Question

Derive H(s) = Vout(s)/Vin(s) for the following circuit:

Assume an Ideal OPAMP with the following parameters:
- R1 = 10 kΩ
- R2 = 20 kΩ
- R3 = 40 kΩ
- C1 = 20 nF

Use any technique you wish to derive the output. Sketch the Bode plot (magnitude and phase) of H(s) using radians per second.

1. Derive the expression for H(s).
2. Sketch the Bode plot (magnitude and phase) of H(s).

Answer 1

The transfer function H(s) is derived using ideal op-amp assumptions and is plotted on a Bode plot by evaluating the logarithmic magnitude and phase across a range of frequencies.

Step-by-step explanation:

To derive the expression for H(s) = Vout(s)/Vin(s) for an ideal operational amplifier (op-amp) with resistors R1, R2, R3, and capacitor C1, we must apply the rules for ideal op-amps and use circuit analysis techniques such as the node voltage method. Given R1 = 10 kΩ, R2 = 20 kΩ, R3 = 40 kΩ, and C1 = 20 nF, we can assume infinite input impedance, zero output impedance, and that the voltage at the inverting and non-inverting terminals of the op-amp are equal.

To sketch the Bode plot, we would calculate the magnitude and phase of H(s) across a range of frequencies. The Bode plot requires us to convert H(s) to a function of jω (where ω = 2πf) and evaluate the logarithmic magnitude (in decibels) and phase (in radians per second). Examples of key points on the Bode plot include the cutoff frequency, where the magnitude drops by 3 dB, and the frequencies where the phase shift is significant.

Answer 2

The transfer function
`\(H(s)\)` for the given circuit with an ideal op-amp,
`\(R_1 = 10 \, \text{k}\Omega\), \(R_2 = 20 \, \text{k}\Omega\)`,
`\(R_3 = 40 \, \text{k}\Omega\)`, and
`\(C_1 = 20 \, \text{nF}\)` is given by
`\(H(s) = -(R2)/(R1) * (1)/(1 + sR3C1)\), resulting in \(H(s) = -2 * (1)/(1 + 2s)\)`.

Step-by-step explanation:

The transfer function of the circuit, denoted as H(s), is derived by analyzing the op-amp circuit. For an ideal op-amp, the virtual short-circuit assumption applies to the input terminals. Considering the inverting amplifier configuration, the voltage at the inverting terminal is the same as the voltage at the non-inverting terminal, which is connected to ground.

Applying the voltage divider rule across
`\(R_2\)` and
`\(R_1\)`, the input to the op-amp can be expressed as
`\((-R2)/(R1) * V_(in)(s)\)`, where ⊅ represents the Laplace transform of the input voltage. Considering the impedance of the capacitor
`\(C_1\)` as
`\(Z_(C1) = (1)/(sC1)\)` and the resistor
`\(R_3\)`, the output is determined by the voltage division between the impedance of
`\(C_1\)`and
`\(R_3\)`, resulting in
`\((1)/(1 + sR3C1)\)`.

Combining these expressions, the transfer function H(s) is obtained as
`\(H(s) = -(R2)/(R1) * (1)/(1 + sR3C1)\)`. Substituting the given resistor and capacitor values into the equation yields
`\(H(s) = -2 * (1)/(1 + 2s)\)`. This transfer function represents the relationship between the input voltage and the output voltage in the Laplace domain for the given circuit.

To sketch the Bode plot (magnitude and phase) of H(s) in radians per second, the magnitude plot indicates the gain in decibels (dB) against frequency, while the phase plot shows the phase shift in degrees. The Bode plot for
`\(H(s) = -2 * (1)/(1 + 2s)\)` illustrates a high-pass filter behavior with a cutoff frequency at
`\(f_c = (1)/(2\pi * 2) \approx 0.08 \, \text{Hz}\)` and a slope of -20 dB/decade. Additionally, the phase plot depicts a linear decrease of -90 degrees per decade.