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)


  • \$\begingroup\$ You can add a schematic using the editor (shortcut: ctrl+M). It's really unclear what circuit you are trying to solve. \$\endgroup\$ Jun 15, 2014 at 5:52
  • \$\begingroup\$ You have absolute values on the left in the equation to prove. Try ((...)^2)^1/2 on you last step. \$\endgroup\$ Jun 15, 2014 at 6:54

1 Answer 1


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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