Finding the initial value based on a transfer function

control theorylaplace transform

There is a system with angular of rotation as output, and its Laplace transform is:

$$
\Theta (s) = \dfrac{4s + 12}{2s^2 + 5s + 1}
$$

Now, I would like to find the following:

  1. Find the initial value of angular of rotation $\theta (t = 0)$
  2. Find the Laplace transform of turning angular velocity of $\Omega (s)$
  3. Find the initial value of turning angular velocity $\omega (t = 0)$

Below is how I attempted to solve the question,

1.

\begin{align}
G(s) &= \dfrac{\Theta (s)}{U(s)} \\
\theta (s) &= \mathscr{L^{-1}}[\Theta (s)] = \mathscr{L^{-1}}\left[\dfrac{4s + 12}{2s^2 + 5s + 1} \right] \\
&= 2e^{-\dfrac{5}{4}t} \cosh\left(\dfrac{\sqrt 17}{4}t\right) + \dfrac{14}{\sqrt 17}e^{\dfrac{5}{4}t} \sinh\left(\dfrac{\sqrt 17}{4}t\right)\\
\theta (0) &= 2
\end{align}

2.
\begin{equation}
\Omega (s) = s * \Theta (s) = \dfrac{s(4s + 12)}{2s^2 + 5s + 1}
\end{equation}

3.
\begin{equation}
\omega (t) = \mathscr{L^{-1}}\left[\dfrac{s(4s + 12)}{2s^2 + 5s + 1}\right] = 2\delta(t) + e^{-\dfrac{5}{4}t} \cosh\left(\dfrac{\sqrt 17}{4}t\right) – \dfrac{9}{\sqrt 17}e^{–\dfrac{5}{4}t} \sinh\left(\dfrac{\sqrt 17}{4}t\right)
\end{equation}

$$
\omega (0) = 2\delta (0) + 1 = ∞
$$

In question three, as far as I know, the Dirac function at t = 0 is ∞, however, I do not think this makes much sense that this system is actually divergent at the beginning. Furthermore, I have attempted to use the initial theorem to write the equation:

$$
\lim_{s\to \infty}s\Omega (s) = \lim_{s\to \infty}s*\dfrac{s(4s + 12)}{2s^2 + 5s +1} = \infty
$$

Is there anything wrong that I missed? Can anyone help me figure this out? Thank you for your time and advice.

Best Answer

The first answer looks correct. The second answer, unfortunately, is not. If you take the derivative of the first answer, you will find,

$$ \omega(t) = \theta'(t) = \frac{1}{17} e^{-5 t/4} \left[ 17\,\cosh(\frac{\sqrt{17}}{4} t) - 9 \sqrt{17}\sinh(\frac{\sqrt{17}}{4} t) \right] $$

whose Laplace transform is,

$$ \Omega(s) = 2 \frac{s - 1}{2 s^2 + 5 s + 1} $$

This obviously doesn't agree with the "multiplying by $s$" rule -- that is, your answer --- so what gives? The theorem you are using there is the following.

Theorem [pp. 387, 1]: Let $\theta(t)$ be continuous on $[0,\infty)$ and $\theta'(t)$ be piecewise continuous on $[0, \infty)$ both of which are of exponential order $\alpha.$ Then, for $\mathfrak{Re}\{s\} > \alpha,$ $$\mathcal{L}\{\theta'\}(s) = s\,\mathcal{L}\{\theta\}(s) - \theta(0)$$

You might have used this theorem when transforming DEs. Compare this with your computation for (2) and note that $\theta(0)$ term is missing. Observe that,

$$ \begin{aligned} s \Theta(s) - \theta(0) &= s \frac{4 s + 12}{2 s^2 + 5 s + 1} - 2\\ &= s \frac{4 s + 12}{2 s^2 + 5 s + 1} - 2 \frac{2 s^2 + 5 s + 1}{2 s^2 + 5s +1}\\ &= \frac{4 s^2 + 12 s}{2 s^2 + 5 s + 1} - 2 \frac{2 s^2 + 5 s + 1}{2 s^2 + 5s +1}\\ &= \frac{2 s - 2}{s^2 + 5 s + 1}\\ &= 2 \frac{s - 1}{2 s^2 + 5 s + 1}\\ \end{aligned} $$

I think you will find your third answer will come out consistent with a direct evaluation of $\lim_{t\to 0^+} \theta'(t)$ once you correct this.

[1]: Nagle, Saff & Snider. Fundamentals of Differential Equations and Boundary Value Problems. 5th Edition.

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