Question

The lower cut-off frequency of \(N\) cascaded stages is given by

• A. \(f_L (2^{1/n} -1)^{1/2}\)
• B. \(\frac{f_L}{\sqrt{2^{1/n}-1}}\)
• C. \(nf_L\)
• D. \(\frac{nf_L}{\sqrt{2^{1/n}-1}}\)

Hint:

Cascading amplifiers modifies bandwidth limits.

Answer 1

The Correct Option is B

Step 1: Write the cascaded gain expression.
For $n$ identical high-pass stages each with lower cutoff $f_L$, the combined voltage gain at any frequency $f$ is $|H(f)|^n = \left[\dfrac{f}{\sqrt{f^2+f_L^2}}\right]^n$. The new lower cutoff $f_L'$ is where the combined gain drops to $1/\sqrt{2}$ of its midband value.

Step 2: Apply the $-3$ dB condition.
Setting $|H(f_L')|^n = 1/\sqrt{2}$: squaring both sides and raising to $1/n$: $\dfrac{f_L'^2}{f_L'^2+f_L^2} = \dfrac{1}{2^{1/n}}$.

Step 3: Solve for the new lower cutoff.
Cross-multiplying: $f_L'^2(2^{1/n}-1) = f_L^2 \Rightarrow f_L' = \dfrac{f_L}{\sqrt{2^{1/n}-1}}$. \[ \boxed{\dfrac{f_L}{\sqrt{2^{1/n}-1}}} \]