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2.2: Introduction to Application of Laplace Transforms

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    The Laplace transform (after French mathematician and celestial mechanician Pierre Simon Laplace, 1749-1827) is a mathematical tool primarily for solving ODEs, but with other important applications in system dynamics that we will study later. In Laplace transformation, we deal with a complex variable denoted as \(s\), which is usually expressed in terms of its real and imaginary parts as

    \[s=\sigma+j \omega\label{eqn:2.9} \]

    in which \(\sigma\) and \(\omega\) both are real. We define a complex function of \(s\), \(F(s)\). The type of function that we will encounter often takes the form of a ratio of two polynomials:

    \[F(s) \equiv \frac{\operatorname{Num}(s)}{\operatorname{Den}(s)}=\frac{b_{1} s^{m}+b_{2} s^{m-1}+\ldots+b_{m+1}}{a_{1} s^{n}+a_{2} s^{n-1}+\ldots+a_{n+1}}=\frac{b_{1}\left(s-z_{1}\right)\left(s-z_{2}\right) \cdots\left(s-z_{m}\right)}{a_{1}\left(s-p_{1}\right)\left(s-p_{2}\right) \cdots\left(s-p_{n}\right)}\label{eqn:2.10} \]

    In the first polynomial form of Equation \(\ref{eqn:2.10}\), \(a_1\), \(\dots\), \(a_{n+1}\) and \(b_1\), \(\dots\), \(b_{m+1}\) are real constants (with the symbols and numbering system keyed to MATLAB notation); numerator \(\operatorname{Num}(s)\) is an \(m\)th degree polynomial in \(s\), and denominator \(\operatorname{Den}(s)\) is an \(n\)th degree polynomial in \(s\), with \(m\) \(\leq\) \(n\) in general. In the second polynomial form, with factored \(\operatorname{Num}(s)\) and \(\operatorname{Den}(s)\), complex constants \(z_1\), \(z_2\), \(\dots\), \(z_m\) are called zeros of \(F(s)\) because \(F(s)\) is zero if \(s\) equals any one of them, and complex constants \(p_1\), \(p_2\), \(\dots\), \(p_n\) are called poles of \(F(s)\) because \(F(s)\) is infinite if \(s\) equals any one of them. If \(m\) < \(n\), \(F(s)\) in Equation \(\ref{eqn:2.10}\) also goes to zero as \(s\rightarrow\inf\).

    Solving a simple ODE problem with Laplace transforms is a gentle introduction to the subject. Consider the 1st order LTI ODE written in standard form: \(\dot{x}-a x=b u(t)\), Equation 1.2.1. Let us solve this ODE with a known IC, \(x(0)=x_{0}\), and with a specific exponential input function, \(u(t)=U e^{-w t}\), \(U\) being a dimensional magnitude; in any physically realistic problem, constant w would be a real number, but for generality here we allow it to be a complex number. So the complete problem statement is:

    \[\dot{x}-a x=b U e^{-w t}, x(0)=x_{0}, \text { solve for } x(t), t>0\label{eqn:2.11} \]

    To begin the solution, we multiply the ODE by a complex exponential, \(e^{-s t}\), then take the definite integral over time from \(t\) = 0 to \(t\) = \(\inf\) of the entire multiplied ODE:

    \[\int_{t=0}^{t=\infty} e^{-s t}\left(\dot{x}-a x=b U e^{-w t}\right) d t \Rightarrow \int_{t=0}^{t=\infty} e^{-s t} \dot{x} d t-a \int_{t=0}^{t=\infty} e^{-s t} x d t=b U \int_{t=0}^{t=\infty} e^{-s t} e^{-w t} d t\label{eqn:2.12} \]

    In Equation \(\ref{eqn:2.12}\), \(s\) is a complex variable that must have values for which the integrals exist. Based upon Equation \(\ref{eqn:2.12}\), we define the forward Laplace transform of dependent variable \(x(t)\)1 :

    \[L[x(t)] \equiv X(s) \equiv \int_{t=0}^{t=\infty} e^{-s t} x(t) d t\label{eqn:2.13} \]

    In Equation \(\ref{eqn:2.13}\), the function \(x\) of time \(t\) is transformed by the definite integration into a function \(X\) of Laplace variable \(s\). Also, the ODE of Equation \(\ref{eqn:2.11}\) is transformed into Equation \(\ref{eqn:2.12}\), which, as we will find, becomes an easily solvable algebraic equation in the unknown \(X(s)\). After we solve that algebraic equation for \(X(s)\), then we will reverse the process to find the original unknown \(x(t)\) by applying the inverse Laplace transform, denoted as \(L^{-1}[X(s)] \equiv x(t)\). For any Laplace-transformable function \(f(t)\), and its transform \(F(s)\), the companion equations \(L[f(t)]=F(s)\) and \(L^{-1}[F(s)]=f(t)\) are called a Laplace transform pair.

    \[\begin{aligned}
    \int_{t=0}^{t=\infty} e^{-s t} e^{-w t} d t &=\int_{t=0}^{t=\infty} e^{-(s+w) t} d t=\frac{1}{-(s+w)} \int_{t=0}^{t=\infty} d\left[e^{-(s+w) t}\right]=\frac{1}{-(s+w)}\left[e^{-(s+w) t}\right]_{t=0}^{t=\infty} \\
    &=\frac{1}{-(s+w)}\left[e^{-(s+w) \infty}-1\right]=\frac{1}{s+w}
    \end{aligned} \nonumber \]

    An important step in this derivation involves the complex exponential, with use of Equations 2.1.5 and 2.1.12: \(e^{z}=e^{(x+j y)}=e^{x}(\cos y+j \sin y)\). Since \(\sin y\) and \(\cos y\) vary periodically as \(y\) varies, the only way for \(e^{z}\) to \(\rightarrow 0 + 0j\) for all values of \(y\) is for \(x \rightarrow-\infty\). Thus, in the last step of the integration, we assume that \(\operatorname{Re}(s+w)>0\), or \(\operatorname{Re}(s)>-\operatorname{Re}(w)\), so that \(e^{-(s+w) \infty}=0+j 0\). The integration above establishes the following Laplace transform pair, which is one of the most important of all pairs for applications:

    \[L\left[e^{-w t}\right]=\frac{1}{s+w} \quad \text { and } \quad L^{-1}\left[\frac{1}{s+w}\right]=e^{-w t}\label{eqn:2.14} \]

    For your convenience, transform pair Equation \(\ref{eqn:2.14}\) and all other fundamental Laplace transform pairs used in this book are tabulated in Appendix A at the end of the book. Appendix A also includes some of the longer but less instructive transform derivations, should you wish to study them.

    Next, let us evaluate the first term on the left-hand side of Equation \(\ref{eqn:2.12}\) by the standard method of integration by parts in the form \(\int_{t=0}^{t=\infty} u d v=[u v]_{t=0}^{t=\infty}-\int_{t=0}^{t=\infty} v d u:\)

    \[\begin{align}
    &\int_{t=0}^{t=\infty} e^{-s t} \cdot \hat{x} d t=\left[e^{-s t} x(t)\right]_{t=0}^{t=\infty}-\int_{t=0}^{t=\infty} x(t) \times(-s) e^{-s t} d t=\left[e^{-s x \infty} x(\infty)-e^{-0} x(0)\right]+s X(s)\\
    &=-x(0)+s X(s)\label{eqn:2.15a}
    \end{align} \nonumber \]

    To obtain this result, we assume that \(e^{-(s \times \infty)} x(\infty)=0\), which generally means \(\operatorname{Re}(s)>0\). This result is the important general Laplace transform of the first derivative of any transformable function \(f(t)\):

    \[L[\dot{f}(t)]=s F(s)-f(0)\label{eqn:2.15b} \]

    Equation \(\ref{eqn:2.15b}\) is the basis for derivation of the Laplace transform of a derivative of any order \(n\), which we will use later (Ogata2, 1998, pp. 25-26):

    \[L\left[\frac{d^{n}}{d t^{n}} f(t)\right]=s^{n} F(s)-s^{n-1} f(0)-s^{n-2} \dot{f}(0)-\cdots-\overset{(n-1)}{f}(0)\label{eqn:2.16} \]

    For example, the Laplace transform of the 2nd derivative is (homework Problem 2.7):

    \[L[\ddot{f}(t)]=s^{2} F(s)-s f(0)-\dot{f}(0)\label{eqn:2.17} \]

    Proceeding with the solution of ODE + IC Equation \(\ref{eqn:2.11}\), we now substitute Equations \(\ref{eqn:2.13}\), \(\ref{eqn:2.14}\), and \(\ref{eqn:2.15a}\) back into Equation \(\ref{eqn:2.12}\) to obtain an algebraic equation for transform \(X(s)\), which is easily solved:

    \[s X(s)-x(0)-a X(s)=\frac{b U}{s+w} \quad \Rightarrow \quad(s-a) X(s)=x_{0}+\frac{b U}{s+w} \nonumber \]

    \[\Rightarrow \quad X(s)=\frac{1}{(s-a)} x_{0}+\frac{b U}{(s-a)(s+w)}\label{eqn:2.18} \]

    To solve for the ultimate unknown \(x(t)\), we now need to find the inverse transform \(X(s)\) of in (2-18). Using Equation \(\ref{eqn:2.14}\) easily gives us the inverse of the first term on the righthand side:

    \[L^{-1}\left[\left(\frac{1}{s-a}\right) x_{0}\right]=x_{0} L^{-1}\left[\frac{1}{s-a}\right]=x_{0} e^{a t}\label{eqn:2.19} \]

    Inverting the second term on the right-hand side of Equation \(\ref{eqn:2.18}\) is a greater challenge, which requires us to use partial-fraction expansion in order to expand that term into two simpler terms, each of which has the easily invertible form of a constant divided by a single factor of the form (\(s\) - \(p\)). We first write the troublesome term as two simpler fractions with unknown constant numerators, \(C_1\) and \(C_2\), which are called residues of the partial-fraction expansion. (A more detailed justification for this form is derived in the next section.):

    \[\frac{1}{(s-a)(s+w)}=\frac{C_{1}}{s-a}+\frac{C_{2}}{s+w} \Rightarrow \frac{C_{1}}{s-a}+\frac{C_{2}}{s+w}=\frac{1}{(s-a)(s+w)}\label{eqn:2.20} \]

    There are two common methods for finding \(C_1\) and \(C_2\), which we shall call the “laborsaving” method and the “brute-force” method.

    Consider first the labor-saving method. Let us go in great detail through the steps required to find \(C_1\), for example. First, multiply both sides of Equation \(\ref{eqn:2.20}\) by , the denominator of the term:

    \[(s-a)\left(\frac{C_{1}}{s-a}+\frac{C_{2}}{s+w}\right)=C_{1}+(s-a)\left(\frac{C_{2}}{s+w}\right)=(s-a)\left(\frac{1}{(s-a)(s+w)}\right) \nonumber \]

    Isolate \(C_1\) on the left-hand side:

    \[C_{1}=-(s-a)\left(\frac{C_{2}}{s+w}\right)+(s-a)\left(\frac{1}{(s-a)(s+w)}\right) \nonumber \]

    Now, in order to eliminate the \(C_2\) term on the right-hand side, set \(s\) = \(a\). Note that this does not eliminate the second right-hand-side term because \((s-1)\) is in both the denominator and the numerator.

    \[\begin{align*} C_{1} &=\left[-(s-a)\left(\frac{C_{2}}{s+w}\right)+(s-a)\left(\frac{1}{(s-a)(s+w)}\right)\right]_{s=a} \\[4pt] &=\left[(s-a)\left(\frac{1}{(s-a)(s+w)}\right)\right]_{s=a} \end{align*} \nonumber \]

    Thus, we obtain the required equation for \(C_1\):

    \[C_{1}=\left[(s-a)\left(\frac{1}{(s-a)(s+w)}\right)\right]_{s=a}=\frac{1}{a+w} \nonumber \]

    Now that we see the logic of the labor-saving method, we can dispense with most of the intermediate steps and quickly write the corresponding equation for \(C_2\) directly from Equation \(\ref{eqn:2.20}\):

    \[C_{2}=\left[(s+w)\left(\frac{1}{(s-a)(s+w)}\right)\right]_{s=-w}=\frac{1}{-w-a}=-\frac{1}{a+w} \nonumber \]

    Next, the brute-force method begins with use of the traditional algebraic method for combining the two fractions of Equation \(\ref{eqn:2.20}\) into a single fraction:

    \[\frac{1}{(s-a)(s+w)}=\frac{C_{1}}{s-a}+\frac{C_{2}}{s+w}=\frac{C_{1}(s+w)+C_{2}(s-a)}{(s-a)(s+w)}=\frac{s\left(C_{1}+C_{2}\right)+\left(C_{1} w-C_{2} a\right)}{(s-a)(s+w)} \nonumber \]

    Equate coefficients of like powers of \(s\) in the numerators:

    \[s^{1}: 0=C_{1}+C_{2} \Rightarrow C_{2}=-C_{1} \nonumber \]

    \[s^{0}: 1=C_{1} w-C_{2} a=C_{1}(w+a) \Rightarrow C_{1}=\frac{1}{a+w}=-C_{2} \nonumber \]

    In this simple problem, the brute-force method is not much more demanding algebraically than the labor-saving method. However, for a slightly more complex original fraction, say, one with three denominator polynomial factors instead of just two, the brute-force method can require orders of magnitude more algebra than the labor-saving method. The point of this discussion: exert the mental energy to understand the laborsaving method, and always use it rather than the brute-force method [see homework Problem 2.9.2].

    Using the partial-fraction expansion developed above, we now express the complete second term on the right-hand side of Equation \(\ref{eqn:2.18}\) in the easily invertible form

    \[F(s)=\frac{b U}{(s-a)(s+w)}=\frac{b U}{a+w}\left(\frac{1}{s-a}-\frac{1}{s+w}\right) \nonumber \]

    Using the fundamental inverse transform from \(\ref{eqn:2.14}\) gives

    \[f(t)=\frac{b U}{a+w}\left(e^{a t}-e^{-w t}\right)\label{eqn:2.21} \]

    Finally, combining Equations \(\ref{eqn:2.19}\) and \(\ref{eqn:2.21}\) gives the inverse of Equation \(\ref{eqn:2.18}\) and the final desired solution of ODE + IC Equation \(\ref{eqn:2.11}\):

    \[x(t)=x_{0} e^{a t}+\frac{b U}{a+w}\left(e^{a t}-e^{-w t}\right), \text { for } t \geq 0\label{eqn:2.22} \]

    The introduction to Laplace transformation in this section is not mathematically rigorous. The focus in introductory system dynamics is more with applying Laplace transforms than with the detailed theory. It is relevant, however, to comment on the existence of Laplace transforms: a transform generally exists (i.e., the defining integral Equation \(\ref{eqn:2.13}\) can be evaluated, in principle) for any function \(f(t)\) for which the product \(e^{-a t}|f(t)| \rightarrow 0\) as \(t \rightarrow \infty\), where \(a\) is some finite, positive, real constant (Hildebrand, 1962, Section 2.2). Practically speaking, this means that any physically realizable function, the type of function that we encounter in engineering, has a Laplace transform.

    1In Section 8.4 we develop a more general definition, Eq. (8-12), in order to accommodate the ideal unit-impulse function, which is important in linear-system theory and applications.

    2Literature sources such as this are described in detail in the References section following Chapter 17.


    This page titled 2.2: Introduction to Application of Laplace Transforms is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by William L. Hallauer Jr. (Virginia Tech Libraries' Open Education Initiative) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.