3.41 in text: A periodic disturbance consists of a sequence of exponentially pulse repeated at intervals \(T\), such that \(Q\relax (t) =Fe^{\frac {-\lambda t}{T}}\) for \(0<t<T\), and \(Q\left (t\pm T\right ) =Q\left ( t\right ) \). The parameter \(\lambda \) is nondimensional. Determine the complex Fourier series representing the force. Evaluate the first \(5\) coefficients when \(\lambda =0.1,1,10\). What does this reveal regarding the influence of \(\lambda \) on the frequency spectrum?
Let \(\tilde {Q}\relax (t) \) be the Fourier series approximation to \(Q\relax (t) \) given by\begin {equation} \tilde {Q}\relax (t) =\frac {1}{2}{\displaystyle \sum \limits _{n=-\infty }^{\infty }} F_{n}e^{in\frac {2\pi }{T}t} \label {eq:1.0} \end {equation}
Where\begin {align*} F_{n} & =\frac {2}{T}{\displaystyle \int \limits _{0}^{T}} Q\relax (t) e^{-in\frac {2\pi }{T}t}dt\\ & =\frac {2}{T}{\displaystyle \int \limits _{0}^{T}} Fe^{\frac {-\lambda t}{T}}e^{-in\frac {2\pi }{T}t}dt=\frac {2F}{T}{\displaystyle \int \limits _{0}^{T}} e^{-t\left (in\frac {2\pi }{T}-\frac {\lambda }{T}\right ) }dt=\frac {2F}{T}\left (\frac {e^{-t\left (in\frac {2\pi }{T}-\frac {\lambda }{T}\right ) }}{in\frac {2\pi }{T}-\frac {\lambda }{T}}\right ) _{0}^{T}\\ & =\frac {2F}{in2\pi -\lambda }\left (e^{-T\left (in\frac {2\pi }{T}-\frac {\lambda }{T}\right ) }-1\right ) \\ & =\frac {2F}{in2\pi -\lambda }\left (e^{-in2\pi }e^{-\lambda }-1\right ) \end {align*}
But \(e^{-in2\pi }=1\), hence \[ F_{n}=\frac {2F}{in2\pi -\lambda }\left (e^{-\lambda }-1\right ) \]
Hence Eq ?? becomes\begin {align*} \tilde {Q}\relax (t) & =\frac {1}{2}{\displaystyle \sum \limits _{n=-\infty }^{\infty }} \frac {2F}{in2\pi -\lambda }\left (e^{-\lambda }-1\right ) e^{in\frac {2\pi }{T}t}\\ & =F{\displaystyle \sum \limits _{n=-\infty }^{\infty }} \frac {\left (e^{-\lambda }-1\right ) }{in2\pi -\lambda }e^{in\frac {2\pi }{T}t}\\ & =F{\displaystyle \sum \limits _{n=-\infty }^{\infty }} \frac {1-e^{-\lambda }}{\lambda +in2\pi }e^{in\frac {2\pi }{T}t} \end {align*}
For \(n=-2,-1,0,1,2\) we obtain\begin {align*} \tilde {Q}\relax (t) & =F{\displaystyle \sum \limits _{n=-2}^{2}} \frac {1-e^{-\lambda }}{\lambda +in2\pi }e^{in\frac {2\pi }{T}t}\\ & =F\left (\frac {1-e^{-\lambda }}{\lambda -i4\pi }e^{-i\frac {4\pi }{T}t}+\frac {1-e^{-\lambda }}{\lambda -i2\pi }e^{-i\frac {2\pi }{T}t}+\frac {1-e^{-\lambda }}{\lambda }+\frac {1-e^{-\lambda }}{\lambda +i2\pi }e^{i\frac {2\pi }{T}t}+\frac {1-e^{-\lambda }}{\lambda +i4\pi }e^{i\frac {4\pi }{T}t}\right ) \end {align*}
For \(\lambda =0.1\)\begin {align*} \tilde {Q}\relax (t) & =F\left (\frac {1-e^{-0.1}}{0.1-i4\pi }e^{-i\frac {4\pi }{T}t}+\frac {1-e^{-0.1}}{0.1-i2\pi }e^{-i\frac {2\pi }{T}t}+\frac {1-e^{-0.1}}{0.1}+\frac {1-e^{-0.1}}{0.1+i2\pi }e^{i\frac {2\pi }{T}t}+\frac {1-e^{-0.1}}{0.1+i4\pi }e^{i\frac {4\pi }{T}t}\right ) \\ & =F\{\left (6.026\times 10^{-5}+7.\allowbreak 572\times 10^{-3}i\right ) e^{-i\frac {4\pi }{T}t}\\ & +\left (2.41\times 10^{-4}+1.514\times 10^{-2}i\right ) e^{-i\frac {2\pi }{T}t}\\ & +0.952\\ & +\left (2.\allowbreak 4099\times 10^{-4}-1.5142\times 10^{-2}i\right ) e^{i\frac {2\pi }{T}t}\\ & +\left (6.026\times 10^{-5}-7.572\times 10^{-3}i\right ) e^{i\frac {4\pi }{T}t}\} \end {align*}
For \(\lambda =1\)\begin {align*} \tilde {Q}\relax (t) & =F\left (\frac {1-e^{-1}}{1-i4\pi }e^{-i\frac {4\pi }{T}t}+\frac {1-e^{-1}}{1-i2\pi }e^{-i\frac {2\pi }{T}t}+\frac {1-e^{-1}}{1}+\frac {1-e^{-1}}{1+i2\pi }e^{i\frac {2\pi }{T}t}+\frac {1-e^{-1}}{1+i4\pi }e^{i\frac {4\pi }{T}t}\right ) \\ & =F\{\left (0.00398+0.05i\right ) e^{-i\frac {4\pi }{T}t}+\left ( 0.016+0.098i\right ) e^{-i\frac {2\pi }{T}t}+0.632+\left (0.016+0.098i\right ) e^{i\frac {2\pi }{T}t}+\left (0.00398+0.05i\right ) e^{i\frac {4\pi }{T}t}\} \end {align*}
For \(\lambda =10\)\begin {align*} \tilde {Q}\relax (t) & =F\left (\frac {1-e^{-10}}{10-i4\pi }e^{-i\frac {4\pi }{T}t}+\frac {1-e^{-10}}{10-i2\pi }e^{-i\frac {2\pi }{T}t}+\frac {1-e^{-10}}{10}+\frac {1-e^{-10}}{10+i2\pi }e^{i\frac {2\pi }{T}t}+\frac {1-e^{-10}}{10+i4\pi }e^{i\frac {4\pi }{T}t}\right ) \\ & =F\{\left (3.877\times 10^{-2}+4.872\times 10^{-2}\allowbreak i\right ) e^{-i\frac {4\pi }{T}t}\\ & +\left (7.169\times 10^{-2}+4.505\times 10^{-2}\allowbreak i\right ) e^{-i\frac {2\pi }{T}t}\\ & +0.1\\ & +\left (7.169\times 10^{-2}-4.505\times 10^{-2}\allowbreak i\right ) e^{i\frac {2\pi }{T}t}\\ & +\left (3.877\times 10^{-2}-4.872\times 10^{-2}\allowbreak i\right ) e^{i\frac {4\pi }{T}t}\} \end {align*}
We notice that as \(\lambda \) became larger, the DC term became smaller. Since the \(DC\) term represents average value of the whole signal, then we can say that as \(\lambda \) gets larger, then the average becomes smaller. This means the energy of the signal becomes smaller as \(\lambda \) becomes larger.
From above, we found for \(\lambda =1\)\begin {align*} F_{n} & =\frac {2F}{in2\pi -\lambda }\left (e^{-\lambda }-1\right ) \\ & =\frac {2F}{in2\pi -1}\left (e^{-1}-1\right ) \end {align*}
and the first \(5\) found to be
| \(n\) | \(F_{n}\) |
| \(-2\) | \(0.00398+0.05i\) |
| \(-1\) | \(0.016+0.098i\) |
| \(0\) | \(0.632\) |
| \(1\) | \(0.016-0.098i\) |
| \(2\) | \(0.00398-0.05i\allowbreak \) |
To verify the result with ffteasy.m using \(\lambda =1\), Using \(F=1\), and using \(T=1\). This below shows the result for \(F_{0},F_{1},F_{2}\) and we see that the DC term \(F_{0}\) agrees, and that complex component of \(F_{1},F_{2}\) also agrees. The real parts are little larger than what I obtained using the above. This might be a scaling issue, and I was not able to determine the reason for it at this time.
EDU>> T=1; del=0.01; t=0:del:T; lambda=1; xt=exp(-lambda*t/T); EDU>> (1/length(t))*fft_easy(xt,t) ans = 0.6326 + 0.0000i 0.0190 - 0.0986i 0.0072 - 0.0502i
We are given that \(m=1200\) kg, \(f=5\) Hz, \(\zeta =0.4\) and\[ z\relax (x) =\left \{ \begin {array} [c]{lll}x-5x^{2} & & 0<x<0.2\\ 0 & & 0.2<x<4 \end {array} \right . \]
A plot of \(z\relax (x) \) for first \(20\) meters is
z[x_] := Piecewise[{{x - 5 x^2, 0 <= x < 0.2}, {0, 0.2 <= x <= 4}}]
z[x_] /; x > 4 := z[Mod[x, 4]];
Table[{x, z[x]}, {x, 0, 21, .1}];
ListLinePlot[%, PlotRange -> {All, {0, .07}}, Frame -> True,
FrameLabel -> {{"z(x) hight or road (mm)", None}, {"meter",
"bumps on road"}}]
We need to be able to express \(z\relax (t) \) as \(\operatorname {Re}\left \{ Ze^{i\frac {2\pi }{T}t}\right \} \) where \(T\) is the period of the function \(z\relax (t) \). Hence we need to represent \(z\relax (x) \) as Fourier series approximation then replace \(x=vt\) and use the result.
The period \(T=4\) meter. Let \(\tilde {z}\relax (x) \) be the Fourier series approximation to \(z\relax (x) \), hence\[ \tilde {z}\relax (x) =\frac {1}{2}F_{0}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{N}} F_{n}e^{in\frac {2\pi }{T}x}\right ) \]
Where\[ F_{n}=\frac {2}{T}{\displaystyle \int \limits _{0}^{T}} z\relax (x) e^{-in\frac {2\pi }{T}x}dx=\frac {1}{2}{\displaystyle \int \limits _{0}^{2/10}} \left (x-5x^{2}\right ) e^{-in\frac {\pi }{2}x}dx=\frac {1}{2}{\displaystyle \int \limits _{0}^{2/10}} xe^{-in\frac {\pi }{2}x}dx-\frac {5}{2}{\displaystyle \int \limits _{0}^{2/10}} x^{2}e^{-in\frac {\pi }{2}x}dx \]
Using integration by parts \(\int udv=uv-\int vdu\), letting \(u=x\) and \(dv=e^{-in\frac {\pi }{2}x}\) then \(v=\int e^{-in\frac {\pi }{2}x}dx=\frac {ie^{-in\frac {\pi }{2}x}}{n\frac {\pi }{2}}\)hence\begin {align*} {\displaystyle \int \limits _{0}^{2/10}} xe^{-in\frac {\pi }{2}x}dx & =\left . x\frac {ie^{-in\frac {\pi }{2}x}}{n\frac {\pi }{2}}\right \vert _{0}^{\frac {2}{10}}-{\displaystyle \int \limits _{0}^{2/10}} \frac {ie^{-in\frac {\pi }{2}x}}{n\frac {\pi }{2}}dx\\ & =\frac {2}{10}\frac {ie^{-in\frac {\pi }{2}\frac {2}{10}}}{n\frac {\pi }{2}}-\frac {2}{n\pi }{\displaystyle \int \limits _{0}^{2/10}} ie^{-in\frac {\pi }{2}x}dx\\ & =\frac {4}{10}\frac {ie^{-in\frac {\pi }{10}}}{n\pi }-\frac {i2}{n\pi }\left ( \frac {e^{-in\frac {\pi }{2}x}}{-in\frac {\pi }{2}}\right ) _{0}^{\frac {2}{10}}\\ & =\frac {4}{10}\frac {ie^{-in\frac {\pi }{10}}}{n\pi }+\frac {4}{n^{2}\pi ^{2}}\left (e^{-in\frac {\pi }{2}x}\right ) _{0}^{\frac {2}{10}}\\ & =\frac {4}{10}\frac {ie^{-in\frac {\pi }{10}}}{n\pi }+\frac {4}{n^{2}\pi ^{2}}\left (e^{-in\frac {\pi }{2}\frac {2}{10}}-1\right ) \\ & =\frac {4i}{10n\pi }e^{-in\frac {\pi }{10}}+\frac {4}{n^{2}\pi ^{2}}e^{-in\frac {\pi }{10}}-\frac {4}{n^{2}\pi ^{2}}\\ & =e^{-in\frac {\pi }{10}}\left (\frac {4}{n^{2}\pi ^{2}}+\frac {2i}{5n\pi }\right ) -\frac {4}{n^{2}\pi ^{2}} \end {align*}
Now we do the second integral \({\displaystyle \int \limits _{0}^{2/10}} x^{2}e^{-in\frac {\pi }{2}x}dx\).
Integration by parts, \(\int udv=uv-\int vdu\), letting \(u=x^{2}\) and \(dv=e^{-in\frac {\pi }{2}x}\) then \(v=\frac {ie^{-in\frac {\pi }{2}x}}{n\frac {\pi }{2}}\) hence\begin {align*} {\displaystyle \int \limits _{0}^{2/10}} x^{2}e^{-in\frac {\pi }{2}x}dx & =\left [ x^{2}\frac {ie^{-in\frac {\pi }{2}x}}{n\frac {\pi }{2}}\right ] _{0}^{\frac {2}{10}}-{\displaystyle \int \limits _{0}^{2/10}} 2x\frac {ie^{-in\frac {\pi }{2}x}}{n\frac {\pi }{2}}dx\\ & =\frac {8}{100}\frac {ie^{-in\frac {\pi }{10}}}{n\pi }-\frac {4i}{n\pi }{\displaystyle \int \limits _{0}^{2/10}} xe^{-in\frac {\pi }{2}x}dx \end {align*}
But \({\displaystyle \int \limits _{0}^{2/10}} xe^{-in\frac {\pi }{2}x}dx\) was solved before and its results is Eq 2.1, hence\begin {align*} {\displaystyle \int \limits _{0}^{2/10}} x^{2}e^{-in\frac {\pi }{2}x}dx & =\frac {4}{100}\frac {ie^{-in\frac {\pi }{10}}}{n\frac {\pi }{2}}-\frac {4i}{n\pi }\left (e^{-in\frac {\pi }{10}}\left (\frac {4}{n^{2}\pi ^{2}}+\frac {2i}{5n\pi }\right ) -\frac {4}{n^{2}\pi ^{2}}\right ) \\ & =\frac {8i}{100n\pi }e^{-in\frac {\pi }{10}}-e^{-in\frac {\pi }{10}}\left ( \frac {16i}{n^{3}\pi ^{3}}-\frac {8}{5n^{2}\pi ^{2}}\right ) +\frac {16i}{n^{3}\pi ^{3}}\\ & =e^{-in\frac {\pi }{10}}\left (\frac {8i}{100n\pi }-\frac {16i}{n^{3}\pi ^{3}}+\frac {8}{5n^{2}\pi ^{2}}\right ) +\frac {16i}{n^{3}\pi ^{3}} \end {align*}
Putting all the above together, we obtain \(F_{n}\) as\begin {align*} F_{n} & =\frac {1}{2}{\displaystyle \int \limits _{0}^{2/10}} xe^{-in\frac {\pi }{2}x}dx-\frac {5}{2}{\displaystyle \int \limits _{0}^{2/10}} x^{2}e^{-in\frac {\pi }{2}x}dx\\ & =\frac {1}{2}\left [ e^{-in\frac {\pi }{10}}\left (\frac {4}{n^{2}\pi ^{2}}+\frac {2i}{5n\pi }\right ) -\frac {4}{n^{2}\pi ^{2}}\right ] -\frac {5}{2}\left [ e^{-in\frac {\pi }{10}}\left (\frac {8i}{100n\pi }-\frac {16i}{n^{3}\pi ^{3}}+\frac {8}{5n^{2}\pi ^{2}}\right ) +\frac {16i}{n^{3}\pi ^{3}}\right ] \\ & =e^{-in\frac {\pi }{10}}\left (\frac {2}{n^{2}\pi ^{2}}+\frac {i}{5n\pi }\right ) -\frac {2}{n^{2}\pi ^{2}}-e^{-in\frac {\pi }{10}}\left (\frac {20i}{100n\pi }-\frac {40i}{n^{3}\pi ^{3}}+\frac {20}{5n^{2}\pi ^{2}}\right ) -\frac {40i}{n^{3}\pi ^{3}}\\ & =e^{-in\frac {\pi }{10}}\left [ \frac {2}{n^{2}\pi ^{2}}+\frac {i}{5n\pi }-\frac {20i}{100n\pi }+\frac {40i}{n^{3}\pi ^{3}}-\frac {4}{n^{2}\pi ^{2}}\right ] -\frac {2}{n^{2}\pi ^{2}}-\frac {40i}{n^{3}\pi ^{3}}\\ & =e^{-in\frac {\pi }{10}}\left (\frac {40i}{n^{3}\pi ^{3}}-\frac {2}{n^{2}\pi ^{2}}\right ) -\frac {2}{n^{2}\pi ^{2}}-\frac {40i}{n^{3}\pi ^{3}} \end {align*}
Now \[ F_{0}=\frac {2}{T}{\displaystyle \int \limits _{0}^{T}} z\relax (x) dx=\frac {1}{2}{\displaystyle \int \limits _{0}^{2/10}} \left (x-5x^{2}\right ) dx=\frac {1}{300}\] Hence\begin {align*} \tilde {z}\relax (x) & =\frac {1}{2}F_{0}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{N}} F_{n}e^{in\frac {2\pi }{T}x}\right ) \\ & =\frac {1}{600}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{N}} \left (e^{-in\frac {\pi }{10}}\left (\frac {40i}{n^{3}\pi ^{3}}-\frac {2}{n^{2}\pi ^{2}}\right ) -\frac {2}{n^{2}\pi ^{2}}-\frac {40i}{n^{3}\pi ^{3}}\right ) e^{in\frac {\pi }{2}x}\right ) \\ & =\frac {1}{600}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{N}} e^{i\left (\frac {n\pi }{2}x-\frac {n\pi }{10}\right ) }\left (\frac {40i}{n^{3}\pi ^{3}}-\frac {2}{n^{2}\pi ^{2}}\right ) -e^{in\frac {\pi }{2}x}\left ( \frac {2}{n^{2}\pi ^{2}}+\frac {40i}{n^{3}\pi ^{3}}\right ) \right ) \\ & =\frac {1}{600}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{N}} \frac {-40}{n^{3}\pi ^{3}}\frac {1}{i}e^{i\left (\frac {n\pi }{2}x-\frac {n\pi }{10}\right ) }-\frac {2}{n^{2}\pi ^{2}}e^{i\left (\frac {n\pi }{2}x-\frac {n\pi }{10}\right ) }-\frac {2}{n^{2}\pi ^{2}}e^{in\frac {\pi }{2}x}+\frac {40}{n^{3}\pi ^{3}}\frac {1}{i}e^{in\frac {\pi }{2}x}\right ) \end {align*}
But \(x=vt\), hence\[ \tilde {z}\relax (t) =\frac {1}{600}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{N}} \frac {-40}{n^{3}\pi ^{3}}\frac {1}{i}e^{i\left (\frac {n\pi v}{2}t-\frac {n\pi }{10}\right ) }-\frac {2}{n^{2}\pi ^{2}}e^{i\left (\frac {n\pi v}{2}t-\frac {n\pi }{10}\right ) }-\frac {2}{n^{2}\pi ^{2}}e^{in\frac {\pi v}{2}t}+\frac {40}{n^{3}\pi ^{3}}\frac {1}{i}e^{in\frac {\pi v}{2}t}\right ) \]
Therefore the forcing frequency is \(n\varpi _{1}=n\frac {\pi v}{2}\) or from \(2\pi f_{1}=\frac {\pi v}{2}\), hence \(f_{1}=\frac {v}{4}\)Hz\(.\)The above can be written as\begin {align*} \tilde {z}\relax (t) & =\frac {1}{600}+{\displaystyle \sum \limits _{n=1}^{N}} \operatorname {Re}\left (\frac {-40}{n^{3}\pi ^{3}}\frac {1}{i}e^{i\left ( \frac {n\pi v}{2}t-\frac {n\pi }{10}\right ) }\right ) -{\displaystyle \sum \limits _{n=1}^{N}} \operatorname {Re}\left (\frac {2}{n^{2}\pi ^{2}}e^{i\left (\frac {n\pi v}{2}t-\frac {n\pi }{10}\right ) }\right ) \\ & -{\displaystyle \sum \limits _{n=1}^{N}} \operatorname {Re}\left (\frac {2}{n^{2}\pi ^{2}}e^{in\frac {\pi v}{2}t}\right ) +{\displaystyle \sum \limits _{n=1}^{N}} \operatorname {Re}\left (\frac {40}{n^{3}\pi ^{3}}\frac {1}{i}e^{in\frac {\pi v}{2}t}\right ) \\ & \\ & =\frac {1}{600}+{\displaystyle \sum \limits _{n=1}^{N}} \frac {-40}{n^{3}\pi ^{3}}\sin \left (n\varpi _{1}t-\frac {n\pi }{10}\right ) -{\displaystyle \sum \limits _{n=1}^{N}} \frac {2}{n^{2}\pi ^{2}}\cos \left (n\varpi _{1}t-\frac {n\pi }{10}\right ) \\ & -{\displaystyle \sum \limits _{n=1}^{N}} \frac {2}{n^{2}\pi ^{2}}\cos \left (n\varpi _{1}t\right ) +{\displaystyle \sum \limits _{n=1}^{N}} \frac {40}{n^{3}\pi ^{3}}\sin \left (n\varpi _{1}t\right ) \\ & \\ & =\frac {1}{600}-\frac {40}{\pi ^{3}}{\displaystyle \sum \limits _{n=1}^{N}} \frac {1}{n^{3}}\sin \left (n\varpi _{1}t-\frac {n\pi }{10}\right ) -\frac {2}{\pi ^{2}}{\displaystyle \sum \limits _{n=1}^{N}} \frac {1}{n^{2}}\cos \left (n\varpi _{1}t-\frac {n\pi }{10}\right ) \\ & -\frac {2}{\pi ^{2}}{\displaystyle \sum \limits _{n=1}^{N}} \frac {1}{n^{2}}\cos \left (n\varpi _{1}t\right ) +\frac {40}{\pi ^{3}}{\displaystyle \sum \limits _{n=1}^{N}} \frac {1}{n^{3}}\sin \left (n\varpi _{1}t\right ) \end {align*}
Where \(\varpi _{1}=\frac {\pi v}{2}\)
To verify the above, here is a plot for different number of fourier series terms showing that approximation improves as \(N\) increases. This was done for \(v=5m/s\) and for \(5\) seconds.
The equation of motion is\begin {align} my^{\prime \prime }+c\left (y^{\prime }-z^{\prime }\right ) +k\left (y-z\right ) & =0\nonumber \\ my^{\prime \prime }+cy^{\prime }+ky & =cz^{\prime }+kz \tag {2.1} \end {align}
From earlier, we found that fourier series approximation to \(z\left ( t\right ) \) is\begin {align*} z\relax (t) & =\frac {1}{600}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{\infty }} \frac {-40}{n^{3}\pi ^{3}}\frac {1}{i}e^{i\left (n\varpi t-\frac {n\pi }{10}\right ) }-\frac {2}{n^{2}\pi ^{2}}e^{i\left (n\varpi t-\frac {n\pi }{10}\right ) }-\frac {2}{n^{2}\pi ^{2}}e^{in\varpi t}+\frac {40}{n^{3}\pi ^{3}}\frac {1}{i}e^{in\varpi t}\right ) \\ & =\frac {1}{600}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{\infty }} \frac {-40}{n^{3}\pi ^{3}}e^{-i\frac {n\pi }{10}}\frac {1}{i}e^{in\varpi t}-\frac {2}{n^{2}\pi ^{2}}e^{-i\frac {n\pi }{10}}e^{in\varpi t}-\frac {2}{n^{2}\pi ^{2}}e^{in\varpi t}+\frac {40}{n^{3}\pi ^{3}}\frac {1}{i}e^{in\varpi t}\right ) \\ & =\frac {1}{600}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{\infty }} e^{in\varpi t}\left [ \frac {-40}{n^{3}\pi ^{3}}e^{-i\left (\frac {n\pi }{10}+\frac {\pi }{2}\right ) }-\frac {2}{n^{2}\pi ^{2}}e^{-i\frac {n\pi }{10}}-\frac {2}{n^{2}\pi ^{2}}+\frac {40}{n^{3}\pi ^{3}}e^{-i\frac {\pi }{2}}\right ] \right ) \end {align*}
Let \[ Z_{n}=\frac {-40}{n^{3}\pi ^{3}}e^{-i\left (\frac {n\pi }{10}+\frac {\pi }{2}\right ) }-\frac {2}{n^{2}\pi ^{2}}e^{-i\frac {n\pi }{10}}-\frac {2}{n^{2}\pi ^{2}}+\frac {40}{n^{3}\pi ^{3}}e^{-i\frac {\pi }{2}}\] Then above can be simplified to\[ z\relax (t) =\frac {1}{600}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{\infty }} e^{in\varpi t}Z_{n}\right ) \]
Where \(\varpi =\frac {\pi v}{2}\), hence\[ z^{\prime }\relax (t) =\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{\infty }} in\varpi e^{in\varpi t}Z_{n}\right ) \]
Hence, let \[ y_{ss}\relax (t) =\operatorname {Re}{\displaystyle \sum \limits _{n=1}^{\infty }} Y_{n}e^{in\varpi t}\]
Hence Eq 2.1 becomes\begin {align*} {\displaystyle \sum \limits _{n=1}^{\infty }} -mn^{2}\varpi ^{2}Y_{n}e^{in\varpi t}+{\displaystyle \sum \limits _{n=1}^{\infty }} icn\varpi Y_{n}e^{in\varpi t}+{\displaystyle \sum \limits _{n=1}^{\infty }} kY_{n}e^{in\varpi t} & ={\displaystyle \sum \limits _{n=1}^{\infty }} icn\varpi e^{in\varpi t}Z_{n}+\frac {k}{600}+{\displaystyle \sum \limits _{n=1}^{\infty }} ke^{in\varpi t}Z_{n}\\{\displaystyle \sum \limits _{n=1}^{\infty }} \left (-mn^{2}\varpi ^{2}+icn\varpi +k\right ) Y_{n}e^{in\varpi t} & ={\displaystyle \sum \limits _{n=1}^{\infty }} \left (icn\varpi +k\right ) Z_{n}e^{in\varpi t}+\frac {k}{600}\\{\displaystyle \sum \limits _{n=1}^{\infty }} \left (-mn^{2}\varpi ^{2}+icn\varpi +k\right ) Y_{n}e^{in\varpi t} & =\frac {k}{600}+{\displaystyle \sum \limits _{n=1}^{\infty }} \left (icn\varpi +k\right ) Z_{n}e^{in\varpi t} \end {align*}
Hence\begin {equation} Y_{n}=\frac {\left (icn\varpi +k\right ) }{-m\left (n\varpi \right ) ^{2}+icn\varpi +k}Z_{n} \label {eq:2.2} \end {equation}
Let \begin {align*} D\left (r_{n},\zeta \right ) & =\frac {icn\varpi +k}{-m\left (n\varpi \right ) ^{2}+icn\varpi +k}\\ & =\frac {i2\zeta m\omega _{nat}n\varpi +\omega _{nat}^{2}m}{-m\left ( n\varpi \right ) ^{2}+i2\zeta m\omega _{nat}n\varpi +\omega _{nat}^{2}m}\\ & =\frac {i2\zeta n\frac {\varpi }{\omega _{nat}}+1}{-\left (n\frac {\varpi }{\omega _{nat}}\right ) ^{2}+i2\zeta n\frac {\varpi }{\omega _{nat}}+1}\\ & =\frac {1+i2\zeta r_{n}}{\left (1-r_{n}^{2}\right ) +i2\zeta r_{n}} \end {align*}
Where in the above \(r_{n}=\frac {n\varpi }{\omega _{nat}}\) where \(\varpi \) is \(\frac {2\pi }{T}\) which means it is the fundamental frequency of the forcing function and \(\omega _{nat}\) is the natural frequency.
Then Eq ?? becomes\[ Y_{n}=D\left (r_{n},\zeta \right ) Z_{n}\]
And the steady state solution \(y_{ss}\relax (t) \) becomes\[ y_{ss}\relax (t) =\frac {k}{600}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{\infty }} D\left (r_{n},\zeta \right ) Z_{n}e^{in\varpi t}\right ) \]
Now we can answer the question. When \(c=0\) then \(D\left (r_{n},\zeta \right ) \) reduces to \(\frac {k}{-m\left (n\varpi \right ) ^{2}+k}=\frac {1}{1-\left ( n\frac {\varpi }{\omega _{nat}}\right ) ^{2}}=\frac {1}{1-r_{n}^{2}}\), hence\[ y_{ss}\relax (t) =\frac {k}{600}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{\infty }} \frac {1}{1-r_{n}^{2}}Z_{n}e^{in\varpi t}\right ) \]
So the displacement \(y_{ss}\relax (t) \) will be resonant when \(r_{n}=1\) or \(\frac {n\pi v}{2\omega _{nat}}=1\) or \(v=\frac {2\omega _{nat}}{n\pi }\)
Hence\[ v=\frac {2\left (2\pi 5\right ) }{n\pi }=\frac {20}{n}\]
Hence \(v=20,10,5,2.5,1.25,\cdots \) meter/sec will each cause resonance. To verify, here is a plot of \(y_{ss}\relax (t) \) with no damper for speed near resonance \(v=19.99\) and comparing this for speeds away from resonance speed. This plot shows that when speed \(v\) is close to any of the above speeds, then the displacement \(y_{ss}\relax (t) \) becomes very large. Once the speed is away from those values, then \(y_{ss}\relax (t) \) quickly comes down to steady state \(F/k\) value.
The function is periodic with period \(T=2\tau \)\[ f\relax (t) =\begin {array} [c]{lll}\frac {P}{\tau }t & & 0<t<\tau \\ 0 & & \tau <t<2\tau \end {array} \]
and \(f\left (t\pm T\right ) =f\relax (t) \). Let \(\tilde {f}\left ( t\right ) \) be the Fourier series approximation to \(f\relax (t) \), hence\begin {equation} \tilde {f}\relax (t) =\frac {1}{2}F_{0}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{N}} F_{n}e^{in\frac {2\pi }{T}x}\right ) \label {eq:3.1} \end {equation}
Where\begin {align*} F_{n} & =\frac {2}{T}{\displaystyle \int \limits _{0}^{T}} f\relax (t) e^{-in\frac {2\pi }{T}t}dt\\ & =\frac {2}{2\tau }{\displaystyle \int \limits _{0}^{\tau }} \frac {P}{\tau }te^{-in\frac {\pi }{\tau }t}dt\\ & =\frac {P}{\tau ^{2}}{\displaystyle \int \limits _{0}^{\tau }} te^{-in\frac {\pi }{\tau }t}dt \end {align*}
Using integration by parts \(\int udv=uv-\int vdu\), letting \(u=t\) and \(dv=e^{-in\frac {\pi }{\tau }t}\) then \(v=\int e^{-in\frac {\pi }{\tau }t}dt=\frac {ie^{-in\frac {\pi }{\tau }t}}{n\frac {\pi }{\tau }}\)hence\begin {align*} F_{n} & =\frac {P}{\tau ^{2}}\left [ \left (t\frac {ie^{-in\frac {\pi }{\tau }t}}{n\frac {\pi }{\tau }}\right ) _{0}^{\tau }-\frac {i}{n\frac {\pi }{\tau }}{\displaystyle \int \limits _{0}^{\tau }} e^{-in\frac {\pi }{\tau }t}dt\right ] \\ & =\frac {P}{\tau ^{2}}\left [ \left (\tau \frac {ie^{-in\frac {\pi }{\tau }\tau }}{n\frac {\pi }{\tau }}\right ) -\frac {i}{n\frac {\pi }{\tau }}\left ( \frac {e^{-in\frac {\pi }{\tau }t}}{-in\frac {\pi }{\tau }}\right ) _{0}^{\tau }\right ] \\ & =\frac {P}{\tau ^{2}}\left [ \left (\tau ^{2}\frac {ie^{-in\pi }}{n\pi }\right ) +\frac {\tau ^{2}}{n^{2}\pi ^{2}}\left (e^{-in\frac {\pi }{\tau }t}\right ) _{0}^{\tau }\right ] \\ & =\frac {P}{\tau ^{2}}\left [ \left (\tau ^{2}\frac {ie^{-in\pi }}{n\pi }\right ) +\frac {\tau ^{2}}{n^{2}\pi ^{2}}\left (e^{-in\pi }-1\right ) \right ] \end {align*}
\(e^{-in\pi }=\cos \left (n\pi \right ) =\left (-1\right ) ^{n}\), hence\[ F_{n}=\frac {P}{\tau ^{2}}\left [ \left (\tau ^{2}\frac {i\left (-1\right ) ^{n}}{n\pi }\right ) +\frac {\tau ^{2}}{n^{2}\pi ^{2}}\left (\left (-1\right ) ^{n}-1\right ) \right ] \]
Hence for even \(n\)\begin {align*} F_{n} & =\frac {P}{\tau ^{2}}\left [ \left (\tau ^{2}\frac {i}{n\pi }\right ) \right ] \\ & =P\frac {i}{n\pi } \end {align*}
and for odd \(n\)\begin {align*} F_{n} & =\frac {P}{\tau ^{2}}\left [ \left (-\tau ^{2}\frac {i}{n\pi }\right ) -2\frac {\tau ^{2}}{n^{2}\pi ^{2}}\right ] \\ & =-\frac {P}{n\pi }\left (\frac {2}{n\pi }+i\right ) \end {align*}
\begin {align*} F_{0} & =\frac {P}{\tau ^{2}}{\displaystyle \int \limits _{0}^{\tau }} tdt\\ & =\frac {P}{\tau ^{2}}\left (\frac {t^{2}}{2}\right ) _{0}^{\tau }=\frac {P}{\tau ^{2}}\left (\frac {\tau ^{2}}{2}\right ) \\ & =\frac {P}{2} \end {align*}
Now Eq ?? becomes\begin {align*} \tilde {f}\relax (t) & =\frac {1}{2}F_{0}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{N}} F_{n}e^{in\frac {2\pi }{T}x}\right ) \\ & =\frac {1}{2}F_{0}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{even\ n}} F_{n}e^{in\frac {2\pi }{T}t}+{\displaystyle \sum \limits _{odd\ n}} F_{n}e^{in\frac {2\pi }{T}t}\right ) \\ & =\frac {p}{4}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{even\ n}} P\frac {i}{n\pi }e^{in\frac {2\pi }{T}t}+{\displaystyle \sum \limits _{odd\ n}} -\frac {P}{n\pi }\left (\frac {2}{n\pi }+i\right ) e^{in\frac {2\pi }{T}t}\right ) \\ & =\frac {p}{4}+\operatorname {Re}\left (\frac {P}{\pi }{\displaystyle \sum \limits _{even\ n}} \frac {i}{n}e^{in\frac {2\pi }{T}t}-\frac {P}{\pi }{\displaystyle \sum \limits _{odd\ n}} \frac {1}{n}\left (\frac {2}{n\pi }+i\right ) e^{in\frac {2\pi }{T}t}\right ) \\ & =\frac {P}{4}+\operatorname {Re}\left (\frac {P}{\pi }{\displaystyle \sum \limits _{even\ n}} \frac {i}{n}e^{in\frac {2\pi }{T}t}-\frac {P}{\pi }{\displaystyle \sum \limits _{odd\ n}} \left (\frac {2}{n^{2}\pi }+\frac {i}{n}\right ) e^{in\frac {2\pi }{T}t}\right ) \end {align*}
To verify, here is a plot of the above, using \(P=1\) and \(\tau =0.5\) sec for \(t=0\cdots 2\) seconds. This shows as more terms are added, the approximation becomes very close to the function. At \(N=40\) the approximation appears very good.
Now we need to write \(f\relax (t) \) as sum of exponential to answer the question. \[ \tilde {f}\relax (t) =\frac {1}{2}F_{0}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{N}} F_{n}e^{in\frac {2\pi }{T}x}\right ) \]
where \(\varpi \) is the fundamental frequency of the force given by \(\frac {2\pi }{T}=\frac {2\pi }{2\tau }=\frac {\pi }{\tau }\)
Hence, let \(y_{ss}={\displaystyle \sum \limits _{n=-\infty }^{\infty }} Y_{n}e^{in\varpi t}\), then\begin {align*} \operatorname {Re}\left (m{\displaystyle \sum \limits _{n=-\infty }^{\infty }} -\left (n\varpi \right ) ^{2}Y_{n}e^{in\varpi t}+c{\displaystyle \sum \limits _{n=-\infty }^{\infty }} in\varpi Y_{n}e^{in\varpi t}+k{\displaystyle \sum \limits _{n=-\infty }^{\infty }} Y_{n}e^{in\varpi t}\right ) & =\frac {1}{2}F_{0}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{N}} F_{n}e^{in\frac {2\pi }{T}x}\right ) \\{\displaystyle \sum \limits _{n=-\infty }^{\infty }} \left (-m\left (n\varpi \right ) ^{2}+icn\varpi +k\right ) Y_{n}e^{in\varpi t} & =\frac {1}{2}F_{0}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{N}} F_{n}e^{in\frac {2\pi }{T}x}\right ) \end {align*}
Hence\begin {align*} Y_{n} & =\frac {F_{n}}{k}\frac {1}{\left (1-\left (n\frac {\varpi }{\omega _{nat}}\right ) ^{2}\right ) +i2\zeta n\frac {\varpi }{\omega _{nat}}}\\ & =\frac {F_{n}}{k}\frac {1}{\left (1-\left (nr\right ) ^{2}\right ) +i2\zeta nr} \end {align*}
Hence\[ y_{ss}=\frac {1}{2}F_{0}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{\infty }} Y_{n}e^{in\varpi t}\right ) \]
Finding \(Y_{n}\) for \(\tau =\frac {\pi }{3\omega _{nat}}\)
where \(r=\frac {\varpi }{\omega _{nat}}\). When \(\zeta =0.04\) and \(\tau =\frac {\pi }{3\omega _{nat}}\), hence now \(r=\frac {2\pi }{\left (2\tau \right ) \omega _{nat}}=\frac {2\pi }{\left (2\frac {\pi }{3\omega _{nat}}\right ) \omega _{nat}}=3\), therefore\begin {align*} Y_{n} & =\frac {F_{n}}{k}\frac {1}{\left (1-\left (3n\right ) ^{2}\right ) +i6\left (0.04\right ) n}\\ & =\frac {F_{n}}{k}\frac {1}{\left (1-9n^{2}\right ) +i0.24n} \end {align*}
The largest \(Y_{n}\) will occur when the denominator of the above is smallest. Plotting the modulus of the denominator \(\sqrt {\left (1-9n^{2}\right ) ^{2}+\left (0.24n\right ) ^{2}}\) for different \(n\) values shows that \(n=1\) is the values which makes it minimum.
This happens since for any \(n>1\) the denominator will become larger due to \(n^{2}\) and hence \(Y_{n}\) will become smaller. So \(n=1\) will be used.
For \(n=1\), we obtain\[ Y_{1}=\frac {F_{1}}{k}\frac {1}{\left (1-9\right ) +i6\left (0.04\right ) }\]
But \(F_{1}=-\frac {P}{\pi }\left (\frac {2}{\pi }+i\right ) \), hence\begin {align*} Y_{1} & =\frac {-\frac {P}{\pi }\left (\frac {2}{\pi }+i\right ) }{k}\frac {1}{\left (1-9\right ) +i6\left (0.04\right ) }=\frac {-P}{\pi k}\frac {\left (\frac {2}{\pi }+i\right ) }{-8+i0.24}\\ & =\frac {P}{\pi k}\frac {\frac {2}{\pi }+i}{8-i0.24}=\frac {P}{\pi k}\frac {\left (\frac {2}{\pi }+i\right ) \left (8+i0.24\right ) }{(8-i0.24)(8+i0.24)}\\ & =\frac {P}{\pi k}\left (0.075759+0.12727\allowbreak i\right ) \end {align*}
Therefore\[ Y_{1}=\frac {P}{k}\left (0.024115+0.0405i\right ) \]
Here is a list of \(Y_{n}\) for \(n=1\cdots 10\) with the phase and magnitude of each (this was done for \(\frac {p}{k}=1\))
From the above we see that most of the energy in the response will be contained in \(Y_{1}\) and adding more terms will not have large effect on the response shape. This is confirmed by the plot that follows.
Plot for the steady state
Since \[ y_{ss}=\frac {1}{2}F_{0}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{\infty }} Y_{n}e^{in\varpi t}\right ) \]
Where now \(r=\frac {\varpi }{\omega _{nat}}\). When \(\zeta =0.04\) and \(\tau =\frac {\pi }{3\omega _{nat}}\), hence now \(r=\frac {2\pi }{\left (2\tau \right ) \omega _{nat}}=\frac {2\pi }{\left (2\frac {\pi }{3\omega _{nat}}\right ) \omega _{nat}}\) therefore \(r=3\)\begin {align*} y_{ss} & =\frac {p}{4}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1,3,5\cdots }^{\infty }} Y_{n}e^{in\varpi t}+{\displaystyle \sum \limits _{n=2,4,6\cdots }^{\infty }} Y_{n}e^{in\varpi t}\right ) \\ & =\frac {p}{4}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1,3,5\cdots }^{\infty }} \frac {F_{n_{odd}}}{k}\frac {1}{\left (1-\left (nr\right ) ^{2}\right ) +i2\zeta nr}e^{in\varpi t}+{\displaystyle \sum \limits _{n=2,4,6\cdots }^{\infty }} \frac {F_{n_{even}}}{k}\frac {1}{\left (1-\left (nr\right ) ^{2}\right ) +i2\zeta nr}e^{in\varpi t}\right ) \\ & =\frac {p}{4}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1,3,5\cdots }^{\infty }} \frac {-\frac {P}{n\pi }\left (\frac {2}{n\pi }+i\right ) }{k}\frac {1}{\left ( 1-\left (nr\right ) ^{2}\right ) +i2\zeta nr}e^{in\varpi t}+{\displaystyle \sum \limits _{n=2,4,6\cdots }^{\infty }} \frac {P\frac {i}{n\pi }}{k}\frac {1}{\left (1-\left (nr\right ) ^{2}\right ) +i2\zeta nr}e^{in\varpi t}\right ) \\ & =\frac {p}{4}+\frac {p}{k}\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1,3,5\cdots }^{\infty }} -\frac {\frac {1}{n\pi }\left (\frac {2}{n\pi }+i\right ) }{\left (1-\left ( nr\right ) ^{2}\right ) +i2\zeta nr}e^{in\varpi t}+{\displaystyle \sum \limits _{n=2,4,6\cdots }^{\infty }} \frac {\frac {i}{n\pi }}{\left (1-\left (nr\right ) ^{2}\right ) +i2\zeta nr}e^{in\varpi t}\right ) \end {align*}
Now let \(r=3\), \(\zeta =0.04\). Normalizing the equation for \(\varpi =1\) which implies \(\tau =\pi \) and \(k=1\) and \(p=1\), then the above becomes\[ y_{ss}=\frac {1}{4}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1,3,5\cdots }^{\infty }} -\frac {\frac {1}{n\pi }\left (\frac {2}{n\pi }+i\right ) }{\left (1-\left ( 3n\right ) ^{2}\right ) +i2\left (0.04\right ) 3n}e^{int}+{\displaystyle \sum \limits _{n=2,4,6\cdots }^{\infty }} \frac {\frac {i}{n\pi }}{\left (1-\left (3n\right ) ^{2}\right ) +i2\left ( 0.04\right ) 3n}e^{int}\right ) \]
Here is a plot of the above for \(t=0\cdots 20\) seconds for different values of \(n\)
We see from the above plot, that \(y_{ss}\relax (t) \) does not change too much as more terms are added, since when \(r=3\), then \(Y_{n}\) for \(n=1\) contains most of the energy, hence adding more terms did not have an effect.
Repeating the calculations for \(\tau =\frac {3\pi }{\omega _{nat}}\)
\(r=\frac {\varpi }{\omega _{nat}}\). When \(\zeta =0.04\) and \(\tau =\frac {3\pi }{\omega _{nat}}\), hence now \(r=\frac {2\pi }{\left (2\tau \right ) \omega _{nat}}=\frac {2\pi }{\left (2\frac {3\pi }{\omega _{nat}}\right ) \omega _{nat}}=\frac {1}{3}\), therefore\begin {align*} Y_{n} & =\frac {F_{n}}{k}\frac {1}{\left (1-\left (nr\right ) ^{2}\right ) +i2\zeta nr}\\ & =\frac {F_{n}}{k}\frac {1}{\left (1-\left (\frac {1}{3}n\right ) ^{2}\right ) +i\frac {2}{3}\left (0.04\right ) n}\\ & =\frac {F_{n}}{k}\frac {1}{\left (1-\frac {n^{2}}{9}\right ) +i0.0267n} \end {align*}
The largest \(Y_{n}\) will occur when the denominator of the above is smallest. Similar to above, we can either find \(n\) which minimizes the denominator (by taking derivative and setting it to zero and solve for \(n\)) or we can make a plot and see how the function behaves. Making a plot shows this
From the above we see that the smallest value of the denominator happens when \(n=3\).
so using \(n=3\) we find\begin {align*} Y_{3} & =\frac {F_{3}}{k}\frac {1}{\left (1-\left (3r\right ) ^{2}\right ) +i2\zeta 3r}\\ & =\frac {F_{3}}{k}\frac {1}{\left (1-\left (3\frac {1}{3}\right ) ^{2}\right ) +i2\left (0.04\right ) 3\frac {1}{3}}\\ & =\frac {F_{3}}{k}\frac {1}{i0.08} \end {align*}
But \(F_{n}=-\frac {P}{n\pi }\left (\frac {2}{n\pi }+i\right ) \), hence \[ F_{3}=-\frac {P}{3\pi }\left (\frac {2}{3\pi }+i\right ) \]
Therefore\[ Y_{3}=\frac {-\frac {P}{3\pi }\left (\frac {2}{3\pi }+i\right ) }{k}\frac {1}{i0.08}\]
Hence\[ Y_{3}=\frac {p}{k}\left (-1.\allowbreak 3263+0.28145\allowbreak i\right ) \]
Here is a list of \(Y_{n}\) for \(n=1\cdots 10\) with the phase and magnitude of each (this was done for \(\frac {p}{k}=1\))
We see from the above that \(\left \vert Y_{3}\right \vert \) is the largest harmonic.
Plot for the steady state
Since \[ y_{ss}=\frac {1}{2}F_{0}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{\infty }} Y_{n}e^{in\varpi t}\right ) \]
Where now \(r=\frac {\varpi }{\omega _{nat}}\). When \(\zeta =0.04\) and \(\tau =\frac {3\pi }{\omega _{nat}}\), hence now \(r=\frac {2\pi }{\left (2\tau \right ) \omega _{nat}}=\frac {2\pi }{\left (2\frac {3\pi }{\omega _{nat}}\right ) \omega _{nat}}=\frac {1}{3}\), therefore from above\[ y_{ss}=\frac {p}{4}+\frac {p}{k}\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1,3,5\cdots }^{\infty }} -\frac {1}{n\pi }\left (\frac {2}{n\pi }+i\right ) \frac {1}{\left (1-\left ( nr\right ) ^{2}\right ) +i2\zeta nr}e^{in\varpi t}+{\displaystyle \sum \limits _{n=2,4,6\cdots }^{\infty }} \frac {i}{n\pi }\frac {1}{\left (1-\left (nr\right ) ^{2}\right ) +i2\zeta nr}e^{in\varpi t}\right ) \]
Now let \(r=\frac {1}{3}\), \(\zeta =0.04\), and assuming \(\tau =0.5\) then \(\varpi =\frac {2\pi }{2\tau }=\frac {\pi }{0.5},\) and assuming \(k=1\), then the above becomes\begin {align*} y_{ss} & =\frac {1}{4}+\frac {1}{k}\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1,3,5\cdots }^{\infty }} -\frac {1}{n\pi }\left (\frac {2}{n\pi }+i\right ) \frac {1}{\left (1-\left ( n\frac {1}{3}\right ) ^{2}\right ) +i2\left (0.04\right ) \frac {1}{3}n}e^{in\frac {\pi }{0.5}t}\right ) \\ & +\frac {1}{k}\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=2,4,6\cdots }^{\infty }} \frac {i}{n\pi }\frac {1}{\left (1-\left (n\frac {1}{3}\right ) ^{2}\right ) +i2\left (0.04\right ) \frac {1}{3}n}e^{in\frac {\pi }{0.5}t}\right ) \end {align*}
Here is a plot of the above for \(t=0\cdots 20\) seconds for different values of \(n\)
We see now that after \(n=3\) that the response did not change much by adding more terms, this is because more of the energy are contained in the first 3 harmonics with \(Y_{n}\) being the the largest.