Non Inverting Operational Amplifier

AI Thread Summary
The discussion focuses on deriving the gain expression Vout/Vin for a non-inverting operational amplifier with a capacitor in series with R1. The initial expression for gain is given as 1 + R2/[R1 + 1/(jwC)], which needs to be converted into the form A + jB. The user struggles with this conversion but eventually rewrites the gain as 1 + R2/[R1 - j/wC]. Guidance is provided on multiplying by the conjugate to simplify the expression, leading to a successful resolution of the problem. The user confirms that they have solved the issue after applying the suggested method.
theuniverse
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Homework Statement


Derive an expression for the gain Vout/Vin. Express your answer in the form of A+jB.

Homework Equations



The Attempt at a Solution


If C is the capacitor in series with R1, gain = 1 + R2/[R1 + 1/(jwC)]. It says to express it in the form of A (real)+ jB (imaginary), but I'm not sure how to make the conversion from my solution of gain to the form of A+jB.
Any help is appreciated.

Edit: Figured that I can rewrite it as gain = 1 + R2/[R1 - j/wC] and I tried multiplying by the conjugate but I still can't get it...
 

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theuniverse said:

The Attempt at a Solution


If C is the capacitor in series with R1, gain = 1 + R2/[R1 + 1/(jwC)]. It says to express it in the form of A (real)+ jB (imaginary), but I'm not sure how to make the conversion from my solution of gain to the form of A+jB.
Any help is appreciated.

Edit: Figured that I can rewrite it as gain = 1 + R2/[R1 - j/wC]
Yes, looks good so far.
and I tried multiplying by the conjugate but I still can't get it...
Multiply by (conjugate/conjugate), where "conjugate" is the conjugate of the denominator.

For example:

\frac{c}{a+jb} = \frac{c(a-jb)}{(a+jb)(a-jb)} = etc.
 
Thanks! yea I ended up doing that by kept forgetting the extra 1 for the gain. All solved now though.
 
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