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Q2. Find the output waveform

If the input to the op-amp circuit below is $A\text{sin}(\omega t) + B\text{cos}(2 \omega t)$ A\text{sin}(\omega t) + B\text{cos}(2 \omega t) where A and B are constants, and $\omega$ \omega is the angular frequency, what is the output waveform? Assume that the op-amp input impedance is infinite but it has a frequency dependent differential gain given by $\frac{G\omega_0}{(\omega_0+j \omega)}$ \frac{G\omega_0}{(\omega_0+j \omega)} , where G is a constant and $\omega_0$ \omega_0 is a constant.

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$V_{out}(t) = H(\omega) A\text{sin}(\omega t + \theta(\omega) )+ H(2\omega) B \text{cos}(2\omega t + \theta(2\omega))$ V_{out}(t) = H(\omega) A\text{sin}(\omega t + \theta(\omega) )+ H(2\omega) B \text{cos}(2\omega t + \theta(2\omega)) given that $H(\omega)e^{j\theta(\omega)} = -\frac{GR_2\omega_0}{R_2(j\omega + (1+G)\omega_0) + R_1(j\omega + \omega_0 + j\omega C_1 R_2 (j \omega + (1+G)\omega_0))}$ H(\omega)e^{j\theta(\omega)} = -\frac{GR_2\omega_0}{R_2(j\omega + (1+G)\omega_0) + R_1(j\omega + \omega_0 + j\omega C_1 R_2 (j \omega + (1+G)\omega_0))}

$V_{out}(t) = H(\omega) A\text{sin}(\omega t + \theta(\omega) )+ H(\omega) B \text{cos}(2\omega t + \theta(\omega))$ V_{out}(t) = H(\omega) A\text{sin}(\omega t + \theta(\omega) )+ H(\omega) B \text{cos}(2\omega t + \theta(\omega)) given that $H(\omega)e^{j\theta(\omega)} = \frac{GR_2\omega_0}{R_2(j\omega + (1+G)\omega_0) + R_1(j\omega + \omega_0 + j\omega C_1 R_2 (j \omega + (1+G)\omega_0))}$ H(\omega)e^{j\theta(\omega)} = \frac{GR_2\omega_0}{R_2(j\omega + (1+G)\omega_0) + R_1(j\omega + \omega_0 + j\omega C_1 R_2 (j \omega + (1+G)\omega_0))}

$V_{out}(t) = H(\omega) A\text{sin}(\omega t + \theta(\omega) )+ H(2\omega) B \text{cos}(2\omega t + \theta(2\omega))$ V_{out}(t) = H(\omega) A\text{sin}(\omega t + \theta(\omega) )+ H(2\omega) B \text{cos}(2\omega t + \theta(2\omega)) given that $H(\omega)e^{j\theta(\omega)} = -\frac{GR_1\omega_0}{R_2(j\omega + (1+G)\omega_0) + R_1(j\omega + \omega_0 + j\omega C_1 R_2 (j \omega + (1+G)\omega_0))}$ H(\omega)e^{j\theta(\omega)} = -\frac{GR_1\omega_0}{R_2(j\omega + (1+G)\omega_0) + R_1(j\omega + \omega_0 + j\omega C_1 R_2 (j \omega + (1+G)\omega_0))}

$V_{out}(t) = H(\omega) A\text{sin}(\omega t + \theta(\omega) )+ H(\omega) B \text{cos}(2\omega t + \theta(\omega))$ V_{out}(t) = H(\omega) A\text{sin}(\omega t + \theta(\omega) )+ H(\omega) B \text{cos}(2\omega t + \theta(\omega)) given that $H(\omega)e^{j\theta(\omega)} = \frac{GR_1\omega_0}{R_2(j\omega + (1+G)\omega_0) + R_1(j\omega + \omega_0 + j\omega C_1 R_2 (j \omega + (1+G)\omega_0))}$ H(\omega)e^{j\theta(\omega)} = \frac{GR_1\omega_0}{R_2(j\omega + (1+G)\omega_0) + R_1(j\omega + \omega_0 + j\omega C_1 R_2 (j \omega + (1+G)\omega_0))}

$V_{out}(t) = H(\omega) A\text{sin}(\omega t + \theta(\omega) )+ H(2\omega) B \text{cos}(2\omega t + \theta(2\omega))$ V_{out}(t) = H(\omega) A\text{sin}(\omega t + \theta(\omega) )+ H(2\omega) B \text{cos}(2\omega t + \theta(2\omega)) given that $H(\omega)e^{j\theta(\omega)} = -\frac{GR_1R_2\omega_0}{R_2(j\omega + (1+G)\omega_0) + R_1(j\omega + \omega_0 + j\omega C_1 R_2 (j \omega + (1+G)\omega_0))}$ H(\omega)e^{j\theta(\omega)} = -\frac{GR_1R_2\omega_0}{R_2(j\omega + (1+G)\omega_0) + R_1(j\omega + \omega_0 + j\omega C_1 R_2 (j \omega + (1+G)\omega_0))}

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ECE3161 Analogue Electronics - MUM S2 2025

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