Determine the output voltage of a network shown in figure if the digital input is 1011 \n\n[Figure needs to be uploaded later]
- A. -3.875 V
- B. -4.875 V
- C. -5.875 V
- D. -6.875 V
Answer: D. -6.875 V
Explanation: [Figure needs to be uploaded later]\n$V_0 = -R_f(\frac{V_A}{800k} + \frac{V_B}{400k} + \frac{V_C}{200k} + \frac{V_D}{100k})$\n$= -(\frac{5}{8} + \frac{5}{4} + \frac{0}{2} + \frac{5}{1})$\n$= -(0.625 + 1.25 + 0 + 5) = -6.875 V$.