EXAMPLE 3.13. (a) Design a second-order band-pass filter with fo BW 100 Hz. What is its resonance gain? (b) Modify the circuit for a resonance gain of 20 dB = 1 kHz and Solution Сд — 10 nF and Ri 3D Rz 3D R3 fo/BW= 10 (К – 1)RA 27.28 V/V (a) Use the equal-component option with C V2/(2л 10° х 10-) — 22.5 kR (use 22.6 kR, 1%). We need Q – /2/10 = 3.858. Pick RA 10.0 kS2, 1%. Then, RB so K = 4 = 28.58 kS2 (use 28.7 k2, 1%). The resonance gain is K/(4 – K (b) Replace R, with two resistances R,A and RB, in the manner of Example 3.9, whose values are found via Eq. (3.63) with Aold = 27.28 V/V and Apew 10 V/V. This gives Ria = 61.9 k, 1%, and RiB = 35.7 k2, 1% 1020/20

 

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