I am learning circuit analysis in s-domain from my textbook "Engineering Circuit Analysis by William Hayt, 8th Ed". I am struggling with the mesh and nodal analysis.

In the book at page 583, there is a small practice question which I tried and my answer and the book solution don't really match. Can somebody help me show where I am getting things wrong?



The book solution to the exercise is: $$ v_{x}(t)=[5+5.657(e^{-1.707t}-e^{-0.2929t})]u(t) $$

My solution to the exercise is: $$ v_{x}(t)=[5+7.07(e^{-1.707t}-e^{-0.293t})]u(t) $$

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    \$\begingroup\$ The constants in the sources are initial conditions and not part of the step inputs. \$\endgroup\$ – Chu Dec 29 '18 at 0:43
  • \$\begingroup\$ @Chu I know. The constant addends are used to find the capacitor voltage at t=0-. But considering an equivalent circuit at t<0, we find that the voltage across the capacitor terminals is zero. Therefore, the capacitor has NO initial energy stored at t=0+. Then, at t>0, we can consider the voltage source to be vs=5*u(t). This is the method used throughout the book. \$\endgroup\$ – billyandriam Dec 29 '18 at 0:48
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    \$\begingroup\$ But there is an initial current through the inductor. \$\endgroup\$ – Chu Dec 29 '18 at 1:03
  • \$\begingroup\$ @Chu OH! you're right! I didn't take that into account. I really welcome your methode since I don't know any alternative strategy to solve this problem. \$\endgroup\$ – billyandriam Dec 29 '18 at 1:07
  • \$\begingroup\$ @Chu You are definitely right. My mistake comes from the fact that I didn't consider the initial current throught the inductor. Thanks a million. \$\endgroup\$ – billyandriam Dec 29 '18 at 13:31

I finally figured out the solution. The key is to firstly find the initial current flowing through the inductor.

For \$t<0\$, the equivalent circuit is the following:


From this circuit, we will calculate the capacitor voltage \$v_c(\infty)\$ and the inductor current \$ i_L(\infty)\$ which will serve as initial stored energy values in the original circuit.

  • It is clear that \$v_c(\infty)=0\text{ Volts}\$
  • Mesh analysis on the two loops will yield \$i_L(\infty)\$

From Mesh 1: $$\begin{align*} \frac{1}{s} &= I - I_L +\frac{2}{s}I\\ \frac{1}{s} &= \left(1+\frac{2}{s}\right)I - I_L \end{align*}$$

From Mesh 2: $$\begin{align*} \frac{1}{s} &= I - I_L -\left(4s\right)I_L\\ \frac{1}{s} &= I - \left(1+4s\right)I_L \end{align*}$$

The linear system is the following:

$$ \begin{cases} \left(1+\frac{2}{s}\right)I - I_L &= \frac{1}{s}, &\text{from Mesh 1}\\ I - \left(1+4s\right)I_L &= \frac{1}{s}, &\text{from Mesh 2} \end{cases} $$

Solving the system, we end up with the expression of \$I_L\$:

$$\begin{align*} I_L &=\frac{{}^{-1}{\mskip -5mu/\mskip -3mu}_2}{s\left(s^2+2s+{}^1{\mskip -5mu/\mskip -3mu}_2\right)}\\ I_L &= \frac{{}^{-1}{\mskip -5mu/\mskip -3mu}_2}{s\left(s+1+\frac{\sqrt{2}}{2}\right)\left(s+1-\frac{\sqrt{2}}{2}\right)} \end{align*}$$

Proceeding with the decomposition of the rational expression:

$$ I_L=\frac{-1}{2}\left(\frac{a}{s}+\frac{b}{\left(s+1+\frac{\sqrt{2}}{2}\right)}+\frac{c}{\left(s+1-\frac{\sqrt{2}}{2}\right)}\right) $$

where \$a,b,c\$ are real numbers. Taking the inverse Laplace trasform of \$I_L\$: $$\begin{align*} \mathcal{L}^{-1}\{I_L\}=i(t)&=\frac{-1}{2}\left(a+b e^{-\left(1+{}^{\sqrt{2}}{\mskip -5mu/\mskip -3mu}_2\right) t} +c e^{-\left(1-{}^{\sqrt{2}}{\mskip -5mu/\mskip -3mu}_2\right) t}\right) u(t)\\ i(\infty)&= \frac{-a}{2} \end{align*}$$ Identifying \$a\$, we find that: $$ a=2 \quad \text{and}\quad i(\infty)=-1 \text{ A} $$

Now, in the second part at \$t\geq 0\$, we return to the intial circuit which becomes as:


Calculating \$V_x\$: $$\begin{align*} \left(\frac{5}{s}-V_x\right)\frac{s}{2} &=V_x+\left(V_x-4-\frac{5}{s}\right)\frac{1}{4s}\\ \left(\frac{5}{s}-V_x\right)\frac{s}{2} &= \left(1+\frac{1}{4s}\right)V_x-\left(\frac{1}{s}+\frac{5}{4s^2}\right)\\ V_x &= \frac{10s^2+4s+5}{s\left(4s+2s^2+1\right)}\\ V_x &= \frac{5s^2+2s+{}^{5}{\mskip -5mu/\mskip -3mu}_2}{s\left(s+1+\frac{\sqrt{2}}{2}\right)\left(s+1-\frac{\sqrt{2}}{2}\right)} \end{align*}$$

Decomposing the rational expression: $$ V_x=\frac{d}{s}+\frac{e}{s+1+\frac{\sqrt{2}}{2}}+\frac{f}{s+1-\frac{\sqrt{2}}{2}} $$

where \$d,e,f\$ are real numbers. Identifying them, we find that: $$ d=5, \qquad e=4\left(\frac{2+\sqrt{2}}{1+\sqrt{2}}\right), \qquad f=4\left(\frac{2-\sqrt{2}}{1-\sqrt{2}}\right) $$

Taking the inverse Laplace transform: $$ \mathcal{L}^{-1}\{V_x\}=v_x(t)=\left[5+4\left(\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)e^{-\left(1+{}^{\sqrt{2}}{\mskip -5mu/\mskip -3mu}_2\right)t}+4\left(\frac{2-\sqrt{2}}{1-\sqrt{2}}\right)e^{-\left(1-{}^{\sqrt{2}}{\mskip -5mu/\mskip -3mu}_2\right)t}\right] $$

Finally, simplifying the expression: $$ v_x(t) =\left(5+5.657e^{-1.707t}-5.657e^{-0.293t}\right)u(t) \text{ V} $$


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