Question E7C02:
What is the frequency response of a T-network with series capacitors and a shunt inductor?  

Choices:  
A: Low-pass  
B: High-pass  
C: Band-pass  
D: Notch    

Correct Answer is  B: High-pass  

A T-network with capacitors in the two series arms and an inductor in the shunt arm is a classic high-pass ladder.

Why the correct answer (High-pass) is correct:
- Reactance vs frequency:
  - Capacitor: $X_C = \dfrac{1}{2\pi f C}$ (decreases with $f$)
  - Inductor: $X_L = 2\pi f L$ (increases with $f$)
- At low frequency: $X_C \to \infty$ (series caps block), $X_L \to 0$ (shunt inductor shorts to ground) → signal is attenuated.
- At high frequency: $X_C \to 0$ (series caps pass), $X_L \to \infty$ (shunt inductor opens) → signal passes.
- Therefore, the network attenuates low frequencies and passes high frequencies → high-pass.

Why the other choices are incorrect:
- Low-pass: Would require series inductors and a shunt capacitor (the dual network). This T has the opposite arrangement.
- Band-pass: Needs both a low-pass and a high-pass effect combined (e.g., series-resonant pass path and parallel-resonant shunt), creating two cutoff edges. This single L–C–C T produces a monotonic high-pass response, not a mid-band pass with two stopbands.
- Notch: Requires a resonant branch that sharply rejects one frequency (e.g., parallel resonance in shunt or series resonance in series). This topology does not create a single narrow rejection notch.

Rules of thumb and useful notes:
- “Capacitors in series, inductors to ground” → high-pass.
- “Inductors in series, capacitors to ground” → low-pass.
- Quick check by limits:
  - $f \to 0$: $X_C \gg R$, $X_L \ll R$ → heavy attenuation.
  - $f \to \infty$: $X_C \ll R$, $X_L \gg R$ → minimal attenuation.
- Order: With two series capacitors and one shunt inductor, this T is a 3rd-order high-pass filter when terminated properly.