Engineering Sinusoidal steady state analysis using Laplace transform

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The discussion focuses on using the Laplace transform to analyze a circuit and reproduce a solution derived through phasor analysis. The user successfully transforms circuit elements and derives expressions for the voltages and currents in the frequency domain. However, upon taking the inverse Laplace transform, the user finds a discrepancy in the final steady-state current expression compared to the professor's result. It is concluded that the phase difference arises from a potential mistake in interpreting the sine and cosine functions in the professor's solution, suggesting that the steady-state answer should align with the source voltage form. This clarification resolves the user's confusion regarding the phase difference between the two approaches.
Wrichik Basu
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Homework Statement
Find ##i_x(t)## in the given circuit at the steady state.
Relevant Equations
Laplace transforms for circuit elements. $$\begin{align}
t\mathrm{-domain} &\rightarrow s\mathrm{-domain}\\
R &\rightarrow R \\
L &\rightarrow sL \\[0.9em]
C &\rightarrow \dfrac{1}{sC}\\[0.9em]
\cos \omega t &\rightarrow \dfrac{s}{s^2 + \omega^2} \\[0.9em]
\sin \omega t &\rightarrow \dfrac{\omega}{s^2 + \omega^2}
\end{align}$$
##\require{physics}##The given circuit is this:

1695896537667.png

The question is taken from this video. The Professor has solved it using Phasor analysis, the final solution being $$\begin{equation}
i_x(t) = 7.59 \sin \qty( 4t + 108.4^\circ )~\mathrm{amps}.
\end{equation}$$My aim, however, is to use Laplace transform to reproduce this solution.

Step 1 is to transform all the circuit elements from the time domain to the frequency domain. The transformed circuit looks like:

1695897060600.png

At node A,$$\begin{align}
&\phantom{implies} \dfrac{\dfrac{20 s}{s^2 + 16} - V_A(s)}{10} - I_x(s) - \dfrac{V_A - V_B}{s} = 0 \nonumber \\[0.8em]
&\implies \dfrac{2s}{s^2 + 16} - \dfrac{V_A}{10} - \dfrac{V_A}{10/s} - \dfrac{V_A - V_B}{s} = 0.\label{eqn1}
\end{align}$$
At node B,$$\begin{align}
\dfrac{V_A - V_B}{s} + 2\dfrac{V_A}{10/s} - \dfrac{V_B}{s/2} = 0.\label{eqn2}
\end{align}$$
Solving equations##~\eqref{eqn1}## and ##\eqref{eqn2}## yield $$\begin{align}
V_A(s) &= \dfrac{60 s^2}{\qty( s^2 + 16 ) \qty( s^2 + 3s + 2)} \\[0.8em]
\implies I_x(s) &= \dfrac{6s^3}{\qty( s^2 + 16 ) \qty( s^2 + 3s + 2)}.
\end{align}$$
Time to take the inverse Laplace transform. $$\begin{align}
I_x(s) &= \dfrac{6s^3}{\qty( s^2 + 16 ) \qty( s^2 + 3s + 20)} \nonumber \\[1em]
&= \dfrac{\dfrac{42}{5}s + 36}{s^2 + 3s + 20} - \dfrac{\dfrac{12}{5}s + \dfrac{144}{5}}{s^2 + 16} \nonumber \\[1em]
&= \dfrac{\dfrac{42}{5}\qty( s + 1.5 )}{\qty( s + 1.5 )^2 + \dfrac{71}{4}}
+ \dfrac{234}{5\sqrt{71}} \dfrac{\sqrt{71}/2}{\qty( s + 1.5 )^2 + \dfrac{71}{4}} - \dfrac{12}{5} \qty[ \dfrac{s}{s^2 + 16} + 3 \dfrac{4}{s^2 + 16} ] \nonumber\\[1em]
\implies i_x (t) &= \dfrac{42}{5} \exp^{-1.5t} \qty[ \cos \qty( \dfrac{\sqrt{71}}{2}t ) + \dfrac{39 \sqrt{71}}{497} \sin \qty( \dfrac{\sqrt{71}}{2}t ) ] - \dfrac{12}{5} \qty[ \cos (4t) + 3 \sin (4t) ].
\end{align}$$
At steady state, i.e. when ##t \rightarrow \infty,## the exponential part vanishes, so I am left with only $$\begin{equation}
i_x (t) = - \dfrac{12}{5} \qty[ \cos (4t) + 3 \sin (4t) ]~\mathrm{amps}. \label{eqn:ixt_final}
\end{equation}$$

I am not sure how to proceed from here to arrive at the same equation that was derived in the video.

If I plot the two values, I get

1695913660053.png

which is just a phase difference. I am not sure where I went wrong. Any leads will be helpful.
 
Last edited:
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Since your source voltage is of the form ##20 \cos(4 t)##, I believe that your steady state answer should also be in that form:
$$ i_x(t) = 7.59 \cos \qty( 4t + 108.4^\circ )~\mathrm{amps}$$
So I think that your professor mistook the sin for cos. This, will I believe solve your phase difference issues between the phasor approach and the Laplace approach.
 
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Likes berkeman and Wrichik Basu
gneill said:
Since your source voltage is of the form ##20 \cos(4 t)##, I believe that your steady state answer should also be in that form:
$$ i_x(t) = 7.59 \cos \qty( 4t + 108.4^\circ )~\mathrm{amps}$$
So I think that your professor mistook the sin for cos. This, will I believe solve your phase difference issues between the phasor approach and the Laplace approach.
That sure does solve the issue. Thank you!
 
You're very welcome!
 

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