FOURIER SERIES AND FOURIER TRANSFORMS 25 1.5 FOURIER SERIES AND FOURIER TRANSFORMS We shall have many occasions where we wish to describe pulses in terms of a frequency- domain representation.We encountered one example in Section 1.4,where we wrote a mode-locked periodic pulse train as a summation over the longitudinal cavity modes.In this section we briefly review Fourier series and Fourier transforms,which are the mathematical tools used to convert between time-and frequency-domain representations.For more detail, see texts such as [11,12].The key point is that any time-dependent signal can be written as a superposition or summation of sines and cosines(or complex exponentials)with different frequencies. 1.5.1 Analytical Aspects If a function f(r)is periodic with period T,we can write it as a Fourier series: f(t)= (1.77 where A@=2/T is the [angular]frequency.The Fn are the Fourier amplitudes or Fourier coefficients and can be obtained from the time-domain signal f(t)as follows3: 1 ft)e-in△aldh (1.78) In the case of an aperiodic time-domain function,we can consider that the period T goes to oo.We then have the frequency spacing A0,and therefore we replace the discrete variable n Aw with a continuous variable w.This results in the Fourier transform: 00 f(t)= F(ejoo do (1.79a) 2J-00 and 00 F()= f(t)e-jo dt (1.79b) F(@)is known as the Fourier transform of f(r),and f(t)is obtained by performing the inverse Fourier transform of F(@). We now review a few useful properties of Fourier transforms,most of which we state without proof.For all of the following we assume that f(t)and F(@)are a Fourier transform pair. Reality condition.If f(t)is a real function,which it will be whenever it represents an actual physical observable,F(-)=F*(@). 3 Instead of integrating fromtoT.one can get the same results by integrating over any other interval of duration T (e.g.-T/2 to T/2)
FOURIER SERIES AND FOURIER TRANSFORMS 25 1.5 FOURIER SERIES AND FOURIER TRANSFORMS We shall have many occasions where we wish to describe pulses in terms of a frequencydomain representation. We encountered one example in Section 1.4, where we wrote a mode-locked periodic pulse train as a summation over the longitudinal cavity modes. In this section we briefly review Fourier series and Fourier transforms, which are the mathematical tools used to convert between time- and frequency-domain representations. For more detail, see texts such as [11,12]. The key point is that any time-dependent signal can be written as a superposition or summation of sines and cosines (or complex exponentials) with different frequencies. 1.5.1 Analytical Aspects If a function f (t) is periodic with period T , we can write it as a Fourier series: f (t) = ∞ n=−∞ Fne jn ωt (1.77) where ω = 2π/T is the [angular] frequency. The Fn are the Fourier amplitudes or Fourier coefficients and can be obtained from the time-domain signal f (t) as follows3 : Fn = 1 T T 0 f (t)e −jn ωt dt (1.78) In the case of an aperiodic time-domain function, we can consider that the period T goes to ∞. We then have the frequency spacing ω → 0, and therefore we replace the discrete variable n ω with a continuous variable ω. This results in the Fourier transform: f (t) = 1 2π ∞ −∞ F(ω)e jωt dω (1.79a) and F(ω) = ∞ −∞ f (t)e −jωt dt (1.79b) F(ω) is known as the Fourier transform of f (t), and f (t) is obtained by performing the inverse Fourier transform of F(ω). We now review a few useful properties of Fourier transforms, most of which we state without proof. For all of the following we assume that f (t) and F(ω) are a Fourier transform pair. Reality condition. If f (t) is a real function, which it will be whenever it represents an actual physical observable, F(−ω) = F ∗ (ω). 3 Instead of integrating from 0 to T , one can get the same results by integrating over any other interval of duration T (e.g., −T/2 to T/2).
26 INTRODUCTION AND REVIEW ·Scaling formula: If h(t)=f(at), then H()() (1.80) ·Time-delay formula: Ifh(t)=f(t-t),then H(@)=F(o)e-joor (1.81) Frequency-offset formula: Ifh(t)=f(t)ejcxo!,then H(@)=F(@-00) (1.82) Comvolution formula.The convolution of two functions f(t)and g(t)is denoted f(t)* g(t)and is defined by f(t)*8(t)= dr'f(r')g(t-t) The convolution formula states that if h(t)=f(t)*g(t) then H(@)=F(@)G(@) (1.83) Parseval's theorem.We know that the intensity I(t)of a pulse whose electric field profile is f(t)is proportional to lf(t)2=f(r)f*(r).Recall that the units of intensity are W/m2.The pulse energy is given by the time-integrated intensity integrated over the cross-sectional area.Parseval's theorem says that 1 f(t)f*(t)dt= F(@)F*()do (1.84) 2元 Therefore,the time-integrated intensity is equal to the frequency-integrated intensity, except for a multiplicative factor.For this reason the quantity F()2=F(@)F*(@) is called the power spectral density. The use of these formulas will be illustrated as they are needed.We do provide a few examples of the Fourier transform: 1.f(t)=8(t),where 8(t)is the delta function or unit impulse function.Recall that ()=0 for(r)=o for =0,andfdr(=1.By substitution into eq.(1.79b),we find that F(@)=1. 2.If f(t)=ejool,then F(@)=278(@-wo).This is verified by plugging into eq.(1.79a)
26 INTRODUCTION AND REVIEW Scaling formula: If h(t) = f (at), then H(ω) = 1 a F ω a (1.80) Time-delay formula: If h(t) = f (t − τ), then H(ω) = F(ω)e −jωτ (1.81) Frequency-offset formula: If h(t) = f (t)e jω0t , then H(ω) = F(ω − ω0) (1.82) Convolution formula. The convolution of two functions f (t) and g(t) is denoted f (t) ∗ g(t) and is defined by f (t) ∗ g(t) = dt′ f (t ′ )g(t − t ′ ) The convolution formula states that if h(t) = f (t) ∗ g(t) then H(ω) = F(ω)G(ω) (1.83) Parseval’s theorem. We know that the intensity I(t) of a pulse whose electric field profile is f (t) is proportional to |f (t)| 2 = f (t)f ∗ (t). Recall that the units of intensity are W/m2 . The pulse energy is given by the time-integrated intensity integrated over the cross-sectional area. Parseval’s theorem says that f (t)f ∗ (t) dt = 1 2π F(ω)F ∗ (ω) dω (1.84) Therefore, the time-integrated intensity is equal to the frequency-integrated intensity, except for a multiplicative factor. For this reason the quantity |F(ω)| 2 = F(ω)F ∗ (ω) is called the power spectral density. The use of these formulas will be illustrated as they are needed. We do provide a few examples of the Fourier transform: 1. f (t) = δ(t), where δ(t) is the delta function or unit impulse function. Recall that δ(t) = 0 for t /= 0, δ(t) = ∞ for t = 0, and ∞ −∞ dt δ(t) = 1. By substitution into eq. (1.79b), we find that F(ω) = 1. 2. If f (t) = e jω0t , then F(ω) = 2πδ(ω − ω0). This is verified by plugging into eq. (1.79a).
FOURIER SERIES AND FOURIER TRANSFORMS 27 3.f(t)is a Gaussian,f(t)=e.Then F()=dte-e-jo= die【+ar/21le-o22/4 Using the substitution=(t+jot2/2)/tp and∫du e-t=√元，we obtain F(o)=tpN元e-w4 The procedure we have used above,called completing the square,is very useful for evaluating Fourier transforms of Gaussian functions. Finally,we frequently write a pulse in terms of a slowly varying envelope function times an optical carrier term: 1 e()=Refa(t)=()+a"()e (1.85) The spectrum E(@)is then given by E@)=2[Aa-o)+A'(-w-oo】 (1.86a) where A(w)= a(t)e-jo dt (1.86b) 0 is the Fourier transform of the envelope function a(t).Equation (1.86a)is sketched in Fig.1.9.Separating the field into envelope and carrier terms is most useful when (as in Fig.1.9)the bandwidth of A(@)is much less than the carrier frequency @o.In the time domain,this means that the envelope is much longer than the optical cycle. E() A*(-0-00) A(0-0) real part real part imaginary part 0 产、0 0=0 00 imaginary part Figure 1.9 Double-sided spectrum corresponding to eq.(1.86a)
FOURIER SERIES AND FOURIER TRANSFORMS 27 3. f (t) is a Gaussian, f (t) = e −t 2 /t2 p . Then F(ω) = dt e−t 2 /t2 p e −jωt = dt e− (t+jωt2 p /2)2 /t2 p e −ω 2 t 2 p /4 Using the substitution u = (t + jωt2 p /2)/tp and ∞ −∞ du e−u 2 = √ π, we obtain F(ω) = tp √ π e−ω 2 t 2 p /4 The procedure we have used above, called completing the square, is very useful for evaluating Fourier transforms of Gaussian functions. Finally, we frequently write a pulse in terms of a slowly varying envelope function times an optical carrier term: e(t) = Re a(t) e jω0t = 1 2 a(t) e jω0t + a ∗ (t) e −jω0t (1.85) The spectrum E(ω) is then given by E(ω) = 1 2 A(ω − ω0) + A ∗ (−ω − ω0) (1.86a) where A(ω) = ∞ −∞ a(t)e −jωt dt (1.86b) is the Fourier transform of the envelope function a(t). Equation (1.86a) is sketched in Fig. 1.9. Separating the field into envelope and carrier terms is most useful when (as in Fig. 1.9) the bandwidth of A(ω) is much less than the carrier frequency ω0. In the time domain, this means that the envelope is much longer than the optical cycle. E 0 0 A A 0 0 Figure 1.9 Double-sided spectrum corresponding to eq. (1.86a)
28 INTRODUCTION AND REVIEW The pulse e(t)can be obtained directly by performing the inverse Fourier transform of the double-sided spectrum described by eq.(1.86a)or from the single-sided spectrum using the formula 40=R{/ (a-oo）ejo do (1.87 1.5.2 Computational Aspects Fourier transform methods are frequently used for computations involving ultrashort pulses Accordingly,we introduce briefly some relevant points relating to numerical computations of Fourier transforms.For detailed discussion,see [13]. Usually computations involving a continuous-time function f(t)involve only samples of f(r)at a set of discrete timesin.Let us assume that we have N evenly spaced samples with time spacing A,so that in =nA.We designate the sample values as fn=f(tn)for n=0,1,...,N-1.The Fourier transform of f(t)is then computed numerically as N-I Fk fne-2zink/N (1.88) Equation (1.88)is formally called the discrete Fourier transform.The inverse discrete Fourier transform is given by 1 N-1 fn=N ∑TFke2rim/N (1.89) k=0 Indices n and k take on integer values in the range 0,1,...,N-1. Discrete Fourier transform routines are widely available in numerical computing soft- ware.Practically,it is preferred to choose the vector length N to be a power of 2.This allows the numerical package to evaluate the discrete Fourier transform using the fast Fourier transform(FFT)algorithm,which dramatically reduces computing time.Although the relationship to the continuous-time Fourier transform may appear to be straightforward, several points must be kept in mind if one is to obtain sensible computational results.These points include the following: Although the f vector contains a sequence of samples of the time-domain function f(t),it does not explicitly provide information on the sampling timestn or the sampling interval.Similarly,the vector of Fourier amplitudes F&does not explicitly stipulate the frequencies at which the spectrum is sampled.Hence,time and frequency vectors must be tracked by the user as auxiliary information.Note,however,that time and frequency vectors are related.If the sampling interval is known to be A,the time vector represents a total time span NA;the frequency vector v&represents a total frequency span 1/A, with vr=k/NA and spacing 1/NA between frequency samples.Here frequency refers to v=@/2 and is measured in hertz(as opposed to angular frequency measured in rad/s)
28 INTRODUCTION AND REVIEW The pulse e(t) can be obtained directly by performing the inverse Fourier transform of the double-sided spectrum described by eq. (1.86a) or from the single-sided spectrum using the formula e(t) = Re 1 2π A(ω − ω0) e jωt dω (1.87) 1.5.2 Computational Aspects Fourier transform methods are frequently used for computations involving ultrashort pulses. Accordingly, we introduce briefly some relevant points relating to numerical computations of Fourier transforms. For detailed discussion, see [13]. Usually computations involving a continuous-time function f (t) involve only samples of f (t) at a set of discrete times tn. Let us assume that we have N evenly spaced samples with time spacing , so that tn = n . We designate the sample values as fn = f (tn) for n = 0, 1, . . . , N − 1. The Fourier transform of f (t) is then computed numerically as Fk = N −1 n=0 fne −2πjnk/N (1.88) Equation (1.88) is formally called the discrete Fourier transform. The inverse discrete Fourier transform is given by fn = 1 N N −1 k=0 Fke 2πjnk/N (1.89) Indices n and k take on integer values in the range 0, 1, . . . , N − 1. Discrete Fourier transform routines are widely available in numerical computing software. Practically, it is preferred to choose the vector length N to be a power of 2. This allows the numerical package to evaluate the discrete Fourier transform using the fast Fourier transform (FFT) algorithm, which dramatically reduces computing time. Although the relationship to the continuous-time Fourier transform may appear to be straightforward, several points must be kept in mind if one is to obtain sensible computational results. These points include the following: Although the fn vector contains a sequence of samples of the time-domain function f (t), it does not explicitly provide information on the sampling timestn or the sampling interval. Similarly, the vector of Fourier amplitudes Fk does not explicitly stipulate the frequencies at which the spectrum is sampled. Hence, time and frequency vectors must be tracked by the user as auxiliary information. Note, however, that time and frequency vectors are related. If the sampling interval is known to be , the time vector represents a total time span N ; the frequency vector νk represents a total frequency span 1/ , with νk = k/N and spacing 1/N between frequency samples. Here frequency refers to ν = ω/2π and is measured in hertz (as opposed to angular frequency ω measured in rad/s).
FOURIER SERIES AND FOURIER TRANSFORMS 29 Inspection of egs.(1.88)and (1.89)reveals that ifindices n andk were allowed to extend beyond the stipulated range,fn and F&would be periodic sequences with period N. fn and F for n and k in the range 0,1.....N-1 would each then represent one period of the corresponding periodic sequences.For signals that are time-limited to duration less than N△，or frequency--limited to bandwidth less than 1/△，meaning that the signal is nonzero only over a range of samples of length less than N,this is of little consequence.However,for signals that are not time-or frequency-limited in this way, signal content appearing at the end of the time (or frequency)vector effectively wraps around to appear at the beginning of the vector. Consider a continuous-time signal f(t)that is frequency-limited such that its spectrum is zero for all frequencies outside the region-B<v<B.A well-known theorem states that all information about f(t)is retained in its sampled representation provided that the sampling interval satisfies A <1/2B.This requirement,called the Nyquist criterion,is equivalent to specifying a minimum of two samples within each period of the highest-frequency component of the frequency-limited signal.Conversely,when the Nyquist criterion is not satisfied,information is lost.In the context of our discussion on the discrete Fourier transform,we note that if Fr is frequency-limited to a bandwidth less than 1/△，we have2B<l/△，and the Nyquist criterion is satisfied.Accordingly, in setting up an F&vector for a computation,it is usually advisable to ensure that the nonzero values of F&fit comfortably within the length of the vector.If necessary,this can be accomplished by zero padding the F vector,which is equivalent in the time domain to decreasing the sample interval (increasing the sample rate). For similar reasons it is usually advisable to make sure that the nonzero values of f fit comfortably within the vector length as well.Again this can be accomplished by zero padding if needed. Inspection of egs.(1.88)and (1.89)also shows that time t=0 of the physical continuous-time function f(t)and frequency v=0 of the continuous-frequency func. tion F(@)correspond to the first point in the fr and F vectors,respectively (i.e., to =0 and vo =0).For a signal with spectrum centered at v=0,the"right"half of the spectrum appears at the beginning of the vector;the"left"half of the spectrum wraps around and appears at the end of the vector.A similar statement is true for a signal centered at t =0 in the time domain If,instead,one inadvertently positions an input spectrum so that it is centered in the frequency vector,then upon performing the inverse transform,one will notice that the expected time-domain signal is multiplied by a mysterious(-1)"function.That is,data points have alternating signs,which is equivalent to a phase shift between points.The explanation is that the discrete Fourier transform "interprets"a shift in the positioning of the input spectrum as an actual frequency shift.Centering the input spectrum amounts to a frequency shift of 1/2A in real units.This corresponds to a period of 2A in time units,equivalent to two sample intervals .Similarly,if one inadvertently positions an input time-domain signal so that it is cen- tered in the time vector,then upon performing the forward transform,one will notice that the expected spectrum is multiplied by a(-1)*function.The explanation here is that a time shift in the input signal gives rise to a sampled linear phase variation in its frequency representation (F).For a time shift of NA/2,the linear spectral phase amounts to a m phase shift per frequency sample
FOURIER SERIES AND FOURIER TRANSFORMS 29 Inspection of eqs. (1.88) and (1.89) reveals that if indices n and k were allowed to extend beyond the stipulated range, fn and Fk would be periodic sequences with period N. fn and Fk for n and k in the range 0, 1, . . . , N − 1 would each then represent one period of the corresponding periodic sequences. For signals that are time-limited to duration less than N , or frequency-limited to bandwidth less than 1/ , meaning that the signal is nonzero only over a range of samples of length less than N, this is of little consequence. However, for signals that are not time- or frequency-limited in this way, signal content appearing at the end of the time (or frequency) vector effectively wraps around to appear at the beginning of the vector. Consider a continuous-time signal f (t) that is frequency-limited such that its spectrum is zero for all frequencies outside the region −B ≤ ν ≤ B. A well-known theorem states that all information about f (t) is retained in its sampled representation provided that the sampling interval satisfies ≤ 1/2B. This requirement, called the Nyquist criterion, is equivalent to specifying a minimum of two samples within each period of the highest-frequency component of the frequency-limited signal. Conversely, when the Nyquist criterion is not satisfied, information is lost. In the context of our discussion on the discrete Fourier transform, we note that if Fk is frequency-limited to a bandwidth less than 1/ , we have 2B < 1/ , and the Nyquist criterion is satisfied. Accordingly, in setting up an Fk vector for a computation, it is usually advisable to ensure that the nonzero values of Fk fit comfortably within the length of the vector. If necessary, this can be accomplished by zero padding the Fk vector, which is equivalent in the time domain to decreasing the sample interval (increasing the sample rate). For similar reasons it is usually advisable to make sure that the nonzero values of fn fit comfortably within the vector length as well. Again this can be accomplished by zero padding if needed. Inspection of eqs. (1.88) and (1.89) also shows that time t = 0 of the physical continuous-time function f (t) and frequency ν = 0 of the continuous-frequency function F(ω) correspond to the first point in the fn and Fk vectors, respectively (i.e., t0 = 0 and ν0 = 0). For a signal with spectrum centered at ν = 0, the “right” half of the spectrum appears at the beginning of the vector; the “left” half of the spectrum wraps around and appears at the end of the vector. A similar statement is true for a signal centered at t = 0 in the time domain. If, instead, one inadvertently positions an input spectrum so that it is centered in the frequency vector, then upon performing the inverse transform, one will notice that the expected time-domain signal is multiplied by a mysterious (−1)n function. That is, data points have alternating signs, which is equivalent to a π phase shift between points. The explanation is that the discrete Fourier transform “interprets” a shift in the positioning of the input spectrum as an actual frequency shift. Centering the input spectrum amounts to a frequency shift of 1/2 in real units. This corresponds to a period of 2 in time units, equivalent to two sample intervals. Similarly, if one inadvertently positions an input time-domain signal so that it is centered in the time vector, then upon performing the forward transform, one will notice that the expected spectrum is multiplied by a (−1)k function. The explanation here is that a time shift in the input signal gives rise to a sampled linear phase variation in its frequency representation (Fk). For a time shift of N /2, the linear spectral phase amounts to a π phase shift per frequency sample.