What is the Rth on the circuit [closed]

Question

I have to find the Resistances R1 & R2 in order to have 0 - 3.3 V input on the ADC. What I cannot find is the Rth on the "Network" part of the circuit.

The text of the original question is:

A sensor produces a voltage between -5 and 3 V. The output resistance of the sensor is 470 Ω. The sensor must be interfaced to an ADC with an input range of 0 to +3.3 V and an input resistance of 3300 Ω. Determine the values of R1 & R2 for the interface network show. Check your work by analyzing the resulting network.

The answer supplied is:

R1 = 9300 930 Ω. R2 = 1400 Ω.

Answer 1

I think the offset should be 2.0625V, not 3.3/2 + 1 (2.65V).

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Using @Tony Stewart Sunnyskyguy EE75 offset calcuated at Vin = -5 and 3V:

\$Vout=Gain*Vin + Offset\$

\$Vout= \dfrac{3.3}{8} * Vin + \dfrac{3.3}{2}+1 \$

\$Vout(Vin=-5V) = \dfrac{3.3}{8} * (-5) + \dfrac{3.3}{2}+1 = 0.5875V\$

\$Vout(Vin=3V) = \dfrac{3.3}{8} * (3) + \dfrac{3.3}{2}+1 = 3.8875V\$

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Both of these endpoints are incorrect by a constant value of 0.5875V. This suggests that the gain is correct and the offset value is incorrect.

Because the gain is multiplying the input, the average value of the input also changes.

\$Avg(Vin) = -1\$

\$Avg(Vin*Gain) = -\dfrac{3.3}{8}\$

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Calculating the offset needed to bring up the scaled signal midpoint to 1.65V:

\$Offset = \dfrac{3.3}{2} - \dfrac{3.3}{8} * \dfrac{-5 + 3}{2}\$

\$Offset = 2.0625V\$

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Checking this with the Vin endpoints again (and Vin midpoint),

\$Vout=Gain*Vin + Offset\$

\$Vout= \dfrac{3.3}{8} * Vin + 2.0625 \$

\$Vout(Vin=-5V) = \dfrac{3.3}{8} * (-5) + 2.0625 = 0.0V\$

\$Vout(Vin=-1V) = \dfrac{3.3}{8} * (-1) + 2.0625 = 1.65V\$

\$Vout(Vin=3V) = \dfrac{3.3}{8} * (3) + 2.0625 = 3.3V\$