OPamp circuit - how to simplify this term?

[altruan23]

Homework Statement

Calculate Uout/Uin

Relevant Equations

Basic op amp eq.

So this is the circuit.

And here i tried to calculate Uout/ Uin , any suggestion how to simplify this term?? I used Uout= Uin * (1+ Z2/Z1)

 

[DaveE]

It looks good so far. The denominator is done; it's in a factored form. For the numerator, you need to expand it again and collect the terms for s into a polynomial. Usually the quadratic terms in s can't be factored (with real numbers); factor it if you can into $(1+\frac{s}{\omega_0})(1+\frac{s}{\omega_1})$, but that never seems to workout. So, then you just leave them in a format like $(as^2 + bs + 1)$ which we would actually express as $({(\frac{s}{\omega_0})}^2 + \frac{s}{Q \omega_0} +1)$ for reasons you'll get to when you study the quadratic response (simple harmonic oscillators and resonant circuits).

 

[DaveE]

It looks good so far. The denominator is done; it's in a factored form. For the numerator, you need to expand it again and collect the terms for s into a polynomial. Usually the quadratic terms in s can't be factored (with real numbers); factor it if you can into $(1+\frac{s}{\omega_0})(1+\frac{s}{\omega_1})$, but that never seems to workout. So, then you just leave them in a format like $(as^2 + bs + 1)$ which we would actually express as $({(\frac{s}{\omega_0})}^2 + \frac{s}{Q \omega_0} +1)$ for reasons you'll get to when you study the quadratic response (simple harmonic oscillators and resonant circuits).

Note that for the quadratic term $({(\frac{s}{\omega_0})}^2 + \frac{s}{Q \omega_0} +1)$, the roots (given by the quadratic formula) are real when $Q \leq \frac{1}{2}$ and the quadratic is factorable with Reals. But, since your analysis is done (correctly, IMO) with variables (like R1, C2, etc.), we don't know what value Q has until we put in the values. So, the general solution isn't factorable, that's why we normally leave it as a quadratic.