- Asked 4/26
- 2080 Bhadra · 4 marks
Non-inverting realization of 4(s+200)/(s+100)
Short answer
Using non-inverting topology with Ra, Rb||C: choose C=1 μF, then Rb=1/(100×1 μF)=10 kΩ, R1=2.5 kΩ, Ra=7.5 kΩ. Yields 4(s+200)/(s+100), HF gain4, DC gain8.
Design the following transfer function using non-inverting op-amp configuration T(s) = 4(s+200)/(s+100)
Given : HF gain 4, DC gain , zero at −200, pole at −100.
Non-inverting configuration: from the inverting input to ground, from output to inverting input:
Because the HF gain is greater than 1, use and :
+--Ra--+--Rb--+
| | |
| +--||--+
| C |
+-----+-|-\ |
| | >--------+--- Vo
R1 Vin---|+/
|
GND
Matching:
Choose F:
Check: DC gain . Correct.
Answer: , , , F.
✓ Answer checked by Claude Sonnet 5.5
Related questions
- Non-inverting realization of (s+8)/(s+2) · Filter Design, chapter 6 · asked 5×
- Active filter; non-inverting (s+8)/(s+2) · Filter Design, chapter 6 · asked 5×
- Inverting realization of 7(s+400)/(s+200) · Filter Design, chapter 6
- Active vs passive; non-inverting (s+8)/(s+2) · Filter Design, chapter 6 · asked 5×
- Non-inverting: zero at -4, pole at -8, HF gain 2 · Filter Design, chapter 6 · asked 4×
- Non-inverting: zero at -5, pole at -3, HF gain 2 · Filter Design, chapter 6 · asked 3×
- Non-inverting filter: zero at 8, pole at 4, gain 2 · Filter Design, chapter 6 · asked 2×
- SC summer, inverting and non-inverting integrators · Filter Design, chapter 9 · asked 4×
- Non-inverting op-amp voltage gain · Fundamental of Electrical and Electronics Engineering, chapter 6
- Transfer function of op-amp phase-shift circuit · Control Systems, chapter 2 · asked 2×
- Transfer function of op-amp with Rs ∥ C feedback · Control Systems, chapter 2
- Ideal op-amp properties; non-inverting gain · Basic Electronics Engineering, chapter 4 · asked 3×