Question

Design an active unity-gain bandpass filter with center frequency 750 Hz and bandwidtg 250 Hz and with 0.1 μF capacitor, R1=6.4kΩ, R2=377Ω, and R3=12.7kΩ.

R. cic HE A R, 1 V. Figure 15.26 4 An active high-Q bandpass filter.

a)Discuss the circuit response with support of a Bode magnitude plot.
b) Assume next that a load R_L is connected to the output of the network at the terminal Vo(s). How does the frequency response of the loaded configuration change?
c) Consider a broadband bandpass op amp filter with center frequency 2.4 kHz and bandwidth 800 Hz. Are the specifications reasonable and would the transfer function poles be real, imaginary, or complex?

Answer 1

To do To design an active unity gain bandpars filter with following specift: fc = 750 Hz 250 uz Bw. во) не R = 3772 = 6.4k 2 R3 = 12.7k-2 The given fig. showes multiple feedback Band pass filter. This circuit uses the full gain of the operational amplifier, with multiple negative feedba back via resistor Rze cap acitor c. Quality factor Q = fe Bw fez conter frequency.

3 Pg & 750 250 here Q is small this implies wide-bandwidth Though given but these can R., R₂ & R₂ values are be calculated relation: following R. H.K R₂ - (20² HK 3 = 20 R where k= antec = 2X3.14 X 7500 0.1x10-6 4 = mid-band gain Vin R, - 3 1 471x10-6 6.36 km R = 6-4 ker similarly R & Rz can be calculated. draw :- Bode magnitude phot

Transfer function 'Als) can be given. ALS) - R₂ R3 Rith, e wm. S 2 2 1+ 22, R2 const R, R₂ R3 2 R, tha RtRa nehere wm = 20 fm azofa 2x 3.14 X 750 c4710 m c.wm 0.1 X10-6 x 4710 o.1 X 10-3 ewa 0.222 X10-S -3 Aus) 12.78103 X377 0.471x10 S (64007377) 1 2X64X10²x377 6400+ 377) X0471x10-3 + 6.4810* 377 x 12.7x103 6400 +377 X0.22L Xlor X&2 4.788 X 0.471 6777 1+ 2.273 X103 st 6.802 68103 6.777 X103 6.777 X103

Pgy A(8) = 3.33x10-1 It 0.3358 + 8 2 This magnitude transfer function has two poles' yo "Ame? АСА) - 3.33 X10-4 (8 to 17+)) (sto17;) is poles at -0.17+j constant - 3. 33 X10-3 &- - 0.17 - note: Conjugate potes will cause a peak at bo frea. peare height = -20logio (26d1abe) where е, а 0.335 = 0.167 2 x 1 0.33 3. pear heigh = 9.6 dB.

Pg. 5 - 20dB f ocs 10-1 10700Hz in mag (dB) when a R load Resistance is connected at the output of (8), this R₂ will come in parallel to Rz due to virtual ground concept). Due to this, the overall feedback resistana decacases. Due to this the quality factor of the filter may change for constant fc. a Bow= 800 H for fc = 2-4 kHz Bod rz Q = be 2-4 X103 = 3 B-W 800 Q factor remain unchanged. R = 2 * fere & 2x3-14 X2 4X10 Xo. 1818-6 R - 1.51x10-3

R, R₂ RO R Values Since valce anges, this will Changes, R, H.K 3 1.51 X10-3 ek z Rz - 3 (2Q2 –H)k (18-1) (:5X103) = 1182 &3 2Q R 2X 3 1.51810-3 a uke ao Denominator of transfer function can be given as: АСА 6² iwan R R₂ R3 + s.c. way 2R, R2 Roth, Roth, 2014 3 2KX 2.48103 = 15-1 x 10 ex wm OLX10-6X15 - 1 X103 = 1.58 10-3 2 (com) 2.25 x 10-6

P4 + expression á مق we become * 2.25X1082 x 1198 4x1097 8.(15X10 37x2x2x10319 (200 000+ 118 200 +118 s² + 0.334 st we compare it with previous transfer function then poles position doesn't change on again conjugate complex poles. me have