2
\$\begingroup\$

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\$

schematic

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!

\$\endgroup\$
2
  • \$\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

4
\$\begingroup\$

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.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.