GATE 2025 EC Q.34: solution and answer

Official answer: 50Source: Official answer key, GATE 2025 EC (IIT Roorkee)

1. Assume an ideal op-amp. The non-inverting input is grounded, so the virtual short holds the inverting input at virtual ground and no current enters the op-amp. The circuit is an inverting amplifier: $H(s)=-\frac{Z_f}{Z_i}$, with $Z_i=R+\frac{1}{s(10C)}$ and $Z_f=2R\parallel \frac{1}{s(0.1C)}$.

2. Substitute the impedances. $Z_i=\frac{1+s10RC}{s10C}$ and $Z_f=\frac{2R}{1+s0.2RC}$, so
   $$H(s)=-\frac{s(20RC)}{\left(1+s10RC\right)\left(1+s0.2RC\right)}$$
   The input network ($R$, $10C$) is a high-pass section with corner ${\omega }_1=\frac{1}{10RC}$. The feedback network ($2R$, $0.1C$) is a low-pass section with corner ${\omega }_2=\frac{1}{0.2RC}=50{\omega }_1$.

3. Midband gain. Between the two corners, $10C$ acts as a short and $0.1C$ acts as an open, so $H=-\frac{2R}{R}=-2$ and the midband gain is $A_M=2$. The lower and upper $-3$ dB frequencies are where the gain falls $3$ dB below $A_M$, which means $|H(j\omega )|=\frac{A_M}{\sqrt{2}}=\sqrt{2}$.

4. Normalise the frequency. Put $x=\omega /{\omega }_1=10RC\omega$. Then $0.2RC\omega =x/50$ and $20RC\omega =2x$, so
   $$|H(j\omega )|=\frac{2x}{\sqrt{\left(1+x^2\right)\left(1+\frac{x^2}{2500}\right)}}$$

5. Apply the $-3$ dB condition $|H{|}^2=2$:
   $$4x^2=2\left(1+x^2\right)\left(1+\frac{x^2}{2500}\right)\Rightarrow 5000x^2=2500+2501x^2+x^4\Rightarrow x^4-2499x^2+2500=0$$

6. Solve the quadratic in $x^2$:
   $$x^2=\frac{2499\pm \sqrt{2499^2-4(2500)}}{2}=\frac{2499\pm 2496.9984}{2}$$
   This gives $x_L^2=1.0008$ and $x_H^2=2497.9992$, so $x_L=1.0004$ and $x_H=49.980$. The product of the roots is $x_L^2{x}_H^2=2500$, so $x_L{x}_H=50$.

7. Ratio of the band edges. Since $x_L=50/x_H$,
   $$\frac{f_H}{f_L}=\frac{x_H}{x_L}=\frac{x_H^2}{50}=\frac{2497.9992}{50}=49.96$$
   The lower edge sits at $1.0004f_1$, almost exactly on the lower corner, and the upper edge sits at $49.98f_1$, almost exactly on the upper corner, because the two corners are a factor of $50$ apart.

8. Put in the data. With $f_L=1\text{MHz}$, $f_H=49.96\times 1\text{MHz}=49.96\text{MHz}$. Rounded to the nearest integer, $f_H=50\text{MHz}$.

9. Note on the reference level. Because the corners are only a factor of $50$ apart, the gain actually peaks at $\omega =\sqrt{{\omega }_1{\omega }_2}$ with value $\frac{2\cdot 50}{51}\approx 1.96$, a little below $A_M=2$. If $-3$ dB is measured from that peak instead of from the midband gain, the edges spread out to $f_H\approx 54\text{MHz}$. The official key, $50$, measures from the midband gain $A_M=2R/R=2$, which is the usual reference for an amplifier or filter whose band is set by separate low- and high-frequency corner networks.