# Reactive impedance question

Discussion in 'Electronic Basics' started by Tim Williams, Oct 3, 2005.

1. ### Rich GriseGuest

At resonance, Z = R. ;-) (these _are_ perfect components, right? ;-) )

Cheers!
Rich

2. ### John PopelishGuest

I believe you are thinking of all 3 components in series.

3. ### Rich GriseGuest

Yeah. Or in parallel. ;-)

You know about the ohnosecond, right? ;-)

Thanks,
Rich

4. ### John PopelishGuest

I missed that.

5. ### The PhantomGuest

Probably you don't have the expressions right. I haven't used the
symbols C, L1, and L2 to represent their reactances, but rather have
used them to represent the values of those components.

The expression for input Z is:

Sqrt[(4*f^2*Pi^2*((L1 + L2)^2*R^2 + 16*C^2*f^4*L1^2*L2^2*Pi^4*R^2 +
4*f^2*L1*L2*Pi^2*(L1*(L2 - 2*C*R^2) - 2*C*L2*R^2)))/
(4*f^2*L2^2*Pi^2 + R^2 - 8*C*f^2*L2*Pi^2*R^2 +
16*C^2*f^4*L2^2*Pi^4*R^2)]

and the voltage you have called Vx is given by:

Vx = Vz*Sqrt[(L2^2*R^2)/((L1 + L2)^2*R^2 +
16*C^2*f^4*L1^2*L2^2*Pi^4*R^2 +
4*f^2*L1*L2*Pi^2*(-2*C*L2*R^2 + L1*(L2 - 2*C*R^2)))]

6. ### Tim WilliamsGuest

I went over the Vx problem twice and got the same result, so there must be
something wrong with my procedure...but it's rock solid as far as I know:
loop current = vector sum of Ir + IL2 + Ic (where Ireactive = IL2 - Ic) and
Vz = vector sum Vx + VL1, where VL1 = XL1 * I. Plug first into second,
switch component currents with expressions in terms of Vx, isolate Vx
outside of the radical and solve for Vx.
Fair enough, but makes for a messier result. I can just as well set Xc = 1
/ 2piFC and so forth when I graph it.
Um.. I'll look at that later... why does it keep growing?...
So that,

Vx = Vz*L2*R / sqrt of:
R^2 L1^2 + R^2 L1 L2 + R^2 L2^2 + 16 pi^4 f^4 L1^2 L2^2 R^2 C^2 +
4 pi^2 f^2 L1^2 L2^2 - 8 pi^2 f^2 L1^2 L2 R^2 C - 8 pi^2 f^2 L1 L2^2 R^2

?

Tim

7. ### Tim WilliamsGuest

Well, that doesn't work the same as the original (if I copied it right...I
copied it twice correctly...), but even so, it gets infinite values for the
range zero to some value of F. The equation I came up with at least shows a
resonance peak, but as I said, it doesn't go to infinity with R, at Fo.

Tim

8. ### The PhantomGuest

The way to do it is to first write the voltage transfer ratio in terms of the
complex variable s. Since it's just a voltage divider, that's pretty easy. The
three components, L2, R, and C form the bottom leg of the divider and L1 forms
the top leg. Let z1 be the impedance of the top leg (just L1), and let z2 be
the impedance of the 3 components of the bottom leg in parallel. Then the
voltage transfer ratio of the divider is given by z2/(z1+z2). Expanding, and
using the complex variable s, we get:

Vz = Vx * [(1/(1/(s L2) + 1/R + s C))/(s L1 +1/(1/(s L2) + 1/R + s C))]
or Vz = Vx * CVTR (CVTR stands for complex voltage transfer ratio).

Replace the variable s in the expression for CVTR with (2 Pi f j), so that
CVTR = (1/(1/(2 Pi f j L2) + 1/R + 2 Pi f j C))/(2 Pi f j L1 +1/(1/(2 Pi f j L2)
+ 1/R + 2 Pi f j C))

Let CONJ[CVTR] be the complex conjugate of CVTR; that is CONJ[CVTR] is just the
same expression as CVTR, but with j (= SQRT[-1]) replaced by -j. Then form the
product: (CVTR * CONJ[CVTR]) and take the square root, so that the (real
magnitude) voltage transfer ratio you want is given as:

Vz = Vx * SQRT[CVTR * CONJ[CVTR]]

This is only the magnitude (or modulus) of the CVTR; the phase information has
been lost with this procedure.
You made a couple of errors in your algebra:
should be ...+ 2 R^2 L1 L2 +...
And this should be ...- 8 pi^2 f^2 L1 L2^2 R^2 C

When I plot this with R=100, C=20uF, L1=50uH, L2=10uH, I get a peak at about
12328 Hz, with a value at the peak of about 25.82

I verify these expressions I've posted by plotting the simpler complex
variable version on the same plot with the complicated real variable version,
and seeing that the two plots are identical (actually, they're on top of each
other).

9. ### Tim WilliamsGuest

We never covered s, and I don't remember how to handle conjugation of i/j.
Ah, ok.
Not a problem, I know phase runs from 0 degrees approaching zero to 180
degrees approaching infinity, with 90 degrees at Fo.
Ok. But how does that help me...

As it turns out I can paste your equation into QBasic verbatim, making minor
corrections (that I can't possibly screw up and it works exactly
perfectly! Well thanks for your help Phantom

Tim

10. ### The PhantomGuest

In an earlier reply of yours to one of my postings, you said:

"And you're certain that the above equation is the real derivation of the
imaginary equation, which is also correct?"

You wanted to know if I was certain, so I thought I'd tell you how I make sure
the complicated real variable equation is correct.
Good! Glad it's working out for you.

11. ### Rich GriseGuest

Ohnosecond:
An ohnosecond is the fraction of time in which you realize you made a big
mistake and you can't undo it. The first documented use of the term was
by Elizabeth Crowe in her book The Electronic Traveler.
-- en.wikipedia.org/wiki/Ohnosecond

Cheers!
Rich