Hence Equ(1) ≥ Equ(2). 12. Determine the transfer function for the practical differentiator
A. Vo(s) /V1(s) = -S×RF×C1/(1+R1×C1)2
B. Vo(s) /V1(s) =-S×RF×C1/(1+RF×C1)2
C. Vo(s) /V1(s) =-S×RF×C1/(1+R1×CF)2
D. None of the mentioned
Answer:A
Hence Equ(1) ≥ Equ(2). 12. Determine the transfer function for the practical differentiator
A. Vo(s) /V1(s) = -S×RF×C1/(1+R1×C1)2 B. Vo(s) /V1(s) =-S×RF×C1/(1+RF×C1)2 C. Vo(s) /V1(s) =-S×RF×C1/(1+R1×CF)2 D. None of the mentioned
Explanation: The transfer function for the circuit,