Problem 13.4-12

Let’s call A the knot for the inverting entry of the amp op. We apply the KCL for knot A:

But we have, as the amp op. is ideal, V_- = V₊ = V_A = 0

Thus we get

So this is:

where p = 1/R₁C₁ = 4000

1+ jω/p can be replaced by 1 when ω < p and by jR₁C₁ω when ω > p

So we get:

H(ω) ≈ jR₁C₂ω when ω < p

           C₂/C₁    when ω > p

So we get

|H(ω)| ≈ R₁C₂ω when ω < p

             C₂/C₁    when ω > p

Finally, as we are looking for the Bode graph, we need G  = 20 log H(ω)

G = 20 log(R₁C₂ω) =20 log (0.05.10-2) + 20 logω when ω < 4000

      20 log (C₂/C₁)                                                when ω > 4000

This is a line of slope 20 dB / dec up to ω = 4000 and then a straight line.