Laplace Error in a Circuit

In summary: The voltage on the capacitor at any given time is the sum of the initial voltage and the constant voltage source.
  • #1
baby_1
159
15
Laplace Error in a Circuit !

Hello
as you see this circuit
1383826100_1352649501.jpg

i want to find the v(t) across the 2ohm resistor.(VC(0-)=0)
i assume A node above the capacitor , so we have (in Laplace)

gif.gif


so i have
gif.gif


so A=0 ! or S=-3/2 ! what can we understand about A=0 or S=-3/2 ? what is their inverse Laplace in time zone?

now how can find the v(t) across the 2ohm resistor? Thanks
 
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  • #2


Hello Again
i have a new question about above circuit. now how can find the voltage across the capacitor?
 
  • #3


What's wrong with A=0? That just means there's no voltage drop across the capacitor, so finding the voltage across the resistor should be pretty straight forward. Also, you shouldn't call s=-3/2 a solution, because your node equations should hold for all values of s. The only way your node equation holds for all values of s is when A=0. Hence, the voltage across the capacitor (A) is zero.

Admittedly this is a bit of an odd circuit. Here's two things you can do to try to understand.

1. Replace the 5/s source with an arbitrary source V1(s) and replace the 10/s source with an arbitrary source V2(s). Solve for A(s) in terms of V1(s) and V2(s). You should see that when V2 = 2*V1 (as is the case here), A(s) goes to zero, but if not A(s) will have an exponential response like you would normally expect. Essentially, in this case the sources are perfectly balanced so that they don't charge up the capacitor. If you had a 5/s source and a 15/s source, the capacitor would charge up to some value and A(s) wouldn't be zero.

2. Look at the steady state response. What would you expect the voltage on the capacitor to be as t goes to ∞? The capacitor is starting at zero volts, and then charging up to this value. Can you see why the voltage on the capacitor would stay constant at zero?
 
  • #4


Hello dear thegreenlaser
Thanks a lot for you explanation
i want to know the laplace give us the steady state value of the circuit?
 
  • #5


Have you studied steady state values at all? In this case you would assume the capacitor acts like an open circuit at t=∞, so you would replace the capacitor with an open circuit and find the voltage across it. If you've never seen anything like that before, then don't worry about it. I was just trying to give you another way to get an intuitive feel for what's going on.

To answer your question about using Laplace, there is indeed a way to find steady state values directly from your Laplace expression. It's called the final value theorem, but the theorem doesn't really offer much in the way of understanding why the final value should be what it is. In this case, I think the final value theorem would just make things more confusing.
 
  • #6


Thanks thegreenlaser again for your best explanation
could you tell me how can we know we have a Jump capacitor voltage with laplace ?for example VC(0-)<> VC(0+)?
 
  • #7


baby_1 said:
Thanks thegreenlaser again for your best explanation
could you tell me how can we know we have a Jump capacitor voltage with laplace ?for example VC(0-)<> VC(0+)?

An impulse would do it.
 
  • #8


thanks dear aralbrec
could you tell my how can find we have a jump in capacitor voltage without impulse source?

Thanks
 
  • #9


baby_1 said:
could you tell my how can find we have a jump in capacitor voltage without impulse source?

You can view an initial capacitor voltage as an impulse. Follow the Laplace transform:

i = C dv/dt

I(s) = sC V(s) - Cv(0-)

This says you can model the current through the capacitor as an uncharged capacitor with an impulse current source in parallel. The impulse source places the initial voltage on the capacitor at t=0. Vi = Q/C = ∫i dt / C = Cv(0-)/C = v(0-)

Normally we manipulate the equation a little further:

V(s) = I(s) /(sC) + v(0-)/s

which says the capacitor can be modeled as an uncharged capacitor in series with a constant voltage source.
 

What is Laplace Error in a Circuit?

Laplace error in a circuit refers to the difference between the actual output of a circuit and the expected output based on theoretical calculations. It is a measure of the accuracy of the circuit's performance.

What causes Laplace Error in a Circuit?

Laplace error can be caused by various factors such as component tolerances, temperature changes, parasitic effects, and manufacturing imperfections. These factors can affect the behavior of the circuit and lead to discrepancies between the expected and actual performance.

How is Laplace Error calculated?

Laplace Error is typically calculated by comparing the actual output of the circuit with the expected output using mathematical models or simulation tools. It is usually expressed as a percentage or in decibels (dB) to quantify the difference between the two values.

How can Laplace Error be minimized?

To minimize Laplace Error in a circuit, engineers can use high-precision components, implement temperature compensation techniques, and perform thorough testing and calibration. It is also important to select appropriate circuit designs and consider potential sources of error during the design phase.

Why is Laplace Error important in circuit design?

Laplace Error is crucial in circuit design as it helps engineers evaluate the accuracy and reliability of a circuit. By understanding the sources of error and minimizing their impact, engineers can improve the performance and functionality of their designs, leading to more efficient and reliable electronic systems.

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