Lecture on Integrator and Differentiator Circuit - Assignment 15

Introduction

• Focus on solving Assignment 15.
• Recommended only for those who have studied integrator and differentiator circuits thoroughly.
• The assignment consists of 5 questions, solved one by one.

Question 1: V(t) Waveform Analysis

• Initial Condition (t = 0): Capacitor shorted, negative feedback present, output is 0V.
• Final Condition (t → ∞): Capacitor open, negative feedback via resistor still present, output is -5V.
• Graph: V_out starts at 0V, exponentially decaying to -5V.
• Equation: V_out(t) = -5(1 - e^(-t/RC))
• Current I(t) Equation:
  • Initial current I(0) = 5/R
  • Final current I(∞) = 0
  • I_C(t) = (5/R) e^(-t/RC)
• V_x: Remains 0V due to virtual short.

Question 2: vx(t) and vout(t) at t = 2 seconds

• Initial Condition (t = 0): Capacitor shorted, negative feedback present, output is 0V.
• Final Condition (t → ∞): Capacitor open, output saturates at -10V.
• Graph: V_out starts at 0V, tries to go to -15V but saturates at -10V.
• Equation: V_out(t) = -10
• V_x(t) Equation:
  • Initially 0V, rises to 2.5V.
  • Equation: V_x(t) = 2.5(1 - e^(-(t-1.09)/RC))
  • At t = 2 seconds: Approx. 1.48V.

Question 3: V_x and V_out Waveform Analysis

• Initial Condition (t = 0): Capacitor shorted, output 10V due to virtual short.
• Final Condition (t → ∞): Capacitor open, output saturates at 12V.
• Graph: V_out starts at 10V, tries to go to 15V but saturates at 12V.
• Equation: V_out(t) = 12
• V_x(t) Equation:
  • Initially 10V, decays to 8.5V.
  • Equation: V_x(t) = 8.5 + 1.5 e^(-(t-0.5)/RC)

Question 4: V_out(t) at t = 0+ and t → ∞

• Initial Condition (t = 0): Capacitor shorted, output is -2.5V due to high frequency approximation of RC.
• Final Condition (t → ∞): Capacitors open, output -5V.
• Transformation Function: First order with time constant 2RC.
• Graph: V_out starts at -2.5V, decays to -5V.
• Equation: V_out(t) = -5(1 - e^(-t/2RC))

Question 5: Analysis with Step Current Source

• Initial Condition (t = 0): Inductor voltage induces step current. Capacitor shorted, output is 0V.
• Final Condition (t → ∞): Capacitor open, output determined by current source and resistor.
• Current i(t) Equation: i(t) = (5/L) u(t)
• Graph: V_out starts at 0V, decays to -5R/L with time constant RC.
• Equation: V_out(t) = -5R/L (1 - e^(-t/RC))

Conclusion

• All problems involved deep understanding of integrator circuit behavior under various initial and final conditions.
• Emphasis on practical applications and complex problem-solving skills.