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## Re: Solve Simple Relationship

 From: Thomas D. Dean Subject: Re: Solve Simple Relationship Date: Fri, 27 Aug 2021 09:33:38 -0700 User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:78.0) Gecko/20100101 Thunderbird/78.11.0

```On 8/27/21 9:04 AM, Doug Stewart wrote:
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On Fri, Aug 27, 2021 at 11:13 AM Thomas D. Dean <tomdean@wavecable.com <mailto:tomdean@wavecable.com>> wrote:
```
On 8/27/21 8:09 AM, Thomas D. Dean wrote:
> I have a voltage divider I want to calibrate. I tried this by
exchanging
> the resistors and measuring the resultant rfesistance.
>
> 1.6516 = 3.3 * R1 / (R1 + R2)
> 1.6484 = 3.3 * R2 / (R1 + R2)
>
> R1 is around 1005 ohms
> R2 is around 1003 ohms
>
> I think this is simple.  But evidently, it is beyond me.
>
> Tom Dean

And, worse, I can not clearly state the problem:  rfesistance = voltage.

I think that you are making it too hard.

Choose r1 say 1000
now calculate r2 from your first eq.
I get 998.06

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I want to calibrate the voltage divider. Two equations, two unknowns. Should have a solution.
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The idea is to have a known value for R1 and R2 so I can measure Vs,
(= 3.3?) in the future.

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Maybe, iterate R1, calculate R2 with the first and see if it satisfies the 2nd equation.
```
I wrote a function
function [y] = eq_r(x)
y(1) = 1.6516 - 3.3 * x(1) / (x(1) + x(2))
y(2) = 1.6484 - 3.3 * x(2) / (x(1) + x(2))
endfunction

```
fsolve returns values that do not satisfy the equations. Very dependent on X0 supplied.
```
Tom Dean

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