# Mind boggling question

Discussion in 'Electronics Homework Help' started by Baws, May 11, 2013.

1. ### Baws

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May 11, 2013
That is making 0 sense to me, can anyone help?

I know how capacitors and inductors work with all the resitance=j/wl and stuff, just this question is like confusing..?

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2. ### The Electrician

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Jul 6, 2012
This is a bridged T notch filter (look up on the web).

They want you to determine the values of Rx and Lx that will make the output zero at some particular frequency.

3. ### Laplace

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Apr 4, 2010
Begin by writing the node equations for the circuit. To keep it simple assume that the source impedance is zero and the load impedance is infinite. (Additional credit may be available accounting for these complicating factors.) From the node equations derive the transfer function then shift it to the steady state frequency domain. The current through load impedance will be zero when the load voltage is zero. Since we are interested only in the zeroes of the transfer function, isolate the numerator and set both the real part and the imaginary part equal to zero. Then solve the equations and see what you get.

In this case the output is zero when the radian frequency equals 1/sqrt(2CLx) and Rx=2Lx/(CR).

Attached is a capture of my MathCad screen.

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4. ### The Electrician

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Jul 6, 2012
Isn't the idea to help the student rather than do all the work for him?

5. ### davennModerator

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Sep 5, 2009
supposed to be

Dave

6. ### Laplace

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Apr 4, 2010
Yet on the positive side there is still a lot of work necessary to account for the effects of non-ideal source and load impedances.

7. ### The Electrician

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Jul 6, 2012
If the source and load impedances are included, they drop out of the final result.

8. ### Laplace

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Apr 4, 2010
There is no reason to suppose that this would be true, and in fact it is demonstrably not true when the load impedance has a reactive component. But now the OP will be robbed of the experience of discovering this for him/herself.

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9. ### The Electrician

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Jul 6, 2012
By "final result", I mean the result required by the original problem, which is the determination of Rx and Lx such that the load current (current in ZL) is zero.

You have found an expression for VL; certainly that expression involves Zs and ZL. Now you need to use that expression to determine Rx and Lx in terms of R, C and the source frequency w such that VL is zero.

10. ### Laplace

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Apr 4, 2010
As before, isolate the numerator of the transfer function, split the load impedance into its real and imaginary parts, set both the real and imaginary parts of the transfer function to zero, and solve for the frequency at which the output voltage is zero. However, in this case the result does not reduce to something simple because the source and load impedances do not drop out of the final result.

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11. ### The Electrician

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Jul 6, 2012
The fact that Mathcad did not reduce its result to something simple doesn't mean that there isn't a simple solution.

Had you made the substitution ZL -> RL + j XL in post #3 before you solved for w, you would have had the same problem of a very large output from Mathcad..

But, it isn't necessary to go to all that trouble anyway. When you have an expression such as:

and the zeroes of the part in parentheses are already known, we can see by inspection that multiplying by ZL simply means that ZL=0 is also a solution; the zeroes of the part in parentheses remain the same. Obviously if ZL is a short, you get no output. That's not a solution the original problem wants anyway.

The solution you obtained in post #3 is still a solution, and Zs and ZL don't appear in it.

Even if Mathcad were unable to simplify the very large output, if we know the solution by means of some other method, we can verify that it's a solution by substitution.

I use Mathematica instead of Mathcad, but with either we can verify the solution by substitution:

If these substitutions were made in the very large output you got in post #10, that output would evaluate to zero (assuming no mistakes in the algebra).

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12. ### Laplace

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Apr 4, 2010
That is a valuable lesson. Of course, I knew it was true at one time in the distant past. But more recently it seemed counter-intuitive and I just did not want to believe. So while adding a reactive load will substantially change the circuit behavior, connecting a point of zero voltage to ground through a passive network will have no immediate effect -- the operating curves of the two different circuits will still intersect at the same zero point.