4 Engineering Example 1

4.1 Charge on a capacitor

In the circuit shown in Figure 19, the switch $S$ is closed at $t=0$ with a capacitor charge $q\left(0\right)=q_0=$ constant and $dq∕dt\left(0\right)=0$ .

Figure 19

Show that $q\left(t\right)=q_0\left(t\right){\text{e}}^{-\alpha t}\left[cos\omega t+\frac{\alpha }{w}sin\omega t\right]$ where $\alpha =\frac{R}{2L}$ and ${\omega }^2=\frac{1}{LC}-{\alpha }^2$

Laplace transform properties required

The following properties are needed to solve this problem.

$F\left(s+a\right)=\mathcal{L}\left\{e^{-at}f\left(t\right)\right\}$ (P1)

$\mathcal{L}\left\{\frac{df\left(t\right)}{dt}\right\}=s\left\{f\left(t\right)\right\}-f\left(0\right)$ (P2)

$\mathcal{L}\left\{\frac{d^{2}f\left(t\right)}{d{t}^2}\right\}=s^2\mathcal{L}\left\{f\left(t\right)\right\}-\frac{df}{dt}\left(0\right)-sf\left(0\right)$ (P3)

$\mathcal{L}\left\{sinkt\right\}=\frac{k}{s^2+k^2}$ with $s>0$ (P4)

$\mathcal{L}\left\{coskt\right\}=\frac{s}{s^2+k^2}$ with $s>0$ (P5)

${\mathcal{L}}^{-1}\left\{\mathcal{L}\left\{f\left(t\right)\right\}\right\}=f\left(t\right)$ (P6)

STEP 1 Establish the differential equation for $q\left(t\right)$ using, for example, Kirchhoff’s law.

Solution

When the switch $S$ is closed, the inductance $L$ , capacitance $C$ and resistance $R$ give rise to a.c. voltages related by

$V_A-V_B=L\frac{di}{dt},V_B-V_D=Ri,V_D-V_F=q∕C\text{ respectively}.$

So since $V_A-V_F=\left(V_A-V_B\right)+\left(V_B-V_D\right)+\left(V_D-V_F\right)=0$ and $i=\frac{dq}{dt}$ we have

$L\frac{d^{2}q}{d{t}^2}+R\frac{dq}{dt}+\frac{q}{C}=0$ (1)

STEP 2 Write the Laplace transform of the differential equation substituting for the initial

conditions:

Solution

Since the Laplace transform is linear, the transform of differential Equation (1) is

$\mathcal{L}\left\{L\frac{d^{2}q}{d{t}^2}+R\frac{dq}{dt}+\frac{q}{C}\right\}=L\mathcal{L}\left\{\frac{d^{2}q}{d{t}^2}\right\}+R\mathcal{L}\left\{\frac{dq}{dt}\right\}+\mathcal{L}\left\{\frac{q}{C}\right\}=0.$ (2)

We deal with each derivative term in turn: Using property (P3),

$\mathcal{L}\left\{\frac{d^{2}q}{d{t}^2}\right\}=s^2\mathcal{L}\left\{q\left(t\right)\right\}-\frac{dq}{dt}\left(0\right)-sq\left(0\right).$

So, using the initial conditions $q\left(0\right)=q_0$ and $\frac{dq}{dt}\left(0\right)=0$

$\mathcal{L}\left\{\frac{d^{2}q}{d{t}^2}\right\}=s^2\mathcal{L}\left\{q\left(t\right)\right\}-s{q}_0.$ (3)

By means of property (2)

$\mathcal{L}\left\{\frac{dq}{dt}\right\}=s\mathcal{L}\left\{q\left(t\right)\right\}-q_0$ (4)

STEP 3 Solve for the function $\mathcal{L}\left\{q\left(t\right)\right\}$ by substituting from (3) and (4) into Equation (2):

Solution

$L\left[s^2\mathcal{L}\left\{q\left(t\right)\right\}-s{q}_0\right]+R\left[s\mathcal{L}\left\{q\left(t\right)\right\}-q_0\right]+\frac{1}{C}\mathcal{L}\left\{q\left(t\right)\right\}=0$

$\Rightarrow \mathcal{L}\left\{q\left(t\right)\right\}\left[L{s}^2+Rs+\frac{1}{C}\right]=Ls{q}_0+R{q}_0$

$\Rightarrow \mathcal{L}\left\{q\left(t\right)\right\}=\frac{\left(Ls+R\right)}{\left(L{s}^2+Rs+\frac{1}{C}\right)}{q}_0$ (5)

Using the definitions $\alpha =\frac{R}{2L}$ and ${\omega }^2=\frac{1}{LC}-{\alpha }^2$ enables the denominator in Equation (5) to be expressed as the sum of two squares,

$L{s}^2+Rs+\frac{1}{C}=L\left[s^2+\frac{Rs}{L}+\frac{1}{LC}\right]=L\left[s^2+2\alpha s+\frac{1}{LC}\right]$

$=L\left[s^2+2\alpha s+{\alpha }^2+{\omega }^2\right]=L\left[{\left\{s+\alpha \right\}}^2+{\omega }^2\right].$

Consequently, with the new expression for the denominator, Equation (5) becomes

$\mathcal{L}\left\{q\left(t\right)\right\}=q_0\left[\frac{s}{{\left(s+\alpha \right)}^2+{\omega }^2}+\frac{R}{L}\frac{1}{{\left(s+\alpha \right)}^2+{\omega }^2}\right].$ (6)

STEP 4 Use the inverse Laplace transform to obtain $q\left(t\right)$ :

Solution

The inverse Laplace transform is used to find $q\left(t\right).$

Taking the inverse Laplace transform of Equation (6) and using the linearity properties

${\mathcal{L}}^{-1}\left\{\mathcal{L}\left\{q\left(t\right)\right\}\right\}=q_0{\mathcal{L}}^{-1}\left\{\frac{s}{{\left(s+\alpha \right)}^2+{\omega }^2}+\frac{R}{L}\frac{1}{{\left(s+\alpha \right)}^2+{\omega }^2}\right\}.$

Using property (P6) this can be written as

$q\left(t\right)=q_0{\mathcal{L}}^{-1}\left\{\frac{s+\alpha }{{\left(s+\alpha \right)}^2+{\omega }^2}+\frac{-\alpha }{{\left(s+\alpha \right)}^2+{\omega }^2}+\frac{R}{L\omega }\frac{\omega }{{\left(s+\alpha \right)}^2+{\omega }^2}\right\}.$

Using the linearity of the Laplace transform again

$q\left(t\right)=q_0{\mathcal{L}}^{-1}\left\{\frac{s+\alpha }{{\left(s+\alpha \right)}^2+{\omega }^2}\right\}+{\mathcal{L}}^{-1}\left\{\frac{-\alpha }{{\left(s+\alpha \right)}^2+{\omega }^2}\right\}+{\mathcal{L}}^{-1}\left\{\frac{R}{L\omega }\frac{\omega }{{\left(s+\alpha \right)}^2+{\omega }^2}\right\}.$ (7)

Using properties (P1) and (P5)

${\mathcal{L}}^{-1}\left\{\frac{s+\alpha }{{\left(s+\alpha \right)}^2+{\omega }^2}\right\}={\text{e}}^{-\alpha t}cos\omega t.$ (8)

Similarly,

${\mathcal{L}}^{-1}\left\{\frac{-\alpha }{{\left(s+\alpha \right)}^2+{\omega }^2}\right\}=-\left(\frac{\alpha }{\omega }\right)\left\{{\text{e}}^{-\alpha t}sin\omega t\right\}$ (9)

and

${\mathcal{L}}^{-1}\left\{\frac{R}{L\omega }\frac{\omega }{{\left(s+a\right)}^2+{\omega }^2}\right\}=\left(\frac{R}{L\omega }\right){\text{e}}^{-\alpha t}sin\omega t.$ (10)

Substituting (8), (9) and (10) in (7) gives

$q\left(t\right)=q_0{\text{e}}^{-\alpha t}\left[cos\omega t+\left\{-\frac{\alpha }{\omega }+\frac{R}{L\omega }\right\}{\text{e}}^{-\alpha t}sin\omega t\right].$ (11)

STEP 5 Finally, show that for $t>0$ the solution is

$q\left(t\right)=q_0{\text{e}}^{-\alpha t}\left[cos\omega t+\left(\frac{\alpha }{\omega }\right)sin\omega t\right]$  where  $\alpha =\frac{R}{2L}$ and ${\omega }^2=\frac{1}{LC}-{\alpha }^2.$

Solution

Substituting $\alpha =\frac{R}{2L}$ in (11) gives

$q\left(t\right)=q_0{\text{e}}^{-\alpha t}\left[cos\omega t+\left[-\frac{\alpha }{\omega }+\frac{2\alpha }{\omega }\right]{\text{e}}^{-\alpha t}sin\omega t\right]$

$=q_0{\text{e}}^{-\alpha t}\left[cos\omega t+\frac{\alpha }{\omega }sin\omega t\right]$