1.3: Solving First-Order ODE Models
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- 17743
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The response of a first-order ODE to a step forcing function with given initial conditions can be obtained with the help of the Laplace Transform. Accordingly, we consider the generic first-order ODE model: \(\tau \frac{dy\left(t\right)}{dt}+y\left(t\right)=u(t)\), and apply Laplace transform assuming an initial condition \(y\left(0\right)=y_0\), to obtain: \[\tau \left(sy\left(s\right)-y_0\right)+y(s)=u(s)\] Next, assuming a unit step input \(u\left(t\right)\), where \(u\left(s\right)=\frac{1}{s}\), the output is solved as: \[y\left(s\right)=\frac{1}{s\left(\tau s+1\right)}+\frac{\tau y_0}{\tau s+1}\] We may use partial fraction expansion (PFE) to express the output as: \[y\left(s\right)=\frac{1}{s}-\frac{\tau }{\tau s+1}+\frac{\tau y_0}{\tau s+1}\] After applying the inverse Laplace transform, we obtain the time-domain solution to the ODE as: \[y\left(t\right)=\left[1+\left(y_0-1\right)e^{-t/\tau }\right]u\left(t\right)\] The \(u\left(t\right)\) in the above expression represents a unit step function, that is used to represent causality, i.e., the output is valid for \(t\ge 0\).
The steady-state value of the system response is denoted as: \(y_{\infty }={\mathop{\mathrm{lim}}_{t\to \infty } y(t)\ }\). In terms of the steady-state output, the step response of the first-order ODE model is given as: \[y\left(t\right)=\left[y_{\infty }+\left(y_0-y_{\infty }\right)e^{-t/\tau }\right]u\left(t\right)\] Assuming zero initial conditions, i.e., \(y_0=0\), the output of the system is expressed as: \[y\left(t\right)=\left(1-e^{-t/\tau }\right)u\left(t\right)\] The output at the selected times, \(t=k\tau ,\ \ k=0,1,\dots\) are compiled in the following table:
| Time | Output value |
|---|---|
| 0 | \(y\left(0\right)=0\) |
| 1\(\tau\) | \(1-e^{-1}\cong 0.632\) |
| 2\(\tau\) | \(1-e^{-2}\cong 0.865\) |
| 3\(\tau\) | \(1-e^{-3}\cong 0.950\) |
| 4\(\tau\) | \(1-e^{-4}\cong 0.982\) |
| 5\(\tau\) | \(1-e^{-5}\cong 0.993\) |
image3 By convention, the model output is assumed to have reached steady-state when the output attains 98% of its final value. Hence, the settling time of the system is given as: \(t_s=4\tau\). The output for \(\tau =1sec\) is plotted in Figure 6.
Figure 6: Output of a first-order system model to a unit-step input.