5.4. Laplace transform

5.4.1. Definition and properties

The Laplace transform is used in signals/controls. It is also a way to solve differential equations using algebra. It is defined as:

(5.59)\[\begin{equation} F(s) = L[f(t)] = \int_0^\infty e^{-st} f(t) \d{t} \end{equation}\]

The inverse Laplace transform of F is f, i.e., \(f(t) = L^{-1}[F(s)]\). The Laplace transform of many functions can be computed using integration by parts. For example, the Laplace transform of \(f(t) = t\) is:

(5.60)\[\begin{align} L[t] &= \int_0^\infty e^{-st} t \d{t} \\ &= \left.-\frac{te^{-st}}{s}\right|_0^\infty + \int_0^\infty \frac{e^{-st}}{s} \d{t} \\ &= -\left.\frac{e^{-st}}{s^2}\right|_0^\infty \\ &= \frac{1}{s^2} \end{align}\]

Importantly, the Laplace transform of the first derivative of an unknown function \(f(t) = y'(t)\) is:

(5.61)\[\begin{align} L[y'(t)] &= \int_0^\infty e^{-st} y'(t) \d{t} \\ &= \left.e^{-st} y(t)\right|_0^\infty - \int_0^\infty y(t) \left(-s e^{-st}\right) \d{t} \\ &= [0 - y (0)] + s \int_0^\infty e^{-st} y \d{t} \\ &= -y(0) + s L[y(t)] \\ &= s Y(s) - y(0) \end{align}\]

using \(u = e^{-st}\) and \(\d{v} = y'(t) \d{t}\). Some common Laplace transforms are:

\(f(t)\)	\(F(s) = L[f(t)]\)
\(t^n\), \(n=0, 1, 2, \cdots\)	\(\dfrac{n!}{s^{n+1}}\)
\(e^{at}\)	\(\dfrac{1}{s - a}\)
\(\sin(at)\)	\(\dfrac{a}{s^2 + a^2}\)
\(\cos(at)\)	\(\dfrac{s}{s^2 + a^2}\)
\(e^{at} \sin(bt)\)	\(\dfrac{b}{(s - a)^2 + b^2}\)
\(e^{at} \cos(bt)\)	\(\dfrac{s - a}{(s - a)^2 + b^2}\)

The Laplace transform is a linear operator, so

(5.62)\[\begin{align} L[kf] &= kL[f] \\ L[f + g] &= L[f] + L[g] \end{align}\]

5.4.2. Solving first-order ODEs

Nonhomogeneous, first-order ODEs with constant coefficients are nice to solve with Laplace transforms:

(5.63)\[\begin{equation} y' + by = r(t), \quad y(0) = y_0 \end{equation}\]

Apply the Laplace transform to both sides of the equation:

(5.64)\[\begin{align} L[y' + by] &= L[r] \\ L[y'] + bL[y] &= L[r]\\ \left(sY - y_0\right) + bY &= R\\ Y &= \frac{y_0 + R}{s+b} \end{align}\]

where \(Y = L[y(t)]\) and \(R = L[r(t)]\). If we can invert \(y = L^{-1}[Y]\) using tables, we have a solution!

Example: Laplace transform

Solve the initial value problem

(5.65)\[\begin{equation} y' =-ky, \quad y(0) =2 \end{equation}\]

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Rearrange, apply Laplace transform \(Y = L[y]\), and solve for \(Y\):

(5.66)\[\begin{align} y'+ ky &= 0 \\ L[y'+ky] &= L[0] \\ L[y']+kL[y] &= L[0] \\ sY - y(0) + k Y &= 0 \\ (s+k) Y - 2 &= 0 \\ Y &= \frac{2}{s+k} \end{align}\]

Invert Laplace transform:

(5.67)\[\begin{equation} y = L^{-1}\left[\frac{2}{s+k}\right] = 2L^{-1}\left[\frac{1}{s+k}\right] = 2 e^{-kt} \end{equation}\]

Example: Laplace transform with partial fractions

Solve the initial value problem:

(5.68)\[\begin{equation} y'-y = t, \quad y(0) = 1 \end{equation}\]

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Rearrange, apply Laplace transform \(Y = L[y]\), and solve for \(Y\):

(5.69)\[\begin{align} L[y'-y] &= L[t]\\ L[y']-L[y] &= L[t]\\ sY - y(0) - Y &= \frac{1}{s^2}\\ (s-1) Y - 1 &= \frac{1}{s^2}\\ Y &= \frac{1}{s-1} + \frac{1}{s^2(s-1)} \end{align}\]

These terms do not immediately correspond to the Laplace transform table, but we can separate them using partial fraction decomposition:

(5.70)\[\begin{equation} \frac{1}{(s-1) s^2} = \frac{A_1}{s-1} + \frac{A_2}{s} + \frac{A_3}{s^2} \end{equation}\]

\(A_1\) and \(A_3\) can be found using the coverup method:

(5.71)\[\begin{align} A_1 &= \frac{1}{1^2} = 1 \\ A_3 &= \frac{1}{0-1} = -1 \end{align}\]

To find \(A_2\), substitute and cross multiply:

(5.72)\[\begin{equation} 1 = s^2 + A_2(s-1)s - (s-1) \end{equation}\]

then compare the coefficient of the \(s^2\) terms on either side:

(5.73)\[\begin{equation} 1 + A_2 = 0 \to A_2 = -1 \end{equation}\]

So, all together:

(5.74)\[\begin{equation} Y = \frac{2}{s-1} - \frac{1}{s} - \frac{1}{s^2} \end{equation}\]

Solve for \(y\) by applying the inverse Laplace transform

(5.75)\[\begin{align} y &=L^{-1}[Y]\\ &=2L^{-1}\left[\frac{1}{s-1}\right]-L^{-1}\left[\frac{1}{s}\right]-L^{-1}\left[\frac{1}{s^2}\right]\\ &=2e^t-1-t \end{align}\]

Example: Hormone level

The concentration of a hormone in the blood c varies due to sinusoidal production by the thyroid and continuous removal according to:

(5.76)\[\begin{equation} c' = A + B\cos\left(\frac{\pi t}{12}\right) - kc \end{equation}\]

The concentration is \(c_0\) at 6 AM (\(t = 0\)). What is the average concentration between 6 PM and 6 AM the same day?

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To solve, rearrange and use the Laplace transform:

(5.77)\[\begin{align} c' + kc &= A + B\cos\left(\frac{\pi t}{12}\right) \\ [sC(s) - c_0] + kC(s) &= L\left[A + B\cos\left(\frac{\pi t}{12}\right)\right] \\ (s+k) C - c_0 &= \frac{A}{s} + \frac{Bs}{s^2 + (\pi/12)^2} \\ C(s) &= \left(\frac{1}{s + k}\right)\left[c_0 + \frac{A}{s} + \frac{Bs}{s^2 + (\pi/12)^2}\right] \\ &= \frac{c_0}{s + k} + \frac{A}{s(s + k)} + \frac{Bs}{[s^2 + (\pi/12)^2](s + k)} \end{align}\]

Use partial fraction decomposition on both fractions. The first one is:

(5.78)\[\begin{equation} \frac{1}{s(s + k)} = \frac{c_1}{s} + \frac{c_2}{s + k} \end{equation}\]

and the cover-up method gives \(c_1 = 1/k\) and \(c_2 = -1/k\). For the next one:

(5.79)\[\begin{equation} \frac{s}{[s^2 + (\pi/12)^2](s + k)} = \frac{c_1 s + c_2}{s^2 + (\pi/12)^2} + \frac{c_3}{s + k} \end{equation}\]

The cover-up method gives

(5.80)\[\begin{equation} c_3 = \frac{-k}{k^2 + (\pi/12)^2} \end{equation}\]

while cross-multiplying and matching coefficients gives

(5.81)\[\begin{align} s &= (c_1 s + c_2)(s + k) + c_3\left[s^2 + (\pi/12)^2\right] \\ s &= (c_1 + c_3)s^2 + (c_1 k + c_2)s + c_2 k + c_3(\pi/12)^2 \end{align}\]

so

(5.82)\[\begin{align} c_1 + c_3 &= 0 \\ c_1 k + c_2 &= 1 \\ c_2 k + c_3 (\pi/12)^2 &= 0 \end{align}\]

Using our solution for \(c_3\) in the first equation gives \(c_1\) and in the third equation gives \(c_2\):

(5.83)\[\begin{equation} c_1 = \frac{k}{k^2 + \left(\frac{\pi}{12}\right)^2}, \quad c_2 = \frac{(\pi/12)^2}{k^2 + (\pi/12)^2} \end{equation}\]

All together,

(5.84)\[\begin{align} C(s) &= \frac{c_0}{s + k} + \frac{A}{k} \left[\frac{1}{s} - \frac{1}{s+k} \right] \\ &+ \frac{B}{k^2 + (\pi/12)^2} \left[\frac{ks}{s^2 + (\pi/12)^2} + \frac{(\pi/12)^2}{s^2 + (\pi/12)^2} - \frac{k}{s+k} \right] \end{align}\]

Invert the Laplace transforms term by term:

(5.85)\[\begin{align} c(t) &= c_0 e^{-kt} + \frac{A}{k} \left(1 - e^{-kt}\right) \\ &+ \frac{B}{k^2 + (\pi/12)^2} \left[k \cos\left(\frac{\pi t}{12}\right) + \frac{\pi}{12} \sin\left(\frac{\pi t}{12}\right) - k e^{-kt}\right] \\ &= \frac{A}{k} + \frac{B}{k^2 + (\pi/12)^2} \left[k \cos\left(\frac{\pi t}{12}\right) + \frac{\pi}{12} \sin\left(\frac{\pi t}{12}\right)\right] \\ &+ \left[c_0 - \frac{A}{k} - \frac{Bk}{k^2 + \left(\frac{\pi}{12}\right)^2}\right] e^{-kt} \end{align}\]

The average concentration is:

(5.86)\[\begin{align} \langle c \rangle &= \frac{1}{t_1-t_0} \int_{t_0}^{t_1} c(t) \d{t} \\ &= \frac{1}{24 - 12} \int_{12}^{24} \Biggl( \frac{A}{k} + \frac{B}{k^2 + (\pi/12)^2} \left[k \cos\left(\frac{\pi t}{12}\right) + \frac{\pi}{12} \sin\left(\frac{\pi t}{12}\right)\right] \\ &+ \left[c_0 - \frac{A}{k} - \frac{Bk}{k^2 + (\pi/12)^2}\right] e^{-kt} \Biggr) \d{t} \\ &= \frac{1}{12} \Biggl[ \frac{A}{k} t + \frac{B}{k^2 + (\pi/12)^2} \left( \frac{12k}{\pi} \sin\left( \frac{\pi t}{12} \right) - \cos\left( \frac{\pi t}{12} \right) \right) \\ &+ \frac{1}{k} \left( \frac{A}{k} + \frac{Bk}{k^2 + (\pi/12)^2} \right) e^{-kt} \Biggr]_{12}^{24} \\ &= \frac{A}{k} - \frac{1}{6} \frac{B}{k^2 + (\pi/12)^2} - \frac{1}{12}\left[\frac{A}{k^2} + \frac{B}{k^2 + (\pi/12)^2}\right] (e^{-12k} - e^{-24k}) \end{align}\]

5.4.3. Skill builder problems

Solve the following IVPs using Laplace transforms:

1. \(y' + 4y = e^{4x}\), \(y(0) = 0\)

   Solution

   (5.87)\[\begin{align} L[y' + 4y] &= L[e^{4x}] \\ sY - y(0) + 4Y &= \frac{1}{s-4} \\ (s+4) Y &= \frac{1}{s-4} \\ Y &= \frac{1}{(s+4)(s-4)} \end{align}\]

   Use partial fractions

   (5.88)\[\begin{equation} \frac{1}{(s+4)(s-4)} = \frac{A_1}{s+4} + \frac{A_2}{s-4} \end{equation}\]

   and cover up to find \(A_1 = -1/8\) and \(A_2 = 1/8\).

   Solve by inverting the Laplace transforms:

   (5.89)\[\begin{align} Y &= \frac{1}{8} \left(\frac{1}{s-4} - \frac{1}{s+4} \right) \\ y = L^{-1}[Y] &= \frac{1}{8}\left( L^{-1}\left[\frac{1}{s-4}\right] - L^{-1}\left[\frac{1}{s+4}\right] \right) \\ &= \frac{1}{8}(e^{4x} - e^{-4x}) \end{align}\]

2. \(y' + 2y = 8\), \(y(0) = 1\)

   Solution

   (5.90)\[\begin{align} L[y' + 2y] &= L[8] \\ sY - y(0) + 2Y &= \frac{8}{s} \\ (s+2) Y - 1 &= \frac{8}{s} \\ Y &= \frac{1}{s+2} + \frac{8}{s(s+2)} \end{align}\]

   Use partial fractions and the cover-up method for the second fraction:

   (5.91)\[\begin{equation} \frac{8}{s(s+2)} = \frac{A_1}{s} + \frac{A_2}{s+2} \end{equation}\]

   to find \(A_1 = 4\) and \(A_2 = -4\). Solve by simplifying and inverting the Laplace transforms:

   (5.92)\[\begin{align} Y &= \frac{4}{s} - \frac{3}{s+2} \\ y = L^{-1}[Y] &= 4L^{-1}\left[\frac{1}{s}\right] - 3L^{-1}\left[\frac{1}{s+2}\right] \\ &= 4 - 3e^{-2x} \end{align}\]

3. \(y' - y = 1 - 2x + \sin(3x), \quad y(0) = -1\)

   Solution

   (5.93)\[\begin{align} L[y' - y] &= L[1 - 2x + \sin 3x ] \\ sY - y(0) - Y &= \frac{1}{s} - \frac{2}{s^2} + \frac{3}{s^2 + 9} \\ (s-1)Y &= -1 + \frac{1}{s} - \frac{2}{s^2} + \frac{3}{s^2 + 9} \\ Y &= -\frac{1}{s - 1} + \frac{1}{s(s - 1)} - \frac{2}{s^2(s - 1)} + \frac{3}{(s^2 + 9)(s - 1)} \end{align}\]

   Use partial fractions:

   (5.94)\[\begin{align} \frac{1}{s(s-1)} &= \frac{A_1}{s} + \frac{A_2}{s-1} \\ \frac{-2}{s^2(s - 1)} &= \frac{A_3}{s} + \frac{A_4}{s^2} + \frac{A_5}{s - 1} \\ \frac{3}{(s^2 + 9)(s - 1)} &= \frac{A_6s + B_6}{s^2 + 9} + \frac{A_7}{s - 1} \end{align}\]

   Cover up to find:

   (5.95)\[\begin{align} A_1 &=\frac{1}{0-1} = -1 & A_4 &= \frac{-2}{0 - 1} = 2 & A_7 &= \frac{3}{1^2 + 9} = \frac{3}{10} \\ A_2 &=\frac{1}{1} = 1 & A_5 &= \frac{-2}{1} = -2 \end{align}\]

   Finish the rest by cross-multiplying or pluggin in values. For the first fraction, plug in \(s=-1\) and known coefficients:

   (5.96)\[\begin{equation} 1 &= -A_3 + 2 + 1 \end{equation}\]

   then solve \(A_3 = 2\). For the third fraction, cross-multiply

   (5.97)\[\begin{equation} 3 = (A_6 s + B_6)(s-1) + \frac{3}{10}(s^2+9) \end{equation}\]

   Then compare like powers of \(s\). Here, I use \(s^2\) and \(s^0\):

   (5.98)\[\begin{align} A_6 + \frac{3}{10} &= 0 \\ -B_6 + \frac{27}{10} &= 3 \end{align}\]

   so \(A_6 = -3/10\) and \(B_6 = -3/10\). Simplify Y

   (5.99)\[\begin{equation} Y = \frac{1}{s} + \frac{2}{s^2} - \frac{17}{10} \frac{1}{s-1} - \frac{3}{10} \frac{1}{s^2+9} - \frac{3}{10} \frac{s}{s^2+9} \end{equation}\]

   Apply inverse Laplace transform across the equation:

   (5.100)\[\begin{equation} y = L^{-1}\left[\frac{1}{s}\right] + 2 L^{-1}\left[\frac{1}{s^2}\right] - \frac{17}{10} L^{-1}\left[\frac{1}{s - 1}\right] - \frac{1}{10} L^{-1}\left[\frac{3}{s^2 + 9}\right] - \frac{3}{10} L^{-1}\left[\frac{s}{s^2 + 9}\right] \end{equation}\]

   So, the solution is

   (5.101)\[\begin{equation} y = 1 + 2x - \frac{17}{10} e^x - \frac{1}{10} \sin 3x - \frac{3}{10} \cos 3x \end{equation}\]