4. The following circuit is a Second Order Active Band Stop filter. Its low and high cut-off frequencies are 13kHz and 18.2kHz respectively. Determine the resistors KR and RA. Vatiaus ta Use! Ans. 40.81k2; 4k2 fo 0.0015µF 2. KR A = R C 1.2k2 0.0015µF VIN O %3D VOUT 18 2472 KR RA RB 68k2

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I need to use these equations to solve for resistors KR & RA. Solve for #4.

4. The following circuit is a Second Order Active Band Stop filter. Its low and high cut-off
frequencies are 13kHz and 18.2kHz respectively. Determine the resistors KR and RA.
Ans. 40.81k2; 4k2
equations do
0.0015µF
2.
KR
オt2
R
1.2k2
0.0015µF
VIN O
%3D
o
VOUT
18.2472
ka KR
RA
RB
68KQ
40.2160
-2.15
2.
2 52
1.2452
Transcribed Image Text:4. The following circuit is a Second Order Active Band Stop filter. Its low and high cut-off frequencies are 13kHz and 18.2kHz respectively. Determine the resistors KR and RA. Ans. 40.81k2; 4k2 equations do 0.0015µF 2. KR オt2 R 1.2k2 0.0015µF VIN O %3D o VOUT 18.2472 ka KR RA RB 68KQ 40.2160 -2.15 2. 2 52 1.2452
Expert Solution
Step 1

Refer to the circuit diagram, the voltage at the non-inverting terminal is determined using the voltage divider rule as given below:

V+=RBRA+RBVin

According to the op-amp characteristic, the voltage at the inverting terminal is given as:

V-=V+=RBRA+RBVin 

The current flowing through the resistor, RK is determined as:

iK=Vout-V-RK=VoutRK-RBRA+RB·VinRK

Step 2

According to the op-amp characteristic, the current flowing through the resistor, RK is equal to the current flowing the capacitor connected to the inverting terminal. Therefore, the voltage drop across the capacitor is determined as:

VC=VoutRK-RBRA+RB·VinRK1sC

The voltage at another node, A of the capacitor is determined as:

VA=V--VC=RBRA+RBVin-VoutRK-RBRA+RB·VinRK1sC

 

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