3.5 Obtain Fourier Series coefficients for a periodic function

3.5.1 Example 1
3.5.2 Example 2
3.5.3 Example 3

3.5.1 Example 1

3.5.1.1 Mathematica

Given a continuous time function \(x\left ( t\right ) =\sin \left (\frac {2 \pi }{T_0} t\right ) \), find its Fourier coefficients \(c_{n}\), where (Using the default definition) \[ x\left ( t\right ) =\sum _{n=-\infty }^{\infty }c_{n}e^{j\omega _{0}nt}\] and \[ c_{n}=\frac {1}{T_{0}}\int _{-\frac {T_{0}}{2}}^{\frac {T_{0}}{2}}x(t)e^{-j\omega _{0}nt}dt \] Where \(T_{0}\) is the fundamental period of \(x\left ( t\right ) \) and \(\omega _{0}=\frac {2\pi }{T_{0}}\).

Notice that in Physics, sometimes the following defintions are used instead of the above

\[ x\left ( t\right ) =\frac {1}{\sqrt {T_0}} \sum _{n=-\infty }^{\infty }c_{n}e^{j\omega _{0}nt}\] and \[ c_{n}= \frac {1}{\sqrt {T_0}} \int _{-\frac {T_{0}}{2}}^{\frac {T_{0}}{2}}x(t)e^{-j\omega _{0}nt}dt \]

3.5.1.1 Mathematica
Clear[T0, w0] 
sol = FourierSeries[Sin[2 Pi/T0 t], t, 3, FourierParameters -> {1, 2 Pi/T0}]
 

\[ \frac {1}{2} i e^{-\frac {2 i \pi t}{\text {T0}}}-\frac {1}{2} i e^{\frac {2 i \pi t}{\text {T0}}} \]

data = Table[{i, 
  FourierCoefficient[Sin[2 Pi/T0 t], t, i, 
  FourierParameters -> {1, 2 Pi/T0}]}, {i, -5, 5}]; 
  head = {"n", "cn"}; 
Grid[Insert[data, head, 1], Frame -> All]
 

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mag = data; 
mag[[All,2]] = Map[Abs[#]&,data[[All,2]]]; 
ListPlot[mag,AxesOrigin->{0,0}, 
         Filling->Axis, 
         FillingStyle->{{Thick,Red}}, 
         AxesLabel->{"n","|Subscript[c, n]|"}]
 

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phase = data; 
phase[[All,2]]=Map[Arg[#]&, 
           data[[All,2]]]* 180/Pi; 
ListPlot[phase,AxesOrigin->{0,0}, 
  Filling->Axis, 
  FillingStyle->{{Thick,Red}}, 
  AxesLabel->{"n","Phase Subscript[c, n] degrees"}, 
  ImageSize->300]
 

pict

3.5.2 Example 2

3.5.2.1 Mathematica

Find Fourier Series for \(f(t) = \cos (\omega _0 t) + \sin ^2(\omega _0 t)\)

3.5.2.1 Mathematica
T0 = 2 Pi; 
w0 = (2 Pi)/T0; 
f[t_] := Cos[w0 t] + Sin[w0 t]^2; 
Plot[f[t], {t, -10 Pi, 10 Pi}, PlotRange -> All, ImageSize -> 300]
 

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sol = FourierSeries[f[t],t,3, FourierParameters->{1,(2 Pi)/T0}]
 

\[ \frac {e^{-i t}}{2}+\frac {e^{i t}}{2}-\frac {1}{4} e^{-2 i t}-\frac {1}{4} e^{2 i t}+\frac {1}{2} \]

data = Table[{i, FourierCoefficient[f[t],t,i, FourierParameters->{1,2 Pi/T0}]},{i,-5,5}]; 
 
head = {"n","cn"}; 
Grid[Insert[data,head,1],Frame->All]
                                                                                    
                                                                                    
 

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mag = data; 
mag[[All,2]] = Map[Abs[#]&,data[[All,2]]]; 
ListPlot[mag,AxesOrigin->{0,0}, 
 Filling->Axis, 
 FillingStyle->{{Thick,Red}}, 
 AxesLabel->{"n","|Subscript[c, n]|"}, 
 ImageSize->300]
 

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Plot the phase

phase = data; 
phase[[All, 2]] = Map[Arg[#] &, 
                  data[[All, 2]]]* 180/Pi; 
ListPlot[phase, AxesOrigin -> {0, 0}, 
 Filling -> Axis, 
 FillingStyle -> {{Thick, Red}}, 
 AxesLabel -> {"n", "Phase cn degrees"}]
 

pict

3.5.3 Example 3

3.5.3.1 Mathematica

Find Fourier series for periodic square wave

3.5.3.1 Mathematica
f[x_] := SquareWave[{0,1},(x+.5)/2] ; 
Plot[f[x],{x,-6,6},Filling->Axis, ImageSize->300]
 

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T0   = 2; 
sol = Chop[FourierSeries[f[t],t,6,  FourierParameters->{1,(2 Pi)/T0}]]
 

\[ 0.5 +0.31831 e^{-i \pi t} + 0.31831 e^{i \pi t} - 0.106103 e^{-3 i \pi t} - 0.106103 e^{3 i \pi t} + 0.063662 e^{-5 i \pi t}+0.063662 e^{5 i \pi t} \]

data = Chop[Table[{i,FourierCoefficient[f[t],t,i, FourierParameters->{1,2 Pi/T}]},{i,-8,8}]]; 
 
head = {"n","cn"}; 
Grid[Insert[data,head,1],Frame->All]
 

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data=Table[{i,FourierCoefficient[f[t],t,i,  FourierParameters->{1,2 Pi/T}]},{i,-20,20}]; 
mag = data; 
mag[[All,2]]=Map[Abs[#]&,data[[All,2]]]; 
ListPlot[mag,AxesOrigin->{0,0}, 
  Filling->Axis, 
  FillingStyle->{{Thick,Red}}, 
  PlotRange->All, 
  AxesLabel->{"n","|Subscript[c, n]|"}, 
  ImageSize->300]
 

pict

Show phase

phase = data; 
phase[[All,2]]=Map[Arg[#]&,data[[All,2]]]* 180/Pi; 
ListPlot[phase,AxesOrigin->{0,0}, 
 Filling->Axis, 
 FillingStyle->{{Thick,Red}}, 
 AxesLabel->{"n","Phase cn degrees"}, 
 ImageSize->300]
 

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