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I know that the transfer function of a low pass filter would be of the form

\$1/(1+s/\omega_c)\$

Two in cascade would have a transfer function equal to:

\$1/(1+s/\omega_c)^2\$

I am still unable to arrive at the correct expression for the -3 dB frequency. Any thoughts?

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    \$\begingroup\$ Hint: If they are identical and cascaded, what would be the attenuation of ONE of the stages when the combination is -3dB? Can you solve for the frequency where one stage is at that attenuation? \$\endgroup\$
    – John D
    Commented Jun 15, 2017 at 22:45
  • \$\begingroup\$ @JohnD Do you mean, when one of the transfer functions decreases by sqrt(2)? \$\endgroup\$
    – peripatein
    Commented Jun 15, 2017 at 22:47
  • \$\begingroup\$ Show how far you've got with the analysis. \$\endgroup\$
    – Chu
    Commented Jun 16, 2017 at 1:10
  • \$\begingroup\$ You might also want to note that making the product of transfer functions of cascaded stages works as long as there are no load effects of the second stage on the first one. The actual transfer function can be very different when you consider load effects (or if you derive the TF directly from circuit analysis). Just sayin'... \$\endgroup\$
    – Sredni Vashtar
    Commented Jun 16, 2017 at 21:12

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no full solution to homeworks, only guidance!

Replace the Laplacian s with j\$\omega\$ . That is =\$ j2 \pi f\$. Your \$\omega_c\$is \$ 2 \pi f_1\$.

You must derive a formula for the absolute value of the complex expression of the cascaded filter transfer function. You must set that to be =\$\frac{1}{\sqrt2}\$ which means -3dB gain. Solving from that equation the frequency gives the wanted result.

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  • \$\begingroup\$ Thank you! I wasn't looking for a complete solution. Your guidance was perfect and I managed arriving at the correct expression :-). Again, thank you! \$\endgroup\$
    – peripatein
    Commented Jun 16, 2017 at 7:27

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