# Method for determining error in an equation

1. Oct 5, 2015

### WK95

1. The problem statement, all variables and given/known data
Ok, so suppose I have an equation like V_out = 0.5(1 + R4/R3)(V1+V2) and I know the R3 and R4 has a tolerance of +-5%. For such an equation and similar ones, how would I estimate the possible range of values V_out? For example, I'd like to find out something like V_out_theoretical * 96% < V_out < V_out_theoretical * 104% or rather, I want to figure out the "tolerance" (no, I don't think that's the proper word) for V_out based on the known tolerances of R3 and R4.

Sorry, I don't know the proper terminology for this sort of stuff.

2. Relevant equations

3. The attempt at a solution

2. Oct 5, 2015

### gleem

First are You assuming V1 and V2 are known exactly? Secondly do You know how to take derivatives?

3. Oct 5, 2015

### WK95

You can assume that V1 and V2 are fixed. And yes, I can take derivatives.

4. Oct 5, 2015

### gleem

the commonly accepted way to determine the uncertainty for V_out is to determine the derivatives of V_out with respect to R3 and R4 to determine the changes in V_out wrt R3 with R4 held fixed and wrt R4 with R3 held fixed. Using dR3 and dR4 as the uncertainty in this case the tolerances substitute into the following equation.

$$dV_{out} = \sqrt{\sum_{i=3}^{4} \left \lfloor \frac{\partial V_{out}}{\partial R_{i}}dR_{i}\right\rfloor^{2}}$$

the two terms in the square root represent the change in V_out due to a change in R3 and R4.

For a more detailed explanation see "Data Reduction and Error Analysis for the Physical Sciences" by Bevington and Robinson

5. Oct 5, 2015

### WK95

So partial derivatives? Finally! I get to use Calculus III stuff!

6. Oct 5, 2015

### WK95

So if R3 and R4 is 10000 Ohms, and V1 is 1.8 and V2 is 3.3, I get 12.8%. Have I done things correctly?

7. Oct 6, 2015

### gleem

I think you are off by a factor of √2 can you find where you missed it?