Laplace plane basic electronic circuit (topic from control engineering)

[NascentOxygen]

Now let's take an example with a capacitor.

We start with 12 volts, and 2 amperes.

Current flows into the capacitor. Electrons. The capacitor, after 5 tau, has reached its steady state.

So now the voltage of the circuit is still 12,
The current is still amperes,

and we have a charged capacitor on top of all that! No energy spent from the circuit to archieve that!

The battery is a bit drained, having supplied the energy to charge the capacitor.

See what I'm saying!

I hope you understand not to say it any further.

 

[Femme_physics]

The battery is a bit drained, having supplied the energy to charge the capacitor.

AHA!

So the initial voltage changes from 12 volts, then! And how can you determined it's been drained a "bit" and not "a lot"? What if we have a capacitor with lots of Farads? And not that much voltage? Wouldn't the voltage be completely drained? or much drained?

 

[NascentOxygen]

The voltage source shown in your schematic is a perfect voltage source. It never changes its voltage no matter how much current is drawn from it, nor for how long you draw that current. If you want to model an imperfect voltage source, you'll have to come up with an accurate electrical model for such a thing.

Whether it's a perfect voltage source or not, does not alter the fact that energy stored in the capacitor has to come from the battery. The energy equation always hold true.

Though using a perfect voltage source makes calculations a lot easier, doesn't it?

 

[Femme_physics]

Ahhhhhhhhhh cheaters ;) now things make a lot more sense :)

Although, do we also consider capacitors are "dormant"? As in, they never discharge as time goes on unless the exercise says so?

PS: As far as the original exercise... I think I'm doing it the hard way if I don't use voltage divider..no?

 

[NascentOxygen]

Although, do we also consider capacitors are "dormant"? As in, they never discharge as time goes on unless the exercise says so?

The symbol for a capacitor is for a perfect capacitor. If you wish to show leakage current then you can include a large value resistance across the plates of the capacitor.

PS: As far as the original exercise... I think I'm doing it the hard way if I don't use voltage divider

The original exercise is so far back that I've forgotten what it was all about. Your threads have a tendency to veer off-course and stray in tangled figures of 8!

 

[Femme_physics]

Sorry :)

But if anyone don't mind going back to look, my question is whether I plug it the value for the coil as 0.2, and C as 4x10^-6

Does it make sense?

 

[NascentOxygen]

But if anyone don't mind going back to look, my question is whether I plug it the value for the coil as 0.2, and C as 4x10^-6

Yes, that's what you will do, but I don't think you ever finished finding the correct[/color] expression for resistor current, I_R(s), did you?

If you think you did present the correct[/color] expression, can you post a link to the message where it's shown?

 

[Femme_physics]

There

http://img515.imageshack.us/img515/1266/vr770.jpg

Wait! It should be R^2, not 2R...

 

[NascentOxygen]

Wait! It should be R^2, not 2R...

Neither, actually.

But you've gone wrong at the first line. How do you arrive at that expression for V_R?

 

[NascentOxygen]

In the first line, shouldn't the numerator be  E · (R || 1/(sC))

 

[Femme_physics]

But I'm using this formulahttp://img844.imageshack.us/img844/9104/certainnagad.jpg

Which is the equation to calculate the voltage on a certain resistor in the circuit

 

[NascentOxygen]

But that "certain resistor" in the equation you quote, corresponds in the circuit here, to a resistor in parallel with a capacitor.

You should be applying the generalised equation: V_Z = E·Z/Z_T

 

[Femme_physics]

True, true,

Tell you what confused me,

My lecturer from some reason appear to have calculated the voltage only despite the fact they're asking for current...or am I seeing something wrong?

(here the question has different values for C, R, L and E )

http://img403.imageshack.us/img403/9259/qqqq1.jpg http://img407.imageshack.us/img407/6528/lec1.jpg

 

[NascentOxygen]

The derivation appears to be V_R(s)/E(s)
then in the last step the decimal point is moved 2 places corresponding to ÷ 100Ω

https://www.physicsforums.com/images/icons/icon2.gif I always use a lower-case handwritten cursive "s" for the Laplace "s" in this work, as I found I would occasionally mistake (5 pages later!) my handwritten printed "s" for a digit 5. The lower-case cursive "s" has the appearance of a robin without legs (that's the bird). Like the one here but without stroke 3.

 

[Ouabache]

Sorry it took me so long to reply...other subjects like Solidworks, Robotics, etc...took my time. OK, back in action!I don't recall us defining "s" as a complex angular frequency. What's so complex about it?

I'm sorry I was not available as you were wading through the mysteries of this problem.
I am glad to see there were others who offered useful input to some of your questions.

Allow me to try and answer your question about "complex angular frequency"

As rude man began to point out, complex refers to the math construct of complex numbers
where such a number may be expressed in the form σ + jω .
Here σ is real and jω an imaginary number. j is the imaginary unit, where j^2 = -1
In a LaPlace Transform of a time-dependent system, upon transformation, s = σ + jω, where ω is the angular frequency
expressed in radians. If you like, we may convert this to frequency expressed in Hertz by substituting ω = 2πf.
In consideration of this complex number that is now frequency dependent in the LaPlace domain;
this is where "complex angular frequency" comes from.

Earlier I stated: as this is a DC circuit (there is no frequency, f = 0 Hz), at steady-state the inductor will have zero impedance
and the capacitor will have infinite impedance (open circuit).

You can determine this for yourself. The impedance for the inductor Z_L = sL = jωL = j[2π(0)L] = 0Ω which is essentially
a short circuit and for a capacitor Z_C = 1/sC = 1/jωC = 1/[j2π(0)C] = 1/(number approaching zero) = ∞ Ω (very large resistance,
essentially an open circuit).

As you found, you may express the transfer function at any time. It is just that at the steady-state,
the Z_L = 0Ω and Z_C = very large resistance, which simplifies your expression.
As you have also noted, voltage division works nicely with impedances of capacitors and inductors
just as it does with the pure resistance of resistors.

 

[Femme_physics]

I'm sorry I was not available as you were wading through the mysteries of this problem.
I am glad to see there were others who offered useful input to some of your questions.

You'll have to excuse me in delaying and I will delay further in replying to the theoretical part-- it's a bit beyond me at this stage. I'll get back to that.

The derivation appears to be VR(s)/E(s)
then in the last step the decimal point is moved 2 places corresponding to ÷ 100Ω

True!

The derivation appears to be VR(s)/E(s)
then in the last step the decimal point is moved 2 places corresponding to ÷ 100Ω

Duly noted, though I'd stay at the side that won't maybe confuse my teacher


Well, it's a bit long but...

http://img593.imageshack.us/img593/4205/vr1x.jpg
http://img812.imageshack.us/img812/9181/vr2i.jpg
http://img11.imageshack.us/img11/9509/vr3u.jpg
http://img217.imageshack.us/img217/1443/vr4bw.jpg
http://img338.imageshack.us/img338/8249/vr5w.jpg
http://img444.imageshack.us/img444/4964/vr6e.jpg
http://img585.imageshack.us/img585/3876/vr7f.jpg
http://img163.imageshack.us/img163/1808/vr8i.jpg
http://img16.imageshack.us/img16/4196/vr9p.jpg

 

[NascentOxygen]

The way this method works is along these lines:

(i) you determine the general transfer function I_R(s) / E(s),
(ii) you consult a table that mathematicians have provided that tells you what E(s) corresponds to the particular E(t) that the circuit is being driven by here,
(iii) you substitute E(s) into the general transfer function from (i), to obtain an expression for I_R(s).
(iv) you arrange I_R(s) in a standard form,
(v) you refer to a table that mathematicians have provided that tells you what exact function of time corresponds to your Laplace expression with its specific coefficients and constants.

What you have erroneously done in your working above is getting to step (iv) and then simply ignoring the s terms in the expression (equivalent to setting s=1). I can't see any way this could be right.

 

[Femme_physics]

Our teacher did not provide us with such a chart, though

 

[NascentOxygen]

Our teacher did not provide us with such a chart, though

Going waaaaaayyy back to the beginning of this thread, to find exactly what you are asked to find:

Write an expression the the laplace plane transfer function, that relates current (that flows through resistor R) to the voltage E.

You are asked for step (i) only. So you will end up with an expression having a polynomial in s in the denominator and numerator. After substituting values for R, C and L, that's where you stop.

Learning about the other steps I outlined will take place in another semester. There is a limit to the amount of excitement they are allowed to include in anyone semester.

 

[Femme_physics]

Ahhh...so therefor this must be the correct final move! :)

Am I right?http://img444.imageshack.us/img444/4964/vr6e.jpg

 

[NascentOxygen]

The s's all seem to have performed a vanishing act! You need to keep them in the expression!

 

[Femme_physics]

There! Without the vanishing act! :)

http://img140.imageshack.us/img140/1628/vsozaperta.jpg

I believe this is it

 

[NascentOxygen]

The most recognizable form is with polynomials in s in the numerator and denominator, just like the last line in this post. Keep it general with L,C and R right up until the last. Only in the final step substitute their values to determine the constants that are the coefficients of the s terms.

You are allowed to cancel s÷s, etc.

 

[NascentOxygen]

When you have finally got it right the expression in s is not as of as little use as my earlier list of steps may have implied. Once you have the expression in s, you can substitute s ← jѠ and the result is the system's frequency response, a very valuable performance parameter to know.