《数字信号处理》教学参考资料（Numerical Recipes in C，The Art of Scientific Computing Second Edition）Chapter 13.1.pdf

538 Chapter 13.Fourier and Spectral Applications 13.1 Convolution and Deconvolution Using the FFT We have defined the comolution of two functions for the continuous case in equation (12.0.8),and have given the comolution theorem as equation(12.0.9).The theorem says that the Fourier transform of the convolution of two functions is equal to the product of their individual Fourier transforms.Now,we want to deal with the discrete case.We will mention first the context in which convolution is a useful procedure,and then discuss how to compute it efficiently using the FFT. The convolution of two functions r(t)and s(t),denoted r*s,is mathematically equal to their convolution in the opposite order,s *r.Nevertheless,in most B applications the two functions have quite different meanings and characters.One of the functions,say s,is typically a signal or data stream,which goes on indefinitely 茶 in time (or in whatever the appropriate independent variable may be).The other function r is a"response function,"typically a peaked function that falls to zero in both directions from its maximum.The effect of convolution is to smear the signal s(t)in time according to the recipe provided by the response function r(t),as shown in Figure 13.1.1.In particular,a spike or delta-function of unit area in s which occurs 9 at some time to is supposed to be smeared into the shape of the response function itself,but translated from time 0 to time to as r(t-to). In the discrete case,the signal s(t)is represented by its sampled values at equal time intervals s;.The response function is also a discrete set of numbersr with the following interpretation:ro tells what multiple of the input signal in one channel (one 含色号分N particular value ofj)is copied into the identical output channel(same value ofj); r tells what multiple of input signal in channelj is additionally copied into output channelj+1;r tells the multiple that is copied into channel j-1:and so on for both positive and negative values of k in rk.Figure 13.1.2 illustrates the situation. Example:a response function with ro 1 and all other r&'s equal to zero is just the identity filter:convolution of a signal with this response function gives identically the signal.Another example is the response function withr14=1.5 and all other rk's equal to zero.This produces convolved output that is the input signal Numerical Recipes 10621 multiplied by 1.5 and delayed by 14 sample intervals. Evidently,we have just described in words the following definition of discrete convolution with a response function of finite duration M: 43106 M/2 (r*s)方≡ sj-k Tk (13.1.1) k=-M/2+1 North If a discrete response function is nonzero only in some range-M/2<k<M/2, where M is a sufficiently large even integer,then the response function is called a finite impulse response (FIR),and its duration is M.(Notice that we are defining M as the number of nonzero values of rk;these values span a time interval of M-1 sampling times.)In most practical circumstances the case of finite M is the case of interest,either because the response really has a finite duration,or because we choose to truncate it at some point and approximate it by a finite-duration response function. The discrete comvolution theorem is this:If a signal sj is periodic with period N,so that it is completely determined by the N values so,...,sN-1,then its

538 Chapter 13. Fourier and Spectral Applications Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). 13.1 Convolution and Deconvolution Using the FFT We have defined the convolution of two functions for the continuous case in equation (12.0.8), and have given the convolution theorem as equation (12.0.9). The theorem says that the Fourier transform of the convolution of two functions is equal to the product of their individual Fourier transforms. Now, we want to deal with the discrete case. We will mention first the context in which convolution is a useful procedure, and then discuss how to compute it efficiently using the FFT. The convolution of two functions r(t) and s(t), denoted r ∗ s, is mathematically equal to their convolution in the opposite order, s ∗ r. Nevertheless, in most applications the two functions have quite different meanings and characters. One of the functions, say s, is typically a signal or data stream, which goes on indefinitely in time (or in whatever the appropriate independent variable may be). The other function r is a “response function,” typically a peaked function that falls to zero in both directions from its maximum. The effect of convolution is to smear the signal s(t) in time according to the recipe provided by the response function r(t), as shown in Figure 13.1.1. In particular, a spike or delta-function of unit area in s which occurs at some time t0 is supposed to be smeared into the shape of the response function itself, but translated from time 0 to time t0 as r(t − t0). In the discrete case, the signal s(t) is represented by its sampled values at equal time intervals sj . The response function is also a discrete set of numbers rk, with the following interpretation: r0 tells what multiple of the input signal in one channel (one particular value of j) is copied into the identical output channel (same value of j); r1 tells what multiple of input signal in channel j is additionally copied into output channel j + 1; r−1 tells the multiple that is copied into channel j − 1; and so on for both positive and negative values of k in rk. Figure 13.1.2 illustrates the situation. Example: a response function with r0 = 1 and all other rk’s equal to zero is just the identity filter: convolution of a signal with this response function gives identically the signal. Another example is the response function with r 14 = 1.5 and all other rk’s equal to zero. This produces convolved output that is the input signal multiplied by 1.5 and delayed by 14 sample intervals. Evidently, we have just described in words the following definition of discrete convolution with a response function of finite duration M: (r ∗ s)j ≡ M/ 2 k=−M/2+1 sj−k rk (13.1.1) If a discrete response function is nonzero only in some range −M/2 < k ≤ M/2, where M is a sufficiently large even integer, then the response function is called a finite impulse response (FIR), and its duration is M. (Notice that we are defining M as the number of nonzero values of rk; these values span a time interval of M − 1 sampling times.) In most practical circumstances the case of finite M is the case of interest, either because the response really has a finite duration, or because we choose to truncate it at some point and approximate it by a finite-duration response function. The discrete convolution theorem is this: If a signal sj is periodic with period N, so that it is completely determined by the N values s0,...,sN−1, then its

540 Chapter 13.Fourier and Spectral Applications discrete convolution with a response function of finite duration N is a member of the discrete Fourier transform pair, N/2 sj-kTk←→SnRm (13.1.2) k=-N/2+1 Here Sn,(n 0,...,N-1)is the discrete Fourier transform of the values sj,(j=0,...,N-1),while Rn,(n =0,...,N-1)is the discrete Fourier transform of the values rk,(=0,...,N-1).These values of r are the same ones as for the range k=-N/2+1,...,N/2,but in wrap-around order,exactly 8 as was described at the end of 812.2. Treatment of End Effects by Zero Padding 茶 The discrete convolution theorem presumes a set of two circumstances that are not universal.First,it assumes that the input signal is periodic,whereas real data often either go forever without repetition or else consist of one nonperiodic stretch of finite length.Second,the convolution theorem takes the duration of the 9 response to be the same as the period of the data;they are both N.We need to work around these two constraints. The second is very straightforward.Almost always,one is interested in a response function whose duration M is much shorter than the length of the data set N.In this case,you simply extend the response function to length N by padding it with zeros,i.e,define rk=0 for M/2 <k N/2 and also for 、屋色 -N/2+1<k<-M/2+1.Dealing with the first constraint is more challenging. OF SCIENTIFIC( Since the convolution theorem rashly assumes that the data are periodic,it will falsely "pollute"the first output channel (r s)o with some wrapped-around data 61 from the far end of the data stream sN-1,sN-2,etc.(See Figure 13.1.3.)So, we need to set up a buffer zone of zero-padded values at the end of the si vector, in order to make this pollution zero.How many zero values do we need in this buffer?Exactly as many as the most negative index for which the response function is nonzero.For example,if r-3 is nonzero,while r-4,r-5,...are all zero,then we 10621 need three zero pads at the end of the data:sN-3 =sN-2 =sN-1=0.These Numerica zeros will protect the first output channel(r s)o from wrap-around pollution.It should be obvious that the second output channel(r*s)1 and subsequent ones will also be protected by these same zeros.Let K denote the number of padding zeros, so that the last actual input data point is sN-K-1. What now about pollution of the very last output channel?Since the data now end with sN-K-1,the last output channel of interest is (r s)N-K-1.This channel can be polluted by wrap-around from input channel so unless the number K is also large enough to take care of the most positive index k for which the response function rk is nonzero.For example,if ro through re are nonzero,while r7,rs...are all zero,then we need at least K=6 padding zeros at the end of the data:sN-6=..=sN-1=0. To summarize-we need to pad the data with a number of zeros on one end equal to the maximum positive duration or maximum negative duration of the response function,whichever is larger.(For a symmetric response function of duration M,you will need only M/2 zero pads.)Combining this operation with the

540 Chapter 13. Fourier and Spectral Applications Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). discrete convolution with a response function of finite duration N is a member of the discrete Fourier transform pair, N/ 2 k=−N/2+1 sj−k rk ⇐⇒ SnRn (13.1.2) Here Sn, (n = 0,...,N − 1) is the discrete Fourier transform of the values sj , (j = 0,...,N − 1), while Rn, (n = 0,...,N − 1) is the discrete Fourier transform of the values rk, (k = 0,...,N − 1). These values of rk are the same ones as for the range k = −N/2+1, . . ., N/2, but in wrap-around order, exactly as was described at the end of §12.2. Treatment of End Effects by Zero Padding The discrete convolution theorem presumes a set of two circumstances that are not universal. First, it assumes that the input signal is periodic, whereas real data often either go forever without repetition or else consist of one nonperiodic stretch of finite length. Second, the convolution theorem takes the duration of the response to be the same as the period of the data; they are both N. We need to work around these two constraints. The second is very straightforward. Almost always, one is interested in a response function whose duration M is much shorter than the length of the data set N. In this case, you simply extend the response function to length N by padding it with zeros, i.e., define rk = 0 for M/2 ≤ k ≤ N/2 and also for −N/2+1 ≤ k ≤ −M/2+1. Dealing with the first constraint is more challenging. Since the convolution theorem rashly assumes that the data are periodic, it will falsely “pollute” the first output channel (r ∗ s)0 with some wrapped-around data from the far end of the data stream sN−1, sN−2, etc. (See Figure 13.1.3.) So, we need to set up a buffer zone of zero-padded values at the end of the s j vector, in order to make this pollution zero. How many zero values do we need in this buffer? Exactly as many as the most negative index for which the response function is nonzero. For example, if r−3 is nonzero, while r−4, r−5,... are all zero, then we need three zero pads at the end of the data: sN−3 = sN−2 = sN−1 = 0. These zeros will protect the first output channel (r ∗ s)0 from wrap-around pollution. It should be obvious that the second output channel (r ∗ s) 1 and subsequent ones will also be protected by these same zeros. Let K denote the number of padding zeros, so that the last actual input data point is sN−K−1. What now about pollution of the very last output channel? Since the data now end with sN−K−1, the last output channel of interest is (r ∗ s)N−K−1. This channel can be polluted by wrap-around from input channel s 0 unless the number K is also large enough to take care of the most positive index k for which the response function rk is nonzero. For example, if r0 through r6 are nonzero, while r7, r8 ... are all zero, then we need at least K = 6 padding zeros at the end of the data: sN−6 = ... = sN−1 = 0. To summarize — we need to pad the data with a number of zeros on one end equal to the maximum positive duration or maximum negative duration of the response function, whichever is larger. (For a symmetric response function of duration M, you will need only M/2 zero pads.) Combining this operation with the

542 Chapter 13.Fourier and Spectral Applications padding of the response rk described above.we effectively insulate the data from artifacts of undesired periodicity.Figure 13.1.4 illustrates matters. Use of FFT for Convolution The data,complete with zero padding,are now a set of real numbers si,j= 0,...,N-1,and the response function is zero padded out to duration N and arranged in wrap-around order.(Generally this means that a large contiguous section of the r's,in the middle of that array,is zero,with nonzero values clustered at 三 the two extreme ends of the array.)You now compute the discrete convolution as follows:Use the FFT algorithm to compute the discrete Fourier transform ofs and of r.Multiply the two transforms together component by component,remembering that the transforms consist of complex numbers.Then use the FFT algorithm to take the inverse discrete Fourier transform of the products.The answer is the convolution r*s. What about decorvolution?Deconvolution is the process of undoing the smearing in a data set that has occurred under the influence of a known response function,for example,because of the known effect of a less-than-perfect measuring apparatus.The defining equation of deconvolution is the same as that for convolution, namely (13.1.1),except now the left-hand side is taken to be known,and (13.1.1)is to be considered as a set of N linear equations for the unknown quantities sj.Solving these simultaneous linear equations in the time domain of(13.1.1)is unrealistic in most cases,but the FFT renders the problem almost trivial.Instead of multiplying the transform of the signal and response to get the transform of the convolution,we just divide the transform of the(known)convolution by the transform of the response 三兽色2∽ to get the transform of the deconvolved signal. OF SCIENTIFIC This procedure can go wrong mathematically if the transform of the response function is exactly zero for some value Rn,so that we can't divide by it.This 6 indicates that the original convolution has truly lost all information at that one frequency,so that a reconstruction of that frequency component is not possible. You should be aware,however,that apart from mathematical problems,the process of deconvolution has other practical shortcomings.The process is generally quite sensitive to noise in the input data,and to the accuracy to which the response function Numerica 10621 rk is known.Perfectly reasonable attempts at deconvolution can sometimes produce nonsense for these reasons.In such cases you may want to make use of the additional 43106 process of optimal filtering,which is discussed in 813.3. Here is our routine for convolution and deconvolution,using the FFT as Recipes implemented in four1 of 812.2.Since the data and response functions are real. not complex,both of their transforms can be taken simultaneously using twofft. Note,however,that two calls to realft should be substituted if data and respns North have very different magnitudes,to minimize roundoff.The data are assumed to be stored in a float array data [1..n],with n an integer power of two.The response function is assumed to be stored in wrap-around order in a sub-array respns [1..m] of the array respns [1..n].The value ofm can be any odd integer less than or equal to n,since the first thing the program does is to recopy the response function into the appropriate wrap-around order in respns [1..n].The answer is provided in ans. #include "nrutil.h" void convlv(float data],unsigned long n,float respns[],unsigned long m

542 Chapter 13. Fourier and Spectral Applications Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machinereadable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). padding of the response rk described above, we effectively insulate the data from artifacts of undesired periodicity. Figure 13.1.4 illustrates matters. Use of FFT for Convolution The data, complete with zero padding, are now a set of real numbers s j, j = 0,...,N − 1, and the response function is zero padded out to duration N and arranged in wrap-around order. (Generally this means that a large contiguous section of the rk’s, in the middle of that array, is zero, with nonzero values clustered at the two extreme ends of the array.) You now compute the discrete convolution as follows: Use the FFT algorithm to compute the discrete Fourier transform of s and of r. Multiply the two transforms together component by component, remembering that the transforms consist of complex numbers. Then use the FFT algorithm to take the inverse discrete Fourier transform of the products. The answer is the convolution r∗s. What about deconvolution? Deconvolution is the process of undoing the smearing in a data set that has occurred under the influence of a known response function, for example, because of the known effect of a less-than-perfect measuring apparatus. The defining equation of deconvolution is the same as that for convolution, namely (13.1.1), except now the left-hand side is taken to be known, and (13.1.1) is to be considered as a set of N linear equations for the unknown quantities s j . Solving these simultaneous linear equations in the time domain of (13.1.1) is unrealistic in most cases, but the FFT renders the problem almost trivial. Instead of multiplying the transform of the signal and response to get the transform of the convolution, we just divide the transform of the (known) convolution by the transform of the response to get the transform of the deconvolved signal. This procedure can go wrong mathematically if the transform of the response function is exactly zero for some value Rn, so that we can’t divide by it. This indicates that the original convolution has truly lost all information at that one frequency, so that a reconstruction of that frequency component is not possible. You should be aware, however, that apart from mathematical problems, the process of deconvolution has other practical shortcomings. The process is generally quite sensitive to noise in the input data, and to the accuracy to which the response function rk is known. Perfectly reasonable attempts at deconvolution can sometimes produce nonsense for these reasons. In such cases you may want to make use of the additional process of optimal filtering, which is discussed in §13.3. Here is our routine for convolution and deconvolution, using the FFT as implemented in four1 of §12.2. Since the data and response functions are real, not complex, both of their transforms can be taken simultaneously using twofft. Note, however, that two calls to realft should be substituted if data and respns have very different magnitudes, to minimize roundoff. The data are assumed to be stored in a float array data[1..n], with n an integer power of two. The response function is assumed to be stored in wrap-around order in a sub-array respns[1..m] of the array respns[1..n]. The value of m can be any odd integer less than or equal to n, since the first thing the program does is to recopy the response function into the appropriate wrap-around order in respns[1..n]. The answer is provided in ans. #include "nrutil.h" void convlv(float data[], unsigned long n, float respns[], unsigned long m