Example: Consider the function \( f(x) = 1-x \) on the interval [0,1]. To define its periodic extension, we use Mathematica command Mod:

To plot half-wave rectifiers, we need to define another piecewise continuous function

Example: Consider the following functions:

• \( f_1 (x) = \mbox{sign}(x)\,: \, (-1,1] \mapsto \, \mathbb{R}) \)
a = Plot[{Sign[t]}, {t, -2, 2}, PlotStyle -> {Thick, Blue}, AspectRatio -> 1/2]
b = Graphics[{Blue, Thick, Circle[{0, 0}, 0.02]}, AspectRatio -> 1]
Show[a, b]
However, it is not what we want because the signum function is defined on the infinite real line \( (-\infty , \infty ) , \) but we need its periodic extension with period 2.
a = Graphics[{Red, Arrowheads[0.05], Arrow[{{-1, -1}, {0, -1}}]}]
b = Graphics[{Red, Arrowheads[0.05], Arrow[{{1, 1}, {0, 1}}]}]
c = Graphics[{Red, Thick, Circle[{0, 0}, 0.05]}, AspectRatio -> 1/2]
b2 = Graphics[{Red, Arrowheads[0.05], Arrow[{{3, 1}, {2, 1}}]}]
a1 = Graphics[{Red, Arrowheads[0.05], Arrow[{{1, -1}, {2, -1}}]}]
c2 = Graphics[{Red, Thick, Circle[{2, 0}, 0.05]}, AspectRatio -> 1/2]
a3 = Graphics[{Red, Arrowheads[0.05], Arrow[{{3, -1}, {4, -1}}]}]
ac = Graphics[{Red, Thick, Circle[{-1, -1}, 0.05]}, AspectRatio -> 1/2]
ac1 = Graphics[{Red, Thick, Circle[{1, -1}, 0.05]}, AspectRatio -> 1/2]
ac3 = Graphics[{Red, Thick, Circle[{3, -1}, 0.05]}, AspectRatio -> 1/2]
c4 = Graphics[{Red, Thick, Circle[{4, 0}, 0.05]}, AspectRatio -> 1/2]
b4 = Graphics[{Red, Arrowheads[0.05], Arrow[{{5, 1}, {4, 1}}]}]
Show[a, b, c, c2, b2, a1, a3, ac, ac1, ac3, c4, b4]
Therefore, we define a periodic extension of the signum function:
fbas[x_] := -1 /; -L <= x < 0 (*square wave*)
fbas[x_] := 1 /; 0 <= x < L
L = 1;
f[x_] := fbas[Mod[x, 2*L, -L]]
SetOptions[Plot, ImageSize -> 300];
Plot[f[x], {x, -3*L, 3*L}, PlotRange -> {-1.2, 1.2}, AxesLabel -> {"x", "f(x)"}, PlotStyle -> Thickness[0.008]]

The function Mod[x, 2*L, -L]
(1) adds L to x,
(2) calculates the remainder on dividing (1) by 2*L,
(3) subtracts L from (2). This has the effect of shifting the original x by ±2L until it falls into the range [-L,L). This modified x is then used as the argument of the original function f[x].

We plot the periodically extended function over three periods -- that is over [-3L, 3L].
Strictly speaking the vertical lines shown on the graph at the discontinuity are not part of the graph. In some ways they make it easier to visualize the discontinuous function, but on those days when we are feeling fussy, we can exclude those segments. This is done by an Option for plot called Exclusions. The plot without the vertical lines is constructed below. The argument of the Option Exclusions is a list of points to be excluded.

Plot[f[x], {x, -3*L, 3*L}, PlotRange -> {-1.2, 1.2}, AxesLabel -> {"x", "f(x)"}, Exclusions -> {-2.0, -1.0, 0.0, 1.0, 2.0}, Epilog -> {Red, PointSize[Medium], Point[{{-2, 0}, {-1.0, 0}, {0, 0}, {1, 0}, {2, 0}, {3, 0}}]}]

Then we plot Fourier approximation with 20 terms:
f = FourierTrigSeries[Sign[x], x, 20, FourierParameters -> {1, Pi}]
A=Plot[f, {x, -5, 5}, PlotStyle -> {Thick, Orange}, AspectRatio -> 2/5]
cm1 = Graphics[{Red, Thick, Disk[{-1, 0}, 0.07]}, AspectRatio -> 1/4]
Show[A, c4, c3, c2, c1, c0, cm1, cm2, cm3, cm4]

because Fourier coefficients are
\[ a_k = \int_{-1}^1 \mbox{sign} (x)\,\cos \left( k\pi x \right) {\text d}x = 0, \qquad b_k = \int_{-1}^1 \mbox{sign} (x)\,\sin \left( k\pi x \right) {\text d}x = \frac{2}{k\pi} \left[ 1- (-1)^k \right] = \frac{4}{k\pi} \times \begin{cases} 1, & \ \mbox{if $k$ is odd}, \\ 0, & \ \mbox{if $k$ is even} . \end{cases} \]
Then
\[ \mbox{sign}(x) = \frac{4}{\pi}\, \sum_{n\ge 1} \frac{1}{2n-1} \,\sin (2n-1)\pi x , \qquad -1 \le x \le 1. \]

 

• \( f_2 (x) = x^2 \,: \, (1,4] \mapsto \, \mathbb{R} \)     We plot its periodic extension:
  fbas[x_] := x^2 /; 1 <= x < 4
  f[x_] := fbas[Mod[x, 3, 1]]
  SetOptions[Plot, ImageSize -> 300];
  Plot[f[x], {x, -3, 7}, PlotRange -> {-1.2, 16.2}, AxesLabel -> {"x", "f(x)"}, PlotStyle -> Thickness[0.01]]
  
  Function f₂(x) is defined on the interval of length 3, so ℓ = 1.5. On symmetrical interval \( (-1.5,1.5) \) the given function can be written as
  \[ f_2 (x) = \begin{cases} (x+3)^2 , & \ \mbox{for } -1.5 < x < 1 , \\ x^2 , & \ \mbox{for } 1 < x < 1.5 . \end{cases} \]
  To find its Fourier coefficients, we calculate the following integrals.
  a0 = 2*Integrate[(x + 3)^2, {x, -3/2, 1/2}]/3 + 2*Integrate[x^2, {x, 1, 3/2}]/3
  335/36
  an = Simplify[ 2*Integrate[(x + 3)^2 * Cos[2*n*Pi*x/3], {x, -3/2, 1/2}]/3 + 2*Integrate[x^2 *Cos[n*Pi*x*2/3], {x, 1, 3/2}]/3]
  (1/(4 n^3 \[Pi]^3))(42 n \[Pi] Cos[(n \[Pi])/3] - 12 n \[Pi] Cos[(2 n \[Pi])/3] - 18 Sin[(n \[Pi])/3] + 49 n^2 \[Pi]^2 Sin[(n \[Pi])/3] + 18 Sin[(2 n \[Pi])/3] - 4 n^2 \[Pi]^2 Sin[(2 n \[Pi])/3] - 36 Sin[n \[Pi]] + 18 n^2 \[Pi]^2 Sin[n \[Pi]])
  bn = Simplify[ 2*Integrate[(x + 3)^2 * Sin[2*n*Pi*x/3], {x, -3/2, 1/2}]/3 + 2*Integrate[x^2 *Sin[n*Pi*x*2/3], {x, 1, 3/2}]/3]
  (1/(4 n^3 \[Pi]^3))((18 - 49 n^2 \[Pi]^2) Cos[(n \[Pi])/3] + 2 ((-9 + 2 n^2 \[Pi]^2) Cos[(2 n \[Pi])/3] + 3 n \[Pi] (7 Sin[(n \[Pi])/3] - 2 Sin[(2 n \[Pi])/3] + 6 Sin[n \[Pi]])))
  Therefore, the required Fourier series for the function f₂(x) becomes
  \[ f_2 (x) = \frac{335}{72} + \frac{1}{4 \pi^3} \sum_{n\ge 1} \frac{1}{n^3} \left\{ 42n\pi \,\cos \frac{n\pi}{3} -12n\pi \,\cos \frac{2n\pi}{3} + \left( 49 n^2 \pi^2 -18 \right) \sin \frac{n\pi}{3} + \left( 18 - 4n^2 \pi^2 \right) \sin \frac{2n\pi}{3} \right\} \cos \frac{2n\pi x}{3} + \frac{1}{4 \pi^3} \sum_{n\ge 1} \frac{1}{n^3} \left\{ \left( 18 -49 n^2 \pi^2 \right) \cos \frac{n\pi}{3} + 2 \left( 2n^2 \pi^2 -9 \right) \cos \frac{2n\pi}{3} + 42 n\pi\,\sin \frac{n\pi}{3} - 4\,\sin \frac{2n\pi}{3} \right\} \sin \frac{2n\pi x}{3} . \]
  Now we build its Fourier approximation
  fourier[m_] = a0/2 + Sum[an*Cos[2*n*Pi*x/3] + bn*Sin[2*n*Pi*x/3] , {n, 1, m}];
  ff10[x_] = 335/72 + Sum[an*Cos[2*n*Pi*x/3] + bn*Sin[2*n*Pi*x/3], {n, 1, 10}];
  ff20[x_] = 335/72 + Sum[an*Cos[2*n*Pi*x/3] + bn*Sin[2*n*Pi*x/3], {n, 1, 20}];
  and plot the corresponding sums with N = 10 and 20 terms:
  Plot[{ff10[x], ff20[x]}, {x, -2, 5}, PlotStyle -> {{Thick, Blue}, {Thick, Orange}}]
  

 

• \( \displaystyle f_3 (x) = \begin{cases} x , & \ \mbox{for } 0 < x < 2 , \\ 1, & \ \mbox{when } -2 < x < 0 . \end{cases} \)   This function maps the finite interval (-2,2) into \( \mathbb{R} . \)

  We calculate the corresponding Fourier coefficients using Mathematica:

  a0 = Integrate[1, {x, -2, 0}]/2 + Integrate[x, {x, 0, 2}]/2
  2
  an = Simplify[ Integrate[Cos[n*Pi*x/2], {x, -2, 0}]/2 + Integrate[x *Cos[n*Pi*x/2], {x, 0, 2}]/2]
  (-2 + 2 Cos[n \[Pi]] + 3 n \[Pi] Sin[n \[Pi]])/(n^2 \[Pi]^2)
  bn = Simplify[ Integrate[Sin[n*Pi*x/2], {x, -2, 0}]/2 + Integrate[x *Sin[n*Pi*x/2], {x, 0, 2}]/2]
  -((n \[Pi] + n \[Pi] Cos[n \[Pi]] - 2 Sin[n \[Pi]])/(n^2 \[Pi]^2))
  Therefore, the Fourier series becomes
  \[ f_3 (x) = 1 - \frac{4}{\pi^2} \sum_{k\ge 0} \frac{1}{(2k+1)^2} \,\cos \frac{(2k+1) \pi x}{2} - \frac{1}{\pi} \sum_{k\ge 1} \frac{1}{k}\,\sin (k\pi x) . \]
  Finally we build partial Fourier sums with 10 and 50 terms
  ff10[x_] = 1 + 4/Pi^2 * Sum[1/(2*k+1)^2 * Cos[(2*k+1)*Pi*x/2], {k,0,10}] - Sum[1/k*Sin[k*Pi*x], {k,1,10}]/Pi; ff50[x_] = 1 + 4/Pi^2 * Sum[1/(2*k+1)^2 * Cos[(2*k+1)*Pi*x/2], {k,0,50}] - Sum[1/k*Sin[k*Pi*x], {k,1,50}]/Pi;
  and plot these approximations
  Plot[{ff10[x], ff50[x]}, {x, -2, 5}, PlotStyle -> {{Thick, Blue}, {Thick, Orange}}]
  
  Since f₃(0) = ½, we get a curious sum:
  \[ \sum_{k\ge 0} \frac{1}{(2k+1)^2} = \frac{\pi^2}{8} \approx 1.2337. \]
  Mathematica also knows this sum:
  Sum[1/(2*k + 1)^2, {k, 0, Infinity}]
  \[Pi]^2/8

 

• \( f_4 (x) = 1-|x|\,: \, (-2,2] \mapsto \, \mathbb{R} . \) We first the corresponding Fourier coefficients
  a0 = Integrate[(1-Abs[x]), {x, -2, 2}]/2
  0
  an = Simplify[ Integrate[(1-Abs[x])*Cos[n*Pi*x/2], {x, -2, 2}]/2]
  (4 - 4 Cos[n \[Pi]] - 2 n \[Pi] Sin[n \[Pi]])/(n^2 \[Pi]^2)
  bn = Simplify[ Integrate[(1-Abs[x])*Sin[n*Pi*x/2], {x, -2, 2}]/2]
  0
  Therefore, the given function f₄ has the following Fourier cosine series
  \[ f_4 (x) = \frac{8}{\pi^2} \sum_{k\ge 0} \frac{1}{(2k+1)^2} \,\cos \frac{(2k+1) \pi x}{2} . \]
  Using Mathematica, we construct partial Fourier cosine sums and the periodic extension of the given function.
  ff10[x_] = 8/Pi^2 * Sum[1/(2*k+1)^2 * Cos[(2*k+1)*Pi*x/2], {k,0,10}];
  ff50[x_] = 8/Pi^2 * Sum[1/(2*k+1)^2 * Cos[(2*k+1)*Pi*x/2], {k,0,50}];
  per[x_] = Which[-2 Then we plot the approximation with N = 10 terms along with the given function:

  Plot[{ff10[x], per[x]}, {x, -2, 5}, PlotStyle -> {{Thick, Blue}, {Thick, Orange}}]

  

 

• \( f_5 (x) = \lfloor x \rfloor \,: \, (-2,2] \mapsto \, \mathbb{R} , \)    where ⌊x⌋ is the floor function of x. Using its Fourier coefficients
  a0 = Integrate[Floor[x], {x, -2, 2}]/2
  -1
  an = Simplify[ Integrate[Floor[x]*Cos[n*Pi*x/2], {x, -2, 2}]/2]
  -(Sin[n \[Pi]]/(n \[Pi]))
  bn = Simplify[ Integrate[Floor[x]*Sin[n*Pi*x/2], {x, -2, 2}]/2]
  (4 (2 + 3 Cos[(n \[Pi])/2]) Sin[(n \[Pi])/4]^2)/(n \[Pi])
  Hence, we get its Fourier series
  \[ f_5 (x) = - \frac{1}{2} + \frac{4}{\pi} \sum_{n\ge 1} \frac{1}{n} \left[ 2 + 3\,\cos \frac{n\pi}{2} \right] \sin^2 \frac{n\pi}{4} \,\sin \frac{n\pi x}{2} . \]
  Since the given function is a piecewise step functions, its Fourier series converges slowly showing Gibbs phenomenon at points of discontinuity (integer values).
  ff10[x_] = -1/2 +4/Pi * Sum[1/n * (2 + 3*Cos[n*Pi/2])*(Sin[n*Pi/4])^2 * Sin[n*Pi*x/2], {n,1,10}];
  ff50[x_] = -1/2 +4/Pi * Sum[1/n * (2 + 3*Cos[n*Pi/2])*(Sin[n*Pi/4])^2 * Sin[n*Pi*x/2], {n,1,50}];
  per[x_] = Which[-2< x < 2,Floor[x], 2 < x < 6,Floor[x-4]]
  Then we plot the approximation with N = 10 (blue) and 50 (orange) terms along with the given function (in black)):
  Plot[{ff10[x], ff50[x], per[x]}, {x, -2, 5}, PlotStyle -> {{Thick, Blue}, {Thick, Orange}, {Thick, Black}}]
  
■

Example: Consider the piecewise continuous function, defined on the interval \( [-2,2] . \)

Other options to define periodic function:

Out[11]= \[ \left\{ \begin{array}{ll} 1 & \ x<0 \\ x^2 & \ x>0 \\ 0 & \ \mbox{True} \end{array} \right. \]

We define a subroutine that provide periodic extension of a function:

Example: Consider the following function that we met previously

We can also define the periodic expansion of any function using Mod command. If we need to define a function with period \( 2*\pi , \) we type:

1. Gregory A. Kriegsmann, A mathematical theory of full-wave rectification, IMA Journal of Applied Mathematics, Volume 36, Issue 1, 1 January 1986, Pages 29--41, https://doi.org/10.1093/imamat/36.1.29
2. C. Marouchos ; M.K. Darwish ; M. El-Habrouk, New mathematical model for analysing three-phase controlled rectifier using switching functions, IET Power Electronics, Volume: 3, Issue: 1, January 2010, 95--110. doi: 10.1049/iet-pel.2008.0328
3. Le Hong Lan and Nguyen Thi Hien, Mathematical model written by the canonical system for some electric rectifier circuits using semiconductor diodes, article
4. Branco Curgus, Fourier periodic extensions of piecewise continuous functions, http://faculty.wwu.edu/curgus/Courses/Math_pages/Math_430/Fourier_periodic_extension.html
5. Mohammed A. Abbas, Falah H.Hanoon, and Lafy F.Al-Badry, Possibility designing half-wave and full-wave molecular rectifiers by using single benzene molecule, Physics Letters A, Volume 382, Issue 8, 27 February 2018, Pages 608--612. https://doi.org/10.1016/j.physleta.2017.12.017