Engineering OPamp circuit - how to simplify this term?

AI Thread Summary
The discussion focuses on simplifying the output-to-input voltage ratio (Uout/Uin) in an OPamp circuit, using the formula Uout = Uin * (1 + Z2/Z1). The denominator is successfully factored, while the numerator requires expansion and collection of terms into a polynomial format. Quadratic terms in 's' typically cannot be factored with real numbers, leading to the suggestion to express them in a standard quadratic form. The roots of the quadratic expression are real when the quality factor (Q) is less than or equal to 0.5, but without specific component values, the general solution remains unfactorable. Understanding these concepts is essential for analyzing the quadratic response in resonant circuits.
altruan23
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Homework Statement
Calculate Uout/Uin
Relevant Equations
Basic op amp eq.
So this is the circuit.
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And here i tried to calculate Uout/ Uin , any suggestion how to simplify this term?? I used Uout= Uin * (1+ Z2/Z1)
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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 said:
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.
 
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