Fourier Series, Fourier Transforms, and Periodic Response to Periodic Forcing

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1 Fourier Series, Fourier ransforms, and Periodic Response to Periodic Forcing CEE 54. Structural Dynamics Department of Civil and Environmental Engineering Duke University Henri P. Gavin Fall, 26 his document describes methods to analyze the steady-state forced-response of single degree of freedom (SDOF) systems to general periodic loading. he analysis is carried out using Fourier series representation of the periodic external forcing and the resulting periodic steady-state response. Fourier Series for Real-Valued Functions Any real-valued function, f(t), that is: periodic, with period, = f(t 2 ) = f(t ) = f(t) = f(t + ) = f(t + 2 ) = square-integrable f 2 (t)dt <. may be represented as a series expansion of sines and cosines, in a Fourier series, ˆf(t; a, b) = 2 a + a q cos 2πqt + b q sin 2πqt, () where the Fourier coefficients, a q and b q are given by the Fourier integrals, a q = 2 b q = 2 f(t) cos 2πqt f(t) sin 2πqt dt, q =, 2,... (2) dt, q =, 2,... (3) he Fourier series ˆf(t; a, b) is a least-squares fit to the function f(t). his may be shown by first defining the error function, e(t) = f(t) ˆf(t; a, b), the quadratic error criterion, J(a, b) = e2 (t)dt, and finding the Fourier coefficients by solving the linear equations resulting from minimizing J(a, b) with respect to the coefficients a and b. (by setting J(a, b)/ a and J(a, b)/ b equal to zero). So doing, J(a, b) = = ( f(t) ˆf(t; a, b) ) 2 dt ( f 2 (t) 2f(t) ˆf(t; a, b) + ˆf 2 (t; a, b) ) dt. (4)

2 2 CEE 54. Structural Dynamics Duke University Fall 26 H.P. Gavin Setting J/ a equal to zero results in = f 2 (t) 2f(t) a ˆf(t; a, b) + ˆf 2 (t; a, b) ] dt = 2 f(t) ˆf(t; a, b)dt + 2 a = 2 f(t) 2 dt a + = f(t)dt + 2 a dt + = f(t)dt + 2 a + ˆf(t; a, b) a ˆf(t; a, b)dt a q cos 2πqt + a q cos 2πqt dt + b q a q + b q b q sin 2πqt 2 dt sin 2πqt dt From which, a = 2 f(t) dt. (5) Note here that (a /2) is the average value of f(t). Proceeding with the rest of the a coefficients, setting J(a, b)/ a k equal to zero results in = f 2 (t) 2f(t) a ˆf(t; a, b) + ˆf 2 (t; a, b) ] dt k = 2 f(t) ˆf(t; a, b)dt + 2 ˆf(t; a, b) ˆf(t; a, b)dt a k a k = f(t) cos 2πkt dt + a 2 + a q cos 2πqt + b q sin 2πqt = f(t) cos 2πkt dt + a cos 2πqt 2 dt + a q cos 2πqt 2πkt cos dt + b q = f(t) cos 2πkt dt + a 2 + a k cos 2 2πkt dt + b q = f(t) cos 2πkt dt + + a k 2 + sin 2πqt cos 2πkt dt 2πkt cos dt From which, a k = 2 f(t) cos 2πkt dt. (6) In a completely similar fashion, J(a, b)/ b k = results in b k = 2 f(t) sin 2πkt dt. (7)

3 Fourier Series and Periodic Response to Periodic Forcing 3 he derivation of the Fourier integrals (equations (5), (6), and (7)) make use of orthogonality properties of sine and cosine functions. ( ) nπt sin 2 dt = 2 ( ) nπt cos 2, n =, 2, dt =, n =, 2, 2 ( ) ( ) nπt mπt sin sin dt =, n m, n, m =, 2, ( ) ( ) nπt mπt cos cos dt =, n m, n, m =, 2, ( ) ( ) nπt mπt sin cos dt =, n m, n, m =, 2,. he complex exponential form of Fourier series Recall the trigonometric identities for complex exponentials, e iθ = cos θ+i sin θ. Defining the complex scalar F as F = (a ib), and its complex conjugate, F as F = (a + ib), it 2 2 is not hard to show that F e iθ + F e iθ = a cos θ + b sin θ. (8) Note that the left hand side of equation (8) is the sum of complex conjugates, and that the right hand side is the sum of real values. So complex conjugates can be used to express the real Fourier series given in equation (). Substituting equation (8) into equation (), ˆf(t; a, b) = 2 a + = 2 a + = 2 a + = 2 a + = F q a q cos 2πqt + b q sin 2πqt ] 2 (a q ib q ) exp i 2πq ] t + 2 (a q + ib q ) exp F q exp i 2πq ] t + F q exp i 2πq ] t exp i 2πq t ] = + Fq exp i 2πq ] t F q exp i 2πq ] t i 2πq t ]] F q e iωqt (9) where the Fourier frequency ω q is 2πq/, F = 2 a, F q = 2 (a q ib q ), and F q = F q. he condition F q = F q holds for the Fourier series of any real-valued function that is periodic

4 4 CEE 54. Structural Dynamics Duke University Fall 26 H.P. Gavin and integrable. Going the other way, ˆf(t; F) = q=+ = F + = F + = F + = F + = F + F q e iωqt q= q=+ F q e iωqt + F q e iωqt + q=+ q=+ F q e iωqt Fq e iωqt + F q e iωqt] F q e iωqt (F q + if q ) (cos ω q t + i sin ω q t) + (F q if q ) (cos ω q t i sin ω q t) ] 2F q cos ω q t 2F q sin ω q t ] So, the real part of F q, F q, is half of a q, the imaginary part of F q, F q, is half of b q, and F q = F q. he real Fourier coefficients, a q, are even about q = and the imaginary Fourier coefficients, b q, are odd about q =. Just as the Fourier expansion may be expressed in terms of complex exponentials, the coefficients F q may also be written in this form. F q = 2 (a q ib q ) = = f(t) f(t) exp cos 2πq t i sin 2πq t ] i 2πq t ] dt () Since the integrand, f(t)e iωqt, of the Fourier integral is periodic in, the limits of integration can be shifted arbitrarily without affecting the resulting Fourier coefficients. f(t)e iωqt dt = τ f(t)e iωqt dt + But, since the integrand is periodic in, τ f(t)e iωqt dt = +τ τ +τ So, the interval of integration can be shifted arbitrarily. f(t)e iωqt dt = +τ +τ f(t)e iωqt dt f(t)e iωqt dt. +τ f(t)e iωqt dt. f(t)e iωqt dt. () On the other hand, shifting the function in time, affects the relative values of a q and b q (i.e., the phase the the complex coefficeint F q ), but does not affect the magnitude, F q = a 2 q + b 2 q. If ˆf(t) = Fq e iωqt, then ˆf(t + τ) = F q e iωq(t+τ) = (F q e iωqτ ) e iωqt, and 2 F q = F q e iωqτ.

5 Fourier Series and Periodic Response to Periodic Forcing 5 2 Fourier Integrals in Maple he Fourier integrals for real valued functions (equations (6) and (7)) can be evaluated using symbolic math software, such as Maple or Mathematica. 2. a periodic square wave function: f(t) = sgn(t) on π < t < π and f(t) = f(t + n(2π)) > assume (k::integer); > f := signum(t); > ak := (2/(2*Pi)) * int(f*cos(2*pi*k*t/(2*pi)),t=-pi..pi); ak := > bk := (2/(2*Pi)) * int(f*sin(2*pi*k*t/(2*pi)),t=-pi..pi); k 2 ((-) - ) bk := Pi k 2.2 a periodic sawtooth function: f(t) = t on π < t < π and f(t) = f(t + n(2π)) > assume (k::integer); > f := t; > ak := (2/(2*Pi)) * int(f*cos(2*pi*k*t/(2*pi)),t=-pi..pi); ak := > bk := (2/(2*Pi)) * int(f*sin(2*pi*k*t/(2*pi)),t=-pi..pi); ( + k ) 2 (-) bk := k 2.3 a periodic triangle function: f(t) = π/2 t sgn(t) on π < t < π and f(t) = f(t + n(2π)) > assume (k::integer); > f := Pi/2 - t * signum(t); > ak := (2/(2*Pi)) * int(f*cos(2*pi*k*t/(2*pi)),t=-pi..pi); k 2 ((-) - ) ak := Pi k > bk := (2/(2*Pi)) * int(f*sin(2*pi*k*t/(2*pi)),t=-pi..pi); bk :=

6 6 CEE 54. Structural Dynamics Duke University Fall 26 H.P. Gavin 3 Fourier ransforms Recall for periodic functions of period,, the Fourier series expansion may be written f(t) = q= F q e iωqt, (2) where the Fourier coefficients, F q, have the same units as f(t), and are given by the Fourier integral, F q = /2 /2 in which the limits of integration have been shifted by τ = /2. f(t) e iωqt dt, (3) Now, consider a change of variables, by defining a frequency increment, ω, and a scaled amplitude, F (ω q ). ω = ω = 2π (ω q = q ω) (4) F (ω q ) = F q = 2π ω F q (5) Where the scaled amplitude, F (ω q ), has units of f(t) t or f(t)/ω. Using these new variables, f(t) = 2π F (ω q ) = q= /2 /2 Finally, taking the limit as, implies ω dω and f(t) = 2π F (ω) = F (ω q ) e iωqt ω, (6) f(t) e iωqt dt. (7) F (ω) e iωt dω, (8) f(t) e iωt dt. (9) hese expressions are the famous Fourier transform pair. Equation (8) is commonly called the inverse Fourier transform and equation (9) is commonly called the forward Fourier transform. hey differ only by the sign of the exponent and the factor of 2π.

7 Fourier Series and Periodic Response to Periodic Forcing 7 4 Fourier Approximation Any periodic function may be approximated as a truncated series expansion of sines and cosines, as a Fourier series, f(t) = 2 a + Q a q cos 2πqt + Q b q sin 2πqt, (2) where the Fourier coefficients, a q and b q may be found by solving the Fourier integrals, a q = 2 b q = 2 /2 /2 /2 /2 f(t) cos 2πqt f(t) sin 2πqt dt, q =, 2,..., Q (2) dt, q =, 2,..., Q (22) he Fourier approximation (2) may also be represented using complex exponential notation. q=q f(t) = F q e iωqt = q= Q q=q q= Q F (ω q ) e iωqt (23) where e iωqt = cos ω q t + i sin ω q t, ω q = 2πq/, F q = 2 (a q ib q ), F q = F q, and F q = /2 /2 f(t) e iωqt dt. (24) he accuracy of Fourier approximations of non-sinusoidal functions increases with the number of terms, Q, in the series. Especially for time series that are discontinuous in time, such as square-waves or saw-tooth waves, Fourier approximations, f(t), will contain a degree of over-shoot at the discontinuity and will oscillate about the approximated function f(t). his is called Gibb s phenomenon. In section 7 we illustrate this effect for square waves and triangle waves.

8 8 CEE 54. Structural Dynamics Duke University Fall 26 H.P. Gavin 5 Discrete Fourier ransform If the function f(t) is represented by N uniformly-spaced points in time, t n = (n)( t), (n =,, N ), the N Fourier coefficients are computed by the discrete Fourier transform (DF), F q = N N n= f(t n ) exp i 2πqn N ], (25) he frequency increment f in the DF is /(N t) Hertz, and the highest frequency is /(2 t) Hertz, the so-called Nyquist frequency. Note that the exponentials in the DF, (25), are periodic in N points. exp i 2πqn ] N = exp i 2πqn ] N + 2π = exp i ] 2π(q N)n N Further, referring to equation (), as long as the set of frequency indices, q, contain N consecutive points, the indices may be shifted arbitrarily. q { N/2 +,,,,, 2,, N/2, N/2 } q {,, N/2 2, N/2, N/2, N/2 +,, 2, } q {,, N/2 2, N/2, N/2, N/2 +,, N 2, N } he N-point discrete Fourier approximation is computed by the inverse discrete Fourier transform, N/2 f(t n ) = F q exp i 2πqn ] N = F q exp i 2πqn ] (26) N N q= N/2+ With the normalization given in equations (25) and (26), the mean square in the time domain is the sum-square of the Fourier amplitudes. N fn 2 = N n= N q= q= F q 2, he number of operations required to compute a DF (equation (25)) increases with N 2. he fast Fourier transform (FF) algorithm computes the DF by simply scaling and re-ordering the sequence f n. he number of operations required to compute a DF using the FF algorithm increases with N log 2 (N): if it takes minute to compute an FF with N = 6, it would take almost 35 days to carry out the same DF using equation (25)! By convention, the fast Fourier transform (FF) computes the DF without the (/N) factor. Also, by convention, the Fourier frequencies, ω q = 2πq/ are sorted as follows: q {,,, N/2, N/2 +,, 2, }. Heideman, Michael.; Johnson, Don H.; Burrus, C. Sidney (985-9-). Gauss and the history of the fast Fourier transform, Archive for History of Exact Sciences, 34(3): doi:.7/bf34843

9 Fourier Series and Periodic Response to Periodic Forcing 9 6 Fourier Series and Periodic Responses of Dynamic Systems he response of a system described by a frequency response function H(ω) to arbitrary periodic forces described by a Fourier series may be found in the frequency domain, or in the time domain. Complex exponential notation allows us to directly determine the steady-state periodic response to general periodic forcing, in terms of both the magnitude of the response and the phase of the response. Recall the relationship between the complex magnitudes X and F for a sinusoidally-driven spring-mass-damper oscillator is X q = H(ω q ) F q = (k mω 2 q) + i(cω q ) F q (27) he function H(ω) is called the frequency response function for the dynamic system relating the input f(t) to the output x(t). Because the oscillator is linear, if the response to f (t) is x (t), and the response to f 2 (t) is x 2 (t), then the response to c f (t) + c 2 f 2 (t) is c x (t) + c 2 x 2 (t). More generally, then, x(t) = q= H(ω q ) F q e iωqt = q= (k mω 2 q) + i(cω q ) F q e iωqt (28) Equivalently, the periodic response can be expressed in terms of the Fourier series coefficients a q and b q (equation ()), x(t) = 2 a /k + = 2 a /k + /k ( Ω 2 q) 2 + (2ζΩ q ) 2 ( aq ( Ω 2 q) b q (2ζΩ q ) ) cos ω q t + ( a q (2ζΩ q ) + b q ( Ω 2 q) ) sin ω q t ] /k ( Ω 2 q ) 2 + (2ζΩ q ) 2 a q cos(ω q t + θ q ) + b q sin(ω q t + θ q ) ] (29) where ω q = 2πq/, Ω q = ω q /ω n, and tan θ q = 2ζΩ q /( Ω 2 q). Note that ω q is not the same symbol as ω n. he natural frequency, k/m, has the symbol ω n. he symbol ω q = 2πq/ is the frequency of a Fourier component of a periodic forcing function of period. For this frequency response function H(ω), the series expansion for the response x(t) converges with fewer terms than the Fourier series for the external forcing, because H(ω q ) decreases as /ω 2 q.

10 CEE 54. Structural Dynamics Duke University Fall 26 H.P. Gavin 7 Examples In the following examples, the external forcing is periodic with a period,, of 2π (seconds). For functions periodic in 2π seconds, the frequency increment in the Fourier series is radian/second. In this case, the frequency index number, q, is also the frequency, ω q, in radians per second, otherwise, ω q = 2πq/. he system is a forced spring-mass-damper oscillator, ẍ(t) + 2ζω n ẋ(t) + ω 2 nx(t) = m f ext (t). (3) he complex-valued frequency response function from the external force, f ext (t), to the response displacement, x(t), is given by H(ω) = where Ω is the frequency ratio, ω/ω n, /ω 2 n Ω 2 + i 2ζΩ, (3) In the following three numerical examples, ω n = rad/s, ζ =., m = kg, and Q = 6 terms. he three examples consider external forcing in the form of a square-wave, a sawtooth-wave, and a triangle-wave. In each example six plots are provided. In the (a) plots, the solid line represents the exact form of f(t), the dashed lines represent the real-valued form of the Fourier approximation and the complex-valued form of the Fourier approximation, and the circles represent 2Q sample points of the function f(t) for use in fast Fourier transform (FF) computations. he two dashed lines are exactly equal. In the (b) plots, the impulse-lines show the values of the Fourier coefficients, F q, found by evaluating the Fourier integral, equation (24), and the circles represent the Fourier coefficients returned by the FF. In the (c) plots, the red solid lines show the cosine terms of the Fourier series and the blue dashed lines show the sine terms of the Fourier series. he (d) and (f) plots show the magnitude and phase of the transfer function, H(ω) as a solid line, the Fourier coefficients, F q, as the green circles, and H(ω q )F q as the blue circles. In the (e) plots, the solid line represents x(t) as computed using equation (28), the circles represent the real part of the result of the inverse FF calculation and the dots represent the imaginary part of the result of the inverse FF calculation. he numerical details involved in the correct representation of periodic functions and frequency indexing for FF computations are provided in the attached Matlab code.

11 Fourier Series and Periodic Response to Periodic Forcing 7. Example : square wave he external forcing is given by f ext (t) = sgn(t) ; π < t < π (32) he Fourier coefficients for the real-valued Fourier series are: a q = and b q = (2/(qπ)) ( q ) function, f(t), (N) Fourier coefficients, F q, (N) (a) (b) cosine and sine terms frequency response.. - (c) (d) e response, x, (m) phase lead, (degrees) (e) (f) Figure. (a): = f ext (t); - - = f ext (t). (b): red solid = R(F q ); blue dashed = I(F q ); *: F q from FF. (c): red solid = component cosine terms, blue dashed = component sine terms. (d): : H(ω) ; *: F q and H(ω q )F q. (e): : x(t); *: x(t) from IFF; : I(x(t)) from IFF. (f): : phase of H(ω); *: phase of F q and H(ω q )F q.

12 2 CEE 54. Structural Dynamics Duke University Fall 26 H.P. Gavin 7.2 Example 2: sawtooth wave he external forcing is given by he Fourier coefficients for the real-valued Fourier series are: a q = and b q = (2/q) ( q ). f ext (t) = t ; π < t < π (33) function, f(t), (N) Fourier coefficients, F q, (N) (a) (b) cosine and sine terms frequency response (c) (d) e response, x, (m) phase lead, (degrees) (e) (f) Figure 2. (a): = f ext (t); - - = f ext (t). (b): red solid = R(F q ); blue dashed = I(F q ); *: F q from FF. (c): red solid = component cosine terms, blue dashed = component sine terms. (d): : H(ω) ; *: F q and H(ω q )F q. (e): : x(t); *: x(t) from IFF; : I(x(t)) from IFF. (f): : phase of H(ω); *: phase of F q and H(ω q )F q.

13 Fourier Series and Periodic Response to Periodic Forcing Example 3: triangle wave he external forcing is given by f ext (t) = π/2 t sgn(t) ; π < t < π (34) he Fourier coefficients for the real-valued Fourier series are: a q = (2/q 2 π)( q ) and b q = function, f(t), (N) Fourier coefficients, F q, (N) (a) (b) cosine and sine terms frequency response.. - e-4 (c) (d) e response, x, (m) phase lead, (degrees) (e) (f) Figure 3. (a): = f ext (t); - - = f ext (t). (b): red solid = R(F q ); blue dashed = I(F q ); *: F q from FF. (c): red solid = component cosine terms, blue dashed = component sine terms. (d): : H(ω) ; *: F q and H(ω q )F q. (e): : x(t); *: x(t) from IFF; : I(x(t)) from IFF. (f): : phase of H(ω); *: phase of F q and H(ω q )F q.

14 4 CEE 54. Structural Dynamics Duke University Fall 26 H.P. Gavin 7.4 Matlab code function Fq,wq,x,t] = Fourier (type,q,wn,z) 2 % F, w, x ] = Fourier ( type,q, wn, z ) 3 % Compute t h e Fourier s e r i e s c o e f f i c i e n t s o f a p e r i o d i c s i g n a l, f ( t ), 4 % in two d i f f e r e n t ways : 5 % complex e x p o n e n t i a l expansion ( i. e., s i n e and c o s i n e expansion ) 6 % f a s t Fourier transform 7 % 8 % he type o f p e r i o d i c s i g n a l may be 9 % square a square wave % sawtooth a saw t o o t h wave % t r i a n g l e a t r i a n g l e wave 2 % 3 % he p e r i o d o f t h e s i g n a l, f ( t ), i s f i x e d at 2 p i ( second ). 4 % 5 % he number o f Fourier s e r i e s c o e f f i c i e n t s i s i n p u t as Q. 6 % his r e s u l t s in a 2 Q Fourier transform c o e f f i c i e n t s. 7 % 8 % he Fourier approximation i s then used to compute t h e steady s t a t e 9 % response o f a s i n g l e degree o f freedom (SDOF) o s c i l l a t o r, d e s c r i b e d by 2 % 2 % 2 22 % x ( t ) + 2 z wn x ( t ) + wn x ( t ) = f ( t ), 23 % 24 % where t h e mass o f t h e system i s f i x e d at ( kg ). 25 % 26 % see : h t t p :// en. w i k i p e d i a. org / w i k i / F o u r i e r s e r i e s 27 % h t t p ://www. jhu. edu / s i g n a l s / f o u r i e r 2 / index. html 28 % 29 3 i f nargin < 4 3 help Fourier 32 return 33 end = 2* pi ; % p e r i o d o f e x t e r n a l f o r c i n g, s a = zeros (,Q); % c o e f f i c i e n t s o f c o s i n e p a r t o f Fourier s e r i e s 39 b = zeros (,Q); % c o e f f i c i e n t s o f s i n e p a r t o f Fourier s e r i e s 4 4 q = : Q]; % p o s i t i v e f r e q u e n c y index v a l u e 42 wq = 2* pi *q/; % p o s i t i v e f r e q u e n c y v a l u e s, rad / sec, eq n (4) dt = /(2* Q); % 2Q d i s c r e t e p o i n t s in time f o r FF sampling 45 ts = -Q:Q -]* dt; % 2Q d i s c r e t e p o i n t s in time f o r FF sampling % s h i f t e d time... /2 <= t < / % t s () = Q dt = /2 = p i... f ( /2) = f (/2) by e x t e n s i o n 49 % t s (Q) = dt = /(2 Q) = p i /Q 5 % t s (Q+) =... f () = f () by e x t e n s i o n 5 % t s (Q+2) = dt = /(2 Q) = p i /Q 52 % t s (2 Q) = (Q ) dt = /2 dt = p i p i /Q P = 28; % number o f p o i n t s in t h e time record ( f o r p l o t t i n g purposes only ) 55 t = -P:P]*/2/ P; % time a x i s % ======================================================================= 58 % Fourier s e r i e s approximation o f g e n e r a l p e r i o d i c f u n c t i o n s 59 6 % s i n ( q p i ) =, cos ( q p i ) = ( ).ˆ q 6 % f t r u e i s used only f o r p l o t t i n g 62 % f samp i s used in FF computations i f strcmp( type, square ) % square wave 66 b = - 2./( q* pi ).* (( -).ˆ q - ); 67 f_true = sign ( t); f_true () = ; f_true (2* P +) = ; 68 f_samp = sign ( ts ); 69 f_samp () = ; % g e t p e r i o d i c i t y r i g h t, f ( p i ) = 7 end

15 Fourier Series and Periodic Response to Periodic Forcing i f strcmp( type, sawtooth ) % sawtooth wave 73 b = -(2./ q).* ( -).ˆ q; 74 f_true = t; f_true () = ; f_true (2* P +) = ; 75 f_samp = ts; 76 f_samp () = ; % g e t p e r i o d i c i t y r i g h t, f ( p i ) = 77 end i f strcmp( type, triangle ) % t r i a n g l e wave 8 a = -(2./( pi *q.ˆ2)).* ( ( -).ˆ q - ); 8 f_true = pi /2 - t.* sign ( t); 82 f_samp = pi /2 - ts.* sign (ts ); 83 end f_approxr = a * cos(q *t) + b * sin (q *t); % r e a l Fourier s e r i e s Fq = (a - i*b )/2; % complex Fourier c o e f f i c i e n t s 89 9 % Complex Fourier s e r i e s ( p o s i t i v e and n e g a t i v e exponents ) 9 f_approxc = Fq * exp( i* wq * t) + conj( Fq) * exp(- i* wq * t); % he imagniary p a r t o f t h e complex Fourier s e r i e s i s e x a c t l y zero! 95 imag_over_real_ = max(abs(imag( f_approxc ))) / max(abs( real ( f_approxc ))) % he f a s t Fourier transform (FF) method 99 % In Matlab, t h e forward Fourier transform has a n e g a t i v e exponent. % According to a convention o f t h e FF method, index number i s f o r time = 2 F_FF = f f t ( f_samp ( Q +:2* Q, :Q ] ) ) / (2* Q); % Fourier c o e f f s % c o e f = Fq F FF ( 2 :Q+) ] % d i s p l a y t h e Fourier c o e f f i c i e n t s 6 7 % P l o t t i n g 8 9 figure (); plot (t,f_true, -r, ts, f_samp, *r, t, f_approxr, -c, t, f_approxc, -b ) xlabel ( time, t, (s) ) 2 ylabel ( function, f(t), (N) ) 3 axis (.5* - pi, pi ]) 4 legend( true, sampled, real Fourier approx, complex Fourier approx ) 5 plot (wq, real (Fq), ˆr, wq,imag(fq), ˆb,... 6 wq, real ( F_FF (2: Q +)), *r, wq,imag( F_FF (2: Q +)), *b ) 7 ylabel ( Fourier coefficients, F( w), ( N) ) 8 xlabel ( frequency, {/ Symbol w}, ( rad /s) ) 9 legend( real (F), imag (F), real ( FF (f)), imag ( FF (f)) ) figure (2); 23 plot (t,a *ones(,2* P +).* cos(q *t), -r, t,b *ones(,2* P +).* sin (q *t), -b ); 24 xlabel ( time, t, (s) ) 25 ylabel ( cosine and sine terms ) 26 axis (.5* - pi, pi ]) % ======================================================================= 3 % Steady s t a t e response o f a SDOF o s c i l l a t o r to g e n e r a l p e r i o d i c f o r c i n g % Complex v a l u e d f r e q u e n c y response f u n c t i o n, H(w), has u n i t s o f m/n] 33 H = (/ wn ˆ2)./ ( - (wq/wn ).ˆ2 + 2*i*z*( wq/wn) ); % Complex Fourier s e r i e s ( p o s i t i v e and n e g a t i v e exponents ) 36 x_approxc = (H.* Fq) * exp(i*wq *t) + (conj(h).* conj(fq )) * exp(-i*wq *t); % In Matlab, t h e i n v e r s e Fourier transform has a p o s i t i v e exponent. 39 %... see t h e t a b l e at t h e end o f t h i s f i l e f o r t h e s o r t i n g convention. 4 x_ff = i f f t ( (/ wn ˆ2), H, conj(h(q -: -:)) ].* F_FF ) * (2* Q); 4

16 6 CEE 54. Structural Dynamics Duke University Fall 26 H.P. Gavin 42 figure (3); % p l o t f r e q u e n c y response f u n c t i o n s 43 semilogy(wq,abs(fq), *g, wq,abs(h), -r, wq,abs(h.* Fq), *b ) 44 xlabel ( frequency, {/ Symbol w}, ( rad /s) ) 45 ylabel ( frequency response ) 46 legend( F, H, H*F ) plot (wq,angle(fq )*8/ pi, *g, wq,angle(h )*8/ pi, -r, wq,angle(h.* Fq )*8/ pi, *b ) 49 xlabel ( frequency, {/ Symbol w}, ( rad /s) ) 5 ylabel ( phase lead, ( degrees ) ) 5 52 figure (4); % p l o t time response 53 ts = ts(q +:2* Q) ts (: Q) ]; % re s o r t t h e time a x i s 54 plot ( t, x_approxc, -r, ts, real ( x_ff ), *b, ts, imag( x_ff ),.b ) 55 axis (.5* - pi, pi ]) 56 xlabel ( time, t, (s) ) 57 ylabel ( response, x, (m) ) 58 legend( x(t) - complex Fourier series, real (x(t)) - FF, imag (x(t)) - FF ) % he imaginary p a r t i s e x a c t l y zero f o r t h e complex exponent s e r i e s method! 62 imag_over_real_2 = max(abs(imag( x_approxc ))) / max(abs( real ( x_approxc ))) % he imaginary p a r t i s p r a c t i c a l l y zero f o r t h e FF method! 65 imag_over_real_3 = max(abs(imag( x_ff ))) / max(abs( real ( x_ff ))) % 68 % SORING o f t h e FF c o e f f i c i e n t s number o f data p o i n t s = N = 2 Q 69 % note : f max = /(2 dt ) ; d f = f max /(N/2) = /(N dt ) = /; 7 % 7 % 72 % index time f r e q u e n c y 73 % ===== ==== ========= 74 % imag (F)= 75 % 2 dt d f 76 % 3 2 dt 2 d f 77 % 4 3 dt 3 d f 78 % : : : 79 % N/2 (N/2 2) dt f max 2 d f = ( N/2 2) d f 8 % N/2 (N/2 ) dt f max d f = ( N/2 ) d f 8 % N/2+ (N/2) dt +/ f max = ( N/2 ) d f imag (F)= 82 % N/2+2 (N/2+) dt f max+d f = ( N/2+) d f 83 % N/2+3 (N/2+2) dt f max+2 d f = ( N/2+2) d f 84 % : : : : 85 % N 2 (N 3) dt 3 d f 86 % N (N 2) dt 2 d f 87 % N (N ) dt d f 88 %