# Impedance function of simple RLC circuit

Discussion in 'Homework Help' started by Fraser_Integration, Nov 3, 2010.

1. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
Hello there. Having problems getting the zeros of this circuit, I believe it should be -1.5 from the answer, but I am not getting a nice quadratic at the end of it. Please see attachment

I start by saying that the impedance function Z(s) = sL + R1 + R2//(1/sC)

= 2s + 2 + [ (1*1/0.5s) / (1+1/0.5s) ]

= 2s + 2 + [ (2/s) / (1 + 2/s) ]

= [ 2s(1 + 2/s) + 2(1 + 2/s) + (2/s) ] / 1 + 2/s

= 2s + 4 + 2 + 4/s + 2/s (ignoring the denominator)

now the zeros of the expression are when the above equals zero,

or 2s^2 + 6s + 6 = 0

however this doesnt give me a nice root, as the answer I am working towards is i(t)=3[1 - e^-1.5t sin(0.866t+90°)]

could someone please tell me where I have gone wrong? many thanks

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2. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
hmmm I think I may have it now.... James if you're reading this.

I think s^2 + 3s + 3 = 0 is the correct equation for the zeros.

then s = -1.5 plus or minus sqrt(3)/2

as s = phi + j-omega, then omega = sqrt(3)/2

and sqrt(3)/2 = 0.866..... as in the answer. I'm getting there bit by bit....

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Your working looks fine so far.

4. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
cheers. still not 100% sure why the final answer is a sine, because the expansion of e^jx = cosx + jsinx no? Or is that irrelevant here?

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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You have an under-damped case so the current will be a damped sinusoid plus DC final value.

6. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
Hey guys, me again.

My lecturer is being devious and making me really work for the final answer.

I just can't seem to mathematically point out where the theta value of 90degrees comes from. It must be a phasor relationship somewhere, but in the examples I've seen it is only the forced response where you have to say i = v/Z and up until now all the natural response answers have been assumed to be exponential, you know?

He says there are two initial conditions that can be worked out. The obvious one is that at t=0, i=0 because the inductor opposes instantaneous changes in current.

He then says the second one can be found when you apply KVL to the circuit, and note that the capacitor voltage is zero at first as that can't instantaneously change. With those two conditions you can soundly prove the two unknowns in the generic: i = Be(-1.5t)sin(0.866t + theta) answer.

7. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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Do you know which device the current leads the voltage by 90°, and which one voltage leads current by 90°?

The two choices are Capacitor and Inductor.

Which (C or L) would give a positive 90° phase shift?

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The generic function you give does not include a non-zero steady state final value for the current, which must exist. At t=∞ the current will be non-zero since there exists a path for current flow and the applied voltage is 9V.

So you have the initial conditions at t=0 as i(0)=0, VL(0)=9V and Vc(0)=0V. You should also be able to deduce the final steady-state current i(∞).

Are you using Laplace transforms or differential calculus to solve the problem?

9. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
The inductor!

Sorry, I really should've said that qualitatively I know what is going on in the circuit, it's just the operation to work out theta, dividing what by what.

10. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
the generic function I gave is for the natural response only. I understand that the final response will be 3A.

my only niggle is working out the operation to define theta. I understand the circuit, it's just theta......

11. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
now i'm thinking maybe there isn't a mathematic operation to find theta, you just accept it is true because intuitively there is only the inductor with a voltage across it, and so there will only be an upwards arrow on the argand diagram because of it?

sorry if this is confusing. I just want to get the example down properly as there will surely be more like this.

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
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There is a consistent mathematical solution to this which will determine all the unknowns - both transient and steady-state. Intuition, whilst sometimes helpful, is not a recommended method of solution - unless you are sufficiently intuitive to solve the problem by inspection.

You didn't say what technique you are using to solve this.

13. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
This is frustrating.

I can't "find" theta, that's all I want to know. A simple mathematical formula, or initail condition.

Look at it this way, the COMPLETE response with unknown constants is:

i(t) = 3 + Be^(-1.5t)sin(0.866t + theta) Amps

Here, there are two unknowns and that would require two initial conditions to solve.

The first is:

i(0) = 0.

The second I don't know? All I do know is that the inductor voltage is 9V, and the capacitor voltage is 0V at t=0+ secs. But I can't see how this information gives me anything I can use to find B and theta?

Hope that clears up my problem.

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
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OK - it would have made things clearer if you had provided the complete generic form of unknown solution. This is after all homework.

You have two unknowns so the only approach using your method of an assumed solution is to find an equation which will involve another of the known initial conditions.

You know VL(0)=9V and VL(0)=L*(di/dt)=9V at t=0.

You might consider making the derivative of the generic current and this will give a second equation which combined with the first (i(0)=0) may be used to solve for the two unknowns.

15. ### The Electrician AAC Fanatic!

Oct 9, 2007
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The answer involves an exponential times a cosine, or it can be expressed as an exponential times a sine + 90°. A cosine is the same as a sine + 90°.

See the attachment.

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16. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I just occurred to me that you may be rolling ideas together about steady state analysis based on complex number theory and transient analysis based on the 's' domain approach. Your thread title lends some weight to that notion where you refer to "impedance". There's essentially nothing wrong with keeping the two approaches in mind when seeking a solution - I guess they are embodied in the complete solution for the variables in a driven circuit. The argand diagram probably only has relevance when analyzing a circuit driven by a fixed frequency sinusoidal source and in which the variables have reached their steady-state condition.

Anyway, just a thought - I may be completely misinterpreting your comments.

Last edited: Nov 7, 2010
17. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Using the "assumed function" method I came up with an answer which agrees with the Electrician's solution.

Hopefully (?) it's clearer why I was asking about your proposed method of solution. The approaches are somewhat different. The Electrician's approach makes no assumptions about the form of the answer - requiring however a knowledge of the use of Laplace transforms.

Using the assumed function approach meant I had to go through the differentiation part which increases the chance of introducing errors.

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