GATE 2024 EC Q.50: solution and answer

Official answer: BSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

1. Amplifier. Take the op-amp as ideal. $R$ runs from the inverting input to ground and $R_1$ runs from the output back to the inverting input, so this is a non-inverting amplifier: $A=1+\frac{R_1}{R}$, a real, positive gain with phase $0^{\circ }$.

2. Feedback network, second section. An ideal op-amp's $+$ input draws no current, so nothing loads node B. The series $R$ and shunt $C$ form a plain voltage divider: $\frac{V_B}{V_A}=\frac{1/(sC)}{R+1/(sC)}=\frac{1}{1+x}$, where $x=sRC$.

3. Impedance at node A. Looking from node A, the shunt $R$ to ground is in parallel with the branch $R+\frac{1}{sC}=R\frac{1+x}{x}$: $$Z_A=\frac{R\cdot R\frac{1+x}{x}}{R+R\frac{1+x}{x}}=\frac{R(1+x)}{1+2x}$$

4. First section. The series capacitor has impedance $\frac{1}{sC}=\frac{R}{x}$, and it forms a divider with $Z_A$: $$\frac{V_A}{V_o}=\frac{Z_A}{Z_A+R/x}=\frac{\frac{1+x}{1+2x}}{\frac{1+x}{1+2x}+\frac{1}{x}}=\frac{x(1+x)}{x(1+x)+(1+2x)}=\frac{x(1+x)}{x^2+3x+1}$$

5. Overall feedback factor. Multiply the two sections: $$\beta (s)=\frac{V_B}{V_o}=\frac{x(1+x)}{x^2+3x+1}\cdot \frac{1}{1+x}=\frac{x}{x^2+3x+1}$$ Now put $x=j\omega RC$: $$\beta (j\omega )=\frac{j\omega RC}{(1-{\omega }^2{R}^2{C}^2)+j3\omega RC}$$

6. Barkhausen phase condition. The amplifier adds $0^{\circ }$, so $\beta (j\omega )$ must also have $0^{\circ }$ phase, which means it must be real and positive. The numerator is purely imaginary, so the denominator must be purely imaginary as well: $1-{\omega }^2{R}^2{C}^2=0$. That gives ${\omega }_0=\frac{1}{RC}$, or $f_0=\frac{1}{2\pi RC}$.

7. Barkhausen magnitude condition. At ${\omega }_0$, $\beta =\frac{j{\omega }_{0}RC}{j3{\omega }_{0}RC}=\frac{1}{3}$, which is real and positive as required. Setting $A\beta =1$ gives $1+\frac{R_1}{R}=3$, so $R_1=2R$.

8. So $R_1=2R$ and $f_0=\frac{1}{2\pi RC}$, which is option (B).