4. Given the following active Filter circuit: RF ww 9K52 Vin 。 sinut + R2 www RT mmm IKUL lokr C 1 •001592µF + + Vo 1. Determine the following: a. The cut off Frequency fc= 9997.17H₂ Fc = 2 Rc 2n (10kr) (.001592 µF) B. The gain of the amplifer at a equal to 100 Hz ONG RF gain = 1+ > H R₁ акл 1KR → frequency gain = 10 Vo/= Gain Vin 10 →Again = 9.999 ₤12 1+ 남기 1 + (100907.17) 2 C. The Gain of the amplifier of the cutoff frequency. √0/= Gain لا 10 → Again = 0.707 Vin 1. 2 1+ $9997.1712 бо 9997.171 The peak-peak amplitude of Vo at a 4. Frequency to 100 H₂ Vop-p = A (100 H₂) • (2 × 0.8) 19.999 (2·0.8) V Vo = 15.998V

Introductory Circuit Analysis (13th Edition)
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Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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4. Given the following active Filter circuit:
RF
ww
9K52
Vin
。 sinut +
R2
www
RT
mmm
IKUL
lokr
C
1
•001592µF
+
+
Vo
1. Determine the following:
a. The cut off Frequency
fc= 9997.17H₂
Fc = 2 Rc
2n (10kr) (.001592 µF)
Transcribed Image Text:4. Given the following active Filter circuit: RF ww 9K52 Vin 。 sinut + R2 www RT mmm IKUL lokr C 1 •001592µF + + Vo 1. Determine the following: a. The cut off Frequency fc= 9997.17H₂ Fc = 2 Rc 2n (10kr) (.001592 µF)
B. The gain of the amplifer at a
equal to 100 Hz
ONG
RF
gain = 1+
> H
R₁
акл
1KR
→
frequency
gain = 10
Vo/= Gain
Vin
10
→Again = 9.999
₤12
1+
남기
1 + (100907.17)
2
C. The Gain of the amplifier of the cutoff
frequency.
√0/= Gain
لا
10
→ Again = 0.707
Vin
1.
2
1+ $9997.1712
бо
9997.171
The peak-peak amplitude of Vo at a
4. Frequency to 100 H₂
Vop-p = A (100 H₂) • (2 × 0.8) 19.999 (2·0.8) V
Vo = 15.998V
Transcribed Image Text:B. The gain of the amplifer at a equal to 100 Hz ONG RF gain = 1+ > H R₁ акл 1KR → frequency gain = 10 Vo/= Gain Vin 10 →Again = 9.999 ₤12 1+ 남기 1 + (100907.17) 2 C. The Gain of the amplifier of the cutoff frequency. √0/= Gain لا 10 → Again = 0.707 Vin 1. 2 1+ $9997.1712 бо 9997.171 The peak-peak amplitude of Vo at a 4. Frequency to 100 H₂ Vop-p = A (100 H₂) • (2 × 0.8) 19.999 (2·0.8) V Vo = 15.998V
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