2.3 HW3

2.3.1 Problem description

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2.3.2 Mathcad initial calculations

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2.3.3 Mathematical model

The equation of motion of the car is$$m{y}^{\prime \prime }+c{y}^{\prime }+ky=c{y}_g^{\prime }+k{y}_g$$

Let $y-y_g=u$ which is the distance between $m$ and the ground. Hence the equation of motion now becomes$$\begin{array}{rl}m(u^{\prime \prime }+y_g^{\prime \prime })+c(u^{\prime }+y_g^{\prime })+k(u+y_g)& =c{y}_g^{\prime }+k{y}_g\\ m{u}^{\prime \prime }+c{u}^{\prime }+ku& =-m{y}_g^{\prime \prime }\end{array}$$

2.3.4 Summary of results found

2.3.4.1 Bridge data

	imperial	SI
span length $\lambda$	$70$ ft	$70\times 0.3048=$ $21.336$ meter
upward camber $\mathrm{\Delta }$	${2.5}^{\prime \prime }=$ $0.20833$ft	$2.5\times 0.0254=$ $0.0635$ meter

2.3.4.2 Car data

	imperial
mass of car	$\frac{1500}{32.2}=$ $46.584$ $\frac{lb.s^2}{ft}$
speed of car	$30$ mile/hr$=44.0$ ft/sec
critical damping ratio is $\zeta$	$0.75$
spring constant $k$	$2400$ lb/ft
natural frequency ${\omega }_n=\sqrt{\frac{k}{m}}$	$\sqrt{\frac{2400}{46.584}}=7.1777$ rad/sec
natural frequency $f_n=\frac{{\omega }_n}{2\pi }$	$\frac{7.1777}{2\pi }=$ $1.1424$ Hz
natural period $T_n=\frac{1}{f_n}$	$\frac{1}{1.1424}=$ $0.87535$ sec
natural damped frequency ${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}$	$7.1777\sqrt{1-{0.75}^2}=$ $4.7476$ rad/sec
natural damped frequency $f_d=\frac{{\omega }_d}{2\pi }$	$\frac{4.7476}{2\pi }=0.7556$ Hz
$T_p$ time to driver over one span $=\frac{\lambda }{v}$	$1.591$sec
$T_a$ time to cross the bridge (duration of loading)	$7\times 1.591=11.137$ sec

2.3.4.3 Results

$a_n$ values found for up to $n=10$

Peak relative displacement of the driver

Maximum relative displacement was $0.24$ inch and it occurred during transient phase.

Peak total displacement of the driver

$0.165$ inch + $2.5$ inch =$2.665$ inch and it occurred during steady state phase at multiples of half the period $T_p$ while on the bridge.

Number of $a_n$ terms used

In addition to $a_0$ term, the next $5$ terms are used for a total of $6$ terms.

2.3.5 Analysis

2.3.5.1 Generate load equation

The equation of the ground is shown in this diagram

Therefore, the equation of span is $$y_g\left(x\right)=\begin{array}{lll}\mathrm{\Delta }\sin \left(\frac{\pi }{\lambda }x\right)& & 0\le x\le \lambda \end{array}$$

Hence, we convert it to be a function of time using $x=vt$, hence$$\begin{array}{rl}{y}_g\left(t\right)& =\begin{array}{lll}\mathrm{\Delta }\sin \left(\frac{\pi }{\lambda }vt\right)& & 0\le t\le \frac{\lambda }{v}\end{array}\\ & =\begin{array}{lll}\mathrm{\Delta }\sin \left(\varpi t\right)& & 0\le t\le T_p\end{array}\\ & =\begin{array}{lll}\mathrm{\Delta }\sin \left(\frac{\pi }{T_p}t\right)& & 0\le t\le T_p\end{array}\end{array}$$

Where in the above $\varpi =\frac{\pi }{T_p}$ is the fundamental frequency of the ground motion. Hence $y_g^{\prime }\left(t\right)=\mathrm{\Delta }\frac{\pi }{T_p}\cos \left(\frac{\pi }{T_p}t\right)$ and $$y_g^{\prime \prime }\left(t\right)=-\mathrm{\Delta }{\left(\frac{\pi }{T_p}\right)}^2\sin \left(\frac{\pi }{T_p}t\right)$$

And $$\beta =\frac{T_n}{T_p}=\frac{0.87535}{1.591}=0.55019$$

Then load in one span $0<t<T_p$ is $$P_a\left(t\right)=m\mathrm{\Delta }{\left(\frac{\pi }{T_p}\right)}^2\sin \left(\frac{\pi }{T_p}t\right)$$

Let$$\begin{array}{rl}{P}_o& =m\mathrm{\Delta }{\left(\frac{\pi }{T_p}\right)}^2=\left(46.584\right)\left(0.20833\right){\left(\frac{\pi }{1.591}\right)}^2\\ & =37.840\text{ lb}\end{array}$$

Then the load becomes$$\begin{array}{}\text{(1)}& P_a\left(t\right)=P_o\sin \left(\frac{\pi }{T_p}t\right)\end{array}$$

2.3.5.2 Convert load to Fourier series

Now we need to convert Eq ?? to Fourier series¹. Let ${\tilde{P}}_a\left(t\right)$ be the Fourier series approximation to $P_a\left(t\right)$, hence$$\begin{array}{rl}{\tilde{P}}_a\left(t\right)& =a_0+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{a}_n\cos \left(n\frac{2\pi }{T_p}t\right)+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{b}_n\sin \left(n\frac{2\pi }{T_p}t\right)\\ a_0& =\frac{1}{T_p}{\displaystyle \underset{0}{\overset{T_p}{\int }}}{P}_a\left(t\right)dt\\ a_n& =\frac{2}{T_p}{\displaystyle \underset{0}{\overset{T_p}{\int }}}{P}_a\left(t\right)\cos \left(n\frac{2\pi }{T_p}t\right)dt\\ b_n& =\frac{2}{T_p}{\displaystyle \underset{0}{\overset{T_p}{\int }}}{P}_a\left(t\right)\sin \left(n\frac{2\pi }{T_p}t\right)dt\end{array}$$

Hence $$\begin{array}{rl}{a}_0& =\frac{1}{T_p}{\displaystyle \underset{0}{\overset{T_p}{\int }}}{P}_a\left(t\right)dt=\frac{P_o}{T_p}{\displaystyle \underset{0}{\overset{T_p}{\int }}}\sin \left(\frac{\pi }{T_p}t\right)dt=\frac{P_o}{T_p}{\left(\frac{-\cos \left(\frac{\pi }{T_p}t\right)}{\frac{\pi }{T_p}}\right)}_0^{T_p}=-\frac{P_o}{\pi }(\cos \left(\pi \right)-1)\\ & =\frac{2P_o}{\pi }=\frac{2\left(37.840\right)}{\pi }\\ & =24.090\text{ lb}\end{array}$$

And$$\begin{array}{rl}{a}_n& =\frac{2}{T_p}{\displaystyle \underset{0}{\overset{T_p}{\int }}}{P}_a\left(t\right)\cos \left(n\frac{2\pi }{T_p}t\right)dt\\ & =\frac{2P_o}{T_p}{\displaystyle \underset{0}{\overset{T_p}{\int }}}\sin \left(\frac{\pi }{T_p}t\right)\cos \left(n\frac{2\pi }{T_p}t\right)dt\\ & =\frac{4P_o}{\pi -4n^2\pi }\cos {\left(n\pi \right)}^2\end{array}$$

But $\cos {\left(n\pi \right)}^2=1$ Hence$$a_n=\frac{4P_o}{\pi -4n^2\pi }$$

and$$\begin{array}{rl}{b}_n& =\frac{1}{T_p}{\displaystyle \underset{0}{\overset{T_p}{\int }}}{P}_a\left(t\right)\sin \left(2\pi n\frac{t}{T_p}\right)dt\\ & =\frac{2P_o}{T_p}{\displaystyle \underset{0}{\overset{T_p}{\int }}}\sin \left(\frac{\pi }{T_p}t\right)\cos \left(n\frac{2\pi }{T_p}t\right)dt\\ & =\frac{2P_o}{\pi -4n^2\pi }\sin \left(2n\pi \right)\end{array}$$

But $\sin \left(2n\pi \right)=0$ for all integer $n$, hence $b_n=0.$Therefore$$\begin{array}{rl}{\tilde{P}}_a\left(t\right)& =a_0+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{a}_n\cos \left(2\pi n\frac{t}{T_p}\right)\\ & =a_0+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}\frac{4P_o}{\pi -4n^2\pi }\cos \left(2\pi n\frac{t}{T_p}\right)\end{array}$$

Using the numerical values found, we obtain$${\tilde{P}}_a\left(t\right)=24.0897+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}\frac{4\left(37.840\right)}{\pi -4n^2\pi }\cos \left(2\pi n\frac{t}{1.591}\right)$$

2.3.5.3 Plot of load and its Fourier series approximation

This plot below shows $P_a\left(t\right)$ and its Fourier series approximation ${\tilde{P}}_a\left(t\right)$ as more terms are added. This was plotted for $t=0\cdots 5\sec$. This was done to verify that the Fourier series approximation is correct before going to the next stage of the analysis. The actual calculations used the first 6 terms of $a_n$.

2.3.5.4 Finding the steady state response

The equation of motion of the car is$$m{y}^{\prime \prime }+c{y}^{\prime }+ky=c{y}_g^{\prime }+k{y}_g$$

Let $y-y_g=u$ which is the distance between $m$ and the ground. Hence the equation of motion now becomes$$\begin{array}{rl}m(u^{\prime \prime }+y_g^{\prime \prime })+c(u^{\prime }+y_g^{\prime })+k(u+y_g)& =c{y}_g^{\prime }+k{y}_g\\ \text{(2)}& m{u}^{\prime \prime }+c{u}^{\prime }+ku& =-m{y}_g^{\prime \prime }\end{array}$$

Hence Eq ?? becomes$$\begin{array}{rl}m{u}^{\prime \prime }+c{u}^{\prime }+ku& =m\mathrm{\Delta }{\left(\frac{\pi }{T_p}\right)}^2\sin \left(\frac{\pi }{T_p}t\right)\\ & =P_a\left(t\right)\\ & ={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n\cos \left(2\pi n\frac{t}{T_p}\right)\\ & =\mathrm{Re}\left\{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{e}^{in\varpi t}\right\}\end{array}$$

Where $\varpi =\frac{2\pi }{T_p}$ is the fundamental loading harmonic. Let $u_{ss}\left(n\right)=\mathrm{Re}\left\{U_n{e}^{in\varpi t}\right\}$ be the response due to the $n$ term in the loading function. Hence the equation of motion now becomes$$\begin{array}{rl}m\mathrm{Re}\{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}-n^2{\varpi }^2{U}_n{e}^{in\varpi t}\}+c\mathrm{Re}\left\{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}i\varpi n{U}_n{e}^{in\varpi t}\right\}+k\mathrm{Re}\left\{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{U}_n{e}^{in\varpi t}\right\}& =\mathrm{Re}\left\{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{e}^{in\varpi t}\right\}\\ (-n^2{\varpi }^{2}m+cin\varpi +k)U_n& =a_n\\ U_n& =\frac{a_n}{-n^2{\varpi }^{2}m+cin\varpi +k}\\ & =\frac{a_n}{k}\frac{1}{(1-n^2{r}^2)+2i\zeta nr}\end{array}$$

Hence the transfer function is

$$\begin{array}{rl}(-n^2{\varpi }^{2}m+cin\varpi +k)U_n& =a_n\\ U_n& =\frac{a_n}{-n^2{\varpi }^{2}m+cin\varpi +k}\\ & =\frac{a_n}{k}\frac{1}{(1-n^2{r}^2)+2i\zeta nr}\end{array}$$

Therefore, steady state response is$$\begin{array}{rl}{y}_{ss}\left(t\right)& =\mathrm{Re}\left\{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{U}_n{e}^{in\varpi t}\right\}\\ & =\mathrm{Re}\left\{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}\frac{a_n}{k}\frac{1}{(1-n^2{r}^2)+2i\zeta nr}{e}^{in\varpi t}\right\}\\ \text{(3)}& & =\mathrm{Re}\left\{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}\stackrel{U_n}{\overbrace{\frac{a_n}{k}D(\zeta ,r,n)}}{e}^{in\varpi t}\right\}\end{array}$$

Where $D(\zeta ,r,n)$ is the $n^{th}$ harmonic dynamic magnification factor $$\left|D(\zeta ,r,n)\right|=\sqrt{\frac{1}{{(1-n^2{r}^2)}^2+{\left(2\zeta nr\right)}^2}}$$

and$$\arg D(\zeta ,r,n)=-{\tan }^{-1}\left(\frac{2\zeta nr}{1-n^2{r}^2}\right)$$

In the above, $$r=\frac{\varpi }{{\omega }_{nat}}=\frac{\frac{2\pi }{T_p}}{{\omega }_{nat}}=\frac{\frac{2\pi }{1.591}}{7.1777}=\boxed{0.5502}$$

This is a list of the magnitude of $U_n$ for different $n$ value to examine the contribution of each harmonic to the steady state response.

2.3.5.5 Find the transient solution

From the steady state solution $u_{ss}\left(t\right)$ we found above, we now find $u_{ss}\left(0\right)$ and $u_{ss}^{\prime }\left(0\right)$ these are the initial conditions, but in opposite sign, that the transient solution have to satisfy.

From above, we found the steady state solution to be$$y_{ss}\left(t\right)=\mathrm{Re}\left\{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{U}_n{e}^{in\varpi t}\right\}$$

Hence$$y_{ss}^{\prime }\left(t\right)=\mathrm{Re}\left\{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}in\varpi U_n{e}^{in\varpi t}\right\}$$

At time $t=0$ the above becomes$$\begin{array}{rl}{y}_{ss}\left(0\right)& =\mathrm{Re}\left\{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{U}_n\right\}\\ y_{ss}^{\prime }\left(0\right)& =\mathrm{Re}\left\{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}in\varpi U_n\right\}\end{array}$$

Now we need to decide on how many harmonics to use in order to determine $y_{ss}\left(0\right)$ and $y_{ss}^{\prime }\left(0\right)$.  From above we see that after $n=5$ then $a_n$ became very small. Hence we will use up to $n=5$ to find the initial conditions from the above 2 equations.$$\begin{array}{rl}{y}_{ss}\left(0\right)& =\mathrm{Re}\left\{{\displaystyle \sum _{n=0}^5}{U}_n\right\}=\mathrm{Re}\left\{{\displaystyle \sum _{n=0}^5}\frac{a_n}{k}\frac{1}{(1-n^2{r}^2)+2i\zeta nr}\right\}\\ & =0.0084435\text{ ft}=0.101322\text{ inch}\end{array}$$

and for the initial velocity we obtain$$\begin{array}{rl}{y}_{ss}^{\prime }\left(0\right)& =\mathrm{Re}\left\{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}in\varpi U_n\right\}\\ & =\mathrm{Re}\left\{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}in\varpi \frac{a_n}{k}\frac{1}{(1-n^2{r}^2)+2i\zeta nr}\right\}\\ & =-0.020207\text{ ft/sec}=-0.242484\text{ inch/sec}\end{array}$$

Now the transient solution for damped system is given by$$u_{tr}\left(t\right)=e^{-\zeta {\omega }_{n}t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)$$

with $$\begin{array}{rl}{u}_{tr}\left(0\right)& =-0.0084435\\ u_{tr}^{\prime }\left(0\right)& =+0.020207\end{array}$$

Hence $$A=u_{tr}\left(0\right)=\boxed{-0.0084435}$$

Taking derivative of $u_{tr}\left(t\right)$ gives$$u_{tr}^{\prime }\left(t\right)=\zeta {\omega }_n{e}^{-\zeta {\omega }_{n}t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)+e^{-\zeta {\omega }_{n}t}(-A{\omega }_d\sin {\omega }_{d}t+B{\omega }_d\cos {\omega }_{d}t)$$

Hence at $t=0$ we obtain$$\begin{array}{rl}{u}_{tr}^{\prime }\left(0\right)& =\zeta {\omega }_{n}A+B{\omega }_d\\ B& =\frac{u_{tr}^{\prime }\left(0\right)-\zeta {\omega }_{n}A}{{\omega }_d}\end{array}$$

But $u_{tr}^{\prime }\left(0\right)=+0.020207$ ft/sec$,A=-0.0084435$ ft$,\zeta =0.75$,${\omega }_d=4.7476$ rad/sec,${\omega }_n=7.1777$ rad/sec, hence$$\begin{array}{rl}B& =\frac{0.020207-0.75\times 7.1777\times (-0.0084435)}{4.7476}\\ & =0.01383\end{array}$$

Therefore $$\begin{array}{rl}{u}_{tr}\left(t\right)& =e^{-\zeta {\omega }_{n}t}(-0.0084435\cos {\omega }_{d}t+0.01383\sin {\omega }_{d}t)\\ & =e^{-0.75\left(7.1777\right)t}(-0.0084435\cos \left(4.7476t\right)+0.01383\sin 4.7476t)\end{array}$$

This solution is now added to the steady state solution.

2.3.5.6 Plot of the absolute total displacement with the bridge for both steady state and transient combined

Zooming on the first $1.8$ seconds shows more clearly the effect of transient solution

The transient solution effect vanishes after about $1.5$ second.

2.3.5.7 Plotting the full relative solution

To better see the solution obtained, we plot the relative displacement. This is the displacement felt by the passenger. First the solution is shown for the whole time to cross the bridge, then we zoom to the first 2 seconds to better see the transient solution

From the above we see that the maximum relative displacement is about $0.24$ inch and it occurs during transient phase. During steady state, the maximum relative displacement is about $0.165$ inch

2.3.6 Appendix

2.3.6.1 Finding Fourier series approximation using complex form

The Fourier series approximation can also be found using the complex representation. This is the derivation using this method which gives the same result as was found earlier.$${\tilde{P}}_a\left(t\right)=\frac{1}{2}{Y}_0+\mathrm{Re}\left({\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{Y}_n{e}^{in\varpi t}\right)$$

Where $$\begin{array}{rl}{Y}_n& =\frac{2}{T_p}{\displaystyle \underset{0}{\overset{T_p}{\int }}}{P}_a\left(t\right)e^{{}^{-in\varpi t}}dt\\ \text{(4)}& & =\frac{2P_o}{T_p}{\displaystyle \underset{0}{\overset{T_p}{\int }}}\sin \left(\frac{\pi }{T_p}t\right)e^{{}^{-in\varpi t}}dt\end{array}$$

Integration by parts, $\int udv=uv-\int vdu$, let $u=\sin \left(\frac{\pi }{T_p}t\right)$, hence $du=\frac{\pi }{T_p}\cos \left(\frac{\pi }{T_p}t\right)$ and $v=\frac{e^{{}^{-in\varpi t}}}{-in\varpi }$, therefore the above becomes$$\begin{array}{rl}{Y}_n& =\frac{2P_o}{T_p}({[\sin \left(\frac{\pi }{T_p}t\right)\frac{e^{{}^{-in\varpi t}}}{-in\varpi }]}_0^{T_p}-{\displaystyle \underset{0}{\overset{T_p}{\int }}}\frac{\pi }{T_p}\cos \left(\frac{\pi }{T_p}t\right)\frac{e^{{}^{-in\varpi t}}}{-in\varpi }dt)\\ & =\frac{2P_o}{T_p}(\sin \left(\frac{\pi }{T_p}{T}_p\right)\frac{i{e}^{{}^{-in\varpi T_p}}}{n\varpi }-\frac{i}{2n}{\displaystyle \underset{0}{\overset{T_p}{\int }}}\cos \left(\frac{\pi }{T_p}t\right)e^{{}^{-in\varpi t}}dt)\\ \text{(5)}& & =-\frac{i{P}_o}{n{T}_p}{\displaystyle \underset{0}{\overset{T_p}{\int }}}\cos \left(\frac{\pi }{T_p}t\right)e^{{}^{-in\varpi t}}dt\end{array}$$

Now integrate by parts again where now $\int udv=uv-\int vdu$, let $u=\cos \left(\frac{\pi }{T_p}t\right)$, hence $du=-\frac{\pi }{T_p}\sin \left(\frac{\pi }{T_p}t\right)$ and $v=\frac{e^{{}^{-in\varpi t}}}{-in\varpi }$, therefore Eq ?? becomes$$\begin{array}{rl}{Y}_n& =-\frac{i{P}_o}{n{T}_p}[{(\cos \left(\frac{\pi }{T_p}t\right)\frac{e^{{}^{-in\varpi t}}}{-in\varpi })}_0^{T_p}-{\displaystyle \underset{0}{\overset{T_p}{\int }}}-\frac{\pi }{T_p}\sin \left(\frac{\pi }{T_p}t\right)\frac{e^{{}^{-in\varpi t}}}{-in\varpi }dt]\\ & =-\frac{i{P}_o}{n{T}_p}[(\cos \left(\frac{\pi }{T_p}{T}_p\right)\frac{e^{{}^{-in\frac{2\pi }{Tp}{T}_p}}}{-in\varpi }-\frac{1}{-in\varpi })+\frac{i}{n2}{\displaystyle \underset{0}{\overset{T_p}{\int }}}\sin \left(\frac{\pi }{T_p}t\right)e^{{}^{-in\varpi t}}dt]\\ & =-\frac{i{P}_o}{n{T}_p}[(-i\frac{e^{{}^{-in2\pi }}}{n\varpi }-\frac{i}{n\varpi })+\frac{i}{n2}{\displaystyle \underset{0}{\overset{T_p}{\int }}}\sin \left(\frac{\pi }{T_p}t\right)e^{{}^{-in\varpi t}}dt]\\ \text{(6)}& & =-\frac{P_o}{n{T}_p}\left(\frac{e^{{}^{-in2\pi }}+1}{n\varpi }\right)+\frac{\mathrm{\Delta }}{2n^2{T}_p}{\displaystyle \underset{0}{\overset{T_p}{\int }}}\sin \left(\frac{\pi }{T_p}t\right)e^{{}^{-in\varpi t}}dt\end{array}$$

Now we see that the term ${\displaystyle \underset{0}{\overset{T_p}{\int }}}\sin \left(\frac{\pi }{T_p}t\right)e^{{}^{-in\varpi t}}dt$ has repeated again. This term is the same as what we started with in Eq ?? , therefore, we write$${\displaystyle \underset{0}{\overset{T_p}{\int }}}\sin \left(\frac{\pi }{T_p}t\right)e^{{}^{-in\varpi t}}dt=\frac{T_p}{2\mathrm{\Delta }}{Y}_n$$

and replace this term back into Eq ?? , hence it becomes$$\begin{array}{rl}{Y}_n& =-\frac{P_o}{n{T}_p}\left(\frac{e^{{}^{-in2\pi }}+1}{n\varpi }\right)+\frac{\mathrm{\Delta }}{2n^2{T}_p}\frac{T_p}{2\mathrm{\Delta }}{Y}_n\\ & =-\frac{P_o}{n{T}_p}\left(\frac{e^{{}^{-in2\pi }}+1}{n\varpi }\right)+\frac{1}{2^2{n}^2}{Y}_n\\ Y_n-\frac{1}{2^2{n}^2}{Y}_n& =-\frac{P_o}{n{T}_p}\left(\frac{e^{{}^{-in2\pi }}+1}{n\varpi }\right)\\ Y_n(1-\frac{1}{{\left(2n\right)}^2})& =-\frac{P_o}{n{T}_p}\left(\frac{e^{{}^{-in2\pi }}+1}{n\varpi }\right)\\ Y_n& =-\frac{2P_o(e^{{}^{-in2\pi }}+1)}{\pi ({\left(2n\right)}^2-1)}=\frac{2P_o(e^{{}^{-in2\pi }}+1)}{\pi -\pi {\left(2n\right)}^2}\\ & =\frac{4P_o}{\pi (1-4n^2)}\end{array}$$

And $$\begin{array}{rl}{Y}_0& =\frac{2}{T_p}{\displaystyle \underset{0}{\overset{T_p}{\int }}}{P}_o\sin \left(\frac{\pi }{T_p}t\right)dt=\frac{2P_o}{T_p}{\displaystyle \underset{0}{\overset{T_p}{\int }}}\sin \left(\frac{\pi }{T_p}t\right)dt=\frac{2P_o}{T_p}{\left(\frac{-\cos \left(\frac{\pi }{T_p}t\right)}{\frac{\pi }{T_p}}\right)}_0^{T_p}=-\frac{2P_o}{\pi }(\cos \left(\frac{\pi }{T_p}{T}_p\right)-1)\\ & =-\frac{2P_o}{\pi }(-1-1)\\ & =\frac{4P_o}{\pi }\end{array}$$

Therefore, the Fourier series approximation for ground motion is now$$\begin{array}{rl}{\tilde{P}}_a\left(t\right)& =\frac{1}{2}{Y}_0+\mathrm{Re}\left({\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{Y}_n{e}^{in\varpi t}\right)\\ & =\frac{4P_o}{2\pi }+\mathrm{Re}\left({\displaystyle \sum _{n=1}^{\mathrm{\infty }}}\frac{4P_o}{\pi (1-4n^2)}{e}^{in\varpi t}\right)\\ & =\frac{2P_o}{\pi }+\mathrm{Re}\left({\displaystyle \sum _{n=1}^{\mathrm{\infty }}}\frac{4P_o}{\pi (1-4n^2)}{e}^{in\varpi t}\right)\end{array}$$

We see that we obtained the same result using the classical Fourier series form.

¹The Fourier series can also be found using complex form. This was done in the appendix.