Classical Fourier Series Introduction: Real Fourier Series

Math 344 May 1, 1 Classical Fourier Series Introduction: Real Fourier Series The orthogonality roerties of the sine and cosine functions make them good candidates as basis functions when orthogonal eansions are needed. In this handout we discuss aroimations associated with the family of functions: { 1, cos(nπ/l, sin(nπ/l } n=1, where L is a ositive number. These functions are orthogonal and comlete on any interval of width L. In articular, any function f in P S[, L] can be aroimated to any desired accuracy with a finite sum of the form S n ( = a n a k cos(kπ/l + b k sin(kπ/l. That is, lim n f( S n ( =. Moreover, it can be shown that f( = lim n S n ( at all oints in the oen interval (, L where f is continuous. At a oint where f has a jum discontinuity, the aroimating sums converge to the average of the left and right limiting values of the function f lim S n( = f( + f( +. n By adoting the convention that f( = (f( + f( + / at all oints of discontinuity, the infinite series formula is valid at all oints in the oen interval (, L. Formulas for the Coefficients (Euler Formulas The inner roducts f( = a a k cos(kπ/l + b k sin(kπ/l (1 cos(kπ/l, cos(kπ/l and sin(kπ/l, sin(kπ/l both evaluate to L in P S[, L]. Therefore, the sine and cosine coefficients in (1 are given by these integral formulas. a k = 1 L b k = 1 L f( cos(kπ/l d ( f( sin(kπ/l d (3 The constant term in (1 is the average value of f on [, L]. Consequently, it can often be determined simly by insection of f s grah. By writing the constant as a /, the integral formula for a k can also be used to find a when necessary. If a comuter algebra system is used to calculate the coefficient formulas it is helful to observe that a = lim k a k. (4 Equations (1 and ( are called Euler formulas in honor of Leonard Euler who was one of the first to use them. Eamle 1. Obtain the formula for the Fourier series aroimation S n ( for the function in P S[ π, π]. Then do the following. f( = {, π < <, < < π 1. Plot f and S 1, S, and S 3 on the same set of aes. Make one lot over the interval π < < π and another on the interval 3π < < 3π. Assume that f( = outside of the interval [ π, π].. Obtain a simle formula for the error associated with the aroimation S n (. 1

3. Use the error formula to find the smallest value of n for which the root square error, RSE, is less than 1% of f. Solution. Let s begin with a lot of the grah of f..5 K K.5.5 The grah of the function f. Since L = π, the n th Fourier aroimation has the simle form S n ( = a n a k cos(k + b k sin(k. By insection of the grah of f, the constant term is a / = ( + π// = π/4. coefficient a k evaluates easily via integration-by-arts (u =, dv = cos(k d. a k = 1 π π π f( cos(k d = 1 cos(k d π π = 1 ( sin(k 1 =π sin(k d π k k = = 1 π cos(kπ 1 = 1 π ( 1k 1 The integral formula for the We have used the fact that cos(kπ = ( 1 k when k is an integer to simlify the formula for a k. Note that the formula for a k can be eressed termwise as follows. a k = π, k = 1, 3, 5,..., k =, 4, 6,... A similar calculation will show that b k = cos(πk k = ( 1k+1 k = { 1/k, k = 1, 3, 5,... 1/k, k =, 4, 6,... Using these formulas, The first term in the sum, S n ( = π n 4 + ( ( 1 k 1 π cos(k + ( 1k+1 k = π 4 π cos( + sin( 1 sin( +. cos( + sin(, π sin(k is called the first harmonic of the function f. It is a sinusoid with amlitude 1 4/π + 1. The second harmonic is 1 sin(, an inverted sine wave with amlitude 1/. The dotted curve in the icture below is the first harmonic shifted uward π/4 units to yield S 1 (, the first-order Fourier aroimation of f. The grah of the second-order aroimation, S ( (dashed curve, is obtained from the grah of S 1 ( by adding the second harmonic. Add the third harmonic to get S 3 (, and so on. 1 An eression of the form a cos(ω+b sin(ω is called a sinusoid. For lotting uroses it can be eressed in the form A cos(ω δ where A = a + b and δ is chosen so that cos(δ = a/a and sin(δ = b/a. See Eercise.

.5 K K.5.5 The dotted curve is S 1 (. The dashed curve is S (. The dash-dot curve is S 3 (. The aroimating curves all begin and end at the same lace, roughly half-way between the oint where the grah of the function f begins and the oint where it ends. This is because classical Fourier sums are eriodic, as we can be see on the grah below. K3 K K 3 Fourier aroimations are eriodic. Because each harmonic has eriod L, it begins and ends at the same lace over any interval of width L, which haens to be π in this eamle. Thus at the endoints of the basic interval, L and, the aroimations tend towards the average of the limiting values of f at the endoints: (f(( + + f(l /. Error Analysis The square of the error associated with the n th aroimation S n ( is f( S n (. The following eression for the squared error in terms of the Fourier coefficients is obtained using the formulas in the Error Analysis handout (May 17. f( S n ( = f( S n ( ( L = f( a n d 4 L + ( a k L + b k L ( L = f( a d L 4 + 1 n (a k + b k Divide both sides by L to obtain the formula for the average squared error associated with S n ( over one eriod. This is also called the Mean Squared Error, denoted MSE. Mean Squared Error Formula MSE = 1 L f( S n( = 1 L f( d ( a 4 + 1 n (a k + b k (5 The fact that S n( = a 4 L + n ( a k L + b k L is simly the equation S n( = a / + n ( ak cos(kπ/l + b k sin(kπ/l in disguise. 3

Using this we have the following simle formula for the n th mean squared error in Eamle 1. MSE = 1 ( π π d π 16 + 1 n ( (( 1 k 1 π k 4 + 1 = π 6 π 16 1 n ( (( 1 k 1 π k 4 + 1 = 5π 48 1 n ( (( 1 k 1 π k 4 + 1 Since RSE = π MSE and f = π 3 /3 (verify, the table below can be used to in down the smallest value of n for which the relative root square error: RSE/ f, is less than 1%. n 1 4 7 3 π MSE.118.111.15.998 π3 /3 Now find the eact value of n by calculating RSE/ f for n = 7, 8, 9, 3. The following icture dislays the grah of f and S 3 (. n 7 8 9 3 πmse.151.13.115.998 π3 /3.5 K K.5.5 The grah of f and S 3 ; relative RSE is less than 1%. Note that almost all of the error associated with S 3 occurs near the two endoints of the basic interval. This is due to the fact that the eriodic etension of the function f has jum discontinuities at π and π. See the net lot. K3 K K 3 The function f etends to this eriodic function. There are jum discontinuities at odd multiles of π. Fourier aroimations imrove dramatically when both f and its eriodic etension are continuous. This is demonstrated in the net eamle which also shows that secial symmetry in the grah of f can be eloited to simlify the calculation of the coefficients. Eamle. Let f be the function in P S[, ] defined as follows. Its grah is shown below. f( = { 1, < 1, otherwise 4

1..5 K K1 1 The function f is continuous, and so is its eriodic etension. Note that this is an even function. Observe that the grah of f is symmetric with resect to the vertical ais. Functions with this roerty are called even. Analytically, this means that, for all, f( = f(. Even functions have the roerty that f( d = f( d. Odd vs Even A function g whose grah is symmetric with resect to the origin is called odd. See the icture on the right. Analytically, this means that, for all, g( = g(. Odd functions have the roerty that g( d =. The grah of g, an odd function. A simle calculation will show that if f is even and g is odd, then their roduct fg is odd. On the other hand, the roduct of two even functions or of two odd functions is always even. Note also that the cosine function is even and the sine function is odd. Because the function f in Eamle is even, its Fourier series aroimations consist of even functions only: a constant term and the cosine terms. This is verified analytically by observing that k th sine coefficient is zero: b k = 1 L f( sin(kπ/l d =. This is because the integrand is odd (even odd. Moreover, the k th cosine coefficient can be calculated by doubling the integral over the right half of the interval: a k = L f( cos(kπ/l d. (6 This is because the integrand, f( cos(kπ/l, is even (even even. Alying these observations to the function in this eamle, its n th Fourier aroimation looks like this. S n ( = a n a k cos(kπ/ By insection of the grah of f, the constant is a / = ( + 1// = 1/4. The coefficient a k evaluates via integration-by-arts: u = 1, dv = cos(kπ/. a k = f( cos(kπ/ d = ( (1 sin(kπ/ = kπ/ = 4 π 1 cos(kπ/ 5 1 (1 cos(kπ/ d + kπ =1 sin(kπ/ d =

The formula for a k cannot be simlified. However, it can be eressed in a simle termwise form. 4 π, k = 1, 3, 5,... a k = 8 π, k =, 6, 1,..., k = 4, 8, 1,... This shows that the k th coefficient decreases to at a rate roortional to 1/, as oosed to the rate of 1/k for the b k coefficient in Eamle 1. This behavior, which is tyical of a function with a continuous eriodic etension, and a discontinuous derivative, means that the Fourier aroimations aroach the limiting function very quickly. The Fourier series reresentation for the function in Eamle is f( = 1 4 + 4 π n 1 cos(kπ/ cos(kπ/ = 1 4 + 4 π ( cos(π/ + 1 cos(π + 1 9 cos(3π/ +. The icture below attests to the quality of the n th aroimation even when n is as small as 3. 1..5 K K1 1 The dotted curve is S 1 (. The dashed curve is S (. The dash-dot curve is S 3 (. In fact the grah of S 3 is so close to the grah of f that it can hardly be seen in the icture above. The net icture shows f and S 3 over the interval 5 < < 5. 1..5 K5 K4 K3 K K1 1 3 4 5 The dotted curve is S 3 (. Observe that S 3 is very good aroimation to the eriodic etension of f. The following error analysis confirms this observation. Error Analysis. Using the mean square error formula on age 3, MSE = 1 ( L f( a d L 4 + 1 n (a k + b k = 1 ( 1 (1 1 d 4 1 16 + 8 n (1 cos(kπ/ π 4 k 4 = 1 6 1 16 8 n (1 cos(kπ/ π 4 k 4 = 5 48 8 π 4 n (1 cos(kπ/ k 4. Since RSE = 4 MSE and f = 6/3 (verify, the following table of relative RSE errors shows how well the low-order aroimations fit the function f. 6

n 1 3 4 5 4 MSE 6/3.364.95.54.54.47 For eamle, the relative error for S 3 is only 5.4% making it an ecellent fit. The error for S 4 is the same because a 4 = imlying that S 3 ( = S 4 (. The net eamle features an odd function with a jum discontinuity at the origin. As was the case in Eamle 1, this causes the Fourier coefficients to aroach at a rate roortional to 1/k so a good aroimation will require several terms in the Fourier aroimation. Eamle 3. Obtain the n th Fourier aroimation for the function f in P S[ π, π] defined as follows. f( = π, π < < π, < < π, elsewhere Although it is not obvious from the defining formulas, the grah of f shown below reveals that it is an odd function. K K K The grah of the function f. It is an odd function. Fourier aroimations for odd functions have only sine terms. Therefore, the Fourier series for an odd function has the form f( = b k sin(kπ/l. This is because the average value of f over [, L] is and f( cos(kπ/l is also odd (odd even so its integral over [, L] evaluates to as well. In addition, because f( sin(kπ/l is even (odd odd, the k th sine coefficient, b k, can be calculated by integrating over the right half of the basic interval and doubling the result. b k = L Secializing to the function f in this eamle, L = π, so S n ( = f( sin(kπ/l d (7 n b k sin(k/. Using integration-by-arts, u = π, dv = sin(k/ d, the calculation of b k roceeds as follows. b k = π π π f( sin(k/ d = 1 (π sin(k/ d π = 1 ( (π cos(k/ =π cos(k/ d π k/ k = = 1 ( π π k sin(kπ/ πk 4 sin(kπ/ = π 7

The formula for b k cannot be simlified. However, termwise, the coefficients have the following form. b k = k 4 π, k = 1, 5, 9,... k, k =, 4, 6,... k + 4 π, k = 3, 7, 11,... This confirms that the k th coefficient goes to as 1/k. The following icture dislays the aroimations corresonding to n = 1, 3, and 6. K K K The dotted curve is S 1 (. The dashed curve is S 3 (. The dash-dot curve is S 6 (. The net icture, showing the grah of f and S (, shows once more that the rincial contribution to the error comes from the aroimation near the jum discontinuity. K K K The function f and S (. The error is analyzed in Eercise 4. Gibbs Phenomenon The American hysicist Josiah Willard Gibbs, was among the first to analyze the behavior of Fourier aroimations near a discontinuity in the target function. Note the overshoot to the right of the oint (, π in the icture above. Gibbs was able to show that all higher-order aroimations make the same overshoot (and undershoot, which is roughly 9% of the jum in function values. The net icture, which zeroes in on (, π, illustrates what is referred to as Gibbs Phenomenon. The aroimations all aear to eak at the dotted line that is lotted.18π units above the grah of f. 8

Half-Range Eansions 1.3 1. 1.1.9.8.7 K1. K.5..5 1. Gibbs Phenomenon: S 1, S, S 3, and S 4 overshoot the target function by aroimately the same amount as from the right. You have robably noticed that formulas (6 and (7, used to calculate Fourier coefficients for even and odd functions in P S[, L], are the same as the formulas for the coefficients in the Fourier sine and cosine series for a function in P S[, L]. Because of this, the aroimations on [, L] are often referred to as Half-Range Eansions. The following table summarizes the essential information needed to obtain the half-range eansions of a function f in P S[, L]. Tye Form Coefficients Cosine Sine a a k cos(kπ/l b k sin(kπ/l a k = L b k = L f( cos(kπ/l d f( sin(kπ/l d Periodic Functions Classical Fourier series are ideally suited to reresent eriodic functions. Here is the formal definition of a eriodic function. Observe that we have switched to the letter t for the indeendent variable. Definition (Periodic Functions A function f, defined for all real t, is called eriodic when there is a ositive number P such that f(t + P = f(t for all t. If P is the smallest ositive number for which this is true, then it is called the eriod of f. The letter t is used to denote the indeendent variable for a eriodic function because we want to think of it as a signal, varying with time. Of course, the best known eriodic functions are sine and cosine, and a quick review of the essential facts about these functions is in order. 1. Both sin(ωt and cos(ωt have eriod P = π/ω. So does any linear combination: A function of this form is called a sinusoid. f(t = a cos(ωt + b sin(ωt.. The recirocal of the eriod of a eriodic function is called its frequency. Thus the sinusoid a cos(ωt+b sin(ωt has frequency ω/π. The number ω is called its circular frequency but we will usually just refer to it as the frequency. It serves as a oint of reference when combining sinusoids of different eriods. 3. By defining A = a + b and choosing δ so that cos(δ = a/a and sin(δ = b/a the sinusiod f(t = a cos(ωt+ b sin(ωt can be written in the form f(t = A cos(ωt δ. 3 A is called the amlitude of the sinusoid and δ is 3 Observe that A cos(ωt δ = A cos(δ cos(ωt + A sin(δ sin(ωt. 9

called the hase angle. We adot the convention that π < δ π so { arctan(b/a when a > δ = arctan(b/a + π when a <. In addition, if a =, then δ = π/ when b > and δ = π/ when b <. A function f of eriod P is called iecewise smooth when it is iecewise smooth over a fundamental eriodic interval [a, a + P ]. Of course, this makes it iecewise smooth over any interval. Such a function can always be reresented as the sum of a Fourier series. The series will have the following form f(t = a a k cos(kωt + b k sin(kωt (8 where ω = π/p. This is the (circular frequency of the first harmonic in the series: a 1 cos(ωt + b 1 sin(ωt. The k th harmonic: a k cos(kωt + b k sin(kωt, oscillates k times faster than the first. Note that the validity of formula (8 deends uon the assumtion that, at a oint t where f is discontinuous, f(t = f(t + f(t+ If we let L = P/, then kω = kπ/l and Equation (8 can be eressed in the familiar form See Equation (1 on age 1. f(t = a a k cos(kπt/l + b k sin(kπt/l.. Imortant Observation. The coefficients in (8 are calculated using the Euler formulas on age 1. However, for a eriodic function the integrals can be evaluated over any eriodic interval [τ, τ + P ]. In this case, a k = P τ+p τ f(t cos(kωt dt, b k = P τ+p τ f(t sin(kωt dt. (9 Eamle 4. Obtain the Fourier series reresentation for the function of eriod P = 1 defined as f(t = t for < t < 1. See the grah below. 1..5 K K1 1 t The eriodic function f with dotted lines at each discontinuity and a circle marking the value of the function. Since P = 1, the fundamental frequency is ω = π. Clearly the average value of f is a / = 1/. Both a k and b k can be calculated using integration-by-arts (u = t. We integrate over the eriodic interval [, 1]. a k = 1 1 ( t sin(kωt t cos(kωt dt = 1 kω kω 1 t=1 sin(kωt dt = t=

and b k = 1 1 ( t sin(kωt dt = t cos(kωt + 1 kω kω = kω = 1 πk t=1 cos(kωt dt t= The Fourier series reresentation is f(t = 1 1 π = 1 1 π sin(kωt k ( sin(ωt + sin(ωt + sin(3ωt + 3 The following icture dislays the grah of f and its third Fourier aroimation.. 1..5 K K1 1 t The function f and its third order Fourier aroimation, S 3 (t. Error Analysis. The mean square error, MSE, is a good measure of the quality of the aroimation when S n ( is used to aroimate a eriodic function. Any eriodic interval can be used in the calculations. In this case, we use [, 1]. The following calculation begins with the formula for MSE for the n th aroimation, then secializes to the function in this eamle. MSE = 1 P = 1 P f(t dt t dt = 1 1 1 π ( 1 4 + 1 π n 1 ( a 4 + 1 n n (a k + b k This can be used to obtain the relative root square error RSE/ f = P MSE/ f. (We are regarding f as a function in [, P ]. In this eamle, since P = 1 and f = 1/3 (verify, the relative error for the aroimation S 3 that is shown above is See Eercise 13. MSE = f 1 1 1 π 3 1 / 1 1/3 =.77. Another Interesting Sum. As n, MSE aroaches. Consequently, the formula for MSE in Eamle 1 imlies that 1 = π 6. (1 11

Hidden Symmetry in a Waveform Surely you noticed that the a k coefficients in Eamle 4 are all zero (ecet for the first one, a. Could we have anticiated this? The answer is yes. Look once more at the grah of f on age 1 and observe that if it is shifted down.5 units, then it will have the symmetry of an odd function. In other words, the eriodic function g defined as g(t = f(t.5 is odd. Consequently, g has only sine terms in its Fourier series reresentation, and so does f. This will be referred to as hidden symmetry in a eriodic waveform. Hidden Symmetry: If a vertical shift makes a erodic function odd, then its Fourier series has no cosine terms, and visa versa. Be on the lookout for hidden symmetry in the eercises because if a waveform has it, then the work involved in the calculation of its coefficients is cut in half. Eamle 5. Obtain the Fourier series reresentation for the function of eriod P = π defined as f(t = sin(t for < t < π. See its grah below. 1..5 K K.5.5 t The function f is even. Because f is even, its series reresentation has only cosines (and a constant. The fact that P = π imlies that the fundamental circular frequency is ω = and the series has the form The k th cosine coefficient is f(t = a a k cos(kωt. a k = π π This integral can be evaluated using the trigonometric identity sin(t cos(kωt dt. sin(t cos(kωt = 1 ( sin((1 + kωt + sin((1 kωt. Substituting into the last integral, and integrating, yields a k = 1 ( cos((1 + kωt + π 1 + kω cos((1 kωt t=π. 1 kω t= This formula looks comlicated, but it simlifies rather easily. First of all, observe that since ω =, when t = π is substituted into the antiderivative both cos((1 + kπ and cos((1 kπ evaluate to 1. When t = is substituted, then the same cosine terms evaluate to 1. Therefore, a k simlifies to a k = 1 π ( 1 + kω + 1 kω = 4 π 1 1 4. Take the limit as k to obtain a = 4/π. This yields the following series reresentation. f(t = π + 4 cos(kt π 1 4 = π 4 ( cos(t + cos(4t π 3 15 + cos(6t 35 + Convergence is very fast. The following icture shows the grah of f along with lots of S 1 (t = π 4 3π cos(t and S (t = π 4 3π cos(t 4 15π cos(4t. 1

1..5 K K.5.5 t f and its Fourier aroimations S 1 (dotted and S (dashed. The One-Sided Amlitude and Power Sectrum The Fourier series reresentation of a eriodic function Equation (8 can be eressed in amlitude and haseangle form as follows. f(t = a A k cos(kωt δ k (11 Here, A k = a k + b k is the amlitude, and δ k = arctan(b k /a k is the hase angle, of the k th harmonic in the series (the formula for the hase angle assumes that the cosine coefficient a k is ositive. A lot of the oints (kω, A k, k =, 1,,... where A = a / is called the One-Sided Amlitude Sectrum of the signal f. It is customary to connect the sectral oints to the horizontal ais with vertical lines to obtain what is called the line sectrum of f. Consider, for eamle, the eriodic function f(t = t, < t < 1 (eriod P = 1 discussed in Eamle 4. In that eamle it was shown that f(t = 1/ 1/π sin(kωt/k where ω = π. Therefore, its one-sided amlitude sectrum consists of the oints (, 1/ and (πk, 1/kπ for k = 1,, 3,.... It is sketched below..5.4.3..1 4 6 8 1 1 14 16 18 The one-sided amlitude sectrum of the function in Eamle 4. The amlitude sectrum for f shows that there are significant contributions to its Fourier aroimations at (and beyond the 1 th harmonic ( π = 1 ω. Consequently, more than 1 harmonics will be needed to obtain a good Fourier aroimation. The One-Sided Power Sectrum is a lot of the oints (, a /4 and (kω, 1 A k, k = 1,, 3.... Its name derives from the fact that in many alications the ower content of the k th harmonic in the function f is roortional to the square of its amlitude. The one-sided ower sectrum of the function in Eamle 4 is shown below..5..15.1.5 4 6 8 1 1 14 16 18 The one-sided ower sectrum of the function in Eamle 4. Observe that almost all of the ower in this signal is concentrated in the constant term and the first three harmonics. This means that, for most alications, the comlete signal f(t can be relaced with its third Fourier aroimation S 3 (t with little or no effect on the analysis of the system. This is in site of the fact that the relative RSE error associated with S 3 is more than %. See the icture on age 11 and the calculation on age 11. 13