Question

Find the inverse Laplace transform of each of the following expressions: (a) \(\frac{4}{s^{2}+4 s+5}\) (b) \(\frac{s+5}{s^{2}+10 s+29}\) (c) \(\frac{2 s-3}{s^{2}+6 s+10}\) (d) \(\frac{s^{2}+6 s-4}{\left(s^{2}+4\right)^{2}}\) (e) \(\frac{1}{4 s^{2}+4 s+1}\)

Short answer

In summary, the inverse Laplace transforms for the given expressions are: (a) \(4e^{-2t}\sin{t}\) (b) \(e^{-5t}\cos{2t}\) (c) \(2e^{-3t}\cos{t}-9e^{-3t}\sin{t}\) (d) \(-\frac{1}{8}(3+t)^{2}t\cos{2t}+\frac{13}{8}t\cos{2t}\) (e) \(2e^{t}\sin{\frac{t}{2}}\)

Step-by-step solution

Recognize known Laplace transform pair

Let's recognize \(\frac{4}{s^{2}+4 s+5}\) as the Laplace transform of some function. Note that the denominator is a quadratic, and we need to complete the square to determine if this Laplace Transform is known. By completing the square, the quadratic can be re-written as \((s+2)^{2}+1\). This expression now looks like the Laplace Transform of the first-order damped sine function, which is given by \(\mathcal{L}\{e^{-at}\sin{(bt)}\}=\frac{b}{(s+a)^{2}+b^{2}}\) or \(\mathcal{L}\{e^{-at}\cos{(bt)}\}=\frac{s+a}{(s+a)^{2}+b^{2}}\).

Apply the inverse Laplace transform

In our case, \(a=2\) and \(b=1\). Applying the inverse Laplace transform, we have \(\mathcal{L}^{-1}\left\{\frac{4}{(s+2)^{2}+1}\right\} = 4 e^{-2t}\sin{(1t)} = 4e^{-2t}\sin{t}\). So, the inverse Laplace transform of \(\frac{4}{s^{2}+4 s+5}\) is \(4e^{-2t}\sin{t}\). # (b) #

Recognize known Laplace transform pair

Let's recognize \(\frac{s+5}{s^{2}+10 s+29}\) as the Laplace Transform of some function. By completing the square for the quadratic in the denominator, we can rewrite it as \((s+5)^{2}+4\). This expression now looks like the Laplace Transform of the first-order damped sine function, which is given by \(\mathcal{L}\{e^{-at}\sin{(bt)}\}=\frac{b}{(s+a)^{2}+b^{2}}\) or \(\mathcal{L}\{e^{-at}\cos{(bt)}\}=\frac{s+a}{(s+a)^{2}+b^{2}}\).

Apply the inverse Laplace transform

In our case, \(a=5\) and \(b=2\). Applying the inverse Laplace transform, we have \(\mathcal{L}^{-1}\left\{\frac{s+5}{(s+5)^{2}+4}\right\} = e^{-5t}\cos{(2t)}\). So, the inverse Laplace transform of \(\frac{s+5}{s^{2}+10 s+29}\) is \(e^{-5t}\cos{2t}\). # (c) #

Recognize known Laplace transform pair

Let's recognize \(\frac{2s-3}{s^{2}+6 s+10}\) as the Laplace Transform of some function. By completing the square for the quadratic in the denominator, we can rewrite it as \((s+3)^{2}+1\). Now, as the Laplace Transform of the first-order damped sine function, this expression looks like either \(\mathcal{L}\{e^{-at}\sin{(bt)}\}=\frac{b}{(s+a)^{2}+b^{2}}\) or \(\mathcal{L}\{e^{-at}\cos{(bt)}\}=\frac{s+a}{(s+a)^{2}+b^{2}}\). In order to align with one of these forms, you need to rewrite \(\frac{2s-3}{(s+3)^{2}+1}\) as the sum/division of two terms, each of which corresponds to one of these known rules.

Reshape numerator

Rewrite the numerator in terms of \((s+3)\). In other words, we rewrite \((2s-3)\) as \(q(s+3)+r\). The variables \(q\) and \(r\) are determined by solving the linear equation system: \(2s-3=q(s+3)+r\). In our case, we have \(q=2\) and \(r=-9\).

Apply the inverse Laplace transform

Now, we can rewrite our expression as \(\frac{2(s+3)-9}{(s+3)^{2}+1}\). This can be further separated into two parts: \(\frac{2(s+3)}{(s+3)^{2}+1}-\frac{9}{(s+3)^{2}+1}\). The first part matches the inverse Laplace transform \(\mathcal{L}\{e^{-at}\cos{(bt)}\}=\frac{s+a}{(s+a)^{2}+b^{2}}\) with \(a=3\) and \(b=1\). The second part matches the inverse Laplace transform \(\mathcal{L}\{e^{-at}\sin{(bt)}\}=\frac{b}{(s+a)^{2}+b^{2}}\) with \(a=3\) and \(b=1\). Therefore, we have the inverse Laplace transform for expression (c) as \(2e^{-3t}\cos{t}-9e^{-3t}\sin{t}\). # (d) #

Recognize known Laplace transform pair

Let's recognize \(\frac{s^{2}+6s-4}{\left(s^{2}+4\right)^{2}}\) as the Laplace Transform of some function. First, rewrite the numerator by completing the square. This results in the expression: \(\frac{(s+3)^{2}-13}{\left(s^{2}+4\right)^{2}}\).

Apply the inverse Laplace transform

From the table of Laplace transforms, we see that this looks like the Laplace transform pair \(\mathcal{L}\{t e^{at} \sin{(bt)}\} =\frac{2a s+b}{\left(s^2+b^2\right)^2}\) or \(\mathcal{L}\{t e^{at} \cos{(bt)}\} =\frac{s^2 - b^2}{\left(s^2+b^2\right)^2}\). We choose the second pair, because the numerator doesn't contain \(s\). In our case, we have \(a=0\) and \(b=2\). Because of the difference in the coefficients, we will find \(\mathcal{L}\{t \cos{2t}\}\) first and then adjust the coefficients. The inverse Laplace transform of \(\frac{s^2 - 4}{\left(s^2+4\right)^{2}}\) is \(t \cos{2t}\). Now, we need to multiply \(t\cos{2t}\) by \(\frac{(s+3)^{2}-13}{s^2-4}\) in the time domain to get the inverse Laplace transform of the expression given.

Multiply by the time domain factors

In the time domain, multiplying \(t\cos{2t}\) by \(\frac{(s+3)^{2}-13}{s^2-4}\) is the same as multiplying \(t\cos{2t}\) by \(\frac{(s+3)^{2}-13}{-8}\). Therefore, the inverse Laplace transform of \(\frac{s^{2}+6s-4}{\left(s^{2}+4\right)^{2}}\) is \(-\frac{1}{8}(3+t)^{2}t\cos{2t}+\frac{13}{8}t\cos{2t}\). # (e) #

Recognize known Laplace transform pair

Let's recognize \(\frac{1}{4 s^{2}+4 s+1}\) as the Laplace Transform of some function. We identify the denominator as the square of a first-order damped sine function, \(\left(\frac{s+1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\).

Apply the inverse Laplace transform

From the table of Laplace transforms, we look for an entry with a denominator in the same form. This pair looks like the Laplace transform of the first-order damped sine function, which is given by \(\mathcal{L}\{e^{-at}\sin{(bt)}\}=\frac{b}{(s+a)^{2}+b^{2}}\). So, let's apply the inverse Laplace transform of this form to our expression with \(a=-1\) and \(b=\frac{1}{2}\). Therefore, we have the inverse Laplace transform of \(\frac{1}{4 s^{2}+4 s+1}\) is \(2e^{t}\sin{\frac{t}{2}}\).

Key concepts

Laplace transform

The Laplace transform is an incredibly useful tool in mathematics and engineering. It's commonly used to transform complex differential equations into simpler algebraic ones, which are easier to manipulate. The Laplace transform \(\mathcal{L}\{f(t)\}\) converts a time-domain function \(f(t)\) into a complex frequency-domain function \(F(s)\). With this transformation, you can analyze systems, especially in control engineering and signal processing.

To visualize the process, imagine you have a messy pile of data (\