Circuit Analysis with Laplace Transforms

In summary, the student attempted to solve a homework equation by transforming it into the 's' domain, but found that they were stuck after calculating the current through the circuit. They then attempted to solve the equation by Laplace transform and found that they were on the right track.
  • #1
gmm
4
0

Homework Statement


circuit1.png


Homework Equations


V=IR
All of them actually

The Attempt at a Solution


So I Started off by transforming the voltage source into the 's' domain
vs(s) = (4/s) -(4/s)*e-.5t

I know the initial conditions are zero, in other words at t=0, the voltage and currents at the capacitors are all 0. which means that that my capacitor 1 can be expressed as an impedance: 106/s. and capacitor 2 as : 3*105/s. All the resistors stay the same.
So now the circuit is transformed into the 's' domain which basically means its comprised of one voltage source and 5 impedences?
This is kind of where I'm stuck... I know that for a Capacitor in the 's' domain the voltage is Ic/sC + vc(0-)/s.. in this case Initial conditions are zero so for both capacitors Vc = Ic/sC ... I now need to find the current I at each capacitor right?
If so I'm not sure how to go about this, all of the examples in my book only show steps for circuits that have components either all in series or all in parallel. Node Voltage Analysis? Current Mesh Analysis? Whats my next step??
 
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  • #2
Node voltage analysis (nodal analysis) looks like a good choice: There's only one essential node.

The Laplace impedance of a capacitor is ##\frac{1}{sC}##.
 
  • #3
Calculate the transfer functions v1(s)/vs(s) and v2(s)/vs(s).

( Calculate the current, I(s), through the 4 rightmost components. Then v(s) = Z(s) * I(s) for both RC pairs. )

Knowing e.g. v2(s)/vs(s) → v2(s) = v2(s)/vs(s) * vs(s).

Then inverse Laplace transform.
 
  • #4
I suggest labeling your components C1, C2, R1 etc rather than working with numbers up front. Then you can also use dimensional analysis for checking the math. Otherwise you're on the right track.
 

FAQ: Circuit Analysis with Laplace Transforms

1. What is a Laplace transform?

A Laplace transform is a mathematical tool used to convert a function of time into a function of complex frequency. It is particularly useful in circuit analysis because it allows for the analysis of time-varying signals in the frequency domain.

2. Why do we use Laplace transforms in circuit analysis?

Laplace transforms are used in circuit analysis because they allow for the simplification of differential equations, making it easier to analyze complex circuits. Additionally, they can help determine the response of a circuit to different inputs, such as sinusoidal or step functions.

3. How do you perform a Laplace transform on a circuit?

To perform a Laplace transform on a circuit, you must first write the circuit's differential equations in terms of voltages and currents. Then, apply the Laplace transform to each term in the equation, using standard transform pairs. Finally, solve for the transformed variables and use inverse Laplace transforms to find the time-domain solution.

4. What is the difference between a Laplace transform and a Fourier transform?

While both Laplace and Fourier transforms are used in circuit analysis, they have different applications. A Fourier transform is used to analyze signals that are periodic in nature, while a Laplace transform is used for non-periodic signals. Additionally, a Fourier transform only considers positive frequencies, while a Laplace transform takes into account both positive and negative frequencies.

5. Are there any limitations to using Laplace transforms in circuit analysis?

Yes, there are some limitations to using Laplace transforms in circuit analysis. For example, they can only be used for linear time-invariant circuits, and they assume that the circuit has reached steady-state. Additionally, Laplace transforms are not as intuitive as other methods of circuit analysis, so they may take some practice to fully understand and apply correctly.

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