Fourier

Fourier Series, Fourier Transforms, andPeriodic Response to Periodic Forcing

CEE 541. Structural DynamicsDepartment of Civil and Environmental Engineering

Duke University

Henri P. GavinFall, 2014

This document describes methods to analyze the steady-state forced-response of singledegree of freedom (SDOF) systems to general periodic loading. The analysis is carried outusing Fourier series representation of the periodic external forcing and the resulting periodicsteady-state response.

1 Fourier Series for Real-Valued Functions

Any real-valued function, f(t), that is:

periodic, with period T ,

= f(t 2T ) = f(t T ) = f(t) = f(t+ T ) = f(t+ 2T ) = square-integrable T

0f 2(t)dt

2 CEE 541. Structural Dynamics Duke University Fall 2014 H.P. Gavin

Setting J/a0 equal to zero results in

0 = T

0

a0

[f 2(t) 2f(t)f(t; a,b) + f 2(t; a,b)

]dt

= 2 T

0f(t)

a0f(t; a,b)dt+ 2

T0f(t; a,b)

a0f(t; a,b)dt

= 2 T

0f(t) 12dt+ 2

T0

12a0 +

q=1

aq cos2piqtT

+q=1

bq sin2piqtT

12dt=

T0f(t)dt+

T0

12a0dt+

q=1

aq

T0

cos 2piqtT

dt+q=1

bq

T0

sin 2piqtT

dt

= T

0f(t)dt+ 12a0T +

q=1

aq 0 +q=1

bq 0

From which,

a0 =2T

T0f(t) dt. (5)

Note here that (a0/2) is the average value of f(t).Proceeding with the rest of the a coefficients, setting J(a,b)/ak equal to zero results in

0 = T

0

ak

[f 2(t) 2f(t)f(t; a,b) + f 2(t; a,b)

]dt

= 2 T

0f(t)

akf(t; a,b)dt+ 2

T0f(t; a,b)

akf(t; a,b)dt

= T

0f(t) cos 2pikt

Tdt+

T0

a02 +

q=1

aq cos2piqtT

+q=1

bq sin2piqtT

cos 2piktT

dt

= T

0f(t) cos 2pikt

Tdt+

a02

T0

cos 2piqtT

dt+q=1

aq

T0

cos 2piqtT

cos 2piktT

dt+q=1

bq

T0

sin 2piqtT

cos 2piktT

dt

= T

0f(t) cos 2pikt

Tdt+ a02 0 + ak

T0

cos2 2piktT

dt+q=1

bq 0

= T

0f(t) cos 2pikt

Tdt+ 0 + ak T2 + 0

From which,

ak =2T

T0f(t) cos 2pikt

Tdt. (6)

In a completely similar fashion, J(a,b)/bk = 0 results in

bk =2T

T0f(t) sin 2pikt

Tdt. (7)

CC BY-NC-ND H.P. Gavin

Fourier Series and Periodic Response to Periodic Forcing 3

The derivation of the Fourier integrals (equations (5), (6), and (7)) make use of orthogonalityproperties of sine and cosine functions.

T0

sin2(npit

T

)dt = T2 , n = 1, 2, T

0cos2

(npit

T

)dt = T2 , n = 1, 2, T

0sin

(npit

T

)sin

(mpit

T

)dt = 0 , n 6= m , n,m = 1, 2, T

0cos

(npit

T

)cos

(mpit

T

)dt = 0 , n 6= m , n,m = 1, 2, T

0sin

(npit

T

)cos

(mpit

T

)dt = 0 , n 6= m , n,m = 1, 2,

1.1 The complex exponential form of Fourier series

Recall the trigonometric identities for complex exponentials, ei = cos +i sin . Definingthe complex scalar F as F = 12(a ib), and its complex conjugate, F as F = 12(a + ib), itis not hard to show that

Fei + F ei = a cos + b sin . (8)

Note that the left hand side of equation (8) is the sum of complex conjugates, and that theright hand side is the sum of real values. So complex conjugates can be used to express thereal Fourier series given in equation (1). Substituting equation (8) into equation (1),

f(t; a,b) = 12a0 +q=1

[aq cos

2piqtT

+ bq sin2piqtT

]

= 12a0 +q=1

[12(aq ibq) exp

[i2piqTt]

+ 12(aq + ibq) exp[i2piq

Tt]]

= 12a0 +q=1

Fq exp[i2piqTt]

+q=1

F q exp[i2piq

Tt]

= 12a0 +q=1

Fq exp[i2piqTt]

+1

q=F q exp

[i2piqTt]

=

q=Fq exp

[i2piqTt]

=

q=Fq e

iqt (9)

where the Fourier frequency q is 2piq/T , F0 = 12a0, Fq =12(aq ibq), and F q = Fq. The

condition F q = Fq holds for the Fourier series of any real-valued function that is periodic



and integrable. Going the other way,

f(t;F) =q=+q=

Fq eiqt

= F0 +q=1q=

Fq eiqt +

q=+q=1

Fq eiqt

= F0 +q=+q=1

Fqeiqt +q=+q=1

Fqeiqt

= F0 +q=1

[Fq e

iqt + Fq eiqt]

= F0 +q=1

[(F q + iF q ) (cosqt+ i sinqt) + (F q iF q ) (cosqt i sinqt)

]

= F0 +q=1

[2F q cosqt 2F q sinqt

]

So, the real part of Fq, F q, is half of aq, the imaginary part of Fq, F q , is half of bq, andFq = F q. The real Fourier coefficients, aq, are even about q = 0 and the imaginary Fouriercoefficients, bq, are odd about q = 0.

Just as the Fourier expansion may be expressed in terms of complex exponentials, thecoefficients Fq may also be written in this form.

Fq =12 (aq ibq) =

1T

T0f(t)

[cos 2piq

Tt i sin 2piq

Tt]

= 1T

T0f(t) exp

[i2piq

Tt]dt (10)

Since the integrand, f(t)eiqt, of the Fourier integral is periodic in T , the limits of integrationcan be shifted arbitrarily without affecting the resulting Fourier coefficients. T

0f(t)eiqt dt =

0f(t)eiqt dt+

T+

f(t)eiqt dt T+T

f(t)eiqt dt.

But, since the integrand is periodic in T , 0f(t)eiqt dt =

T+T

f(t)eiqt dt .

So, the interval of integration can be shifted arbitrarily. T0f(t)eiqt dt =

T+0+

f(t)eiqt dt.

On the other hand, shifting the function in time, affects the relative values of aq and bq(i.e., the phase the the complex coefficeint Fq), but does not affect the magnitude, |Fq| =12

a2q + b2q. If f(t) =

Fqe

iqt, then f(t + ) = Fqeiq(t+) = (Fqeiq ) eiqt, and|Fq| = |Fqeiq |.



2 Fourier Integrals in Maple

The Fourier integrals for real valued functions (equations (6) and (7)) can be evaluatedusing symbolic math software, such as Maple or Mathematica.

2.1 a periodic square wave function: f(t) = sgn(t) on pi < t < pi and f(t) = f(t+ n(2pi))

> assume (k::integer);> f := signum(t);> ak := (2/(2*Pi)) * int(f*cos(2*Pi*k*t/(2*Pi)),t=-Pi..Pi);

ak := 0

> bk := (2/(2*Pi)) * int(f*sin(2*Pi*k*t/(2*Pi)),t=-Pi..Pi);k

2 ((-1) - 1)bk := - --------------

Pi k

2.2 a periodic sawtooth function: f(t) = t on pi < t < pi and f(t) = f(t+ n(2pi))

> assume (k::integer);> f := t;> ak := (2/(2*Pi)) * int(f*cos(2*Pi*k*t/(2*Pi)),t=-Pi..Pi);

ak := 0

> bk := (2/(2*Pi)) * int(f*sin(2*Pi*k*t/(2*Pi)),t=-Pi..Pi);(1 + k)

2 (-1)bk := --------------

k

2.3 a periodic triangle function: f(t) = pi/2 t sgn(t) on pi < t < pi and f(t) = f(t+n(2pi))

> 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);

k2 ((-1) - 1)

ak := - --------------2

Pi k

> bk := (2/(2*Pi)) * int(f*sin(2*Pi*k*t/(2*Pi)),t=-Pi..Pi);bk := 0



3 Fourier Transforms

Recall for periodic functions of period, T , the Fourier series expansion may be written

f(t) =q=q=

Fq eiqt , (11)

where the Fourier coefficients, Fq, have the same units as f(t), and are given by the Fourierintegral,

Fq =1T

T/2T/2

f(t) eiqt dt , (12)

in which the limits of integration have been shifted by = T/2.

Now, consider a change of variables, by defining a frequency increment, , and a scaledamplitude, F (q).

= 1 =2piT

(q = q ) (13)

F (q) = TFq =2piFq (14)

Where the scaled amplitude, F (q), has units of f(t) t or f(t)/.

Using these new variables,

f(t) = 12pi

q=q=

F (q) eiqt , (15)

F (q) = T/2T/2

f(t) eiqt dt . (16)

Finally, taking the limit as T , implies d and f(t) = 12pi

F () eit d , (17)

F () =

f(t) eit dt . (18)

These expressions are the famous Fourier transform pair. Equation (17) is commonly calledthe inverse Fourier transform and equation (18) is commonly called the forward Fouriertransform. They differ only by the sign of the exponent and the factor of 2pi.



4 Fourier Approximation

Any periodic function may be approximated as a truncated series expansion of sines andcosines, as a Fourier series,

f(t) = 12a0 +Qq=1

aq cos2piqtT

+Qq=1

bq sin2piqtT

, (19)

where the Fourier coefficients, aq and bq may be found by solving the Fourier integrals,

aq =2T

T/2T/2

f(t) cos 2piqtT

dt , q = 1, 2, . . . , Q (20)

bq =2T

T/2T/2

f(t) sin 2piqtT

dt , q = 1, 2, . . . , Q (21)

The Fourier approximation (19) may also be represented using complex exponential notation.

f(t) =q=Qq=Q

Fq eiqt = 1

T

q=Qq=Q

F (q) eiqt (22)

where eiqt = cosqt+ i sinqt, q = 2piq/T , Fq = 12(aq ibq), Fq = F q, and

Fq =1T

T/2T/2

f(t) eiqt dt . (23)

The accuracy of Fourier approximations of non-sinusoidal functions increases with thenumber 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 degreeof over-shoot at the discontinuity and will oscillate about the approximated function f(t).This is called Gibbs phenomenon. In section 7 we illustrate this effect for square waves andtriangle waves.

5 Discrete Fourier Transform

If the function f(t) is represented by N uniformly-spaced points in time, tn = (n)(t),(n = 0, , N1), the N Fourier coefficients are computed by the discrete Fourier transform,

Fq =1N

N1n=0

f(tn) exp[i2piqn

N

], q = [N/2 + 1, ,2,1, 0, 1, , N/2] (24)

and the N -point discrete Fourier approximation is computed by the inverse discrete Fouriertransform,

f(tn) =N/2

q=N/2+1Fq exp

[i2piqnN

](25)

With the normalization given in equations (24) and (25), 1N

f 2n =

|Fq|2. By convention,in the fast Fourier transform (FFT) the factor (1/N) is applied to equation (25) instead ofequation (24). Also, by convention, the Fourier frequencies, q = 2piq/T are sorted as follows:q = 0, 1, , N/2,N/2 + 1, ,2,1. The lowest positve frequency in the FFT is 1/THertz, and the highest frequency is 1/(2t) Hertz, which is called the Nyquist frequency.



6 Fourier Series and Periodic Responses of Dynamic Systems

The response of a system described by a frequency response function H() to arbitraryperiodic forces described by a Fourier series may be found in the frequency domain, or in thetime domain.

Complex exponential notation allows us to directly determine the steady-state periodicresponse to general periodic forcing, in terms of both the magnitude of the response and thephase of the response. Recall the relationship between the complex magnitudes X and F fora sinusoidally-driven spring-mass-damper oscillator is

Xq = H(q) Fq =1

(k m2q ) + i(cq)Fq (26)

The function H() is called the frequency response function for the dynamic system relatingthe input f(t) to the output x(t).

Because the oscillator is linear, if the response to f1(t) is x1(t), and the response to f2(t)is x2(t), then the response to c1f1(t) + c2f2(t) is c1x1(t) + c2x2(t). More generally, then,

x(t) =q=q=

H(q) Fq eiqt =q=q=

1(k m2q ) + i(cq)

Fq eiqt (27)

where q = 2piq/T .

Note that q is not the same symbol as n. The natural frequency,k/m, has the symbol

n. The symbol q = 2piq/T is the frequency of a component of a periodic forcing functionof period T .

For this frequency response function H(), the series expansion for the response x(t)converges with fewer terms than the Fourier series for the external forcing, because |H(q)|decreases as 1/2q .



7 Examples

In the following examples, the external forcing is periodic with a period, T , of 2pi (sec-onds). For functions periodic in 2pi seconds, the frequency increment in the Fourier series is1 radian/second. In this case, the frequency index number, q, is also the frequency, q, inradians per second, otherwise, q = 2piq/T .

The system is a forced spring-mass-damper oscillator,

x(t) + 2nx(t) + 2nx(t) =1mf ext(t) . (28)

The complex-valued frequency response function from the external force, f ext(t), to the re-sponse displacement, x(t), is given by

H() = 1/2n

1 2 + i 2 , (29)

where is the frequency ratio, /n,

In the following three numerical examples, n = 10 rad/s, = 0.1, m = 1 kg, andQ = 16 terms. The three examples consider external forcing in the form of a square-wave, asawtooth-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 representthe real-valued form of the Fourier approximation and the complex-valued form of the Fourierapproximation, and the circles represent 2Q sample points of the function f(t) for use in fastFourier transform (FFT) computations. The two dashed lines are exactly equal.

In the (b) plots, the impulse-lines show the values of the Fourier coefficients, Fq, found byevaluating the Fourier integral, equation (23), and the circles represent the Fourier coefficientsreturned by the FFT.

In the (c) plots, the red solid lines show the cosine terms of the Fourier series and theblue dashed lines show the sine terms of the Fourier series.

The (d) and (f) plots show the magnitude and phase of the transfer function, H() as asolid line, the Fourier coefficients, Fq, as the green circles, and H(q)Fq as the blue circles.

In the (e) plots, the solid line represents x(t) as computed using equation (27), the circlesrepresent the real part of the result of the inverse FFT calculation and the dots representthe imaginary part of the result of the inverse FFT calculation.

The numerical details involved in the correct representation of periodic functions andfrequency indexing for FFT computations are provided in the attached Matlab code.



7.1 Example 1: square wave

The external forcing is given by

f ext(t) = sgn(t) ; pi < t < pi (30)

The Fourier coefficients for the real-valued Fourier series are:aq = 0 and bq = (2/(qpi)) (1q 1).

(a)-1.5

-1

-0.5

0

0.5

1

1.5

-3 -2 -1 0 1 2 3

func

tion,

f(t), (

N)

time, t, (s) (b)-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0

0 2 4 6 8 10 12 14 16Fo

urie

r coe

fficie

nts,

Fq,

(N

)frequency, , (rad/s)

(c)-1.5

-1

-0.5

0

0.5

1

1.5

-3 -2 -1 0 1 2 3

cosi

ne a

nd s

ine

term

s

time, t, (s) (d) 1e-04

0.001

0.01

0.1

1

0 2 4 6 8 10 12 14 16

frequ

ency

resp

onse

frequency, , (rad/s)

(e)-0.025

-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

0.015

0.02

0.025

-3 -2 -1 0 1 2 3

resp

onse

, x, (m

)

time, t, (s) (f)-200

-150

-100

-50

0

50

100

150

200

0 2 4 6 8 10 12 14 16

phas

e le

ad, (d

egree

s)


Figure 1. (a): = fext(t); - - = f ext(t). (b): red solid =


7.2 Example 2: sawtooth wave


f ext(t) = t ; pi < t < pi (31)

The Fourier coefficients for the real-valued Fourier series are:aq = 0 and bq = (2/q) (1q).

(a)-4

-3

-2

-1

0

1

2

3

4

-3 -2 -1 0 1 2 3

func

tion,

f(t), (

N)

time, t, (s) (b)-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0 2 4 6 8 10 12 14 16

Four

ier c

oeffi

cient

s, F

q, (N


(c)-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

-3 -2 -1 0 1 2 3

cosi

ne a

nd s

ine

term

s

time, t, (s) (d) 1e-04

0.001

0.01

0.1

1

0 2 4 6 8 10 12 14 16

frequ

ency

resp

onse


(e)-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

-3 -2 -1 0 1 2 3

resp

onse

, x, (m

)

time, t, (s) (f)-200

-150

-100

-50

0

50

100

150

0 2 4 6 8 10 12 14 16

phas

e le

ad, (d

egree

s)




7.3 Example 3: triangle wave


f ext(t) = pi/2 t sgn(t) ; pi < t < pi (32)

The Fourier coefficients for the real-valued Fourier series are:aq = (2/q2pi)(1q 1) and bq = 0.

(a)-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

-3 -2 -1 0 1 2 3

func

tion,

f(t), (

N)

time, t, (s) (b)-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 2 4 6 8 10 12 14 16

Four

ier c

oeffi

cient

s, F

q, (N


(c)-1.5

-1

-0.5

0

0.5

1

1.5

-3 -2 -1 0 1 2 3

cosi

ne a

nd s

ine

term

s

time, t, (s) (d) 1e-05

1e-04

0.001

0.01

0.1

1

0 2 4 6 8 10 12 14 16

frequ

ency

resp

onse


(e)-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

0.015

0.02

-3 -2 -1 0 1 2 3

resp

onse

, x, (m

)

time, t, (s) (f)-200

-150

-100

-50

0

50

100

150

200

0 2 4 6 8 10 12 14 16

phas

e le

ad, (d

egree

s)




7.4 Matlab code1 function [Fq ,wq,x,t] = Fourier(type,Q,wn,z)2 % [F,w, x ] = Fourier ( type ,Q,wn, z )3 % Compute the Fourier s e r i e s c o e f f i c i e n t s o f a pe r i od i c s i gna l , f ( t ) ,4 % in two d i f f e r e n t ways :5 % complex exponen t i a l expansion ( i . e . , s ine and cos ine expansion )6 % f a s t Fourier transform7 %8 % The type o f p e r i od i c s i g n a l may be9 % square a square wave

10 % sawtooth a sawt oo th wave11 % t r i an g l e a t r i a n g l e wave12 %13 % The per iod o f the s i gna l , f ( t ) , i s f i x e d at 2 p i ( second ) .14 %15 % The number o f Fourier s e r i e s c o e f f i c i e n t s i s input as Q.16 % This r e s u l t s in a 2Q Fourier transform c o e f f i c i e n t s .17 %18 % The Fourier approximation i s then used to compute the steadys t a t e19 % 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 ed by20 %21 % 222 % x( t ) + 2 z wn x ( t ) + wn x ( t ) = f ( t ) ,23 %24 % where the mass o f the system i s f i x e d at 1 ( kg ) .25 %26 % see : h t t p :// en . w ik iped ia . org/wik i / Fou r i e r s e r i e s27 % ht tp ://www. jhu . edu/ s i g n a l s / f ou r i e r 2 / index . html28 %2930 i f nargin < 431 help Fourier32 return33 end343536 T = 2*pi; % per iod o f e x t e rna l forc ing , s3738 a = zeros(1,Q); % co e f f i c i e n t s o f cos ine par t o f Fourier s e r i e s39 b = zeros(1,Q); % co e f f i c i e n t s o f s ine par t o f Fourier s e r i e s4041 q = [1:Q]; % po s i t i v e f requency index va lue42 wq = 2*pi*q/T; % po s i t i v e f requency va lues , rad/sec , eq n (13)4344 dt = T/(2*Q); % 2Q d i s c r e t e po in t s in time fo r FFT sampling45 ts = [-Q:Q-1]*dt; % 2Q d i s c r e t e po in t s in time fo r FFT sampling4647 % s h i f t e d time . . . T/2


71 end7273 i f strcmp( type,sawtooth ) % sawtooth wave74 b = -(2./q) .* (-1).q;75 f_true = t; f_true (1) = 0; f_true (2*P+1) = 0;76 f_samp = ts;77 f_samp (1) = 0; % ge t p e r i o d i c i t y r i gh t , f (p i ) = 078 end7980 i f strcmp( type,triangle ) % t r i a n g l e wave81 a = -(2./(pi*q.2)) .* ( ( -1).q - 1 );82 f_true = pi/2 - t .* sign(t);83 f_samp = pi/2 - ts.* sign(ts);84 end858687 f_approxR = a * cos(q*t) + b * sin(q*t); % rea l Fourier s e r i e s8889 Fq = (a - i*b)/2; % complex Fourier c o e f f i c i e n t s9091 % Complex Fourier s e r i e s ( p o s i t i v e and nega t i v e exponents )92 f_approxC = Fq * exp(i*wq *t) + conj(Fq) * exp(-i*wq *t);939495 % The imagniary par t o f the complex Fourier s e r i e s i s e x a c t l y zero !96 imag_over_real_1 = max(abs(imag(f_approxC ))) / max(abs( real (f_approxC )))979899 % The f a s t Fourier transform (FFT) method

100 % In Matlab , the forward Fourier transform has a nega t i v e exponent .101 % According to a convent ion o f the FFT method , index number 1 i s f o r time = dt102 % t s (1) = Q dt = T/2 ; t s (Q) = dt ; t s (Q+1)=0 ; t s (2Q) = (Q1)dt = T/2dt103 % . . . see the t a b l e at the end o f t h i s f i l e .104105 F_FFT = f f t ( [ f_samp(Q+1:2*Q) f_samp (1:Q) ] ) / (2*Q); % Fourier coe f f s106107108 % coe f = [ Fq F FFT(2 :Q+1) ] % d i s p l a y the Fourier c o e f f i c i e n t s109110 % P lo t t i n g 111112 figure (1);113 plot(t,f_true ,-r , ts ,f_samp ,*r , t,f_approxR ,-c , t,f_approxC ,-b )114 xlabel(time , t, (s))115 ylabel(function , f(t), (N))116 axis (1.05*[ -pi ,pi])117 legend(true,sampled ,real Fourier approx ,complex Fourier approx )118 plot(wq , real (Fq),r, wq,imag(Fq),b , ...119 wq, real (F_FFT (2:Q+1)),*r, wq,imag(F_FFT (2:Q+1)),*b )120 ylabel(Fourier coefficients , F(w), (N))121 xlabel(frequency , {/ Symbol w}, (rad/s))122 legend(real(F), imag(F), real(FFT(f)), imag(FFT(f)))123124125 figure (2);126 plot(t,a*ones(1,2*P+1).*cos(q*t),-r , t,b*ones(1,2*P+1).* sin(q*t),-b);127 xlabel(time , t, (s))128 ylabel(cosine and sine terms)129 axis (1.05*[ -pi ,pi])130131132 % =======================================================================133 % Steadys t a t e response o f a SDOF o s c i l l a t o r to genera l p e r i od i c f o r c i n g .134135 % Complexva lued frequency response funct ion , H(w) , has un i t s o f [m/N]136 H = (1/wn2) ./ ( 1 - (wq/wn).2 + 2*i*z*(wq/wn) );137138 % Complex Fourier s e r i e s ( p o s i t i v e and nega t i v e exponents )139 x_approxC = (H .* Fq) * exp(i*wq *t) + (conj(H) .* conj(Fq)) * exp(-i*wq *t);140141 % In Matlab , the inve r s e Fourier transform has a p o s i t i v e exponent .



142 % . . . see the t a b l e at the end o f t h i s f i l e f o r the s o r t i n g convent ion .143 x_FFT = i f f t ( [ (1/wn2) , H , conj(H(Q-1: -1:1)) ] .* F_FFT ) * (2*Q);144145 figure (3); % p l o t f requency response func t i ons146 semilogy(wq ,abs(Fq),*g, wq,abs(H),-r, wq ,abs(H.*Fq),*b )147 xlabel(frequency , {/ Symbol w}, (rad/s))148 ylabel(frequency response )149 legend(|F|, |H|, |H*F|)150151 plot (wq ,angle(Fq )*180/pi ,*g, wq,angle(H)*180/pi ,-r, wq,angle(H.*Fq )*180/pi ,*b)152 xlabel(frequency , {/ Symbol w}, (rad/s))153 ylabel(phase lead , (degrees))154155 figure (4); % p l o t time response156 ts = [ ts(Q+1:2*Q) ts(1:Q) ]; % res o r t the time ax i s157 plot ( t,x_approxC ,-r , ts , real (x_FFT),*b , ts, imag(x_FFT),.b )158 axis (1.05*[ -pi ,pi])159 xlabel(time , t, (s))160 ylabel(response , x, (m))161 legend(x(t) - complex Fourier series , real(x(t)) - FFT, imag(x(t)) - FFT)162163164 % The imaginary par t i s e x a c t l y zero f o r the complex exponent s e r i e s method !165 imag_over_real_2 = max(abs(imag(x_approxC ))) / max(abs( real (x_approxC )))166167 % The imaginary par t i s p r a c t i c a l l y zero f o r the FFT method !168 imag_over_real_3 = max(abs(imag(x_FFT ))) / max(abs( real (x_FFT )))169170 % 171 % SORTING of the FFT c o e f f i c i e n t s number o f data po in t s = N = 2Q172 % note : f max = 1/(2 dt ) ; d f = f max /(N/2) = 1/(N dt ) = 1/T;173 %174 %175 % index time frequency176 % ===== ==== =========177 % 1 0 0178 % 2 dt d f179 % 3 2 dt 2 d f180 % 4 3 dt 3 d f181 % : : :182 % N/21 (N/22) dt f max2d f = ( N/22) d f183 % N/2 (N/21) dt f maxd f = ( N/21) d f184 % N/2+1 (N/2) dt +/ f max = ( N/2 ) d f185 % N/2+2 (N/2+1) dt f max+df = (N/2+1) d f186 % N/2+3 (N/2+2) dt f max+2d f = (N/2+2) d f187 % : : : :188 % N2 (N3)dt 3d f189 % N1 (N2)dt 2d f190 % N (N1)dt d f191 %


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