Consider the circuit on the left with two resistors, a capacitor and an ideal OP-AMP. You may make simplifying assumptions about the OP-AMP but you must state your assumptions clearly. (a) By making a suitable assumption for the capacitor, find the DC gain of this circuit, namely, (vo/v₁)DC (Hint: How does the capacitor behave for steady- state DC voltages?)
Consider the circuit on the left with two resistors, a capacitor and an ideal OP-AMP. You may make simplifying assumptions about the OP-AMP but you must state your assumptions clearly. (a) By making a suitable assumption for the capacitor, find the DC gain of this circuit, namely, (vo/v₁)DC (Hint: How does the capacitor behave for steady- state DC voltages?)
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I solved it in frequency-domain way. And answer is correct. But didn't get point, Can you explain why should I use this way to solve it? More details. Thank you! As soon as possible!
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- **R1**: Resistor 1
- **R2**: Resistor 2
- **C**: Capacitor
#### Assumptions:
1. The OP-AMP is ideal.
2. For steady-state DC voltages, the capacitor behaves as an open circuit.
### (a) Determining DC Gain
To find the DC gain (\( \frac{v_o}{v_I} \)) of the circuit, we assume that for steady-state DC voltages, the capacitor acts as an open circuit. Therefore:
\[ \left( \frac{v_O}{v_I} \right)_{DC} = -\frac{R_1}{R_2} \]
### (b) General Expression Relating \( v_I \) to \( v_O \)
We set up the general expression for the relationship:
\[ v^- = v^+ = 0 \]
Applying Kirchhoff's Current Law (KCL) at the node:
\[ \frac{v_I}{R_2} = \frac{v_O - v_{C}}{R_1} + C \frac{dv_O}{dt} \]
Substituting \( v_L = \frac{dv_O}{dt} \):
\[ \frac{v_I}{R_2} = \frac{-v_O + v_{C}}{R_1} + C \frac{dv_O}{dt} \]
### (c) Step Function Input
Assuming \( v_I \) is a step function that is zero before \( t < 0 \) and \( v_I \) is \( V_I \) after \( t > 0 \):
1. \( v_I(t) = V_I u(t) \)
2. Considering \( v_O = 0 \) at \( t = 0 \):
\[ V_I(s) = \frac{VI}{s} \]
\[ \frac{V_I}{sR_2} = -\frac{V_O(s)}{R_1} + sCV_O(s)\]
Rearranging and simplifying:
\[ V_O(s) = -\frac{R_1 V_I}{R_2 (1 + s R_1 C](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a579a7f-99c6-465f-8352-06e32f8837ce%2F2f7c9b2c-851c-458f-a74c-e83b36143e22%2Fap4rig9_processed.png&w=3840&q=75)
Transcribed Image Text:# Operational Amplifier Circuit Analysis
### Consider the following circuit:
The circuit contains two resistors, a capacitor, and an ideal operational amplifier (OP-AMP).
- **R1**: Resistor 1
- **R2**: Resistor 2
- **C**: Capacitor
#### Assumptions:
1. The OP-AMP is ideal.
2. For steady-state DC voltages, the capacitor behaves as an open circuit.
### (a) Determining DC Gain
To find the DC gain (\( \frac{v_o}{v_I} \)) of the circuit, we assume that for steady-state DC voltages, the capacitor acts as an open circuit. Therefore:
\[ \left( \frac{v_O}{v_I} \right)_{DC} = -\frac{R_1}{R_2} \]
### (b) General Expression Relating \( v_I \) to \( v_O \)
We set up the general expression for the relationship:
\[ v^- = v^+ = 0 \]
Applying Kirchhoff's Current Law (KCL) at the node:
\[ \frac{v_I}{R_2} = \frac{v_O - v_{C}}{R_1} + C \frac{dv_O}{dt} \]
Substituting \( v_L = \frac{dv_O}{dt} \):
\[ \frac{v_I}{R_2} = \frac{-v_O + v_{C}}{R_1} + C \frac{dv_O}{dt} \]
### (c) Step Function Input
Assuming \( v_I \) is a step function that is zero before \( t < 0 \) and \( v_I \) is \( V_I \) after \( t > 0 \):
1. \( v_I(t) = V_I u(t) \)
2. Considering \( v_O = 0 \) at \( t = 0 \):
\[ V_I(s) = \frac{VI}{s} \]
\[ \frac{V_I}{sR_2} = -\frac{V_O(s)}{R_1} + sCV_O(s)\]
Rearranging and simplifying:
\[ V_O(s) = -\frac{R_1 V_I}{R_2 (1 + s R_1 C
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