Solving a circuit in Freq Domain

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  • Thread starter phsyics_197
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  • #1

Homework Statement



Question as in image:

rjnqx1.jpg


Homework Equations





The Attempt at a Solution



When I did Mesh, I got this:
-20j-20Ia+(Ia-Ib)2j=0
(Ib-Ia)2j-(Ib-5)j-(Ib-Ic)50=0
-(Ic-Ib)50-(Ic-5)8j-2(Ia-Ib)=0
2(Ia-Ib)-100Id=0
Ia-Ib=Ix
**This is done from Left to Right**

One thing I am not sure about is the Sin voltage source. I believe you convert it to Cos first? Then it should give you -20j.

Well, when I solved the above equations in Matlab, I got Ix = -6.4622 - 0.5310i.

I was wondering if that is correct, and how could I verify it?

Thanks
 

Answers and Replies

  • #2
gneill
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The polarity of your controlled voltage source is a bit vague, and it will affect the operation of the circuit. Could you clarify it?
 
  • #3
Yes, sorry about that. It should (+) on top and (-) on bottom.
 
  • #4
gneill
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Yes, sorry about that. It should (+) on top and (-) on bottom.
Hmm. In that case I think that the circuit won't reach a steady state; the current Ix is going to grow without bound as the controlled voltage grows while Ix does -- there's positive feedback occurring. (I just confirmed this with a Spice simulation).
 
  • #5
But can't you still model the current sinusoidally?
 
  • #6
gneill
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But can't you still model the current sinusoidally?
Sure, but you'll need to use a slightly different analysis method; Solve for the current Ix as a function of time using Laplace transforms. This is mostly the same as the usual mesh method, only no assumptions about the circuit going to steady state are made. The voltage and current sources are replace with Laplace transform counterparts for sine and cosine driving functions. The expression for Ix(s) is found and a reverse Laplace transformation performed on it.

The result will be something like a sum of sine and cosine terms multiplied by an exponential that INCREASES with time, like [itex] e^{A t} [/itex].

[EDIT] BTW, the Spice simulation indicated that Ix would reach an amplitude of about 150 Million Amps in about 216 seconds.
[EDIT] Also note that this problem does NOT occur if the controlled voltage source has its polarity reversed (+ down, - up).
 
Last edited:
  • #7
Are you 100% positive it does not reach steady state? Cause everything we have done in class has been for steady state problems. I find it hard to believe he is giving us a problem that we have never seen before.
 
  • #8
gneill
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Are you 100% positive it does not reach steady state? Cause everything we have done in class has been for steady state problems. I find it hard to believe he is giving us a problem that we have never seen before.
I believe what the simulation is telling me. Note that the problem could also be alleviated if the controlled source gain were -2 rather than +2. You might want to check that.
 
  • #9
He specifically said that the (+) is on top. And when I worked it out by hand following his steps, I got what seems to be an appropriate answer.
 
  • #10
gneill
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He specifically said that the (+) is on top. And when I worked it out by hand following his steps, I got what seems to be an appropriate answer.
Sometimes there is a world of difference between "seems" and "is" :smile:

If you have LTSpice on your PC (It's a free download, and practically an industry standard) I'd be happy to post the wirelist for you to run the simulation for yourself. You can see the difference if you change the gain from +2 to -2, effectively reversing the polarity of the controlled source.
 
  • #11
Sorry, I was playing around with PSpice and I know what you are saying.
 
  • #12
gneill
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Sorry, I was playing around with PSpice and I know what you are saying.
So it looks like a misprint in the problem statement, or perhaps the person who created it didn't check for the positive feedback possibility.

The steady-state analysis method WILL yield a result in both cases, but it lacks the finesse to properly handle circuits that don't in fact have a steady state.
 
  • #13
Do you know the equation?

I was getting an amplitude of about 6, not sure how to extract the phase shift from the plot.
 
  • #14
gneill
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I calculate a magnitude of 5.996A for Ix, with a phase shift (with respect to the current source 5cos(2t) ) of -144.107°. This is for steady-state case, of course.
 

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