# Homework Help: Laplace plane basic electronic circuit (topic from control engineering)

1. Apr 28, 2012

### Femme_physics

1. The problem statement, all variables and given/known data

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 [Broken]

3. The attempt at a solution

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

Last edited by a moderator: May 5, 2017
2. Apr 28, 2012

### 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.

Last edited: Apr 28, 2012
3. Apr 28, 2012

### rude man

That's wrong, no.

In computing Vx you have neglected the effect of R.

4. Apr 28, 2012

### 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. )

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Last edited: Apr 28, 2012
5. Apr 28, 2012

### Ouabache

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

6. Apr 28, 2012

### I like Serena

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.

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

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.

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.

7. Apr 28, 2012

### 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?

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 $\frac{E}{R}$

Thanks I will check that out..

Last edited: Apr 28, 2012
8. May 5, 2012

### Femme_physics

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

Yes, that was a mistype, sorry!

I do remember that!

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

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

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.

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

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

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

And yes my exercise is like the one you drew.

9. May 5, 2012

### I like Serena

I don't know.

Anyway, I preferred to assume that you were right!

10. May 5, 2012

### Femme_physics

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

11. May 5, 2012

### 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.

12. May 5, 2012

### rude man

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?

13. May 5, 2012

### 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.

14. May 5, 2012

### Femme_physics

Excuse me for not answering the advanced replies-- I'm still trying to understand basic stuff.....I wanna 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?

15. May 5, 2012

### 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.

Last edited: May 5, 2012
16. May 5, 2012

### rude man

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 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 gotta give you the value of R sometime ...

17. May 5, 2012

### RoshanBBQ

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.

18. May 5, 2012

### rude man

19. May 5, 2012

### RoshanBBQ

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.

20. May 8, 2012

### Femme_physics

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 [Broken]

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?

Perfect explanation.

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

Last edited by a moderator: May 6, 2017
21. May 8, 2012

### 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?

22. May 8, 2012

### 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?

23. May 8, 2012

### Femme_physics

Last edited by a moderator: May 6, 2017
24. May 8, 2012

### I like Serena

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

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. ;)

25. May 12, 2012

### Femme_physics

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

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 [Broken]

Last edited by a moderator: May 6, 2017