represents a continuous, periodic function in time by a discrete sequence of equidistant δ-functions in the fruency domain. Consider now the effect of sampling in the time domain {h n } = ĥ(t) = h(n )δ(t n )=h(t)comb (t) n= Clearly, in the Fourier Domain, the sampling process is a convolution ] Ĥ(ω) =F [ĥ(t) = ω 0 H(ω) Comb ω 0 (ω) As the sampling interval increases, the seperation of the δ-functions in the frequency domain decreases. Eventually the replicated functions in the frequency domain overlap leading to aliasing he Sampling heorem he Shannon Sampling heorem [Shannon, 1949] is fundamental in information theory. It states If the Frequency ransform of a function h(t) is zero above a certain frequency ω c, i.e. H(ω) =0 for ω ω c (22) then h(t) can be uniquely determined from its values ( ) nπ h n = h 14 ω c

13 i.e. samples of h(t) at a sequence of equidistant points, π ω c apart. In fact h(t) is given by h(t) = n= h n sin(ω c t nπ) (ω c t nπ) (23) Proof Applying the Inverse Fourier ransform to Eq. 22 gives h(t) = 1 so that the sampled function is given by ( ) nπ h n = h ω c ωc ω c H(w)e iωt dω = 1 ωc H(w)e inπω ωc ω c If we expand H(ω) into a Fourier Series in the interval ( ω c,ω c ) dω H(ω) = n= A n e inω 2ωc ω c <ω<ω c where A n = 1 ωc H(w)e inπω ωc dω 2ω c ω c which gives immediately that A n = π ω c h n. However, the Fourier Series represents a replicated function H(ω) = n= π h n e inπω ωc ω c where H(ω) =H(ω) within the interval ( ω c,ω c ). o recover H(ω) fully requires masking it outside the ( ω c,ω c )interval given the Fourier ransform pair we have the result Eq. 23. H(ω) =Rect ωc (ω) ω c π sin(ω c t nπ) (ω c t nπ) n= π h n e inπω ωc ω c F Rect ω c (ω)e inπω ωc F 1 he frequency 2ω c is called the Nyquist Sampling Rate. If time-domain samples are taken at an interval less than Δ c = π ω c then the original function is exactly reproduced by Sinc interpolation Eq. 23. If Δ >Δ c then aliasing occurs. 15

14 2.3.3 Frequency Sampling heorem he corresponding sampling theorem in the frequency domain is as follows If a function h(t) is zero outside a certain period, i.e. h(t) =0 for t (24) then its Fourier ransform H(ω) can be uniquely determined from its values H ( ) nπ at a sequence of equidistant points, π apart in frequency. In fact H(ω) is given by H(ω) = n= H ( nπ ) sin(ω nπ) (ω nπ) (25) 16

15 3 Interpolation of Functions 3.1 Interpolation through Shannon Reconstruction he Shannon Sampling heorem Eq. 23 tells us that a function can be exactly reconstructed by convolution with a Sinc function and summing to infinity, provided that it is sampled above the Nyquist rate. Let s look at an example Figure 1: op Left : he function cos(5t)+sin3t 1 sampled at different rates : + 4samples in period, 8 samples per period, 10 samples per period (i.e. the Nyquist rate), 16 samples per period. op Right : Reconstruction of the function at 128 sampling points from Eq. 23 using n =1; reconstruction from 4 samples in period, reconstruction from 8 samples per period, reconstruction from 10 samples per period (i.e. the Nyquist rate), reconstruction from 16 samples per period. Bottom Left : As op Right with n = 3. Bottom Right : As op Right with n =

16 In figure 1 we show the function cos(5t)+sin3t 1 with different sampling rates. By construction the maximum frequency in the signal is 5 (units of radians) which means that 10 samples per period are required to meet the Nyquist sampling rate. We show samples with {4, 8, 10, 16} per period. When using these samples in Eq. 23 we see two things 1. When sampling below the Nyquist rate, the function cannot be reconstructed adequately 2. When sampling at or above the Nyquist rate, the function can be well reconstructed, but the accuracy still depends on how large a number of periods we reconstruct from; i.e. how close n is to the ideal value of infinity. 3.2 Other Interpolation Schemes he Shannon reconstruction is the ideal interpolant because convolving with the Sinc (Sinc(t) := sin(t) t ) corresponds to truncation in the Fourier domain of all the higher harmonic copies of the original function which were induced by the sampling process. However, in the spatial domain, it is interpolation with a Sinc function. Since the Sinc function has infinite support, to interpolate to between-sample-points amounts to finding a weighted sum of all the other samples, not just those taken, but those implicitly taken by replicating the sampled period out to infinity in both directions; (recall this is an assumption of the Fourier ransform : the function being sampled is periodic and so the samples repeat infinitely). Clearly this becomes computationally heavy, especially in multiple dimensions. Other interpolation schemes use interpolants of finite support which means that the summation of weighted samples is limited to a small number. Nearest Neighbour Interpolation simply takes the between-sample value to be that of the nearest value. his is equivalent to convolution with Rect Δt (t) : a rectangle function of width Δt. In the frequency domain this is therefore multiplication with a Sinc with period Nω 0, i.e. the one whose zeroes correspond to the maximum sampling frequency and multiples thereof. We can see that the resultant function is not the original one but one with multiple harmonics, going out to infinity, decreasing at a rate of 1/ω. hushigh frequencies are generated in the original function which should not be there Linear Interpolation is the next simplest scheme. Here the between-sample value is the linear interpolation of the two nearest samples. his is equivalent to convolution with riangle 2Δt (t) : a triangle function of width 2Δt. he riangle function is in fact the convolution of two Rect functions. From this we see that its Fourier ransform is Sinc 2. he Sinc 2 function has zeros at the same places as the Sinc function, but its decay rate is 1/ω 2. hus the generation of false high frequency is less intrusive than for Nearest Neighbor interpolation. Cubic Interpolation. By convolving a riangle with itself we get a cubic function. his weights four sample points to give a between-sample value. Its Fourier ransform is Sinc 4. 18

17 hus the false high frequencies are killed off even faster; but so too are the true high frequencies up to the Nyquist. So this interpolant has a smoothing effect but might still not be all that is desired. 3.3 Fourier Zero-Fill Method Figure 2: Left : he Fourier ransform of the original function and the samples from figure 1. Right : the result of adding N M zeros to the DF of the sampled data followed by IDF (Fourier zero-fill interpolation). he above interpolation schemes give an analytic function from a set of discrete samples, i.e. they are discrete-to-continuous schemes. If we only require an increased number of samples, e.g. from the original N samples to a larger number M, but still in the same period, we can instead exploit the Fourier ransform as a consequence of the Sampling heorem. Recall that the Sampling heorem not only tells us that we can reconstruct exactly given samples taken at or above the Nyquist rate, but, in fact, that the Nyquist Rate is all that is needed and going at a higher rate adds no extra information. o see this recall that the original function is bandlimited at maximum frequency ω c. his means that its Fourier ransform is zero for ω >ω c. When sampling in an interval of length,thelowest frequency obtained is ω 0 =,knownas the fundamental frequency. If we take N samples in the time domain they lie at the points 2 = Nπ = π Δt { N/2ω 0, (1 N/2)ω 0..., ω 0, 0,ω 0,...(N/2 1)ω 0 } where Nω0 Nω0.husif 2 >ω c it will have a value of zero, because we are just taking samples of a function which is zero above the critical frequency. We therefore see that in order to synthesise samples in the time domain that are at a higher rate, we only need to add the extra frquencies in the Fourier domain - which are all zero. his is the basis of Fourier Zero-Fill interpolation which is summarised like this 19

18 ake N samples of the function h(t) at spacing Δ : {h n = h(nδ ); n =0,...,N 1} ake the DF : {H n } = DF({h n }) Add M N zeros into the list {H n } at the most positive and most negative frequencies { H n } = {0, 0,...,0, { H n }, 0,...0}. ake the inverse DF : { h n } = DF 1 ({ H n }) Now h n = h ( N M nδ ) ; i.e. the sample interval has decreased from Δ to N M Δ 20