Our R's are computable,but...

I neither said they were the same nor did i say they were different. I just gave the defining equations. On inspection it is obvious that they are, and MUST be different.

Hmmm...do not think so, but i am going to sit down and use the old-fashioned computer implements...paper & pencil.

Bummer..who sells those? In 0.1 percent..

There appear to be a couple of errors here:

Phantom schrieb:

Hello,

I agree with the corrected values -.0699 and .2083.

Bye

I thought you were asking an engineering question not an algebra question. Solve for S1 in equation one and substitute that into the other equations. It's really only a series of six, two-variable simultaneous equations, since there is only one variable difference between equation #1 and each of the others.

Well, since Mouser refuses to carry 0.1% negative resistors, the errors are moot. What was really disturbing, is that (via Google) i found a site that "solved" the 7 equations in the 7 unknowns, getting.... .... ... R1 positive, and all other resistors negative!!!!!

Now boys and girls, positive or negative, the resistors and the equations are LINEAR!! So, how in the G-D h*ll can there be _two_ solutions?

Robert Baer schrieb:

Hello,

it is easy to test a solution of the equations, simply set the values in the equations and calculate if both sides of each equation are equal. Did you do that with both solutions and what was the result?

Testing a solution is much easier than finding a solution, so what are you waiting for?

The theory says that a system of seven linear equations with seven unknowns has only one solution if a solution exists.

Bye

There doesn't have to be a solution.

R2 has to be 1K for the first two lines to be true.

However, you see R7 altering the impedance of the parallel network by a factor of more than a magnitude.

With R2 present, the last three conditions are physically impossible.

What do you actually need?

RL

Huh?

Convert to conductance (k-Ohms):

5 = G1 + G2 + G3 + G4 + G5 + G6 + G7 4 = G1 + G3 + G4 + G5 + G6 + G7 3.333 = G1 + G2 + G4 + G5 + G6 + G7 1.667 = G1 + G2 + G3 + G5 + G6 + G7 1 = G1 + G2 + G3 + G4 + G6 + G7 0.5 = G1 + G2 + G3 + G4 + G5 + G7 0.2 = G1 + G2 + G3 + G4 + G5 + G6

Then solution is trivial (if you don't mind negative conductance/resistance)... G1 = -14.3

G2 = 1

G3 = 1.667

G4 = 3.333

G5 = 4

G6 = 4.5

G7 = 4.8 ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |

      Remember: Once you go over the hill, you pick up speed

My recollection is that a set of 7 equations with 7 unknowns has either one, zero, or infinity many solutions?

In fact, in practice, sets of many equations with many unknowns usually have zero solutions (due to noise in measurements and a finite number of digits) :-), so people resort to things like least-squared fitting to get a "reasonable" answer; the condition number of the matrix of the coefficients gives one some idea as to just how "reasonable" that answer might be.

Here's a new set of equations that is solvable for the values you seem to be searching for:

200 = R1||R2||R3||R4||R5||R6||R7 250 = R2||R3||R4||R5||R6||R7 300 = R3||R4||R5||R6||R7 600 = R4||R5||R6||R7 1000 = R5||R6||R7 2000 = R6||R7 5000 = R7

How do you want to skin the cat?

RL

A real solution is needed for real resistor values.

RL

You've never seen a negative resistance ?:-) ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |

      Remember: Once you go over the hill, you pick up speed

I could see Baer working with a bipolar supply, non-digital source, or two inputs from a diff amp, but judging from the point of the original post and responses to it, I'm not holding my breath.

RL

[snip]

Baer was the OP? Yep, don't hold your breath :-) ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |

      Remember: Once you go over the hill, you pick up speed

Where do you think they got the idea for the Mobius resistor? ;-)

--
You can't fix stupid. You can't even put a band-aid on it, because it's
Teflon coated.

Ever seen a Klein bottle ?:-) ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |

      Remember: Once you go over the hill, you pick up speed

Yes, that's where you store your Mobius resistors. ;-)

--
You can't fix stupid. You can't even put a band-aid on it, because it's
Teflon coated.