Elementos Constituyentes de un sistema de eDemocracia
2.2 El concepto del Gobierno Abierto
In the previous section Fourier series were determined for a number of standard functions. In the same way one can, in principle, determine the Fourier series for many more periodic functions. This, however, is quite cumbersome. By using a number of properties of Fourier series one can determine in a relatively simple way the Fourier series of a large number of periodic functions. These properties have also been included in table 2 at the back of the book.
3.5.1 Linearity
Fourier coefficients of linear combinations of functions are equal to the same linear combination of the Fourier coefficients of the individual functions. This property is formulated in the following theorem.
When the complex Fourier coefficients of f(t) and g(t) are fnand gnrespectively, THEOREM 3.1
Linearity of the Fourier
transform then one has for a, b ∈ C:
a f(t) + bg(t) ↔ a fn+ bgn.
Proof
The proof of this theorem is a straightforward application of the linearity of integra-tion. When cndenotes the Fourier coefficients of a f(t) + bg(t), then
cn = 1 T
T/2
−T/2(a f (t) + bg(t))e−inω0tdt
= a T
T/2
−T/2f(t)e−inω0tdt+ b T
T/2
−T/2g(t)e−inω0tdt= a fn+ bgn.
With the linearity property one can easily determine the Fourier coefficients of linear EXAMPLE 3.5
combinations of functions whose individual Fourier coefficients are already known.
Let f be the periodic function with period 6 as sketched in figure 3.11. The function
t 2
0 3
–3
–1
f
FIGURE 3.11
Periodic function as a combination of periodic block functions.
f is then equal to 2g− 3h, where g and h are periodic block functions, as defined in section 3.4.1, with period 6 and, respectively, a = 4 and a = 2. Using (3.14) or table 1 and applying theorem 3.1, it then follows that the Fourier coefficients are given by
cn = 4 6
sin(nω04/2) nω0
−6 6
sin(nω02/2) nω0
= 2 3
sin(n2π/3)
n2π/6 −sin(nπ/3)
n2π/6 = 2 sin(2nπ/3) − 3 sin(nπ/3)
nπ .
3.5.2 Conjugation
The Fourier coefficients of the complex conjugate of f can be derived from the Fourier coefficients of the function itself. How this can be done is the subject of our next theorem.
When the Fourier coefficients of f(t) are equal to cn, then THEOREM 3.2
Fourier coefficients of a
conjugate f(t) ↔ c−n.
Proof
Since ei nω0t = e−inω0t, it follows by direct calculation of the Fourier coefficients of f(t) that
1 T
T/2
−T/2 f(t)e−inω0tdt = 1 T
T/2
−T/2f(t)ei nω0tdt
= 1 T
T/2
−T/2f(t)e−i(−n)ω0tdt = c−n.
This property has a special consequence when f(t) is real. This is because we then have f(t) = f (t). The Fourier coefficients of f (t) and f (t), whenever these exist at least, must then also be equal and hence cn = c−n. This result has been derived before, see (3.12). Furthermore, one has for the moduli that| cn| = | c−n| and since the moduli of complex conjugates are the same (see (2.3)), this in turn equals| c−n|. For a real function it thus follows that | cn| = | c−n|, which means that the amplitude spectrum is even. We also know that the arguments of complex conjugates are each other’s opposite, and so arg(cn) = arg(c−n) = − arg(c−n).
Hence, the phase spectrum is odd.
The standard functions treated in section 3.4 are all real. One can easily check that EXAMPLE
the amplitude spectra are indeed even. For the sawtooth function one can check moreover that the phase spectrum is odd, while the phase spectra of the periodic block and triangle functions are zero, and so odd as well.
3.5.3 Shift in time
The standard functions treated in section 3.4 were all neatly ‘centred’ around t= 0.
From these one can, by a shift in time, obtain functions that are, of course, no longer centred around t= 0. When the shift equals t0, then the new function will be given by f(t − t0). The Fourier coefficients of the shifted function can immediately be obtained from the Fourier coefficients of the original function.
When cnare the Fourier coefficients of f(t), then THEOREM 3.3
Shift in time
f(t − t0) ↔ e−inω0t0cn. Proof
The Fourier coefficients of f(t − t0) can be calculated using the definition. In this calculation we introduce the new variableτ = t − t0and we integrate over (−T/2, T/2) instead of ((−T/2) + t0, (T/2) + t0), since this gives the same result:
1 T
T/2
−T/2 f(t − t0)e−inω0tdt
= e−inω0t01 T
(T/2)+t0 (−T/2)+t0
f(t − t0)e−inω0(t−t0)d(t − t0)
= e−inω0t01 T
T/2
−T/2f(τ)e−inω0τdτ = e−inω0t0· cn.
It follows immediately from theorem 3.3 that the amplitude spectra of f(t) and f(t − t0) are the same: |e−inω0t0cn| = | cn|. Hence, the amplitude spectrum of
a function does not change when the function is shifted in time. For the phase spectrum one has
arg
e−inω0t0cn
= arg e−inω0t0
+ arg(cn) = n arg e−iω0t0
+ arg(cn).
The phase spectrum thus changes linearly with n, apart from the fact that the argu-ment can always be reduced to a value in the interval [−π, π].
The periodic block function from example 3.4 is centred around t= 0. The Fourier EXAMPLE
coefficients of the periodic function with f(t) = 1 for 0 ≤ t < T/2 and f (t) = 0 for−T/2 < t < 0, see figure 3.12, can easily be derived from this. For this periodic
t 1
0 T/2 T
–T/2 –T
FIGURE 3.12
The shifted periodic block function.
block function one has t0= T/4 and a = T/2, so the Fourier coefficients are equal to
cn = e−inω0t0 1
nπ sin(nπ/2) = e−inπ/2 1 nπ
ei nπ/2− e−inπ/2 2i
= 1− e−inπ
2i nπ = (−1)n− 1 2nπ i.
Hence, for even n and n= 0 one has cn= 0 and for odd n one has cn= −i/nπ.
Furthermore, c0= 1/2. The amplitude spectrum is thus equal to that of the periodic
block function in example 3.4.
3.5.4 Time reversal
The process of changing from the variable t to the variable−t is called time reversal.
In this case there is again a simple relationship between the Fourier coefficients of the functions.
When cnare the Fourier coefficients of f(t), then THEOREM 3.4
Time reversal
f(−t) ↔ c−n. Proof
A direct application of the definition and changing from the variable−t to τ gives the proof:
1 T
T/2
−T/2 f(−t)e−inω0tdt = 1 T
−T/2
T/2 f(τ)e−inω0(−τ)d(−τ)
= 1 T
T/2
−T/2 f(τ)e−i(−n)ω0τdτ = c−n.
A direct consequence of this theorem is that if f(t) is even, so f (t) = f (−t) for all t, then cn = c−n. In this case the amplitude spectrum as well as the phase spectrum are even. Furthermore, it follows from (3.8) that the coefficients bn of the ordinary Fourier series are all 0. Thus, the Fourier series of an even function contains only cosine terms. This result is easily understood: the sines are odd, while the cosines are even. A series containing sine functions will never be even. When, moreover, f(t) is real, then cn and c−nare each other’s complex conjugate (see (3.12)) and in that case the coefficients will be real as well.
The periodic block function and the periodic triangle function from sections 3.4.1 EXAMPLE
and 3.4.2 are even and real. The spectra are also even and real. When f(t) is odd, so f (t) = − f (−t), it follows that cn = −c−nand so the spectrum is odd. Since c−n = cnfor f(t) real, the Fourier coefficients are purely imaginary. The spectrum of a real and odd function is thus odd and purely imagi-nary. Moreover, in the case of an odd function it follows from (3.8) that the coeffi-cients anare 0 and that the Fourier series consists of sine functions only.
The periodic sawtooth function is a real and odd function. The complex Fourier EXAMPLE
coefficients are odd and purely imaginary, while the Fourier series contains only
sine functions.
EXERCISES
The periodic function f with period 4 is given by f(t) = 1 + | t | for | t | ≤ 1 and 3.14
f(t) = 0 for 1 < | t | < 2. Sketch the graph of the function and determine its Fourier coefficients.
Determine the Fourier coefficients of the periodic function with period T defined by 3.15
f(t) = t on the interval (0, T ).
Let the complex-valued function f(t) = u(t) + iv(t) be given, where u(t) and v(t) 3.16
are real functions with Fourier coefficients unand vn. a Determine the Fourier coefficients of f(t) and of f (t).
b Suppose that f(t) is even, but not real. Will the Fourier coefficients of f (t) be even and real then?
The amplitude spectrum of a function does not change when a shift in time is 3.17
applied. For which shifts does the phase spectrum remains unchanged as well?
In section 3.5.4 we derived that for even functions the ordinary Fourier series con-3.18
tains only cosine terms. Show that this also follows directly from (3.5) and (3.6).