# Op Amp magnitude of gain proof

I've been working through a proof but I'm stuck on one of the last steps. Consider an inverting op amp with a feedback resistor $R_f$ in series with a capacitor and resistor $R_1$

simulate this circuit – Schematic created using CircuitLab

I must prove that: $$\frac {|V_o|}{|V_i|} = \frac {R_f}{R_1} \frac {1}{\sqrt{1+\frac{f_1^2}{f^2}}}$$

My steps so far: $$\frac {V_o}{V_i} = \frac {R_f}{R_1 - \frac {j}{\omega C}}$$ $$= \frac {1}{\frac{R_1}{R_f} - \frac{j}{\omega RC}}$$ $$=\frac{1}{\frac{R_1}{R_f}-j\frac{\omega_c}{\omega}}$$ $$=\frac{1}{\frac{R_1}{R_f}-j\frac{f_c}{f}}$$ Now, how would I go about bringing the resistances out? (a little rusty on my algebra haha)

Thanks!

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You can add a schematic using the editor (shortcut: ctrl+M). It's really unclear what circuit you are trying to solve. –  helloworld922 Jun 15 at 5:52
You have absolute values on the left in the equation to prove. Try ((...)^2)^1/2 on you last step. –  Szymon Bęczkowski Jun 15 at 6:54

Bringing the resistances out

Starting from your final equation, $$\frac{V_o}{V_i} = \frac{1}{\frac{R_1}{R_f}-j\frac{f_c}{f}}\tag{1}$$ where $f_c =\dfrac{1}{2\pi R_f C}$.

Let $f_1 = \dfrac{1}{2\pi R_1 C}$ then, $$f_c = \frac{R_1}{R_f}\times f_1$$ Applying this in $(1)$, $$\frac{V_o}{V_i} = \frac{1}{\frac{R_1}{R_f}-j\frac{f_1}{f}\times \frac{R_1}{R_f}}$$ $$\left|\frac{V_o}{V_i}\right| = \left|\frac{R_f}{R_1}\times\frac{1}{1-j\frac{f_1}{f}}\right|$$ $$= \frac{R_f}{R_1}\frac{1}{\sqrt{1+\frac{f_1^2}{f^2}}}$$

Alternate method

You should have started this way. Especially when you had the final answer with you. :)

$$\frac {V_o}{V_i} = \frac {R_f}{R_1 - \frac {j}{\omega C}}$$ $$= \frac{R_f}{R_1}\frac{1}{1-j\frac{1}{\omega R_1 C}}$$ $$= \frac{R_f}{R_1}\frac{1}{1-j\frac{f_1}{f}}$$

Taking absolute value results in the required result.

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Thank you, I can see where I was stuck! –  Paldan Jun 15 at 9:03