Sinusoidal steady state analysis using Laplace transform

  • #1
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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  • #2
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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  • #3
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!
 
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  • #4
You're very welcome!
 

Related to Sinusoidal steady state analysis using Laplace transform

What is sinusoidal steady state analysis?

Sinusoidal steady state analysis is a method used to determine the behavior of electrical circuits when they are driven by sinusoidal sources. In this state, all transient effects have died out, and the system's response is purely sinusoidal at the same frequency as the input signal.

How does the Laplace transform help in sinusoidal steady state analysis?

The Laplace transform simplifies the process of analyzing circuits by converting differential equations into algebraic equations. For sinusoidal steady state analysis, it allows us to handle complex impedance and phasors systematically, making it easier to solve for voltages and currents in the circuit.

What is the significance of the s-domain in this analysis?

The s-domain, or complex frequency domain, is used in the Laplace transform to represent circuit elements and their behaviors. In sinusoidal steady state analysis, the s-domain helps in transforming complex time-domain differential equations into simpler algebraic forms, facilitating easier manipulation and solution.

How do you convert a sinusoidal source to the s-domain using Laplace transform?

A sinusoidal source can be represented in the s-domain using the Laplace transform. For example, a sinusoidal function like \( V(t) = V_0 \sin(\omega t + \phi) \) can be transformed into the s-domain using \( V(s) = \frac{V_0 \omega e^{j\phi}}{s^2 + \omega^2} \), where \( \omega \) is the angular frequency and \( \phi \) is the phase angle.

What are the common steps in performing sinusoidal steady state analysis using Laplace transform?

The common steps include: (1) Transforming the circuit elements and sources into the s-domain, (2) Writing the algebraic equations based on Kirchhoff's laws, (3) Solving the algebraic equations for the desired quantities, (4) Transforming the solutions back to the time domain if necessary, and (5) Analyzing the steady-state behavior of the circuit.

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