Problem 14.9-13

Let’s call node 1, the node on the left, between R1 and R2.

Let’s call node 2, the node in the middle, between R2 and R3, corresponding to the inverting entry of the op amp. The voltage there, V2 is equal to zero.

Let’s call node 3, the node on the right, between R3 and R4.

We apply the KCL at those three nodes.

At node 1, we have

(V₁ – V_S)/R₁ + (V₁ – V₃)*Cs +  (V₁ – V₂)/R₂ = 0   (1)

and V₂ = 0

At node 2, we have

(V₂ – V₁)/R₂ +  (V₂ – V₃)/R₃ = 0   (2)

so we find V₃/R₃  = – V₁/R₂ (2b)

At node 3, we have

(V₃ – V₂)/R₃ + (V₃ – V₁)*Cs +  (V₃ – V₀)/R₄ = 0   (3)

We take (2b) and replace it in equation (1), to get:

(V₁ – V_S)/R₁ + (V₁ + R₃V₁/R₂ )*Cs +  V₁/R₂ = 0

From this we find V₁ as a function of V_S:

Now, we take (2b) and replace it in equation (3) as well, to get:

– V₁/R₂ – (R₃V₁/R₂ + V₁)*Cs – (R₃V₁/R₂ + V₀)/R₄ = 0

which is exactly V₁/R₂ + (R₃V₁/R₂ + V₁)*Cs + (R₃V₁/R₂ + V₀)/R₄ = 0

From this we find V₁ as a function of V₀:

So finally the transfer function is: