I still cant find the equation using laplace transform

 

Have you re-drawn the circuit in its Laplace form?

For instance with the source the Laplace equivalent becomes

\(V_{in}(s)=\frac{\omega}{\( s^2+\omega^2 \)}\)

What working can you demonstrate thus far?

 

Hey, t_n_k, does that problem look a bit off to you? It says that the v(t) is a pure sinusoid which should result in a sinusoidal response with a 90deg phase shift at the cap (so a peak at t=0s), yet the answer it says you are trying to show has an exponential component and is zero at t=0s. I'm thinking they've gotten sloppy and meant to say that

v_in(t) = 1V sin(wt)u(t)

which is what you gave the Laplace transform of (ignoring the absence of the 1V).

What do you think?

 

I take your point - although without a specific statement to the contrary, most people familiar with the Laplace method would assume the circuit has no initial or steady-state condition and that a transient solution is required in that context. Many authors seem to pass over the distinction without much concern.

The Laplace domain deals only with positive time values. I always wonder at such concepts as t=0- and t=0+. Presumably this is basis for consideration of the 'special case' of initial conditions when using the Laplace method.

The introduction of the Heaviside u(t) factor as your indicate, probably helps to remove any ambiguity as to the initial state of the circuit.

Perhaps steveb might help us out.

 

Perhaps steveb might help us out.

I think I would agree with both of you. It seems most people would consider the single sided Laplace transform (integral from zero to infinity, not negative infinity to infinity). This is equivalent to the full Laplace transform when multiplying by the Heaviside function which zeros out the other half of the full integral. Also, most would assume initial condition of zero on the state variable, unless otherwise specified.

The answer given seems in order because the initial value of the output voltage is zero, and the transient portion of the solution is needed to offset the initial value from the steady state portion of the solution. Also, the Laplace transform you give for the input signal is the single sided Laplace transform of sin(wt), or the full Laplace transform of sin(wt) u(t).

I think from a mathematician's point of view, engineers are always sloppy when describing these situations, but I think a better way to say it is that there is an assumed convention in place when engineers are talking about these problems.

 

Thanks for the help guys. But i still dont get how to arrive at the equation with cos, sin and exp in there!

 

You've not shown any attempt at working even a partial solution - we are waiting for that key piece of the puzzle.

 

I think from a mathematician's point of view, engineers are always sloppy when describing these situations, but I think a better way to say it is that there is an assumed convention in place when engineers are talking about these problems.

I would agree with this statement. The main problem I have is that most engineers don't realize they are assuming a convention in a large fraction of situations. Either they never learned it in anything resembling a rigorous way or they have long since forgotten it. I suspect it is a mixture of both. I think teachers at all levels make an unconsious mistake that if you teach something with rigor when it is first presented that the student will forever be aware of the assumptions and limitations. I know, in my case as a student, this was a really bad assumption. Like most students, I had a hard time caring about regions of convergence and whatnot in part because it seemed like extraneous stuff that was only of theoretical significance. But also because few courses do more than just brush over the stuff in passing. I think it would be much better if courses routinely ensured that problems in which such details are important constitute a noticeable portion of the assignments and exam problems. In addition, on the handful of occasions in which a follow-on course spent the first class doing a quick but thorough review of the entire preceding course to be extremely useful because I could then see how some of the subtle points of the foundational work from the beginning of the course tied in with other parts of the course in behind-the-scenes ways.

Thanks for the help guys. But i still dont get how to arrive at the equation with cos, sin and exp in there!

Yeah, we did kinda get off on a tangent. But we're expecting to see something from you showing what you've done so far and where you are having problems. It's hard to give good help without knowing where the problem lies.

1) What are the Laplace transforms for a resistor and a capacitor?

2) What's the output voltage in the Laplace (complex frequency) domain?

3) What pieces of that voltage look like elements in standard Laplace transform pairs?

 

Here are my workings

 

It seems you may not be familiar with the Laplace method. Have you studied it in your classes?

 

we are doing self learning for this chap. so can you show us your solutions that can prove it to the equation abv???

 

Self learning has its place but I'm not sure going it alone on Laplace methods is a sensible course of action. I'm surprised your professor would leave you in this unfortunate situation.

Frankly, I would not think simply providing you with a worked solution has any merit. Maybe others will be happy to give you a short course here, but that's not my role - it's your teacher's responsibility.

Sorry I can't be of assistance - I feel [without knowing all the facts] that you are possibly being treated unfairly.

 

I agree. Transform Methods is one of the more difficult courses for most EE undergrads and there are a lot of important, but subtle, things that most people doing any kind of self study are going to gloss right past.

Looking at your work, I will offer the following suggestions:

1) In your first line, you are using the Fourier transform (i.e., jw) which is only valid for sinusoidal steady state solutions. In this case you are looking for the full response including both the steady state and the transient, so you have to use the full Laplace transform (i.e. s where 's' is a complex variable). The Fourier transform is a special case of the Laplace transform that is obtain by setting s equal to jw.

2) When you are using transforms, you either transform everything or you transform nothing. You can't mix the two, like you do starting with your third line in which you have the transform version of the impedances but the time domain version of the voltage.

3) In your fifth line, it looks like you are saying

\(
V_R=V_{out}
\)

Are you sure about this? Is the output voltage the voltage across the resistor?

4) I would recommend trying to me a little neater with your handwriting. It isn't too bad and I have certainly seen far worse, but there is a sloppiness in places that make it hard to tell, for instance, what some of the subscripts are. Sooner or later you will make mistakes by misreading your own writing.


5) There are some very useful conventions commonly employed when working with transform methods: time domain signals are expressed using lower case variables and transform (i.e., complex frequency domain) signals are expressed using upper case variable. This allows you to quickly see if your expressions are consistent and in a single domain.

With these in mind, go back to your book and see it things make a bit more sense to you and then answer the following:

1) What are the Laplace transforms for a resistor and a capacitor (and an inductor, even though it isn't used here)?

2) What's the input voltage in the Laplace (complex frequency) domain?

3) What's the output voltage in the Laplace (complex frequency) domain?


Let's get this far and then we will worry about getting the answer back into the time domain.

 

okay! last question, what do you mean by "Is the output voltage the voltage across the resistor?" Is'nt that what the question stated? Is it the voltage across the capacitor instead?

 

okay! last question, what do you mean by "Is the output voltage the voltage across the resistor?" Is'nt that what the question stated? Is it the voltage across the capacitor instead?

Read the question again, carefully, and ask yourself why you think Vout is the voltage across the resistor. Remember, "voltage across" means the voltage on one side relative to the voltage on the other. The output voltage is the voltage on the top right node compared to the voltage on the ground node. Which component can be described as having that same voltage across it?

 

Here is a pdf that may answer some of the OPs questions.

http://staff.najah.edu/sites/default/files/signal8.pdf

But, as stated above, it is not easy to learn this without a formal class setting and a live teacher to question.

It is easy for us to answer specific questions and provide guidance. But, providing the broader context required to know this subject well is awkward to communicate in a forum. A textbook is the next best thing, compared to a teacher, and as far as value for money, as well as suitability to a beginner trying to learn solo, the Schaum's Outline of Signals and Systems is very good, in my opinion.

http://www.amazon.com/Schaums-Outli...words=Schaum's+Outline+on+Signals+and+Systems

 

Read the question again, carefully, and ask yourself why you think Vout is the voltage across the resistor. Remember, "voltage across" means the voltage on one side relative to the voltage on the other. The output voltage is the voltage on the top right node compared to the voltage on the ground node. Which component can be described as having that same voltage across it?

Okay, I went back and read the question again, carefully, and I see where the confusion is coming from. The statement in the hint (which I overlooked) does state that the output voltage is the voltage across the resistor. But the diagram clearly shows it being the voltage across the capacitor and this is the voltage that the given answer represents.

 

The hint is wrong. For the laplace problem 2 it should be Vout= Vc. Confirmed with teacher.

 

That's good to have confirmed. Thanks for doing that and reporting back. Hopefully the teacher made an announcement to the entire class. Actually, I would have hoped that at least some subset of the students would have caught the discrepancy and brought it to the teacher's attention already.