Finding the initial value based on a transfer function

Question

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.

Answer 1

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.