Fall 2009/JonathanS

Problem

A simple pendulum with a length L = 0.5m is pulled back and released from an initial angle $\theta _{0}=12^{o}$. Then it is run with a forcing function of cos(3$\theta$) Find an equation that gives the angle of the pendulum at any given time t.

Solution

Assuming no damping and a small angle($\theta <15^{o}$), the equation for the motion of a simple pendulum can be written as

${\mathrm {d} ^{2}\theta \over \mathrm {d} t^{2}}+{g \over \ell }\theta =cos(3\theta ).$

Substituting values we get

${\mathrm {d} ^{2}\theta \over \mathrm {d} t^{2}}+{9.81 \over 0.5}\theta =cos(3\theta ).$

$\Rightarrow {\mathrm {d} ^{2}\theta \over \mathrm {d} t^{2}}+{19.62}\theta =cos(3\theta ).$

Remember the identities

${\mathcal {L}}\{f(t)\}=F(s)=\int _{0}^{\infty }e^{-st}f(t)\,dt.$

${\mathcal {L}}\{f^{''}(t)\}=s^{2}F(s)-sf(0)-f^{'}(0)$

Now we can take the Laplace Transform to change the second order differential equation, from the t domain, into a simple linear equation, from the s domain, that's much easier to work with

${\mathcal {L}}{\bigg \{}{\mathrm {d} ^{2}\theta \over \mathrm {d} t^{2}}+{19.62}\theta {\bigg \}}=s^{2}F(s)-sf(0)-f^{'}(0)+{19.62}\theta ={\mathcal {L}}\{cos(3\theta )\}$

$\Rightarrow$ $s^{2}{\boldsymbol {\theta }}-s\theta (0)-\theta ^{'}(0)+{19.62}{\boldsymbol {\theta }}={s \over {s^{2}+9}}\,$

Since we know that $\theta (0)=12^{o}$ and the initial velocity $\theta ^{'}(0)=0$ we get

$s^{2}{\boldsymbol {\theta }}-12s+19.62{\boldsymbol {\theta }}={s \over {s^{2}+9}}\,$

$\Rightarrow$ ${\boldsymbol {\theta }}(s^{2}+19.62)={s \over {s^{2}+9}}+12s\,$

$\Rightarrow$ ${\boldsymbol {\theta }}(s^{2}+19.62)={s \over {s^{2}+9}}+{12s(s^{2}+9) \over {s^{2}+9}}\,$

$\Rightarrow$ ${\boldsymbol {\theta }}(s^{2}+19.62)={(12s^{3}+109s) \over {s^{2}+9}}\,$

$\Rightarrow$ ${\boldsymbol {\theta }}={(12s^{3}+109s) \over {(s^{2}+9)(s^{2}+19.62)}}$

$\Rightarrow$ ${\boldsymbol {\theta }}={12s^{3}+109s \over {(s^{4}+28.62s^{2}+176.58)}}$

Now we can take the inverse Laplace Transform to convert our equation back into the time domain

${\mathcal {L}}^{-1}{\bigg \{}{12s^{3}+109s \over {(s^{4}+28.62s^{2}+176.58)}}{\bigg \}}=0.094162cos(3t)+11.9058cos(4.42945t)$

This will give us the angle (in degrees) of the pendulum at any given time t.

Initial Value Theorem

We can use the Initial Value Theorem as a check that our initial values for the problem are valid.

${\mathcal {L}}\{f^{'}(t)\}=sF(s)-f(0^{-})=\int _{0^{-}}^{\infty }f^{'}(t)e^{-st}\,dt$

${\begin{alignedat}{3}\lim _{n\to \infty }{\mathcal {L}}\{f^{'}(t)\}&=\lim _{n\to \infty }sF(s)-f(0^{-})\\&=\lim _{n\to \infty }sF(s)-f(0^{-})=0\\&=\lim _{n\to \infty }sF(s)=f(0^{-})\\\end{alignedat}}$

Below we will use this theorem to check the values for our problem.

${\begin{alignedat}{3}\lim _{n\to \infty }{\mathcal {L}}\{f^{'}(t)\}&=\lim _{n\to \infty }sF(s)-f(0^{-})=\lim _{n\to \infty }\int _{0^{-}}^{\infty }f^{'}(t)e^{-st}\,dt\\&=\lim _{n\to \infty }s{\bigg (}{12s^{3}+109s \over {(s^{4}+28.62s^{2}+176.58)}}{\bigg )}-f(0^{-})=0\\&=\lim _{n\to \infty }s{\bigg (}{12s^{3}+109s \over {(s^{4}+28.62s^{2}+176.58)}}{\bigg )}=f(0^{-})\\12&=f(0^{-})\\\end{alignedat}}$

This value $f(0^{-})=12$ is the initial angle we gave the pendulum so it checks out.

Final Value Theorem

We can use the Final Value Theorem as a check that our final values for the problem are valid.

${\mathcal {L}}\{f^{'}(t)\}=sF(s)-f(\infty )=\int _{0^{-}}^{\infty }f^{'}(t)e^{-st}\,dt$

${\begin{alignedat}{3}\lim _{n\to 0}{\mathcal {L}}\{f^{'}(t)\}&=\lim _{n\to 0}sF(s)-f(\infty )\\&=\lim _{n\to 0}sF(s)-f(\infty )=0\\&=\lim _{n\to 0}sF(s)=f(\infty )\\\end{alignedat}}$

Below we will use this theorem to check the values for our problem.

${\begin{alignedat}{3}\lim _{n\to 0}{\mathcal {L}}\{f^{'}(t)\}&=\lim _{n\to 0}sF(s)-f(\infty )=\lim _{n\to 0}\int _{0^{-}}^{\infty }f^{'}(t)e^{-st}\,dt\\&=\lim _{n\to 0}s{\bigg (}{12s^{3}+109s \over {(s^{4}+28.62s^{2}+176.58)}}{\bigg )}-f(\infty )=0\\&=\lim _{n\to 0}s{\bigg (}{12s^{3}+109s \over {(s^{4}+28.62s^{2}+176.58)}}{\bigg )}=f(\infty )\\0&=f(\infty )\\\end{alignedat}}$

This is zero because the average angle as time goes to infinity will be zero (halfway between -12 and 12 degrees).

Bode Plots

To find the bode plot we use the function

$H(s)={12s^{3}+109s \over {(s^{4}+28.62s^{2}+176.58)}}$

Entering this into a program such as Maple, Octave, or MATLAB will give you a plot that looks like this

Asymptotes & Break Points

In this section I will describe how you can use the transfer function to quickly estimate what the bode plot will look like.