- Asked 2/26
- 2075 Asoj · 1+4 marks
Band pass (2000 rad/s, 400 rad/s) from 4th order
Short answer
For ω₀ = 2000 rad/s, B = 400 rad/s and fourth‑order Butterworth prototype (0.7654 H, 1.848 F, 1.848 H, 0.7654 F): Z1: 1.9135 mH series with 130.65 µF; Y2: 4.62 mF parallel with 54.113 µH; Z3: 4.62 mH series with 54.113 µF; Y4: 1.9135 mF parallel with 130.65 µH. All resonate at 2000 rad/s; band edges are 1810 and 2210 rad/s.
What is the importance of frequency transformation? Obtain a bandpass filter having ω0 = 2000 rad/s and B = 400 rad/s from fourth order Butterworth lowpass filter. [Refer table 2]
Importance of frequency transformation
Frequency transformation converts a normalized low-pass prototype into HP, BP or BS filters by a change of variable, so:
- Only one set of approximation tables (low-pass poles and ladder element values) is needed; HP, BP and BS filters are derived from it.
- The difficult approximation step (finding for given , ) is done once, for the simpler low-pass case; specifications of the other types are converted to an equivalent LP specification.
- It can be applied either to the transfer function (substitute for ) or directly to each element of an LC ladder, so the circuit is obtained without re-deriving it.
- It reduces design time and errors, and keeps the properties of the prototype (equiripple, maximally flat, etc.).
Band-pass filter from the fourth-order Butterworth low-pass filter
LP → BP: , rad/s (), rad/s.
Prototype (Table 2, , , starting with a series inductor): H, F, H, F.
Rs=1 [ Z1 ] [ Z3 ]
o-/\/\-[ ]--+-----[ ]--+------+---o
| | | | +
Vs [ Y2 ] [ Y4 ] RL=1 Vo
| | | | -
o--------------+-------------+------+---o
| Arm | Band-pass element |
|---|---|
| Z1 (series) | 1.9135 mH in series with 130.65 μF |
| Y2 (shunt) | 4.62 mF ∥ 54.113 μH |
| Z3 (series) | 4.62 mH in series with 54.113 μF |
| Y4 (shunt) | 1.9135 mF ∥ 130.65 μH |
All LC pairs resonate at 2000 rad/s; the 3 dB band edges are 1810.0 and 2210.0 rad/s. These values are for terminations. For practical values, magnitude-scale by (e.g. for 1 kΩ terminations): multiply every L by and divide every C by .
✓ Answer checked by Claude Sonnet 5.5
Related questions
- Band pass (2000 rad/s, 400 rad/s) from 3rd order · Filter Design, chapter 3 · asked 2×
- Band stop filter (2000 rad/s, 400 rad/s) from 3rd order · Filter Design, chapter 3 · asked 4×
- Band pass (1 krad/s, 100 rad/s) from 4th order ladder · Filter Design, chapter 3
- Band pass (1500 rad/s, 300 rad/s) from 4th order ladder · Filter Design, chapter 3 · asked 4×
- Band pass (2000 rad/s, 400 rad/s) from given ladder · Filter Design, chapter 3
- Bruton transformation; FDNR 4th order LPF at 2000 rad/s · Filter Design, chapter 8 · asked 3×
- Bruton transformation; FDNR 4th order LPF at 20,000 rad/s · Filter Design, chapter 8 · asked 5×