Question 1 of 6: Direct Laplace Transforms of Differential and Integral Equations

Question 1: Direct Laplace Transforms of Differential and Integral Equations (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Three lumped-parameter models: (a) a series $RLC$ loop driven by a voltage $e(t)$, with inductance $L$, resistance $R$ and capacitance $C$; (b) a mass–damper–spring translational system with mass $M$, viscous coefficient $B$ and stiffness $K$ driven by a ramp force $3t$; (c) a rotational inertia–damper–spring system with inertia $J$, viscous coefficient $B$ and torsional stiffness $K$ driven by a sinusoidal torque $10\sin\omega t$.

Find. The transformed algebraic equation for each system, and hence the transform of the dependent variable, $I(s)$, $X(s)$ and $\Theta(s)$.

Approach. Transform each equation term by term with the derivative, integration and forcing-function pairs from the supplied table, carrying the initial conditions explicitly and then setting them to zero to obtain the transfer relationship.

1. State the three operational pairs the question turns on. From the supplied table, for a function $f(t)$ with transform $F(s)$, $$\mathcal{L}\{f^{\prime}(t)\} = sF(s)-f(0), \qquad \mathcal{L}\{f^{\prime\prime}(t)\} = s^{2}F(s)-s\,f(0)-f^{\prime}(0), \qquad \mathcal{L}\left\{\int_{0}^{t} f(\tau)\,d\tau\right\} = \frac{F(s)}{s}.$$ Differentiation multiplies by $s$ and subtracts the initial state; integration divides by $s$. Each equation below is transformed one term at a time using only these rules plus the table entry for the right-hand side.
2. (a) Transform the series $RLC$ loop. With $i(0)$ the initial inductor current and $q(0)=\int i\,dt\big|_{t=0}$ the initial capacitor charge, $$L\left[sI(s)-i(0)\right] + R\,I(s) + \frac{1}{C}\left[\frac{I(s)}{s}+\frac{q(0)}{s}\right] = E(s).$$ Grouping the terms in $I(s)$ gives the transformed loop equation.
3. Reduce (a) to the quiescent-start form. Setting $i(0)=0$ and $q(0)=0$, as is standard when a transfer function is wanted, $$\boxed{\;\left(Ls + R + \frac{1}{Cs}\right) I(s) = E(s) \quad\Longrightarrow\quad \frac{I(s)}{E(s)} = \frac{Cs}{LCs^{2}+RCs+1}\;}$$ The loop impedance in the $s$-domain is exactly $Ls+R+1/(Cs)$, which is why the transformed circuit can be written down directly from the element impedances without ever writing the integro-differential equation.
4. (b) Transform the mechanical system and its ramp input. Table entry (11) with $n=1$ gives $\mathcal{L}\{3t\}=3/s^{2}$, so with $x(0)$ and $\dot{x}(0)$ retained $$M\left[s^{2}X(s)-s\,x(0)-\dot{x}(0)\right] + B\left[sX(s)-x(0)\right] + K X(s) = \frac{3}{s^{2}}.$$
5. Reduce (b) to the quiescent-start form. With $x(0)=\dot{x}(0)=0$, $$\boxed{\;\left(Ms^{2}+Bs+K\right)X(s) = \frac{3}{s^{2}} \quad\Longrightarrow\quad X(s) = \frac{3}{s^{2}\left(Ms^{2}+Bs+K\right)}\;}$$ The double pole at the origin is the ramp, not the plant; the plant contributes only the quadratic factor.
6. (c) Transform the rotational system and its sinusoidal torque. Table entry (13) gives $\mathcal{L}\{10\sin\omega t\}=10\omega/(s^{2}+\omega^{2})$, so with $\Theta(0)$ and $\dot{\Theta}(0)$ retained $$J\left[s^{2}\Theta(s)-s\,\Theta(0)-\dot{\Theta}(0)\right] + B\left[s\Theta(s)-\Theta(0)\right] + K\,\Theta(s) = \frac{10\omega}{s^{2}+\omega^{2}}.$$
7. Reduce (c) to the quiescent-start form. With $\Theta(0)=\dot{\Theta}(0)=0$, $$\boxed{\;\left(Js^{2}+Bs+K\right)\Theta(s) = \frac{10\omega}{s^{2}+\omega^{2}} \quad\Longrightarrow\quad \Theta(s) = \frac{10\omega}{\left(s^{2}+\omega^{2}\right)\left(Js^{2}+Bs+K\right)}\;}$$ The pole pair at $s=\pm j\omega$ carries the steady sinusoidal response while the plant quadratic carries the decaying transient.

All three results share the same structure: a polynomial in $s$ built from the element constants multiplies the transform of the dependent variable, and the transform of the driving function stands alone on the right. This is the whole point of the transform — calculus in $t$ has become algebra in $s$.