Problem 14.9-10

Let’s call node 1, the node on the left, corresponding to the non-inverting entry of the left op amp. Let’s call node 2, the node on the right, corresponding to the inverting entry of the right op amp.

 

The KCL gives us, for node 1:

(V₁ – V)*C₁s + (V₁ – V₀)/R₁ = 0   (1)

The KCL gives us, for node 2:

(V₂ – V₁)/R₂ + (V₂ – V₀)*C₂s = 0   (2)

But we have V₂ = 0 thus we get from (2)

V₁  = – V₀ R₂C₂s

We replace that in equation (1) and we get