EC.23)For the non-inverting integrating amplifier in the figure,the ideal closed-loop gain $\frac{V_0(s)}{V_i(s)}$ equals:
(a) $ \frac {1}{(R1C1s+1)} .\frac{( RCs+1)}{RCs}$
(b) $\frac{1}{(R1C1s+1)}.\frac{RCs}{(RCs + 1)}$
(c) $\frac{1}{(R1C1s)}.\frac{(RCs+1)}{RCs}$
(d) none of above
Answer is (a)
To find the solution, let's go step by step , $\frac{V_o(s)}{V_a(s)} = \frac{R + 1/Cs}{R}$ (Explained above for EC.21)...Eqn 1
Next,the low pass filter at the front , $\frac{V_a(s)}{V_i(s)} = \frac{1}{R1C1s+1}$ ..Eqn(2) Combining Eqns(1)&(2),$\frac{V_o(s)}{V_i(s)} = \frac {1}{(R1C1s+1)} .\frac{( RCs+1)}{RCs}$
