Laplace plane basic electronic circuit (topic from control engineering)

In summary: OK, thanks.I'm not entirely sure. I guess what confuses me is that why when f = 0 then s necessarily = 0.In summary, the conversation discusses a circuit with a capacitor, inductor, and resistor, and the goal is to write an expression for the Laplace plane transfer function that relates current to voltage. The circuit is assumed to be a DC circuit with no internal resistance. The conversation also touches on the concept of impedance and how it extends the concept of resistance to AC circuits. It is mentioned that for DC circuits with no frequency component, the parameter s representing complex angular frequency is equal to 0. There is also discussion about the effect of R on computing Vx and the implications of steady
  • #36
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?
 
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  • #37
Femme_physics said:
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 expression for resistor current, IR(s), did you?

If you think you did present the correct expression, can you post a link to the message where it's shown?
 
  • #39
Wait! It should be R^2, not 2R...
Neither, actually. :cry:

But you've gone wrong at the first line. How do you arrive at that expression for VR?
 
  • #40
In the first line, shouldn't the numerator be  E · (R || 1/(sC))
 
  • #42
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: VZ = E·Z/ZT
 
  • #44
The derivation appears to be VR(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.
 
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  • #45
Femme_physics said:
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 [itex]j^2 = -1 [/itex]
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 [itex]Z_L[/itex] = sL = jωL = j[2π(0)L] = 0Ω which is essentially
a short circuit and for a capacitor [itex]Z_C[/itex] = 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 [itex] Z_L[/itex] = 0Ω and [itex]Z_C[/itex] = 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.
 
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  • #46
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
 
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  • #47
The way this method works is along these lines:

(i) you determine the general transfer function IR(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 IR(s).
(iv) you arrange IR(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.
 
  • #48
Our teacher did not provide us with such a chart, though
 
  • #49
Femme_physics said:
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. :biggrin:
 
  • #50
Ahhh...so therefor this must be the correct final move! :)

Am I right?http://img444.imageshack.us/img444/4964/vr6e.jpg
 
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  • #51
The s's all seem to have performed a vanishing act! :smile: You need to keep them in the expression!
 
  • #53
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. :smile:
 
  • #54
When you have finally got it right :wink: 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.
 

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