Inverse Laplace transforming

First, define a symbolic s.

syms s 'complex'

Now we can define F, a symbolic expression for F(s).

F = (s^2 + 2*s + 2)/(s^2 + 6*s + 36)*6/(s+6);

Now all that remains is the apply partfrac.

F_pf = partfrac(F)

F_pf =
13/(3*(s + 6)) + ((5*s)/3 - 24)/(s^2 + 6*s + 36)

Now consider the Laplace transform table. The first term can easily be inverted: $$\begin{align} \mathcal{L}^{-1}\left(\frac{13}{3}\cdot\frac{1}{s+6}\right) &= \frac{13}{3}\mathcal{L}^{-1}\frac{1}{s+6} \tag{linearity} \\ \frac{13}{3} e^{-6 t}. \tag{table} \end{align}$$

The second term, call it F₂, is not quite as obvious, but the preimage $$\begin{aligned} \frac{s-a}{(s-a)^2 + \omega^2} \end{aligned}$$ is close. Let’s first make the numerator match: $$\begin{aligned} \frac{5}{3} s - 24 = \frac{5}{3}\left( s - \frac{72}{5} \right), \end{aligned}$$ so a₁ = 72/5. Now we need the term (s−a₁)² in the denominator. Asserting the equality $$\begin{aligned} s^2 + 6 s + 36 &= (s - a_2)^2 + \omega^2 \\ &= s^2 - 2 a_2 s + a_2^2 + \omega^2. \end{aligned}$$ Equating the s⁰ coefficents yields ω² = 36 − a₂² and equating the s coefficient yields a₂ =  − 3 ≠ a₁ = 72/5, so no cigar! What if we “force” the rule by using a new a₁′ = a₂, which can be achieved by adding a term (and subtracting it elsewhere)? We need a₁′ =  − 3, so if we add (and subtract) a term $$\begin{aligned} \frac{\frac{5}{3}(a_1 - a_1')}{(s - a_2)^2 + \omega^2}, \end{aligned}$$ like $$\begin{aligned} F_2 = \frac{\frac{5}{3}(s - a_1)}{(s - a_2)^2 + \omega^2} + \frac{\frac{5}{3}(a_1 - a_1')}{(s - a_2)^2 + \omega^2} - \frac{\frac{5}{3}(a_1 - a_1')}{(s - a_2)^2 + \omega^2} \end{aligned}$$ we can combine the first two terms to yield $$\begin{aligned} F_2 = \frac{\frac{5}{3}(s - a_1')}{(s - a_2)^2 + \omega^2} - \frac{\frac{5}{3}(a_1 - a_1')}{(s - a_2)^2 + \omega^2} \end{aligned}$$ where we recall that a₁′ = a₂ by construction.

Now the expression is $$\begin{aligned} F_2 = \frac{\frac{5}{3}(s - a_2)}{(s - a_2)^2 + \omega^2} - \frac{\frac{5}{3}(a_1 - a_2)}{(s - a_2)^2 + \omega^2} \end{aligned}$$

The first term is, by construction, in the Laplace transform table. The second term is close to $$\begin{aligned} \frac{\omega}{(s - a)^2 + \omega^2} \end{aligned}$$ for which we must make the numerator equal ω. Our ω² = 36 − a₂² = 27, so $\omega = \pm \sqrt{27}$. The current numerator is $$\begin{aligned} \frac{5}{3} (a_1 - a_2) &= \frac{5}{3}\left(\frac{72}{5} + 3\right) \\ &= 29. \end{aligned}$$ So we factor out $29/\sqrt{27}$ to yield $$\begin{aligned} \frac{\frac{29}{\sqrt{27}} \omega}{(s - a_2)^2 + \omega^2} \end{aligned}$$

Returning to F₂, we have arrived at $$\begin{aligned} F_2 = \frac{\frac{5}{3}(s - a_2)}{(s - a_2)^2 + \omega^2} - \frac{\frac{29}{\sqrt{27}} \omega}{(s - a_2)^2 + \omega^2} \end{aligned}$$

Now the inverse transform is $$\begin{align} \mathcal{L}^{-1} F_2 &= \frac{5}{3} \mathcal{L}^{-1} \frac{(s - a_2)}{(s - a_2)^2 + \omega^2} - \frac{29}{\sqrt{27}} \mathcal{L}^{-1} \frac{\omega}{(s - a_2)^2 + \omega^2} \tag{linearity} \\ &= \frac{5}{3} e^{a_2 t} \cos\omega t - \frac{29}{\sqrt{27}} e^{a_2 t} \sin\omega t. \end{align}$$ Simple! Putting it all together, then, $$\begin{align} F(s) &= \frac{13}{3} e^{-6 t} + \frac{5}{3} e^{-3 t} \cos(3\sqrt{3} t) - \frac{29}{3\sqrt{3}} e^{-3 t} \sin(3\sqrt{3} t). \end{align}$$

First, define a symbolic s.

syms s 'complex'

Now we can define F, a symbolic expression for F(s).

F = (s^2 + 2*s + 2)/(s^2 + 6*s + 36)*6/(s+6);

Now all that remains is the apply ilaplace.

F_pf = ilaplace(F)

F_pf =
(13*exp(-6*t))/3 + (5*exp(-3*t)*(cos(3*3^(1/2)*t) - (29*3^(1/2)*sin(3*3^(1/2)*t))/15))/3

This is easily seen to be equivalent to our previous result $$\begin{aligned} F(s) &= \frac{13}{3} e^{-6 t} + \frac{5}{3} e^{-3 t} \cos(3\sqrt{3} t) - \frac{29}{3\sqrt{3}} e^{-3 t} \sin(3\sqrt{3} t). \end{aligned}$$