Question

Applying Kirchoff's current law (KCL) at the blue input node and the red output node, we can write Op amp has infinite input impedance and zero output impedance!! 0 Blue input node: lin(s) If(s) + H(s) = (will not "steal" current from your circuitry on the input side, nor will it steal anything from the output side! Red output node: lour(s)+ lats) Excellent! Now, since we don't care about the out node for now (because we're just focused what's happening between the input and the output), we can just focus on the blue input node for the rest of the problem) From the blue input node, we can plug in lin(s) and () as V1(s) - Vi(s)Vin(s) - Vout (s) Blue input node equation: The basic input - output relationship Vout (s)A Vin(s) Your tasks start on the next page!!

Answer 1

Given that: 1/R_1C = K rightarrow rightarrow now let inverting terminal voltage = ''v' rightarrow as open loop gain of op-amp = (very High) TildeEqual infinity Therefore V_non-inverting = V_inverting = > V = V_2(s) = 0 = > V = 0 rightarrow (2) rightarrow now applying Nodal Analysis at Node(V): i_1 = i_2 + i_3 rightarrow (111) as we know: for op-amp: R_in = (i/p) impedance TildeEqual 'infinity' Therefore current Entering TildeEqual 0 = > (i_2 = 0) Therefore i_1 = i_3 = > v_1(s) - 0/R_1 = 0 - v_o(s)/ (1/sc) = > v_1(s) /R_1 = 0 - v_o (s)/ (1/sc) = > v_1(s) /R_1 = -v_o (s)/(1/sc) = > v_o(s)/ v_1(s) = -(1/sc)/R_1 = > T(s) = V_o(s)/ v_1(s) = [-1/ scR_1) Therefore T(s) = -[k/s] (Therefore k = 1/ R_1 c) Given Closed loop Transfer function = T(s) = (-k/s) T(s) = v_o (s)/v_1(s) = (-k/ s) = > v_o (s) = (- k/ s) v_1(s) Therefore equivalent system diagram: v_1 (s) rightarrow T(s) = (-k/s) rightarrow v_o(s) Therefore for the above op-amp circuit: T(s) = (-k/s) Therefore T(s) has one pole at origin (s = 0) Therefore it is an