Laplace plane basic electronic circuit (topic from control engineering)

[Femme_physics]

Homework Statement

C = 4 μF
L = 0.2 Hy
E is DC current. No internal resistance.

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

(There is another question. I will just address the first one for now)http://img7.imageshack.us/img7/6659/circuitttttt.jpg

The Attempt at a Solution

I tried reaching the expression without simplifying or plugging in numbers. That's right, yes?

http://img819.imageshack.us/img819/9473/circuittttttanswer.jpg

 

[Ouabache]

E is at the schematic symbol for voltage potential, I suspect they meant to say, voltage E instead of current. The current flowing out of E and through L you might call I.

Do you remember from one of your earlier questions, how we represent impedance (Z) of a component? (for a resistor Z = R, for a capacitor Z= 1/sC, for an inductor Z = sL)
Impedance extends the concept of resistance to AC circuits. It contains both magnitude and phase information, whereas resistance has only magnitude. In the case of a DC circuit,
the impedance of C and of L are special cases. Did your instructor discuss complex angular frequency? The same parameter s is a complex angular frequency. For DC circuits there is no frequency component (f=0) and therefore s=0. How would that affect the impedance of a capacitor? an inductor?

Because this is a DC circuit, analysis becomes trivial.
Now if you are considering the instantaneous current when switch S has just closed (at t=0+) versus steady state (switch has been closed for long time) (t=long time), that makes the problem more interesting.

 

[rude man]

Homework Statement

I tried reaching the expression without simplifying or plugging in numbers. That's right, yes?

[

That's wrong, no.

In computing V_x you have neglected the effect of R.

 

[I like Serena]

Can it be that with E you meant the DC voltage?

I'm having a little trouble interpreting your problem statement.
I suspect that something like this drawing is intended.
If that is the case, then your expression for the Laplace transfer function is correct, assuming it applies to the circuit in the middle.
Your current through R would follow from that, but it would only be correct if we can assume that this current is much smaller than the other currents.

(Btw, I just got a new drawing program on my computer, so I just had to make a drawing. )

.

 

[Ouabache]

Can it be that with E you meant the DC voltage?

Hmmm, I thought I just mentioned that. :uhh:

I'm having a little trouble interpreting your problem statement.
I suspect that something like this drawing is intended.

Are you attempting to back-engineer the circuit from the information stated?
(Write an expression for the Laplace plane transfer function)
What happens to this transfer function under steady-state condition?

Cool drawing program.. What is it you are using?

 

[I like Serena]

Hmmm, I thought I just mentioned that.

I know. I just decided to repeat it to make sure we can agree on the definitions of the symbols, which is rather important IMO.

Are you attempting to back-engineer the circuit from the information stated?
(Write an expression for the Laplace plane transfer function)

Yes.
The fact that they ask for a transfer function suggests that a variable load is not supposed to be part of it.

What happens to this transfer function under steady state condition?

What should happen to it?

If you rewrite the transfer function, you get a nice characteristic time from it.
An inverse Laplace transform will show the expected loading curve for the capacitor.

Cool drawing program.. What is it you are using?

Visio.
Where I work it's the preferred choice to make all the different types of diagrams and pictures we need in our Word documents.
This is the first time I made an electronic circuit with it though.

 

[Ouabache]

Just for politeness if you had stated something earlier that I wished to re-emphasize, I might say; As ILS also discussed, can it be that with E you meant the DC voltage?

What should happen to it? (transfer function)

Since this is a DC circuit, at steady-state the inductor will have zero impedance (short circuit), the capacitor will have infinite impedance (open circuit) and all of the supply current will flow through the resistor. $I_R$ reduces to $\frac{E}{R}$

Visio.
Where I work it's the preferred choice to make all the different types of diagrams..

Thanks I will check that out..

 

[Femme_physics]

Sorry it took me so long to reply...other subjects like Solidworks, Robotics, etc...took my time. OK, back in action!

E is at the schematic symbol for voltage potential, I suspect they meant to say, voltage E instead of current. The current flowing out of E and through L you might call I.

Yes, that was a mistype, sorry!

Do you remember from one of your earlier questions, how we represent impedance (Z) of a component? (for a resistor Z = R, for a capacitor Z= 1/sC, for an inductor Z = sL)
Impedance extends the concept of resistance to AC circuits. It contains both magnitude and phase information, whereas resistance has only magnitude. In the case of a DC circuit,

I do remember that!

the impedance of C and of L are special cases. Did your instructor discuss complex angular frequency? The same parameter s is a complex angular frequency.

I don't recall us defining "s" as a complex angular frequency. What's so complex about it?

For DC circuits there is no frequency component (f=0) and therefore s=0. How would that affect the impedance of a capacitor? an inductor?

I'm not entirely sure. I guess what confuses me is that why when f = 0 then s necessarily = 0.

Now if you are considering the instantaneous current when switch S has just closed (at t=0+) versus steady state (switch has been closed for long time) (t=long time), that makes the problem more interesting.

The problem does not define time, so we can't tell whether it's the steady state or not, and yet there is a capacitor in the picture...that kind of makes it impossible to solve without time defined.

That's wrong, no.

In computing Vx you have neglected the effect of R.

You're right--I can't use voltage divider this way, I forgot.

Since this is a DC circuit, at steady-state the inductor will have zero impedance (short circuit), the capacitor will have...

Again, time isn't define so how can we tell it's a steady-state?

I'm having a little trouble interpreting your problem statement.
I suspect that something like this drawing is intended.
If that is the case, then your expression for the Laplace transfer function is correct, assuming it applies to the circuit in the middle.
Your current through R would follow from that, but it would only be correct if we can assume that this current is much smaller than the other currents.

So rude man was wrong, and I did use voltage divider correctly?

And yes my exercise is like the one you drew.

 

[I like Serena]

So rude man was wrong, and I did use voltage divider correctly?

And yes my exercise is like the one you drew.

I don't know.
If you don't know either, you should ask your teacher.

Anyway, I preferred to assume that you were right!

 

[Femme_physics]

:) Heh...I'll have a class tuesday about it...I'll see about my voltage divider usage.

 

[RoshanBBQ]

First off, a transfer function is a ratio of output to input in the Laplace domain. Your problem does not state whether to treat the current as input or the voltage as input. We will assume the problem meant for E to be the input.

The problem does not state to assume initial conditions are zero, so these will be part of the solution. We can do mesh analysis:

$$-E(s)+LsI_1(s)-Li_1(0^-)+\frac{I_1(s) -I_2(s)+v_C(0^-)}{Cs}=0$$

$$\frac{I_2(s)-I_1(s)-v_C(0^-)}{Cs}+I_2(s)\left(R+A \right)=0$$

Use these coupled equations to solve for the ratio

$$\frac{I_2(s)}{E(s)}$$

I have no idea why people are talking about zero frequencies and whatnot. The transfer function doesn't care what frequency E is. And yes, your answer is wrong.

 

[rude man]

Sorry it took me so long to reply...other subjects like Solidworks, Robotics, etc...took my time. OK, back in action!

So rude man was wrong, and I did use voltage divider correctly?

.

No, rude man was right.
Signed,
rude man.

(R is negligible in your voltage divider only if R = ∞.)

And oh yes, s is complex because s = σ + jω, and j = √(-1). Isn't that complex?

 

[Femme_physics]

Thanks for the replies. I don't recal having done mesh analysis, unless it just means to use KVL and KCL--although I've never applied KVL and KCL with a capacitor and coil in the picture.

I'll try to see how to figure it all out, so far I feel like I got myself into a mess.

 

[Femme_physics]

Excuse me for not answering the advanced replies-- I'm still trying to understand basic stuff...I want to get some things clear.

1) How come at t = 0 the capacitor is shortciruited and the inductor is a disconnection?

2) How come at the steady state it's the other way around?

 

[I like Serena]

The basic property of a capacitor is that it stores current.

When a capacitor is empty (t=0), current can freely flow into the capacitor, filling up the capacitor with charge.
Just like current can flow freely through a shortcircuit.

When a capacitor is full (t=∞), no current can flow into it anymore.
Just like there is an actual break in the wire, which is a disconnection (hence the symbol).



The basic property of an inductor is that it resists a change in current.

When the switch is closed (t=0), the current suddenly starts flowing, and the inductor resists this change, hardly allowing any current through at first.

When things even out (t=∞), and the current becomes constant, the inductor has nothing to resist anymore and the (constant) current can flow freely through it.

 

[rude man]

Thanks for the replies. I don't recal having done mesh analysis, unless it just means to use KVL and KCL--although I've never applied KVL and KCL with a capacitor and coil in the picture.

I'll try to see how to figure it all out, so far I feel like I got myself into a mess.

No KCL or KVL with L's and C's? Oh dear.
But, all is not lost:

1. calulate Z1 = parallel combination of your R and C.
2. Now V_x = E*Z1/(Z1+ sL).
3. Then i = V_x/R.

You did not state the problem completely at all. If S is closed at t=0, what is the Laplace representation of E? And oh yeah, someone's got to give you the value of R sometime ...

 

[RoshanBBQ]

Thanks for the replies. I don't recal having done mesh analysis, unless it just means to use KVL and KCL--although I've never applied KVL and KCL with a capacitor and coil in the picture.

I'll try to see how to figure it all out, so far I feel like I got myself into a mess.

The two equations comes from doing KVL for each loop, using two currents we assign to each loop (I_1 and I_2). But if you are not comfortable with that method, the method posted above me will bring about the same answer.

And I'll say it again, I have no idea why people keep talking about what exactly E is ("You did not state the problem completely at all. If S is closed at t=0, what is the Laplace representation of E?"). The transfer function is the same no matter what nonzero signal you apply, which is why you should just keep E symbolic in your work.

As to why an uncharged capacitor is a short if a DC voltage is just applied to it and an open after some time, it comes from its current/voltage characteristics. For zero initial charge on the capacitor, we can write (for a voltage applied at t = t_0, thereby creating current to integrate)

$$v_c(t) = \frac{1}{C} \int \limits_{t_0}^{t} i_c(t)dt$$

This states the voltage across a capacitor is equal to the integral of the current through it. So let's just think about this. The current is some smooth function that moves around a bit as the capacitor charges. Since it is smooth, the integral cannot instantaneously capture area. So at the moment t = t_0, we have

$$v_c(t) = \frac{1}{C} \int \limits_{t_0}^{t_0} i_c(t)dt = 0$$

And if the voltage across an element is zero, it is a short circuit.

As for the open circuit as t -> inf, consider that a current flowing into the capacitor makes the voltage across it bigger (since its voltage equals the integral (i.e. summation) of current). This growth in voltage will continue (for a DC signal) until the voltage across the capacitor is so great that no current flows through it. This condition is known as "open circuit" for any element.

Examining inductor equations brings about similar intuition.

 

[rude man]

And I'll say it again, I have no idea why people keep talking about what exactly E is ("You did not state the problem completely at all. If S is closed at t=0, what is the Laplace representation of E?"). The transfer function is the same no matter what nonzero signal you apply, which is why you should just keep E symbolic in your work.
/QUOTE]

Here's an idea: I anticipated the second part of her question, which undoubtedly is to determine i(t) in the resistor. Seeing as E is shown to be a dc source with a switch S ... make sense?

The OP was badly misled from the start by her leaving out the effect of R on her voltage divider. That should be the thing for her to correct, not start her on a treatise on how to write KVL and KCL equations using transformed impedances. She does that one thing, she' practically home free.

 

[RoshanBBQ]

Here's an idea: I anticipated the second part of her question, which undoubtedly is to determine i(t) in the resistor. Seeing as E is shown to be a dc source with a switch S ... make sense?

The OP was badly misled from the start by her leaving out the effect of R on her voltage divider. That should be the thing for her to correct, not start her on a treatise on how to write KVL and KCL equations using transformed impedances. She does that one thing, she' practically home free.

I am assuming the OP knows how to solve DC circuits. Laplace analysis of circuits is exactly the same except you deal with a symbolic s thrown into the mix. She obviously understands what KVL is, and I think she can handle writing two equations by looking at two loops. Plus, that is beyond the point. The question does not tell her to calculate the response. It merely asks for the transfer function, which is independent of the input E and of its form in the Laplace domain.

 

[Femme_physics]

No KCL or KVL with L's and C's? Oh dear.
But, all is not lost:

1. calulate Z1 = parallel combination of your R and C.
2. Now Vx = E*Z1/(Z1+ sL).
3. Then i = Vx/R.

You say parellel combination of R and C...

well look, according to my solution manual...here are two different cases and their solutions...

http://img715.imageshack.us/img715/6774/comparisonxv.jpg

I am aware that for all cases we use I = E/R
BUT, it appears when a capacitor and coil are in parallel, they are added to the voltage part of the equation in pluses, whereas you have a resistor and a capacitor, it's added in parallel.

What gives? Why is it different?

The basic property of a capacitor is that it stores current.

When a capacitor is empty (t=0), current can freely flow into the capacitor, filling up the capacitor with charge.
Just like current can flow freely through a shortcircuit.

When a capacitor is full (t=∞), no current can flow into it anymore.
Just like there is an actual break in the wire, which is a disconnection (hence the symbol).
The basic property of an inductor is that it resists a change in current.

When the switch is closed (t=0), the current suddenly starts flowing, and the inductor resists this change, hardly allowing any current through at first.

When things even out (t=∞), and the current becomes constant, the inductor has nothing to resist anymore and the (constant) current can flow freely through it.

Perfect explanation.

Thanks Roshan, as well. Yes, I master KVL and KCL well :)

 

[rude man]

For the second diagram, you made a typo in the denominator - it should be (1/sC)||R + sL.

Then, I = ({(1/sC)||R}/{(1/sC||R + sL})(E/R).

I can give you the correct expression for the first diagram, but you'll have to figure out if it's the same as yours: E/I = sL/(s²LC + 1) + R

So have you figured out what R||(1/sC) is yet?

 

[Femme_physics]

You're right-- I just have issues with my math :) I figured it out now-- thanks.

On a different note, when speaking about the steady state...what supplies the voltage? The capacitor only?

 

[Femme_physics]

OK-- so this is the equation I need to simplify...I just applied KVL after turning the circuit into a single loop by combining the capacitor and the resistor in parallel.

http://img696.imageshack.us/img696/6036/transss.jpg

 

[I like Serena]

On a different note, when speaking about the steady state...what supplies the voltage? The capacitor only?

In steady state the voltage is supplied by the battery.
The capacitor is full at that time and does not do anything anymore.

OK-- so this is the equation I need to simplify...I just applied KVL after turning the circuit into a single loop by combining the capacitor and the resistor in parallel.

That's the right equation in the 1st and 2nd line.

But there is a math issue between the 2nd and the 3rd line.
Perhaps you should try it with wolframalpha. ;)

 

[Femme_physics]

In steady state the voltage is supplied by the battery.
The capacitor is full at that time and does not do anything anymore.

Doesn't it mean that the voltage is reduced by a certain amount?

In steady state the voltage is supplied by the battery.
The capacitor is full at that time and does not do anything anymore.

But won't these equations only find me the current over Ls?

In order to find the current at R I need to do this:

http://img17.imageshack.us/img17/497/transfuncccc.jpg

 

[I like Serena]

Doesn't it mean that the voltage is reduced by a certain amount?

Huh?

Are you are talking about a possible reduction by the inductor?
Well, in steady state the current is constant, meaning the inductor does not do anything.
So there is no voltage drop over the inductor in steady state.

But won't these equations only find me the current over Ls?

In order to find the current at R I need to do this:

Yup. :)

(Don't forget the s after L in the first line.)

 

[Femme_physics]

Are you are talking about a possible reduction by the inductor?
Well, in steady state the current is constant, meaning the inductor does not do anything.
So there is no voltage drop over the inductor in steady state.

No, I'm not talking about cases where there are inductors. My question is more simplified to understand the idea behind a capacitor.

http://img825.imageshack.us/img825/1451/1st2ndk.jpg

I'm going to go with case 1.

Initially...the capacitor gets charged up. Once it's finished charging, all of the power, and voltage, and current, whatever you want to call it in the circuit passed on to the capacitor. That means that from this point on the capacitor supplies the voltage! But you say it's not the case. Fine. So the capacitor takes "some" of the voltage of the circuit. Therefor there is less voltage on the entire circuit. No? You get what I'm saying?

Yup. :)

(Don't forget the s after L in the first line.)

Ah...dooly noted :) I'll work on it now

 

[NascentOxygen]

No, I'm not talking about cases where there are inductors. My question is more simplified to understand the idea behind a capacitor.

I'm going to go with case 1.

Initially...the capacitor gets charged up.

Hi FP! That's right...so far.

Once it's finished charging, all of the power, and voltage, and current, whatever you want to call it in the circuit passed on to the capacitor.

That is a carelessly thrown together sentence! I'm sure you know the difference between current and voltage. I can't conceive of what you could mean by something being "passed on the the capacitor" here. The fundamental operation of a capacitor involves adding (or removing) charge on its plates, and this directly affects the voltage between its plates.

That means that from this point on the capacitor supplies the voltage!

Supplying the voltage would imply "supplying current" I guess. But if the capacitor is supplying current then it would be losing charge, resulting in it losing voltage, would it not? This conflicts with the assertion that in the steady state the capacitor voltage is unchanging.

But you say it's not the case. Fine. So the capacitor takes "some" of the voltage of the circuit.

If the capacitor voltage is unchanging, then it is neither gaining nor losing charge, so it can't very well be supplying or taking anything. If it is neither supplying nor accepting charge, then it's doing nothing so it may as well be invisible to the operation of the circuit in the steady state.

Therefor there is less voltage on the entire circuit. No?

No.

You get what I'm saying?

I do. But I hope you'll understand otherwise and stop saying it. :shy:

 

[RoshanBBQ]

No, I'm not talking about cases where there are inductors. My question is more simplified to understand the idea behind a capacitor.

http://img825.imageshack.us/img825/1451/1st2ndk.jpg

I'm going to go with case 1.

Initially...the capacitor gets charged up. Once it's finished charging, all of the power, and voltage, and current, whatever you want to call it in the circuit passed on to the capacitor. That means that from this point on the capacitor supplies the voltage! But you say it's not the case. Fine. So the capacitor takes "some" of the voltage of the circuit. Therefor there is less voltage on the entire circuit. No? You get what I'm saying?
Ah...dooly noted :) I'll work on it now

In steady-state, the capacitor doesn't supply a thing for circuit one. Can we agree that the voltage is a constant value at steady-state? Well, then by this equation:

$$i_c(t) = c\frac{dv_c(t)}{dt}$$

we see that the current 'through' the capacitor is zero. How can it supply a thing without current flowing 'through' it? Its effect becomes naught. You just remove it from the circuit. The battery still delivers the v_battery^2/R power and all of the v_battery/R current to the resistor R.

 

[Femme_physics]

If the capacitor voltage is unchanging, then it is neither gaining nor losing charge, so it can't very well be supplying or taking anything. If it is neither supplying nor accepting charge, then it's doing nothing so it may as well be invisible to the operation of the circuit in the steady state.

Your very premise contradicts the first rule of physics: Energy isn't gained or loss, it only transforms shape.

In your case, energy is gained.

Let's start with case 1.

Supposed we have a voltage of 12, and a resistor of 6. Imagine that the line that connects the capacitor doesn't exist.

So,

Voltage: 12
Current: 12/6 = 2 Amperes

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! We don't need the sun anymore! We figured out how to harness energy out of nothing! (should I keep going? I'm rather enjoying it... :) ) See what I'm saying!

 

[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!