Re: Tanks, impedance, the j operator and excel

On May 25, 8:58 pm, "Bob" <nob...@xxxxxxxx> wrote:
Hi gang,
This one is directed at you math whizzes..... I have a formula for tank
circuit impedance that I am attempting to analyze/plot in MS Excel. The tank
consists of a resistor in series with the coil, a cap in parallel with the
coil resistor combination and a another resistor in parallel with the cap
and coil/resistor.
The formula consists of two quotients; One has a numerator that consists of
several terms multiplied, added, squared, etc and the denominator is
similar. The second quotient is preceded by the "j" operator. (That is the
entire quotient appears to be "j"ed rather than any single term or factor)..
This quotient has a numerator and denominator similar to the first).
 I assume that the first quotient is the real part of the impedance, while
the second one is the imaginary part, yielding the rectangular Z =R +/-j
form. Does this sound correct?

Yes, sounds right.

Assuming the math is done correctly, I believe I can use the IMABS function
on the result to get the magnitude of the impedance. Yes?

http://office.microsoft.com/en-us/excel/HP052091191033.aspx

It appears that, if you want to form a complex number from its real
and imaginary components, you need to synthesize the complex number
with the COMPLEX function before given that synthesized number to
IMABS.

In Excel, I have entered the first quotient as ordinary numbers, i.e.
"=A1*(C2+B3)..../D4*6.28*C3......"
I'm getting numbers that seem to make sense.
The second quotient has me scratching my head a bit. I have tried using the
same process on this part, and then using "=IMABS(first quotient
results,+second quotient results)" to get the magnitude of the impedance but
I think this may be incorrect.
Any thoughts? If someone wants it, I can post my Excel file to them.....

You probably already know that it is not necessary to use IMABS or
COMPLEX.

Just plug in omega for first quotient, omega for second quotient. You
will get two numbers, one for each quotient, X, and Y, respectively
Compute the total impedance Z from them using Pythagorean theorem.

Z = sqrt (X^2+Y^2).

Naturally, you ignore the j on second quotient.

-Le Chaud Lapin-
.

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